Chemical and Statistical Thermodynamic Chemistry

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PREFACE TO T,IJE SECOND 'EDITION .
The . book "Chemical & Statisdcal Thermodynamics~' is wriijen with the .objective of
providing description. ofpriµcfples. irivo lyed iJi Chemical an4 ·S.Jatistical Theqnpdyriamics in
Physical (:hemi~try inclu~irig their appHcation at a levei Sllitabl~ for post-mdu~te students of .
Chemistry in different ·universitie~.. This bo~k. will oefhighly. useful for the candidates ·
preparing for various competitive examinations at natio,nal artd state levels. such as CSIR'." . ·
UGC-NET, GATE, TIFR, SLET, etc. ·
: ,1
In all the competitive examinations such as. CSIR-:UGC-NET/JRF, yATE, TIFR, SLET, etc.
. . . . .. ·. . . ! ./. : . l .
the section of Chemical and.Statistical Thermodynamics cal!Y appro~itnately 12-15 percent
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' •• · / . • . ' • '1 . ..:/. • '
weightage of total marks. ' /
• In every year CSIR-UGC-NET includes at least J6-20 tfurks from this section....
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• In every year·GATE paper.includes at least 8-10 marks from this section out of 100
marks.
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. . . . . 1. 11 ' .
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Apart from NET, GATE and PSUs this bo9k will also be very usefu\ to candidates ~ppearing
fot various Ph.D entrance examinations.
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The book contains a brief description on important topic of Chemical and Statistical
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Thermodynamics. Students preparing for competitive examinations are advised to go through

every topic for their understanding and conceptual clarity .and then get into solving the
is
problems.
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Much care has been taken to minimize the typographical errors, yet if any noticed,
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then your suggestions are invited with a view to improve the book in future editions.
We would like to thank our respected faculty members for their valuable inputs in the
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book. Thanks to Mr. Neeraj Panwar, Mr. Vinay Dwivedi and Mr. Habib Ali for their
constructive support in doing corrections ·and proofreading. A note of thanks also to
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Mohammad Firoz for formatting, designing and type-setting the book.

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CONTENTSQUZZES MORE NOTES ON www.ChemistryABC.com
Chapter 1: Introduction to Thermodynamics •- .l

01-25
Chapter 2: Zeroth & First Law of Thermodynall)icS 26-60
Chapter 3: Second Law of Thermodynamics! :' 61-104

·,
Chapter 4: Equilibrium Criteria Free Energy F_unction
·,)
105-114
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Chapter 5: Third Law of Thermody-nam,ics 115-120
Chapter 6: Partial Molar Quantity 121-133
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Chapter 7: Thermochemistry '..
'·' 13'4..:140
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Chapter 8: Statistical Thermodynamics l.41-163
Chapter 9: Kinetic Theory of Gases 164-178

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Chapter 10: Chemical Equilibria 179-194

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Chapter 11: Phase Rule 195-214

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Chapter 12: Solution & ColligativeProperties 215-226

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[ Chapter 1]
Introduction to Thermo��namics
J
·.\
Introduction .)
>
'?
. . . . . I .
The subjectofthennodynamics deals basically with the.interaction of or,i,
. 'body with another in tenns of the
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quantities ofheat and work. The entire fonnulation ofthennodynamics)s ba&ed on the two fundamental laws
which have been-established on the b3&is,ofthe experimental behavioty 9f rrlacroscopic aggregates ofm�tter
collected over a long period oftime. Wrth the help ofthermodynamics principles the experimental criteria for
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equilibrium or for the spontaneity ofprocess·are readily established
C
Mathematics Involved in Chemical Thermodynamics ,,
\
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Differential Formulas: A
d(c) d(xn )
(i) =0 (ii) -·-· = �
y
I
dx dx
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·.··
.. ''' ·d(cxn ) - d(xn) - ....fn_: ) )J(sin:X), , ·. · X. . ·.·
(lll)
is
- C-- -Cru'
I
(iv} = COS
dx dx dx
m
d(cosx) d(ax )
(v) =-smx (vi)�= ff In a
dx
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... d(lnx) 1
(vm) .. dx = � x>O
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x
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[t d ( x)
(x)J g(x) [f ] -f(x).
d[g(x) ]
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d
(x) - -- � dx dx
dx g(x)
[g(x)]2
Integration Formulas:
J J
( i) dx = X + C, dt = t + C (ii) Jx ndx = -+c
n+I
Xn+I
Xn+I
(iii) Jkx dx = k Jx dx = k--+c
n n
. (n + 1)
(v) J!dx = f!ne x + c _(vi) Jsinxdx =-cos�+c

_ HAND MADE NOTES www.ChemistryABC.com
(vii) Jcosdx = smx+c
cvm) 1u cx)g(x>; dx :f(x> Jg(x)dx _{ ! J

f(x). g(x )dx ]dx
Logarithms:
(i) log10 2 ""0.3010 (ii) log10 3 = 0.4771 (iii) log 10 4 0.6020
(iv) log10 5 =0.699 (v) log 10 6=0:7781. ·(vi) log1-0 7 =·0'.8451
( vii) log10 8 := 0.903 0 (viii) log 10 -9 = 0.954 . (ix) log!O 10 =1

,)
Important Rules of Logarithms:
(i)ln~ = 7.303xlog 10 2 = 2.303x0.3010 =0.693
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.. . m
(ii) log 10 xm = m log 10 .x ; log h = l9g m -,:log n
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(iii)log(mx n) = logm+ logn
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1
(iv) log 10 x =-2 => x = 10.,.2 =-- =0.01
100
log 10 (a I b) = log 10 a -log 10 b
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(v)
Selected SI units
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Quantity Unit
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Basic Units Name S~bol ·

m
Lengths Metre m
Mass Kilogram kg
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Tnne Second s
Electric current Ampere A
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Temperature Kelvin K
Luminous intensity Candela cd
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Amount ofsubstance mole rool

Derived Units:
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Vohnne cubic metre mJ

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Density kilogram per kg/rrr

cubic metre
Force newton N(kgm/s2)
Pressure pascal Pa(N/m2)
Energy, work, heat joule J(N-m)

. ·liiiiiiiiillii)
Conversion of Selected Non-system Units to SI units:

Quantity Unit . Com·ersion factor to SI
centimetre lx10-21m
· micrometre(µm) lx10-6m
nanometre (nm) ;'lxl0-9 m
Length angstrom( A) lx10- 10 m
· inch 2.54x 10-2 m =2.54 cm
Mass gram · lxl0-31$/
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Volwne ,,litre c:Jx1of~3 1dm3 =103 cm3
millilitre' . 1 cny
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. bar .105 Pa= 100 kPa
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atmosphere( atm) l.01325xJ05 P9-,,y= 760 mmHg = 76 cmHg
Pressure
133.322 Pd
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, miUinietre of Hg s?
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or torr
1.60219xl0-19 J
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electron volt( eV)

calorie( cal) 4.1868J
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Energy, work, amount ofheat kilocalorie (kcal) 4186.8J

m
erg 10-1 J
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Values of Selected Fundamental Physical Constants

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Quantity · Symbol _ Value

=6.6262 X 10-27 erg• S
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i Planck constant
= 6.6262x10_ J ·s(SI)
1
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.:1
23
Avogadro number N = 6.022x 10 moi-' (SI)
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(Avogadro constant)
Boltzmann constant k =l.3807xl0-16 ergdeg- 1 molecule-'
= 1.3807 X 10-23 JK-l (SI)
Gas constant R = 8.314x 107 ergdeg-1 mole-1
= 8.314JK- 1mor1 (SI)

-
=0.0821L·atmdeg-1 mole-1
=l.987caldeg- 1 mole-1
Molar volume ofgas at NTP ;V HAND MADE NOTES
== 22.414Lmole- www.ChemistryABC.com
1
=22.414xl0-3 m3mole-1 (SI)

Velocity oflight c ·= 2.998xl0 10 cms-:
'
· =2;9~f~xlQ~m(l(~J) "
Electron mass me =9, 109 X 10-28 g, · _1
31
_ = 9.109xl0- kg(SI)
Electron charge e = 4.80,,3 X 1o-lO eSU.
=r6p2xl0- 19 coulomb(SI).
Faraday constant F = 96485 coulomb eq-1
_96484 coulomp mor 1 (SI)
/
/
2
Acceleration due to gravity g = 980.66$ cms-
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./ J
9.80665 m s-2{SI)
1 ·.
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Atomic mass unit amu l.66xl0- kg(SI)
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.,,"""""'5 _____ ,,. -- --~---.~-.---..,,---
Function:
Function is a rule that relates two or more variables. .
If z =3x + y, where zisafunctionof x and y. Inidealgasequa,tionP isafuncti~nofTandV, Tisa function
of P and V and Vis a function ofT and P.
Operator;.
An oper~tor is a symbol for a rule for transforming a given ma,thematical function into another function. It has no
physical meaning ifwritten alone. It is represented by cap (A).
Example:
(i) !
,
is an operator which transfunns a function into its first derivative ~iylfrespect to x.
I
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d
(ii) dx transforms the ~ction sin x into the function cos x.
. f '
In generaL if A denotes an operator which transforms the function J(x) into the function g (x), then we
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write Af (x) = g(x). ·
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B
Example:
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(a)Let A be dxd and /(x)=ax2 ;then 1/(x)=. dx
d'(ax2 )2axie. g(x)=2ax .
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(b) Let A. be a (i.e. multiplication by a) and / ( x) = x2 +c, then Af (x) =a.(x 2 +c) = at£+ ac 1.e.
is
g(x) = ax2 +ac.

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Differentiation:
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In calculus (a branch of mathematics) the derivative is a measure of how a function changes as its input
_changes. Derivative can be thought of as how much one quantity is changing'inresponse to changes insome
other quantity; for example, the derivative of the position ofa vehicle with respect to time is the velocity at
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which the vehicle is traveling. The process offinding a derivative is called differentiation. The reverse process
is called antidifferentiation. The fundamental theorem ofcalculus states that antidifferentiation is the same as
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integration.
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Differentiation is a method to compute the rate at which a dependent out;put y changes with respect to the
change in the independent input x. This rate of change is called the derivative of y with respect to x. The
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dependence ofyuponxmeansthatyis a function ofx. This functional relationship is oftendenotedy= f(x),

where f denotes the function. If infinitesimal change inx is denoted by dx, and the derivative ofy with respect
. . dy
to x ts wntten as dx .
In mathematics, a partial derivative ofa function ofseveral variables is its derivative with respect to one of
those variables, with the others held constant as opposed to the total derivative.
For a function ofseveral variables, for example f(x,y,z), the partial derivatives are simply usual derivative in
one of the variables when you keep the other variables fixed. For example, take
f =3xy+yz2
In this example f is a function ofx,y and z and we can calculate the value ofpartial derivative in all form
Total derivative
QUZZES
is just the usual derivative (with respect to x) ofthe function
MORE NOTES ON
f, It can be expressed in terms of
www.ChemistryABC.com
partial differentiation as
df =(!) dx+(of)
y,z f}y x,z
dy+(!). dz .
. x,y .
Max-Born thermodynamics Square: Given by Gennan chemist Max-Born with the help ofthis square we
find out nearly all thermodynamics equation related withtherm9dynarnic potential connecting internalenergy
(U), enthalpy (H), Hehnholtz free energy (A) and Gibbs free' energy (G), with relevant parameters sudi as
entropy, pressure, temperature and volume, may be put as ·
:[SJ:'
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V·' A' ·· T
1
(where, positive sign because Sand. Pat tail ofarrow) .
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dH=TdS+VdP . - . '
Similarly, dG=VdP-SdT (where, negative sigti because T at head ofarrow)
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dA = -PdV - SdT (where, negative sign because V and ::J,tat head ofarrow)
B
dU =TdS-PdV (where, negative sign because V\tt he~ ofarrow)
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These equations are known as Maxwell f!quation.
Maxwell's relationship : From the various expansions
tr
IfV is constant so that dV: is zero, then,

is
(au) =T ... (1)

m
as v
If S constant /so that dS is zero,

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i
.
(au)
av
=-P ... (2)
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1 . s ·
Differentiating equation (1) with respect to V keeping S constant and differentiationg equation (2) with
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respect to S keeeping V constant, we get ·
(ar)
w
2
au ... (3)
(as)(av) = av s
w
a u (aP)
2
And ... (4)

(av)(as) = as v
It follows from equation (3) and( 4) that
(:n =-(:1 ... (5)
Following the same mathematical procedure as above, the following expressions can be easily derived.
(:a=-(:a, HAND MADE NOTES

... (6)
as 1· (aP1
r av L= ar L ... (7)
QUZZES(!!1 =-(:;), MORE NOTES ON ... www.ChemistryABC.com
(8)
Equation (5) and (8) are known as Maxwell's relationships: ....

Another set ofM-axwell's relationships
(:l =-(!;), ... (9)
(!~), =-(: ), ... (10)
(!~t =-(!~t ... (fl)

?
(::), ~-(:n
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. ·!
\ / /
. ... ('12)
1
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PROBLEMS
C
=T-P(:;),
.,
,,
B
t
Showfuat ( : ) ,
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oln. From the combined form of the first and the second laws ofthermodynamics, we know that in a system
undergoing expansion,
tr
dU;; TdS- PdV

Dividing by dS, keeping Pconstant, we have
is
( au) =T-P(av)
m
asp asp
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Since, (:ar(:;), [From Maxwell's relation]

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·(au)· =T-P(ar)
w
as p aP s
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Using the combined form of the first and second law of thermodynamics and appropriate Maxwell rela-
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tions, derive the two thermodynamics equations of state.

oln. dU=TdS-PdV :
Dividing by dV, keeping T constant, we have
( au) =T(as) 1-P

av r av r
Since, (!:t=(:;t [From Maxwell's relation]
(au) =T(aP) -P
av r ar v
HAND
This is the first thermodynamic equation of MADE NOTES
state. www.ChemistryABC.com
Again, dH =TdS+VdP
Dividing by dP, -keeping T constant, we have
( aH)
aP r
··=r(as) +v
aP r,
Since, (aPas) · ·(av)
r =-ar I'
[From Maxwell's relation]
(a.I}{) =v-r(av)
aP r . . ar p
This is the second thermodynamics equation of state.

'
.)
Indicate which one ofthe following relations is NOT correc~. [NET June 2012]
av =(ap)
-(BT) (b)-(:), =(~)! .·
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(a)
sas v
i '
-(as)
av =-(aP) -(as) (av)
.c
(c) (d) =
T BT v . o.P
T BT p
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Jin. The correct option is (b)
B
The Maxwell's relationship derived :from the equation dG = VdP-SdT is
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The correct option is (c)

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Cyclic Rule :
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The triple product rule, known variously as the cyclic chain rule, cyclic relation, or Euler~ chain rule, is
a formula which relates partial derivatives of three interdependent variables. The rule finds application in
thermodynamics.
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The triple product rule for such interrelated variables x, y, and z

comes :from using a reciprocity relation is given by
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Cyclic Rule in the form of P, V & T:

Let us consider one function and any two variables then we proof that cyclic rule
P= f(T,V), we have
dP=(aP). dT+(aP) dV
ar v av r
Since, P is constant, so dP = 0
(:a (dl')P +(!;), (dV)P =0
dividing by ( av )p' we have . HAND MADE NOTES www.ChemistryABC.com

BP)
QUZZES
( oTvBVp(of)·
+(BP) -O
BVT.
MORE NOTES ON www.ChemistryABC.com
Or, of (of)·
( OP) v BV p =- (BP)
BV T ·'
Or, BP) (of)

( ofvBVp'c)PT+l=O
·
(oV)
Or,
oP) (of)
(ofv (BY) __1
fJVp fJP T
Thermodynamic Coefficient:
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(i)-l~bar.icthermalexpansion coefficient (a);
The degree ofexpansion divided by the change in temperature at C\i)nstant pressure is called the material's
coefficient ofthermal_expansion and generally varies with temperature · ·
.c
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(ii) Isothermal Compressibility factor (f:J):
B
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In thermodynamics and fluid mechanics;compressibllityis a measureoftherelativevolume change ofa fluid
or so lid as.a response to a pressure (or mean stress) change at constant temperature.
tr
~ _ _!_(fJV)
is
V fJP T
Note: most textbooks use the notation K for this quantity.
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where, Vis volurrie andP is pressure

The above statement is incomplete, because for any object or system the magnitude ofthe compressibility
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. . . .
depends strongly on whether the process is adiabatic or isothermal. Accor-dinglyisothermal compressibility is
defined as
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A = __!_(fJV)
t-'T V fJP
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where the subscript T indicates that the partial differential is to be taken at constant temperature
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Adiabatic compressibility is defined:

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~. =- ~(:),
(iii) Isochoric Thermal Pressure Expansion Coefficeint (r):

The relative change in pressure with respect changing temperature il,t constant volume known as isochoric
thermal pressure expansion coefficient. · ·
r=!(:l
QUZZES (PROBLEMS)
I. Using the cyclic rule, show that the thermal coefficients a. and p are related as a. =J3 · yP. Where y is the
pressure coefficient.
Solo. B y ~ a= ~(~a;p ~(~a and !(!;),

=- r=
··(av) -av· (aP)· ___ (a~)/ 1 =-1 ... (i)

Hence, aT p- ' av /JV
T -aT yP and V
'
.)
. .
Fromthecyclicrule,
(av)
aT p(aT)
ap V (aP)
av T -__1 . ...(ii)
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Substituting equation, (i) in equatio; (ii), we have
.c
C
a
B
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Or, -=1
/JyP
Or, a =/JrP
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is
2. Show that for a gaseous substance, ( : ; ). = ;

m
(!H!~H!l
he
Som. Fromthecyclicrule, =-I

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We have,
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Multiplying the numerator and the denominator on the RHS by IN, we obtain
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aP) =_(11 Y)( av IaT)p = a

(aT v (11 v)( av IaP)r /1
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3. Show that fora gaseous substance, (;) = ( V h ~

Som. Wehaveshownabovethat ( ; )=(!l
Fora vanderwaals gas, (P+ a I v 2 )(v -b) = RT
RT a
Or, P=---2
V-b V
.· Show that (
Let .
!;l ~ -{
QUZZES :a
V = f (T,P)
av=(BvJaT+(Bv) dP ... (i)

BT BP T
Since, bydefinition, a =.1V (BVJ . (BVJ

BT P' hence BT P =aV ... (ii)
Also,since
__ _!_(BvJ
p - V aP T'
(BvJ __~~·
hence BP T - ... (m) '
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Substituting equation (ii) and (iii) in equation (i), we have .·
dV = a dT.;_; Pndrp
dV =a VdT - {JVdP or V
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Or, dlnV=adT-PdP
Since V is a state function, therefore, using the Euler reciprocal relation, we get
11
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tr
State Function and Exact differentials :

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In thermodynamics, a state function, state quantity, or aJunction ofstate, is a property of a system that
m
depends only on the current state of the system, not on the way in which the system acquired that state. For
example, internal energy, enthalpy and entropy are state quantities .In contrast, mechanical work and heat are
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process quantities and state functions gives exact differentiables. As E is a state function and dE is an exact
differential
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e.g. dE = 8q + 8w can be exact differential

To prove: State function we use Euler's theorem. This theorem applies on linear equation (Not zig-zag
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equation according to Euler)

Ifz is function ofx and y variable.
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z=z(x,y)
dz=(:), dx+(:J. dy
Jut, M=(:J,, =(:J. N
So, equation change from zig-zag to linear
dz=Mdx+Ndy
Than for z being a state function it must follow this relation and dz would be exactdifferential if

z can be considered as any thermodynamics function E, H, G etc. and x~ y may be the thermodynamic
variables T, P, V etc. are also state functions.
PROBLEM~ :
1. Verify whether dz = xdy + ydx is an exact diflerential or not.
.;
Soln. Here, M ( x, y) = y and N ( x, y) x
Hence, (aM) (aN) .;. I

fy " =1 and ax Y -
.Hence, dz is an exact aifferential.

.
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' . . ."/ :
2. The exact d~erential df ofa state function f ( x, y) among the f~n/wmi is [NET Dec. 2014]
I X
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(a) xdy (b) dx ~dy (c) ydx-xdy (d) -dx--2 dy
y y y
C
I X
Soln. If df= -dx- - 2 dy
B
y y
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· 1 X
Here, M ( x, y) =- and N ( x, y) = --:--2
y y
tr
(a:). J, 1
is
Hence, = - :, and ( : y2
m
:1 =(:),
he
ThJs, (
Hence, df is an exact differential

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Essential Criteria of a state function :

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The essential criteria ofa state function are as follows :

• The change in the value ofa state function depends only the initial and fj:i;ia1 states and not on the path ofth1
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process carried out in going ii-om initial state to final state.

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• The cyclic integration involving a state function is zero

• The state function has an exact differential, i.e. if P = f ( T, V) is a state function theri
0 0
dP=( PJ
ar v
ar+( av
PJ dV
r
with the condition that

.·Iii!!!!!!!!!!!!
• All thermodynamics properties satisfy the requirements of state functiomA few ofthem are,,,'.. ·..
!iUQUZZES
=q + w MORE NOTES ON
c~e·in thermodynamic~reriergy www.ChemistryABC.com
.:;;~~;,:
S = qrev entropy
T
H=U+PV enthalpy
G=H-TS Gibb's free energy
A=U-TS Helmohltz free energy
chemical potential
·,\
. . .
(PROBLEMS).
om
Verify whether dz = (5 lx y + 47 y ) dx + (17 x + l 88xy3) czy is an exa'fdnre;ential or not.
2 4 3
n. Here, M(x,yr=5Ix 2y'+47y4 and N(x,y) = 17x3 + 188xy3 f
.c
Hence,
·(aM) . +188y and ·(oN)
ay =5lx 2 · + 188y
3
ox =5lx 2 ··· :s
C
x Y
B
yA
Hence, dz is an exact differential.
tr
Writing Vas a function ofTandP, show that for an ideal gas, dVis an exact difrerential.,
is
n. V = f(T,P)
m
dV =(av) dT +(av) dP ... (i) .

he
aT p aP r
For an ideal gas, PV = RT so that V = RT IP. Hence,
.C
(:a=: aoo(:a =:(9, and(:a =;; ... (ii)

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/ Substituting equation (ii) in equation we get

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dV=(:)dT (;)ell'
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According to the Euler reciprocal relation, dV would be an exact differential if

2
a(R!P)]' =-[a(RT/ P )] .
[ aP · aT ·
T P
R R
Or, if -=
.. -p2 which is true.
System and Surroundings .
A system is that part ofuniverse which is under thermodynamic study and therestofuniverseis caled surroundings.
A system is made up of one or more than one phase and is called homogenous and heterogenous systems
respectively.
A phase is defin~d as homogenous, physically djstinct and mechanically seperable,portion of system
·..::-=......,=-------__;------------
·-------=
'.l~.'!~'!_J ----~......,_...,;__ _ _ _ _ _
Types of system:
QUZZES
Depending
:
MORE
upon nature ofboundary between NOTES
system
.
ONsurroundmgs: .www.ChemistryABC.com
and .
(a) Isolated system: If the boundary prevents any interaction with the surroundings, the system is called
isolated system
OR .
A system is said to be isolated when it can neither exchange energy nor matter with its surroundings..
e.g. - Boiling water, A cup oftea in thermus flask.
(b) Open system: Iftµe matter and energy both can pass apross the boundary, the system is called open
system
e.g. Tea kept in cup.
(c) Closed system: If matter can't exchange but energy exchange it is called closep system,
1
e.g. - Tea kept in kettle. · ,:
E:ttensive Property or variable: That properties which depend Oij fu.ass or size are known as extensive
om
property. e.g. volume, energy, heat capacity enthalpy, entropy, free Jriergy, length, mass etc. .
Intensive Property or variable :,That properties which does Jl,Ot depend on mass or size are knonw as
intensive property e.g. temperature, pressure, concentration, density, dipole moment, viscosity, smface tension,
.c
molar volume, dielectric constant, colour etc.
C
Thermodynamic process: , /t
It is the path or operation by which a system changes from one state to anothJr.
1.
B
Isothermal process: Temperature of system remains constant during each step.
yA
T =constant. => dT = O
Details:
tr
(i) For such change system should be contained in a perfectly co ducting container.
(ii) Perfect isothermal change is impossible but when a change is carried out very slowly approximate isothermal
is
change occurs.
(iii) It follows Boyle's law.
m
(iv) Work done in isothermal process is graphically given by area under P-V curve.
(v) Llli=nCp.8T and 8E=nCy,8T
he
In isothermal process 8 T = O
.C
8H = 0 and 8E = 0
(vi) Specific heat at constant Tis infinitely great in isothermal process.
w
2. Adiabatic process: There is no heat exchange between system and surroundings.

q =constant.
w
dq=O
Perfectly adiabatic change is impossible but when a process is carried out very rapidly fairly approximab
w
adiabatic change occurs.

3. Isobaric process: It is the process in which pressure ofthe system remains constant during each step.
P = constant ·
dP 0
4. Polytropic process: In this process heat capacity of the body remains constant.
· CP constant and Cv = constant
dCP = 0 and dCv = 0
5. Quasistatic process: The processs in which the deviation from thermodynamic eguihbrium is infinitesimal an
all the states through which the system passes can be considered as equilibrium states.
6. Isochoric process: When these is no change in the volume ofthe system during various operations the chang
is said to be isochoric.
V = constant.
=> dV=O
. ·~- ;-) {,
Cyclic process: The process which brings back a system to its qriginalstate afterta;s.efies ofchanges is called
a cyclic process.
QUZZES MORE NOTES ON ·
As the E andH depends only on their state.
: . · E and H are constant
So, dE =0 and dH =0 . ;
Reversible process: A thermodynamically reve,rsible process is one in which all changes occuring in any part
ofthe process are exactly reversed when it is carried out in opposite direction.
Characteristics of reversible process:
It has to be carried out in infmitesimal amounts and hence require infinite time. Therefore changes in this process
are very slow. : .
At all time, during the process, the driving force for the change is opposed by a rpstraining force infinitesimally
smaller than the driving force. ' '. r ' / ~ i .•
It can be reversed by infinitesimal increase in opposing fo~ce. · . .. . (. . . .
om
1
.· The wor~ produced in a reversible pr~cess is maximum.···· .> · . ,\ // .. [ . . ; • . •• ,
Irreversible process: The process which occurs suddenly or spontaneously without the restnct1on ofoccunng
in successive stages ofinfinitesimal quantities. ··· · f· ; ·
.c
For such process driving force is very nmch greater than opposing force; so the~ can't be reversed.
These are also called spontaneous process.
C
1
Characteristics of irreversible proc.ess: \ 1!
B
Ifthe changes be reversed, the work in the forward direction and in the backward process would be tmequal.
yA
Ifinitial and final stages be specified, the internal energy change would always be same, whether the process
has been affected reversibly or irreversibly. · · · ·
tr
In an irreversible process, since workterms(dw) in two opposite directions are uriequa,4 the heattransfers
(dq) would also be unequal
is
All natural processes are irreversible.

e.g. => (i) flow ofheat from high temperature to low temperature;
m
(ii) Exapansion ofgas from higher pressure to lower pressure..

he
Internal Energy:
Inernal energy is the energy associated with a system by virtue ofits molecular consitution and motion of its
molecules.
.C
It depends upon internal potential energy and internal kinetic energy.

ETota1 = Internal potential+ Internal kinetic enegy
w
I ETotal =Et +Er+ Ev + Eb +Ee+ Eo + VT + Enuc I

w
where E1 TranslationalK.E. • Ee =Electronic energy·

Er Rotational KE. E0 = Zero point energy
w
Ev= Vibrational K.E. ·Enuc = Energy due to interaction between nuclei.

Eb= Bond energy VT= Intermolecular interaction energy.
=> It is an extensive property ( E and t:\E both).
=> Internal energy is a state function.
dE = E2 El =Efinal - Einitial
It follows Euler's theorem ofExactness.

Sign of AE:
If E1 > E2 :. Lill = -ve
i.e. energy will be absorbed by the sytem from its surroundings.

· Hence reaction will be exothennic.
=> For cyclic process HAND MADE NOTES www.ChemistryABC.com
LLill = 0; cf dE =0
t
p
Work: v~ , .·
Any quantity that flows across the boundary ofa system during a change ipits state and is completely convertibl«
into the lifting ofa weight in the surroundings. . . ./
om
1. Pressure - Volume work or work of expan;ion of gas:' ( ·
W = -[(Fe"~ xdistance through wliich piston moves ( LlR)] = +[Fb.1 xAreax(t.d)J
.c
I W=-Pext.AV I [;. Areaxllf =LlV]
C
Or I W =-PeKt (V2 -VI) I
(a) for expansion V2 > Yi , thus W = negative
B
yA
(b) for compression V1 > V2 , thus W = positive
tr
2. Isothermal reversible· expansion work of ideal gas

is
m
he
.C
Expansion will takeplaceonlywhen

w
pint> pext
Where, Pint= Pressure exerted by gas (Internal pressure)
w
Pext= External pressure or atmospheric pressure

w
v,
Work done, W= -JdW= ·JPdV
V,
v,
W= JPdV (w=-PdV) ... (I)
V,
For an ideal gas; we know that

PV=nRT
nRT
P=v
Putting this value of Pin equation (1 ), we get
W= r 2
I
nRT
--dV => W=-nRT
V
('
IV
dV MADE NOTES
HAND · (v/l
v. · www.ChemistryABC.com
-=> W=-nRT[lnVJv21 =-nRTlnl-v.)
1
W= -2.303nRTlog10 ( ~ )
Since V2 > V1 (due to expansion)
V2 (V2)
-v1 > 1 => log 10 -V1 > l ,· W< 0
i.e wisothemal < 0

i.e. Work is done by the system
Isothermal irreversible expansion work of an ideal gas ' f

(i) Free expansion (In vacuum): . . '· .· ., · ./
om
When an ideal gas is expanded isothermally in vacuum from a v0Iu19i\r; Jb V2 irreversibly, Then
Work done =-_Pext (V2-:V 1) 0 ·
since in vacuum there isno external pressure on the gas V'
.c
(ii) When the ideal gas is expanded isothermally from a volume V1 to V2 irreversible upto a constant
external pressure Pext and if Pext < Pgas' then
C
!Work done =-Pex1 (V2 -V1)j
Reversible adiabatic expansion of an ideal gas:

B
yA
Starting with the proofof TV1 - 1 = constant [Geo.:.chemist exam 2015]
tr
Proof: From the 1st law ofthermodynamics, we know that,

- dQ=dU+dW ... (1)
is
But for an adiabatic process, Q is constant i.e dQ 0

m
O=dU+dW
dU+dW=O
he
dU + dW:;;Q
Sine, dU :;; nCvdT and dW = PdV
.C
=> nCvdT+PdV 0 ... (2)

Foronemoln= l
w
=> CV dT + PdV = 0
=> CvdT+ PdV 0 ... (3)
w
Now for one mole ofan ideal gas, we know that

vRT
w
PV=RT => P=
Now putting this value of Pin equation (3), we get
RT) dT dV
CvdT+ ( V dV=O =>Cv T+Rv 0
dT dV
=> Cv +(Cp-Cv)-=O [:. R=CP-Cv]
T V
dT (Cp )dV dT dV
T+ Cv -l v= O => T+(y-l)v= 0
Integrating both sides, we get,
J dT
T+(1-1)
JdV
v =o
=> lnT + (y-1) In V=ln C => In T + In y(l-i)= lnC
=> 1n Tvh-1) Inc
Taking antilog, we get TV( r.:_1} =C
i.e. ITV(y-1) = constant I ... (4)
Now, for 1 mol of an ideal gas, we know that,
PV= RT => T= PV
R
Putting this value ofT in (4) , we get
·,,)
( -PV) y<r-I) =constant
R
[ :. R is also a constant] . , .
om
I => !?V 7
=constant I ... (5)
/
.c
Again, for one mol of ari ideal gas, we have
C
RT
PV RT => V= p
Now putting this value of Vin equation (4). we get
B
·A¥r
yA
=constant
tr
I Trp(I-r) == constant j
is
m
Problem. A given mass ofa gas at 0°C is compressed reversiblly and adiabatically to a pressure 20 times
initial value what is the final temp ofthe gas. [Given: y= 1.42]
he
(a) 66.21 K (b) 77.21 K (c) 60.21 K (d) 70.21 K

Soln. We know that,
.C
TY p(l-y) = Constant
=> TY p(l-y) = TY p(J-y)
w
I I 2 2
w
=> ... (1)

w
Here,
T 1 = 0°C 273 K, T2 = ?, P1 = P 1, P2 20P 1 & y = 1.42
Putting all there values in Equation (1 ), we get
42 42 42
273
( -T2
l!. (20P
- -1 ll-!. => (273
- lt. = (20t04.2
P1 T2
. (273\
Taking log, we get, l.42 log 10 T J =-0.42 logwC20) l 2
=> l .42 log 10 (273) l .42 log 10 T2 = - 0.42 log 10(2x 10)
=> l.42 log 10 T2 = l.42 log 10 273 + 0.42 [~0 2+!~0 10]
=> lq_\/1; = 4.00577/1.42 = 2.820
Taking antilog, we get
T2= 102 820
QUZZES
· = 66.21 K MORE NOTES ON www.ChemistryABC.com
Correct option is (a)

Work done in reversible adiabatic expansion for an ideal gas: ·
,/
fdw - ( PdV
2
.. : (i)
=> w=- ( PdV. .•. (ii)

l
Now we know that for adiabatic process, PV1 =constaJtt

PV1 =K
Also P1Vl = P2V/. = K
om
Now puttingthisvalueofPmequation{ii), we get
' .
t' Kv-, dV => w~ -K £' v-,dV => w ~ -K[~-r+f]v

2
.c
w - ;.
1 ,y+Lv ...
I
C
= -K [vi-r
. 2 _,.,14']
B
W
=> 1-y
Nl ·
yA
W-
- _:!__[Kv
(1-y) 2
1
-r -KvI1-rJ
tr
l n v.rv,t-y -,cl'
nv.rul-y]
is
c·
W := . - - C:i · ·2. 2 y'¥1 . [Using (iii)]
=> (y-1)
m
... (iv) .
he
=>
for
Now an ideal gas, we know that, PV = nRT
.C
=> P1V1 nRT 1 ... (v)

And P2V2 =nRT2 ••• (vi)
w
Equation, (vi)-(v), we get

=> P2V2 P1V2 = nR(T2 -T 1) •
w
Now making this substitution in equation (iv), we get

w
1
w~--nR(T2 -T1)
(y-1)
Work done in irreversible adiabatic expansion for an ideal gas:

,(i) Free expansion (in vacuum):
When an ideal gas is expanded adiabatically (in vacuum) from a volume V1 to V2 irreversibly, Then
workdone = Pext(V2-V 1)=0
Since in vacuum there is no external pressure on the gas
(ii) When the ideal gas is expanded adiabatically from a volume V1to V2 irreversibly upto a constant external
pressure pext and if Pext< pgas' Then HAND MADE NOTES www.ChemistryABC.com
Work done= - Pext (V2-V 1) ••• (0
Now :from the 1st low ofthermodynamics, we know that ·
dQ=dU+dW
O= dU + dW [ :. For adiabatic Process Q = constant]
=> O=nCvdT+PdV => O=nCy(T2 -T 1)+P(V2 -V 1)
.·· ·nC
=> -P(V2 -V 1)=nCy(T2 -T 1) => (V2 -V 1)=- / (T2 -T 1)
Now making this substitution in equation(i), we get
p
Work done ( w) = nCv _EL ( I;_ - 1'i) [Where, P is presslilre ofthe gas]
p ' , \' ,}
(iii) We know that, work done in reversible l;ldiabatic expaps~ti'(w) is given as:
om
nR. (T -T) n(c;-cv1(T i.T) . .
w=
~
(y- l) (T2 -T 1) =
(Cc: -1
J 2 1 (C -CJJ .
Cv .
2 1 nCvR . _
= ~ (T2 -T 1) - nCvdT
.
.c
P
Since, W =dU (Change in internal energy)
C
IdU=nCvdI' I
B
So,
i.e. in reversible adiabatic expansion, the work done is equal to the change in internal energy.
yA
[PROBLEMS)
tr
1. Prove that from cycle rule for P Vm = RT

is
Soln. Differentiating the given equation, P Vm = RT , we have

m
. PdVm + VmdP = RdT

Dividing the equation by dT and introducing the condition ofconstant volume, we get
he
PdVm + VmdP = RdT

.C
Dividing the equation by dT and introducing the condition ofconstant volume, we get
l.. :.
w
V. ( : ),. ;R i.e. ( :
w
Similarly, we have
w
Now substituting these in the cyclic rule of ( :; H::H:;), + I ; 0 , we get
(8P)
8T
·(~]
avm
v.,
(avm) +l=(!ll(P)(-Vm)+1=-l+l=O
8P p vm R P r
2. Test the cyclic rule for
.(P+ ~ )rv.);RT HAND MADE NOTES www.ChemistryABC.com

equation, (PV. +:. )= RT and
,In. From givenQUZZES thenNOTES
MORE diflerentialin)l,
ON we have.www.ChemistryABC.com
a·
PdVm+ VmdP- 2
dVm =RdT
Vm
Dividing by dT introducing the condition ofconstant volume, i.e. dVm =0 , we get
. v.(:a =R
,.:. (:a. =RdT
Similarly, we have
a;;1
.i
(:;. ) = P-;
v; =
and (
t
, P-a/Vm2
om
Substituting these in the cyclic rule, we get
(
ap
)
(j!__)· avm +l= ·�� P->Cl !V,; ·
( ) ( ( ·) (
vm
)+1=-1+1=0
(
.c
BT vm avm p BP T ' v m) R P-a/V,;
C
•,
Show that the pressure is a state function for a gas obeying

\,
B
yA
(P+� Jrv.) = RT
tr
RT_!!_
)In. Given that, P=
is
Vm Vm
2
m
Therefore, (-
0Vm
ap) =-
RT 2a
-.2+-
Vm Vm
;
(:;t R
=-
Vm
he
3
T
a2 P R aip
=- 2 and =--2
.C
R
aT avm vm avmar vm
w
a2p a2 P
Hence, =
aravm o VmoT
w
Therefore, dP is an exact differential and P is a state function.

w
From the following thermodynamics relation,

G = H-TS; H = u+P V; dqrev =TdS; dU = dqrev - PdV
Showthat,(:), =-S; (:), =V and(:a =-( ),

!
>In. We start with,
G = H-TS
Since, H=U +PV
Therefore, G = U+PV -TS
Differential ofthis expression is, HAND MADE NOTES www.ChemistryABC.com
. dG=dU+PdV+VdP-TdS- SdT
TdS =dq,ev
Now, QUZZES =dU + PdV, MORE NOTES ON www.ChemistryABC.com
Therefore, dG =VdP- SdT ... (i)

Thus, we establish that G is a function ofT and P. Moreover since G is a state function, therefore, we have
dG=(aG) dP+(aG) dT ... (ii)

. aP aT p
.T
Comparing equation (i) and (ii), we get
(aG)
ap
=V and (aG) =-S
r ar p
Applying Euler's reciprocity relation to equation (i), we get
( av) (as)
om
ar p =- aP r
5. Considering U as a function ofanytwo ofthe variables P, V and?l~ prove that
.c
C
B
Soln. (i) Given that, U = f ( T, P) we get
yA
du=(au) dr+(au) dP
ar p ap r
tr
Dividing by dP and introducing the conditiion of constant V, we have

is
(~~l =(~~u:a +(~a

m
(~~t-(~~l =(~~u:~1
he
Or, ... (i)
Now, U= f(P,V), we get

.C
w
du =(au) dP+(au) dV
aP v av p
w
Divindg by dP and introducing t;he condition of constant T, we have
(~~l =(~a +(:~u~:1

w
Or, (~~l-(!~l =(~~H~a

Comparing equation (i) and (ii), we have the required relation,
... (ii)
(ii) Since U = f(T,V), we get
du =(au) dr+(au) dVHAND MADE NOTES www.ChemistryABC.com
ar v av r
Diving by dP and introducing the condition ofconsta.nt temperature.( dT =0) , we 4ave
(!�l �(:�u:;t
1\vo moles ofa monoatomic gas expand adiabatically in a reversiple mannefon heating 27°C to 327°C.Then
calculate total change in internal energy ofthe entire process.
n: We know that, chnage in internal energy (dU)is given as
dU=nCvdT
� dU=nCv(T2 -T 1 ) ••• (i)
Here, dU= change in internal energy and n =2 moles
For monoatomic gas, Cv= R f

Given, T 1= 27°C =(27 + 273) K = 300 K
om
T2= 327°C =(327 + 273) K 600 K
Now putting all these values in equation(i), we get change in internal e �ergy
.c
� dU= 2molxixR(600-300)K=3 x2 x �x300 mol K
2 k.mol
dU= 1800 cal
C
�
\•.,
B
The temperature of54g ofwater is raised from l 5 C to 75 C at constant pressure. The change in the enthalpy
° °
ofthe system (given that Cp.m of water 75 JK.- 1 mor')is · [GATE 2007]
yA
(a)4.5kJ (b)l3.5kJ (b)9.0kJ (d)I8.0kJ
n. Since, weknow that,
tr
, dH=nCpdT
is
wt. in gm
Since number ofmole (n) = gm
m
molecular wt.x-
mole
he
54
n=-=3 mole
18
.C
And Ti= 273+ 15 =288K and r;_ =23+ 75 =348K

w
dT =T;_-fi =348-288 =60K

w
dH =3x75JK- 1 moZ- 1 x60K =13500J =13.5kl

Correct option is (b)
w

( PRACTICE SET)
I. What is a cyclic rule ? Derive this rule for a function Z = j ( x, y)
2. Show that the volume ofthe following equations is a state function: 1 '
(i) Ideal gas equation
(ii) Real gas at low pressure for which van der waals equati6n reduces to ( P + ;; ) (V) nRT
3. An arbitraryvariable tfJ is found to have the following relationship

T
dt/J = R -ap - RdT
p i
om
Are the following statements correct? Justify your answers '/
I
(a) tfJ is a state function
(
(b) dtfJ is an exact differential.
.c
(c)4d¢i*O
C
(d) It is possible to wrie down tfJ explicitly in terms ofT andP as the intieperidentvari ables, i.e. tfJ = f (T,P
4. From cyclic rule,
B
HZ)J:),
yA
show that, (: =-l [GATE1991]
tr
5. For an ideal gas [GATE2010]

is
aP of
( ) ( ) ( ) 0 av _
av8T) (av)
m
=_1
(a) oT V av p ap T -
(b) ( aP ) (
BT v aPp T
he
aP BT
( ) ( ) ( ) +I av -
(c) of V av p ap T - (d) (:a(:). (!a +2
.C
6. Among W ( work), Q (heat), U(intemal energy) and S(entropy) [GATE2010]

w
(a) Wand Uare path functions but Q and Sare state functions.
(b) Wand Sare path functions but Q and Uare state functions.
w
(c) Sand Uare path functions but Q and Ware state functions.
(d) Wand Q are path functions but Uand Sare state functions.
w
7. Match the following: [GATE2004)
p{:a I.A
au)
Q.(av II.-S
s
R_(:\ III.T
s.(�). HAND MADE NOTES

IV.-P
V. H
VI.V
(a) P-III Q-IV R-VI S-II (b) P-III Q-I R-Ii S-V
(c) P-I , .Q-III R-V S-II (d) P-IV Q-III R-VI S-V
Indicate which one ofthe following relations is NOT correct. .: '· [NET June 2012]
(a)-(!), =(:l
(c) -(as)
av =-(aP)
ofT v
From the following thermodynamic relations

A=U-TS; dq=TdS; dU=dq-pdV
om
Showthat
(i)
aA)r =-P
(av (ii)
(,aA)
(ar v =-S
., .
(iii)
·(aP) (as)
fir v = av r '
.
.c
The Maxwell's relationship derived from the equation dG =VdP-SdT is [NET Dec. 2013]
C
(av)
of =(as)· (avaP ). =(of)
as . Y:(ov)
of =-(as) av ·(of,)
·('BP)·· lf!
<a) ap
p T (b) T aP
p (c p
B T <d) Tas =- p
yA
tr
is
m
he
.C
w
ANSWER KEY
w
Questions 5 • 6 1 8 10
Option (bl (d) (b) (b) (c)
w

[ Chapter�
Zeroth & First law of Thermodynamics

Introduction / ..
The Zeroth Law of Thermodynamics also known as the law {hegnal equilibrium was put forward much�
om
the establishment of the first and second laws of thermodynami&. .c'
in
Two-systems 'in thermal equilibrium with a third system ar¢ also thermal equilibrium with each otlxr
.c
Zeroth Law of Thermodynamics:
Ifbody A is in equilibrium with body C and body B is also in equilibrium w.ith body C, then bodies A and B
C
in equilibrium with each other. In other words, if systems A and B 'are pliced in contact with each other
B
exchange ofheat will take. place. Recording oftemperature ofa system by a thermometer is also basal on
law. When a thermometer is placed in the system, it comes to thermal equilibrium with the latter and 1
yA
records a constant value.
tr
First law of thermodynamics:

It is the law of conversation ofenergy and can be stated in various ways.
is
(i) "The total mass and euc:::rgy of an isolated system remains constant"
(ii) ''Energy can be transformed from one fonn to another form but energy can neither be created nor destroy
m
(iii) It is impossible to construct a perpetual motion machines which could produce work without consur
he
any energy."
..-------,
Mathematically, I�-------'
dU = dq + dw I
.C
State A ---+ State B

w
J J J
B B B
dU= dq+ dw • ::::> U B - UA =q+w
w
j AU=q+w I
A A A
w
Or ... (1)
It may be noted that the values of both q and w depend upon the path followed in going from state A to:
B, but the algebraic sum o fthese two quantities is independent of the path. Thus ( q+w) is state function
At constant volurne
dV=O
W=-Pex dV 0
from equation ( 1)
I AU=q I v
Thus heat absorbed at constant volume increases the internal energy.
At constant temperature
AU=O HAND MADE NOTES www.ChemistryABC.com
q+w =0 ::::> q -w
Thus amount ofheat absorbed bythe system from the surroundings is equal to the work done bythe system on
the surroundings.
QUZZES · MORE NOTES ON ·
:=;, For cyclic process
AU=O
q+w=O
:=;, At constant pressure
q H = enthalpy
·ffi'st law becomes
AU=H+W
:=;, · For adiabatic process
q 0
AU= O+ w :=;,AU= w :=;, AU =-Pext (av)
om
Thus work is equal to the loss in internal energy.
Enthalpy or heat content (H):. · .
Thennal changes occuring at constant pressure is called as enthalpy.'
.c
H=U+PV .... (1)
... (2)
C
:=;, dH = dU + PdV + VdP
where, H is a state function and dH is an exact cµfferential.
B
from equation ( 1)
yA
HI= ul + P1"1 and· H2 = U2 + P.2v.2
AH =H 2 -H1 =(U 2 +P2V2 )-(U 1 +P1V1 ) =(U 2 -U1 )-(P2 V2 -P1V1 )
tr
AH= AU+ P(V2 -Yi) [At constant pressure P2 = P1 = P]

is
j AH=AU +PAV I ... (3)

m
Or I AH= AU+ W I ... (4)

he
st
Froml law
AU=q w
.C
equation (4) becomes.

I
AH = q - w + w ; AH = qp l
w
If AH= +ve, process will be endothermic

w
If AH = -ve, process be ~ill exothermic.

w
Relation between AH & AU:

From first law,
AU =q-w :=;, AU =q PAV
at constant volume
AV=O
Again we know,
AH=qp
Therefore, from equation (3)
AH=AU+PAV
Let n1 moles of gaseous reactant changes into 11i mole of gaseous product::under constant T and P.
PV1 = n1RT; PV2 = n2RT
PV - PV = n_ -n.RT
.._....,..,~~l"-"'lrl 11;;;;;,,,v........,.~,r _ , •••-,.•··--.J••Mlfll~
PLiV. LingRT where, Ang= (mole ofgaseous product) (molesofgaseousreactant)

J m = LiE + LingRT I ... (5)
Application of First Law of Thermodynamics: .
1. Heat capacity: The heat required to raise the temperature by one degree.
It is oftwo types: ·
(a) Specific heat capacity: It is the amount ofheat absorbed by 1 gm ofa substance to raise its temperature
~·~------ C= q
m(T2 -T1 )
/
. ,)
(b) Molar heat capacity: It is the amount ofheat absorbed by 1 mole or~ substance to raise its temperature
by 1°c. , Ii
om
- . q
C- n(T2 Ti) ~ Cxn= (Tz~'fi) V'
.c
C q ~C
dq
Lm
C
M (T2 -Ti) ~
B
From :first law dq dU + PdV
yA
[ C= ctu:;ctv [
tr
(i) At constant volume: dV= 0
~ Cv =(~l ~1 dU=C,dT I
is
Cv =(~~)v
m
Fornmoles I dU = nCvdT I
he
(ii) At constant pressure:

.C
( dq)P ( dH)P
dH=dU+PdV
w
... ~ I dH" C,dT I

w
c, =( ~~),
w
Fornmole ! dH = nCPdT I
( SOLVED PROBLEMS)
1. For the combustion ofl mole ofliquid benzene at 25°C, the heat ofreaction at constant pressure is given by
1
C6H 6 (I)+ 7-0 2(g)~6C0 2 (g)+3H 20( £); LiH =-780980 cal
2
What would be the heat ofreaction at constant volume?
(a)-780 kcal (b) 780 kcal (c)-580 kcal (d) 580 kcal
Soln. We have, Lill= Li£+ LingRT
Here, Ling= 6-7.5 =-LS
Thus, LiE = Lill -LingRT = 780980-(-1.5)x2x 298 =-780090 calories =-780.090 kcal
~·~-· .l - - _ ..... -· . ,.._ ..... __ _...., .._.. ....... _ _.
When 1 mole ofice melts at 0°C and at constant pressure of 1 atm, 1440 'calories ofheat areabsdrbed by the
system. The mo Jar volumes ofice and water are 0.0196
QUZZES MORE NOTES ON
and O.0180 www.ChemistryABC.com and. AE\
litre respectively:Calculati"\{i-r
Since, Mi =qP
= heat absorbed by the system at constant pressure
= 1440 calories.
In the equation, MI = AE + pl\V
pAV = 76x13.6x98I(18 19.6) ergs
76xI3.6x98lxl.6 l .
= .1 x · ca ones
4 8 107
= -0.039 calories
(p= 1 atm 76xl3.6x98I dynes/cm2, V2 = 18 cc, V1 = 19.6 cc)
Since, pAV is very small compared to AE, pt\V can be neglected.
om
Thus, MI = AE = 1440 calories
Show that in an isothermal expansion of ru:i ideal gas (a) AU= O and (bf MI= 0.
(au) ·
.c
We know that for one mo le of anideal gas, Cv = of · v , hence, dU = CvdT
C
)
B
Forafinitechange, AU=CvAT '·
yA
For an isothermal process, Tis constant so that AT = 0 . Hence, AU= O
(b) We know that, MI= AU +A(PV) ·
tr
Since, foranidealgas PV =RT

MI= AU+ A(RT) =AU+ RAT= RAT (·: AU= 0)
is
Since, T is constant, AT = 0 . Hence, MI = 0 .

m
Calculate q, w, AU and L\H for the reversible isothermal expansion of one mole of an ideal gas 27°C from a
he
volume oflO dm3 to a volumeof20dm3•

T = 27°C 27 + 273 300 K
.C
Since the process is isothermaL

AU =Ml 0
w
w =-nRTln[V2 J
v1 .
w
=-(1 mol)(8.314 JK-1moi-1 )(300K)

w
in (20 dm3/10 dm3) = 1729 J

From the first law, AU= q +w
Since, AU= 0, q = -w = l 729J
Calculate q, w, AU and L\H for the isothermal expansion ofone mole of an ideal gas of27°C from a volume
, -0fl Odm3 to a volume of20 dm3, against a constant external pressure ofl atm.
Since, the process is isothermaL AU = MI = 0
But, in this.case, w = P(Vi - V1 ) =-(1 atm)(20 dm 3 -10 dm 3 )
= -10 dm3 atm =-10131
Since, AU= q + w and AU= O, hence q = -w = 10131
A gas expands isothermally against a constant external pressure
HAND MADE NOTES volume oflO dm1 to a volume
ofl atmfromawww.ChemistryABC.com
. of20 dm • In this process it absorbs 8001 ofthermal energy from its surroundings. Calculate the AU for the
3
process in ·oules.
. .
. .
V2 . . . . ; .
3 -10dm3 ) =-10dm3atin
S0.In. w =-JPdV
QUZZES
~
=~P(V2-V1 ) =-(l atm)(20dm
'
=-I0dm3x 8.314JK.-lmorl' =-1013J

0.08206 dm3 K-1mor1
From the first law ofthermodynamics.. ·
~u = q +\\' = sooJ +(-10nJ) =-213J
7. Show that
(QC, -Cv r(!l(:J.

= (it) c,-cv.,,y:r~tgas.
,)
om
.· 2TV:'
. . .a
(fu) C, Cv =R (1+ R~ Jfofi:eal gas (iv) Cp -Cv =1-~-
.c
C
(v)
.p
c -cv =-T(aPJ2(avJ
aT aP ·
B
V T
c, -c;, ~r(%i}(~J.
yA
soln. (Q
tr
Let S = f ( V, T)
is
On complete differential
m
cts = ( as J .dv + ( as J .dT

av T aT v
he
TdS=T(asJ .dV+T(asJ .dT

. avT aTy
.C
WeknowthaS [( r(:)}cv]
w
w
av
TdS=T(asJ .dV+Cy,dT
T .
w
Divide by dT at constant pressure
( TdSJ
dT p
=T(asJ
av T
(avJ
ar p
+C (dTJ
v dT p
c,=r(!H:J. +Cv
c, Cv=r(!l(:J.
From thermodynamic square
(!l =(!l HAND MADE NOTES www.ChemistryABC.com

... QUZZES MORE NOTES ON www.ChemistryABC.com
· Therefore, the above relation is called as standard equation various relations can be derived by using this rue.
(it) . I CP -Cv = R I for ideal gas for ideal gas
From standard equation. PV=RT ·
c,-Cv =T(:H:a P==·RT

. V.
;:d V RT
·an
. =p-
C -C =T(~ RTJ (~ RT)
.
p V 8T V V 8T p p
C -C = TR(oTJ x R(8TJ =TR R =P.x R,
om
p V V 8T V p 8T p . V ·p ' p
I Cp-Cy~R I . f
.c
Fo~nmole I Cp-Cv =nR I Proved.
C
2a J·
(iii) CP -Cv = R ( 1+ RTV for real gas:
From standard equation,
B
yA
cp -cv =T(aPJ
oT .(av)
or ... (1)
tr
V P
Now we put value ofP and V from vander waal's equation
is
(P+ ; 2 )(v-b)=RT
m
... (2)
P-[
he
( P+~J=
V2
RT
(V b);
RT -~]
- (V b) V2
.C
oP) ( a RT J ( a a ) .
(or v = oT(V-b) - orv v
w
2
V U
w
0
w
... (3)
Again differentiation equation (2) w.r.t. temperature at constant pressure

I II
,{!(P+ :, )(v-b)l =(!RT),

=>(P+-;)[~(V-b)] +(V-b)(~P+-;J =R(8T) +T(aR)
V 8T p 8T V p 8T p 8T p
u
' 0

~·--· _ _.. ..... - .. __.. .... _.. --- .. -~·. -~-- ,.-- ..
(. Va)(av)
=>. P+-2 (ab) +(V-b)
Mp BTp
- · . (BP)
- - _+ (-8
-·
Mp BTV p
-;a)
~ =R
. u u
0 0
~ (p<,)(:l +(V-b)[ t:(:)}R

RT
=> (V-b) M P V3
(av) _ (v )(av) =
2a -h
8T P
R
(:). =[ RT -~a(V-b)] \ I ./·/
om
3
.· .:... (4)
_ (V-b) V ....
Putting value ofequation (3) and(4) in equation (1)
.c
R R
C -C = T X - - X = - - - - - - - - =
C
P v. (V-b) [ RT 2a(V. -b)]
V3
B
(V-b)
yA
Cp-C _ RT R ·
v-(V-b)x RT [ _ 2a(V-b)2]
1
tr
(V-b) . V3RT
is
So, V > b, ( V - b )2 ~ V2
m
Cp - Cv = R 2 =R [1 2a Let (V-b= V)
he
1 2aV RTV
V3RT
.C
Proved.
w
w
a?TV ·
(iv) CP -Cv =-~- foridealgas:
w
...
a<(:); ~~-t(:a
C """C
[J_(av) ] xTxV _l
= V \. 8T P
2
= v2 aT
(av) 2
xTxV
(av) (av)
P ··-
P v 1 1
V aP T v. ap T

From cyclic rule.
(:n:u:aQUZZES
=-l . ~ (:u:a MORE NOTES ON
= (~J
8P y
c, =-r(:):-(!). =rf:H:t
-Cv
Thus standard rule proves validity ofgiven relation.
(v) c,-Cv =-r(:H:a

From fundamental rule.
om
'.f
/
/
.c
C
'
\
B
yA
=(oV)
tr
Using Euler's theorem, 1/ ( oP )

av T BP T
is
=T[-(1HHH](!l
m
c,-Cv
he
C -C =-T(oV)-.(oP)2 Proved.
P y 8P 8T
.C
T Y
Calculate CP -Cy for nitrogen at 298K and 100 bar pressure, the van derwaals constant 'a' is 141 dm.6 kPa
w
mot2•
• Wehave,
w
p = 100 bar= 104 kPa

a= 141 dm6 kPa mo1-2
w
For a gas obeying van der Waals equation, we have
Cp - Cy = R ( 1+ R~; 2 P)
Substituting the given values, we get
2(14ldm kPamor )
6 2
J
c, -Cv = ( 8.314 JK_-'mol-1) X ( 1+ (8.314 dm' kPaK-'mor' )' ( 298K )' .
= (8.314 JK-1mor 1 )(1 +0459) =12.13 JK-1mor 1

L.t::oru,.,.-1 ~-:- I(~"' l,,,,,c:;I~ V'I I··-:--· 1 ·--J• •c-11 •.••cl?
~ i·
First Thermodynamic Equation of State · ···~·;

QUZZES
Change in MORE NOTES
internal energy with respect to volume ON www.ChemistryABC.com
at constanttemperature is known as first thermodynamicl '
' ~- .
equation ofstate i.e. ( !~ ),

Dimensionally it was seen that its dimension is the dimension ofpres;ure. Thus it is also called as internal
pressure ( rcT )
From thermodynamic square
dU=TdS-PdV
( au) -T(as) -P(av) u G

avT avT avT
( au) =T(as) -P
om
T
avT avT
Fromthermodynamicsquare
(:a =(:t /
.c
C
B
( au)
av T
=T(aP) -P
8T v
yA
Or ... (1)
tr
is
For ideal gas internal pressure (1tT) is zero.

From equation (l)
m
a RT)
nT =T (- R(oT)
- -P =T- - RT
-P =--P .
=P-P=O ~ nT =0 foridealgas
he
oTVv VoTv V
a
.C
For real gas internal pressure (1tT) is y .:

2
(P<, )tv :-b)=RT P=[( VR~J ;, ]

w
For real gas; =>

w
So, from equation (1), we get.

w
~ =T(! (:~b)l-(! y, l-P =T {V~bi(:l-P

0
= (VR~b) p =(P+ ;2 )-P
I~<, I
Fornrn:oles,

1 & Firetl.aw of Therrnoaynamice
Second Thermodynamic Equation of State
Change in enthalpy w.r. t pressure at consant temperature is knowri as second thermodynamic equation of
state.
dH=TdS+VdP
( OH)
8P r
=T(8S) +v(8P) => (aH) =T(8S), +V..
8P r 8P :r 8P r ~oP r
From thermodynamic square,
(~l =-(:), ·,)
:. (1H ~[-T( ), +v]
om
l
.c
Forideal gas ( : =O
C
(Zl =-T(! ~T), V=-T ~(:), + V=-V + V~ (:t=O
B
+
yA
Prob.,m: Show that fur fixed amount ofideal gi,s ( ~ )r =O
tr
(~l=[!(:U
is
m
Changing the operator and U is a state function.
(~; l=[!(!~)1 8U)

he
but ( o furidealgas
av r
.C
·(acav )
_v =0 proved.
w
Variation of heat of reaction temperature:

w
For low range oftemperature, two relations are

w
.!\Cp 8H2 .!\HI
T2 T1
Consider the equation, aA + bB -t cC + dD

(Where, a, b, c and dare stoichiometric Coefficients)
.!\H DIP DIR
=> 8H (cHC + dHrJ product - ( aHA + bHB)reactant
[ o(L\H)J
8T P
[c(8Hc)
8T
+d(8HD)
8T
P
] , -[a(8HA)
8T
+b(8HB )• ] : .
HAND MADE NOTES
8T ·
P Product P
P Reactant
-.-.- - - -- .., . ----~--·~
.
' .
o(Aff))
QUZZES = [ c Cp(C)+d Cp(D) J·NOTES·-[ONaCp(A)+.b Cp(B)J
MORE ., . .
.( _of p . . Specific heat Capacities ofProduct Product Specific,heat Capacities ofReac!ant Reactant
8(AH)J .[8(AH)] ·· .
[ of . = (CP)product - (CP)Reacmnt; of = ACP;
p p d(AFI) ACP dT
(2 d(AH) =;. f ACpdT ... (i)
Ifat T T 1, AH AH 1 and T = T2 , AH AH2

then from equation (i)
AH2 -AHt= ACP (T2 - T 1)
jAH 2 = AH,+ ACp (T2 -T1)j ... (ii)
om
Similarly, j~ 2 = AE 1+ L\9v (T2 - T1) I ... (m)
Equation (ii) and (iii) are known as 'Kirchoff's equations'.
.c
9. The enthalpy ofreaction AH for the formation ofammonia according to the reaction .
C
N2 + 38i ~2NH3
at 27°C was found to be 91.94 kJ.
B
yA
What willbe the enthalpy ofreaction AH at 50°C?
Molar heat capacities at constant pressure and at 27°C forN 2, H2 and NH3 are28.45, 28.32 & 37.0
tr
Joules/ K respectively,
Soln: From Kirchoff's equation we have
is
AH2= AHi + ACP (T2 -Tl) ... (i)

m
Here T 1= 27°C (27 + 273) K 300K ... (ii)

T2 50°C = (50 + 273) K = 323 K ... (m)
he
AH 1 -91.94 K.J =-91.94 x 10 J 3

... (iv)
ACP= 2CP(NH3) [Cp(N) + 3 cp (H2)] ... (v)
.C
=> ACP= 2 x 37.07 J/K-[28.45 + 3 x 28.32] J/K ... (v)

=> ACP= 39.27 /J/K and AH2=? ... (vi)
w
Now making substitution from the equation (ii), (iii) (iv) and ( v) in equation (i), we get
w
AH2=-9l.94 x 103J +(-39.~7) [323-300)] -91.94 x 103 J-39.27 x 23 J

= 91.94 X 103 J -903.21 J -92843.21 J = -92.843 kJ
w
10. Prove that for reversible process w = 0 1

Soln. Let a gas is expanding from an initial volume ofV1 to a final volume V2•
Thus reversible work
V
w = -2.303RTlog-2
1
vi
Now the gas is compressed from V2 -+ V1
V V
W2 -2.303RTlog-1 =2.303RTlog-.2
v2 v1
Total cyclic work.
4w =WI+ w 2
HAND MADE NOTES
= -:2.303RTlog Vz + 2.303RTlog
v1
v2 = 0
v1
So, gw=O ·
Prove thatFor irreversible process cJ w * 0
. .
ht, Let a gas expanding irreversibly from V1 ~ v2 · against external pressure P2• ,.
· wexp =-P2 ext(V2 -Yi)

cfw=w 1 +w~
=-P2 (V2 -Yi)-P1 (V, -V2 )
=-P2 (V2 - V, )+ Pt (V2 - V,)
C}w =(V2 -V1 )+(P1 -P2 ) V,,:'
/.
w.llT =+ive value
om
So, cf,w:#0
Showthatforanidealgas {
.c
(a) (fJU)
av r
=0 (b) (aH)·
ap r
=0 (c)(!a=O (d) (fJH)
av r
=0
C
(a) Foran idealgas, PV =nRTor P=;nRTJV· ·
B
ln.
yA
aP) =nRIV=PIT
( ar v
tr
· Sub;tituting this value in the first thermodynamics equation ofstate, we have
(!~l =r(:a-P=r(;)~P=O
is
m
'
(b) Foran ideal gas, PV =nRT or V=nRT
--
he
. p
(:a=~=;
.C
w
Substituting this value in the Second thermodynamics equation of state, we have
=V-r( 0i;:) =V-T(v)=o

w
( oH)
aP r ar p r
w
(c)Since, H=U+PV, hence, U=H-PV=H-nRT
Or, ( au)
aP r
=(o(H-nRT)) =(aH) -(a(nRT)) =O-O=O
aP r ap r ap r
' (d) Since, ·H=U+PV=U+nRT
( aH)
av r
=(a(u +nRT)) =(au) +(a(nRT)) =O+O=O
av r av r av r

···--v··-···'"""g"'1
---------------------------------'-----------·,
Thermodynamics changes in isothermal variation in volume of an ideal gas :
- - · ---·· - • _•• - - - - · · _.a .. · · - ·
QUZZES
Expression for q, w, AH and AH: MORE NOTES ON www.ChemistryABC.com
We shall use the first law of thermodynamics to calculate the changes in thermodynamics properties when an
· ideal gas undergoes the process ofexpansion or. compression. ·
In an isothennal expansion or compression process, ~he tempemfure ofthe system remains constant
throughout the expansion qr compression process. Since for an ideal gas, U depends only on temperature
(Joule's law, ( W), ~p ), it follows that
I dU=OorAU=O J .
Substituting the above expression in the first law ofthermodynamics, we get

dq =-dw or q=-w
Hence, in an isothermal expansion or compression process, heat is con:y6rted into work and vice versa Thus,
om
ifheat is supplied to the sytsem (q positive), the equivalent amount bf}Vbrk}s done by the system(w negative),
and ifsome work is done on the system. (w positive), the equivalent1amo~nt of heat is given out (q negative)
The enthalpy change ofthe sytsem is also zero as f
.c
AH =A(U +pV)=AU +A(pV)
C
=AU +A(nRT)=AU +nR(AT) =0+0=0 / /f
The magnitude ofw (or q) depens on how the ·expansion or compression'proctfss is carried out. Two different
B
types ofprocesses may be distinguished, namely, reversible and irreversible.
yA
Reversible Expansion or Compression Process :
The expansion for the work involved in. an isothermal expansion or compression from volume V1 to V2
tr
can be worked out as follows:

Sine, dw= -p PPdV, therefore,
is
V2
f
m
w=- PoppdV
he
In the reversible change ofvolume, the opposite pressure differs from the pressure ofthe gas by an infinitesimal
ammmt,ie.
.C
Popp= Ant± dp
where +ve and -ve signs are meant for compression and expansion processes, respectively.
w
Substituting this is the previous expression, we get

~ ~ ~
w
w= J(Ant ±dp) dV :::!- JAntdV+ JdpdV

w
Vi
The second integration can be neglected, since it is product oftwo infinitesimal small differentialie. For an ideal
gas, P(nt is given by
nRT
An1=V
with this the previous expression becomes
w=-)nRT dV
v.I V
Since temperature remains constant in an isothermal process, we have
dV V:
w = -nRT J-
V2
V .
v.
= -nRTln-1.
Yi HAND MADE NOTES www.ChemistryABC.com
I
In terms ofpressure; we have
w=-nRTin.!i
Pi
Hence, for an isothennal reversible expansion or compression, we have
_;
q=-w=nRTln V2 =nRTln.!i
v; P2
[ PROBLEMS)
: .. - 'i \
WhaUs the work which could be obtained from an irothermal reversibleexpaqsiph of t mol of Cl2 from 1 dm3
to 50 dm3 at 273K using ideal gas behaviour. · , · · ;
om
1 1
( a)-8.88 kJ mo1- · (b) 8.88 kJ mol- ( c )-88.8 kJ mo1-
1
Cd_),,88.8-'kJ mol- 1
For an ideal gas I
w=-RTin V2 i
.c
v;
l
C
3
1
= -( 8.314JK-'mol- )(273K)x 2.303x log(SOd~
B
.ldm .
yA
= -8.880J mol-4 = -8.88 kl mo!-'
Correct!option is (a)
tr
___ ,, __
One dm3 of an ideal gas at a pressure ofl .013 3 MPa expands reversibly and isothermally from its volume to
is
10 dm3• How much of heat is absorbed and work is done. in expansion?

1
,
For an i~eal gas undergoing reversible volume change, we have ·
m
V:
q =-w= nRTJn-1.
v;
he
The temperature in the above expression can be replaced in terms ofp1 and V I by usingthe ideal gas equation.
Thus, ·
.C
q=(Piv;)ln v;
. v;
w
Substituting the value ofp" V1 and V2, we have

w
1
q-(1.013 3Mpa)(l dm')x2.io3xlog( ~:;']
w
= 2.333 6 Mpa dm 3 = 2.333 6kf

Irreversible Expansion or Compression process;
Two types of irreversible change in volume may be distinguished, namely, expansion against a zero pressure
, (free expansion) and ·expansion or compression against a constant pressure (intermediate expansion or
compression). The change in q, w, !1U and Ml for these processes are given below.
For free expansion, p opp =0
Therefore, J J
w = dw = . .:. . p appdV = 0
1bus, q=-w=O
!1U =Ml =0
Intermediate expansion or compression :
Here, QUZZES
the work is done against a constantMORE
external
NOTESpressure.
ON www.ChemistryABC.com
V2
Therefore, w=-J PoppdV =-A,pp(V2 -v;)

Pj
Hence, in the present case, we have
q = -w Popp (J?; -v;)
l!iU :f Ml = 0
Comparison between reversible isothermal and adiabatic expan,sion of an ideal gas:

Comparison of the Isothermal curve and Adiabatic curve : . ·
'It is clear from the curve that the work done in isothern:pl process is gryater than that in a<liabat:bprocess ii;
· greater than that in adiabatic process since the area (ABDE) u�der t9�curve (AB) for the isothermalproces
om
�s more than the area (ACDE) under the curve (AC) for the adi�b,ffc pr,ocess. ·
I.e, WABDE ::::: WABC + WACDE
W.!SO= WABC + Wad1.a
.c
C
Adiabatic
ip
B
y A
v�
tr
is
Prove that ifthe final volume of the gas in isothermal and adiabatic process is same, then the final pressure c
the gas in isothermal process will be more than that in adiabatic process.
m
Le P.!SO > Padta. or (Pl'r).!SO > (Pl'r)•d.ta

he
Proof: Let Pi be the initial pressure of the gas in both the process (viz isothermal and adiabatic) and let� be the initi:
volume of the gas in both the processes (viz isothermal and adiabatic).
.C
Let V1be the final volume of the gas in both the processes (viz isothermal and adiabatic). Let Piso be the firu
pre�ure of the gas in isothermal process and let Padia be the final pressure of the gas in adiabatic process.
w
Isothermal Process : For isothermal process, we know that PV= Constant

=> P,V 1= P2V2 => pyi= p isoVr
w
w
... (1)
Adiabatic Process:
For adiabatic Process we know that
PVY = constant
=> PV.
1 t
Y= pad'1a vfr
p (V r
P.�a = � J HAND MADE NOTES
... (2)
pi P. Jy
QUZZES
From(l)and(2), we have, padia =( pi:o
-·· (3)
Since V1 > v, =:, ~ > I =:, ( ;J > I and aIBo yis always grei\ler than F(i.e. y> I).
Therefore, we have from equation (3)

P. P.I
_I > _
1 1
·_>_
padia piso ' padia piso
=> Ipiso > padiaI

i.e The final pressure ofthe gas in isothermal process is more thanthat in adiapatic process.
om
' . ) .I
Prove that if the final pressure of the ,gas inisothernialandad,.iaba.¢:pr~cess is same, then the fmal
volume of the gas in isothermal proce~s will be more than that in adiabatic process.
1
i.e Viso >Vadia OR (V}iso > p adia . . f
.c
In this case also I wiso > w adia I
C
Let Pi be the initial pressure ofthe gas in both the processes (viz isothermal anl{ adi~tic) & Let P1be the final
B
pressure ofthe gas in both the process (viz isothertnal and adiabatic).
yA
Let Vi be the initial volume ofthe gas in both the processes (viz isothermal and adiabatic).
Let Viso be the final volume ofthe gas in isothermal process and letVadia be the final volume ofthe gas in
adiabatic process.
tr
Isothermal Process :
is
For isothermal process, we know that PV= constant

m
PIVI= P2V2 => pyi= PJViso

he
viso pi
-=-
=> ... (i)
vi P1
.C
Adiabatic Process:
For adiabatic process, we know that
w
PV1 constant
w
=> ... (ii)

w
From (i) and (ii), we have -V:-

V,dia (v,~ );
0
Taking log on both the sides, we get
log" ( V;.• J- ; log"(~' J. =:, log" ( :7) ylog10 ( v;,,,) ... (iii)
Now since, V.!SO > V.I => V

vi > 1 => log (V
ISO
~ J> 0
!SO

QUZZES
Vadia
vi > MORE
·( Vadia
. vi
NOTES
1 => log ON
J.> 0 www.ChemistryABC.com
And i Also, y > l I

Therefore it is clear from the Equation (iii)
log(Yt}1og(v;;;,J~ :';> v:,

=> jviso > vadia I [:.vi> O]
i.e The final volume ofthe gas in isothermal process is more than that:fu adiabatic pro~ess.
om
Combined Result of (6), (7), (8) and (9) \ ;·
!(Slope)adiabatic = 'Y (Slope)isothermal I
.c
wisothermal > wadiabati~
C
When final volume ofthe gas in same in both the process (viz isothe$1 anl adiabatic), Then
I > (Pfinal) adiabatic I

B
yA
(Pfmal)isothermal
When final pressure ofthe gas .is same· in both the processes (viz isothermal and adiabatic), then
tr
is
Joule-Thomson Efrect
m
Porus Plug Experiment:

he
When any gas is subjected from a higher pressure region to a lower pressure region through a p01
plug adiabatically, cooling takes place except for I1z and He. At the same conditions of temp and presst
when these gases are subjected from a higher pressure region to a lower pressure region through a poi
.C
plug adiabatically, heating takes place.

w
This effect is known as 'Joule-Thomson effect'. It is an isoenthalpic and adiabatic process.

w
A G B
w
--+ I --+ I
I I
I I
I I
------,
I
P1 r-----
1 I
...,._v--+'I
I ...,._Vl--+
I I
I I
I I I
Figure: The porous plus experiment
Since Joule-Thomson effects is an adiabatic Process, therefore,

dQ=dU+dW
0 = dU + dW;-dW= dU (': dQ=O)
HAND MADE NOTES
... (1)
i.e. the work done by the g~ is equal to. the change in internal enetgy;
PdV =dU
- P(V2 - VI )= U2 - U I
-PV2 + PV1=U2 -U1
u 1 + PV 1= u 2 + Pu2
Ht=H 2
H-H=O
2 I
Aff=O => dH=O
constant
Joule-Thomson effect is an isoentha]pic pmcess ,\
Joule Thomson Coefficient:
om
H=H(T, P)
i
dH = ( oH) dT + ( oI:I) dP .•·
.c
.or p oP .r
C
(aH) dr+(aH)
l
~.
B
O= dP [ :. ]nJoule-Thomson effect dH = O]
of p OP T
yA
tr
is
m
(:l
he
is known as Joule-Thomson coefficient and is denoted by µ,_y_ (Joule Thomson Coefficient)

.C
µJT =- ~p (!l .. (J)

w
From thermodynamic square.

w
dH TdS+VdP
(!l =T(:l +v(!l

w
Aga~ (:l =-(:a

µ,y=-½[-T(~), +VJ
=½[T(~),-V]
µ1 y ... (2)

Joule-Thomson Coefficient and Ideal gas equation of state :'
The expression for Jqule-Thomsen coefficient as given by
µJ.T =-1
Cp
[T(av)
fJT p
-v] =-1
Cp
[T(~ RT) -v] =_!_[R
fJT P P Cp P
T(fJT)
fJT P
-v]
=[_!_
CP
R_.:!__]
PCP
=~-~=-1
Cr Cr Cp ·
(v-v)
/ µ 1.r = 0 j for ideal gas ·
Joule-Thomson Coefficient and vander Waal equation of state :

The expression for Joule-Thomsen coefficient as given by
=-1 [T(av) -v] \ .. (1)
om
µJ.T
. Cp fJT p
We have to find val~e of ( : ),fu, real gas and put it into abo te eq;ation
.c
I II III
C
![(r+ :, ){v-b)J. =!(R T),
B
yA
(r+ :, )!(v-b), +(V-b) !(r+;,), =R(:), +T(:),
tr
(r+ :, )(:),-(:), +(V-b)[(:), +(! :, ),

is
=R
m
(r+ ;, )[(:),-(:)J+(v-bi[(!), +(!, )JR

he
(r+ ;, )[(:),-o]+(v-bi(o+ ~:(:)JR

.C
w
(r+ ;, )(:), +(V-b)- ~~(:l =>(:),[(r+ ;,}~~(V-b)]=R

w
w
(av)
fJT
[( (V-b)
RT J-~(V-b)]=R
p V 3
(av)
fJT
[_!!_
V-b V
p
2a(v b)]=R 3
(:), =-R-T____,[==-1_-~a-(V---b)-= 2]
(V-b) VRT 3
... (2)
Putting value ofequation (2) in equation (1).

{V-b=V]
1
1
µ1 T = - [(v-b)(1-~J- -v]
i · Cp RTV
i .
2ab , .
I
1
[so,(1-~. J-I =(1+_2_.a )]
RTV . RTV ( RTV · is verysrt:tallmay be ~eglected)
I .
~
}
.;
om
...Jh.T _1.[_.2a
C
-blj \
I
/
.· p RT /
1
.c
Inversion temperature (Ti):
C
The temperature at which µ 1.T, becomes zero i.e. neither heating nor cooling tak;es place ;md gas behaves
ideally is called inversion temperature. \ /p
=(:a=~p[~~ -b] B
yA
µJT
If µJ.T =0, T = Ti
tr
=2_[ 2a · b] 2a -b-O 2a _
is
:::::;> 0 - :::::;> .. - :::::;>--b

Cp R½ RTi RTi
m
~
he
... (3)
Now following experimental condition indicates heating or cooling.
During J. T. effect. ·
.C
=(!a
w
dP -ve µJT
w
If µ 1.T positive then of should be negative, T2 < T1

Thus cooling takes place.
w
In experimental condition:
µJT = ~[:-b] If µn = +ve means µ 1,r > 0

1 2
, , :::::;> 0<- [~-b]. ( ~ bJ>o. (~-bJ>o. a >T
CP RT ' RT ' RT ' Rb
Thus experimental temperature should be lesser than the inversion temperature ofa gas for cooling.
If µ 1.T = negative:
Thus of should be positive HAND MADE NOTES www.ChemistryABC.com
T2 - T1 > 0 ; T2 > T1 (heating talces place)

In experiemental
QUZZES condition: MORE NOTES ON www.ChemistryABC.com
1 [2a
0>- ---b 2a
-<b J
2a
-<T ·
T <T [T = experimental temperature]
CP RT => RT => Rb => i
Ifexperimetnaltemperature is greater than ¾heating takes during J. T. experiement.
Note: f1i and He gases have a very low Ti (-80°C and -240°C). This temperature is always below t
experiemental condition temperature, thus heating happen.
Joule-Thomson Coefficient (or a Real gas : ,

Joule-Thomson coefficient may be positive or negative bfzero depending upon tp.e temperature and prt'Sst
ofthe gas. Since the change in pressure, dp, in Joule-Thomson experiment is always negative, it folows that i
is negative when µrr is positive, and vice-versa, the e:ffects:are s ~ in table below.
om
Experimental result (aH)
aP r Comments
.c
dT negative, i.e. (+) (-) Most gases, H2and He
T2 <T1 at low temperature.
C
Cooling
B
dT zero, i.e. Zero Zero Ideal gas, real gas at inversion
yA
T2 = T1 . temperature
Neither heating nor cooling
tr
dT positive i.e., (-) (+) Most gases at high temperature

T2 >T 1 H2and He at ordinary temperatm
is
Heating
m
3. Calculate the Joule-Thomson coefficient of carbon monoxide at 298K and 40.53 MPa pressure, given tl
he
( ~) (::), is 0.984, the molar volume is 76 cm' moJ-1 and C, = 37.28 JK:-1 mot'.
.C
:a-I]
Soln. Substituting the given values in the expression
~~
w
µ JT = [ (
.
w
( 76 X 10-6 m3mor 1 ) .
w
We get, µ =-'-------'-(0.984-1)
n ( 37.28JK-1mor')
r'
=--0.032 6 X 10-6 KF 1m3 =-0.032 6 X 10-6 K (Pam 3 m3
= --0.032 6xl0-o K(l atm/101.325 x 103 f =-0.003 3 K atm- 1
1
4. Show that inversion temperature, Ti =- for ideal gas.
ai
Soln.
(lfT = Ti) than µJ,T becomes zero.

r1?!'J-"'- r •• ;;;,..,_........,...,,.._ - · .•·•·-· ···--J·-·\~··--
0=~,[r;( ),-v] T;(:),=v

QUZZES
0
V =
1 1
=-
T; =(:), _!_(BV)
V 8T P
ai
~ proved.
~
Note: (1) IsothennalJ.T. coefficient ( q>)
'
,)
( <!>) =(BH)
BP /;
om
T
\
(2) Isoenthalpic J.T. coeffic!ent ( µ)
~,(:1
.c
. µ,T=-
C
Evaluation. of (BH)
B
yA
- from µ.JTof a van der Waals gas :
' BP T '
tr
We know that, ( ':), =-CPµJT ... (i)

is
m
And µJT= ;J!;-b) ... (ii)

he
From equation (i) and (ii), we get

.C
(BHmJ =b-2::_
w
BP T RT
w
For a gas undergoing isothermal expansion or compression, we will have

w
Mi.m = (·b-2::_)
RT M
Problem : Calcu1ate the value of Ml for the isothermal compression at 300K ofl mol ofnitrogen from 1()5
Pa to 500xl05 Pa. Given: a 135.78 (dm3)2 kPa mo1-2 and b =0.039 dm3 mo1- 1•
. ;Ne have,
Ap =500xl05 Pa-Ix 10 5 Pa= 499x 105 Pa= 499xl02 kPa
a= 135.78( dm 3 )2 kPa mor2

b =0.039 dm 3mol-1
Substituting these values in the expression,
Mfm=(b- ~; )Ap
6
QUZZES
We get, m . .
MORE
Mi =[0.039dm 3mar 1
-
NOTES
3 1
xl3
2 ON 5.73 dm kPamal.-
1
(8.314dm kPaK- mar )(300K) .
](499xl02 kPa)
2
=( 0.039 dm 3
maz- 1 -0.1089 dni3 mar 1 )( 4'99xl0 2 kPaf
=-3 488 dm 3 kPa mar' =-3.488 kl mar1

Calculation of various thermodynamic relation for van derwaal's relation:
1. Expression ofW:
dw =-Pext·dv
For reversible change in volume, Pext =~nt ± dP = P ~1.
rv2( na) . \
om
2
nRT
dw -Pint _dv => Jdw =- Jv1 . V -.nb v2 dV
'i
.c
W =-nRT .en..v2 -nb -n2a ( -1- -1· J
C
V1 -nb V2 V1
2. Expressed for AU :
B
yA
(:l ;T(!l-P
tr
is
m
2 2 2
au) _ nRT -P _ nRT -( nRT n aJ= n a (au) = n a
he
( av T -v nb -(V-nb) (V-nb) V 2
V2 => av T.
U=f(T,V)
.C
w
ctu =(au) .dT +(au) .ctv

av v aT v
.
w
For Isothennal change dT = 0

w
dU =( -au) .dV =>

avT
f dU = Jv,v ;y
f'2 n : dV
2
=> AU=-n
2
a(-
1
V
2
1
--
VI
J
Similarly, q = AU -w
q nRTln V2 -nb
=> VI -nb
·Expression for AH :
·: AH=QUZZES
AU +A(PV) MORE NOTES ON www.ChemistryABC.com
2
2
AH=-n a( 1--
- 1 ) +A [( -nRT
- - -n2a) V ]
V2 V1 V-nb V
. '/
om
'/
l
(sotVED PROllLEM:sJ f
.c
When229 Jofenergy is supplied to3 moles ofa gas at constantpressure,theternperature ofthe gas increased
C
by2.55 K Calculate Cr and Cv for the gas, assuming ideal behaviour. ( 11 [GATE 1996]
Given, dqP = 229 J, n = 3 mole and dt = 2.55K
B
\ ,·
Since, we know that
yA
dqp nCpdt
tr
229J=3 molxCP x2.55K

is
CP = 229J
3 molx2.55K
m
~Cf'= 29.9 J moZ- 1K- 1

he
Since, CP Cv =R (foridealbehaviour)
.C
29.9 J moZ-1 K- 1 -Cv = 8.314 J K- 1 moZ-1

Cv =(29.9-8.3I4)JK-1moZ- 1
w
Cv = 21.58 J K- 1moZ- 1
w
One mole ofan ideal gas ( CP = 29 .234 JK- 1moZ-1 ) is expanded reversibly and adiabatically from 1 dm3 to
w
10 dm3• Ifthe initial temperature is 750 K, the final temperature will be [GATE 1996]
(a) 1000K (b) 750 K (c) 300K (d) 100 K
Given, T; =750, ~ =1 dm 3 , V2 =10 dm 3 , T2 ? and CP = 29 .234 JK- moZ-
1 1
We know that, CP- Cv R

29.234 JK- 1moZ- 1 Cv = 8.314 JK- 1moZ-1
Cv 20.92 JK-'moZ- 1
_ CP _ 29.234 JK-'moZ- 1
r - Cv - 20.92 JK- 1moZ-1
r=I.4
"c::;rv.....-, O!, r,n::o ... i,..i,:IW VI I IIC>I II · ~ l l c : l f l l l " ' l l
.
Smce, r,v;r-1
I I
_ ,.,.. rn-t
-.L2Y2
.
. . .·
. . 1··. ...
.
\'
QUZZES MORE NOTES ON www.ChemistryABC.com ·3l
.~
1'.2 = 750K =300K

2.51
Correct option is (cj.
3. The relationship between volume change in an isothe~l process ( 6. V;) and an apiabatic I?rocess ( 6. r:,) fo:
~
a pressure change from P I to P2 is , [GATE 1998]
(a) 6.V; > 6.Va (b) 6.V; < 6.f:, (c), 6.V:I =6.Va
om
Soln. IsothermalProcess:
For"isothermal process, wdmow that PV= constant v'
=>
.c
PtV,= P2V2; pyi= P/Viso
V;"° = P;
C
=> vi P1 . ... (i)
Adiabatic Process:
B
yA
PV1 = constant
tr
is
=> ... (ii)

m
From (i) & (ii), we have V; = ~

Vadia (V ); 0
he
Taking log on both the sides, we get

.C
logw(V¼ia )= ~logw(~·); I0g 0

10 ( ~ )= ylog 10 (V~ia) ..• (111)
w
· Vi,, (Vi,,o)
w
0
Now since,Viso > Vi => • V; > 1 => log V; > 0
w
And V_A. >V vadia> l => log (vadia)

- - >O
outa 1
vi vi
And I Also, y > 1 I

Therefore it is clear from the Equation (iii)
log(~o) > log(V¼ia]; ~o > V;

'"' ..
_.. .. -~
For an ideal gas foll0wing adiabatic reversible expansio~ plot oflog p versusJog,\{ is linear witha slope equal
;o(r= ~) <}{GATE 2000]
1
(b) - - (c) r (d) -r
r
pvr =constant =k
P=kv-r
Taldng log both side we get
log P = log k-"- r log V
om
Or, log P = -y log V + log k ... (i)
Since, y = mx+c ... (ii)
· From equation (i) and(ii),,we get
.c
slope= m = tan 0 = -r
C
So, correct option is (a)
The internal pressure of one mole.ofa van der Waru gas is equal to {GATE1996]
a
B a
yA
(a) zero (b) b2 . (c)7 {<i)h--·
RT
tr
is
m
1t =T( 8 · RT ) (~-~) -P _ R (of)

T of(V-b) v of v 2 v -T(v-b) of v -P
he
u .
0
=(:~b) r=(r<, )-P

.C
w
Ire,=;, I
w
The internal pressure of a van der Waals gas is [GATE2003]

w
(a) independentofthemolarvolume
(b) inverselyproportional to square ofthe molar volume
(c) inversely proportional to square ofthe molar volume
(d) directlyproportionalto themolarvolume.
a
. , -Since internal pressure Kr = 2
V
Internal pressure is inversely proportional to square ofthe molar volume
Correct option is (c)
In an adiabatic system, the work done to change its state fromA to B is [GATE2005]
(a) dependent on path fromAto B
(b) independent ofpath fromAto B
(c) path dependence is related to particulars ofHAND
statesMADE
A and B
NOTES www.ChemistryABC.com
(d) path dependence is related to both states A and B and choice ofpath.
. Correct option is (b)
8. QUZZES pressure, ", =
. The internal r{ :;),- MORE
P NOTES ON www.ChemistryABC.com
forone mole a yander Waals gas is [GATE2005]
~
;
a RT
(a) v2 (c) zero (d) V -b.
Soln. For vander Waal gas, for one mole,
(P+ ; )(V-b) =RT

2
p RT a
(v b) v2 ... (i)
(!;). V~b
om
Since, gi~ intema'l pressure, ir;. =T ( :; ), - P =:, "r = 'Q (lb) - p
.c
a
C
From(i), 7rr =P+ -P
B
a
1rr =-2
yA
V
tr
Common Data for Q. 9 and Q.10. [GATE2006]

is
Consider the following P-V diagram for an ideal gas that follows the diagonal path fromA to B.
m
(T=300K)
B
he
20
P(atm)
.C
1
w
·C A(T 300K)
w
0.5 10
V(L)
9.
w
The work done (in atm-L) on the gas in the process is

(a) 9.5 (b) 99.75 (c) 190 (d) 10 ln(20)
V
Soln. Since, w = -nRT ln r: = -Pi~ ln r,r
I' I rl
5
w = -lOln 0. = 101n(20)
10
Correct option is (d)
10. For the above process,
(a) Afl=W (b) Afl=Q (c) Ml 11G (d) Ml= till
Soln. Ml (pdT) = 0
till= CvdT 0 HAND MADE NOTES www.ChemistryABC.com
Ml= till
Correct o tion is (d)
} . . ( CP - Cv ) . ;
QUZZES
rt,:'·Lfagas obeys the equation of state P(V-nb)MORE NOTES
=nRT; theON
ratio (·. C -C )www.ChemistryABC.com
· }8 , [GATE2008]
· · . •. P V ideal :
(a)>l (b) <l (c) 1 (d){l.:,.-b)
1ln•. Given, P(V-nb)=nRT ... (i)
P= nRT
V-nb
oP) =( V-nb
nR )
( oT V and pV-pnb=nRT
om
\
nRT
pV=nRT+pnb => V=--+nb
P·
.c
... (ii)
C
cp -c B
yA
=T(aP) (av)
. v ar . ar V P
tr
Or, CP-Cv Tx ( -nR- ) x-.

nR =nR .
(from(i))
is
V-nb P
m
we know that, ( CP Cv )ideal = nR

he
=> Cp Cv =1
. ( Cp -C)
.C
v· ideal
Correct option is (e)

w
. R
w
Show that Cv - and Cp =(. __I_)R

. y-1 ly-1
w
. Cp -1-~ R
iln. CP Cv =R => C - C => Y l =Cy
-
V V
Cv=(~J
y-1
R
Again, Cp--=R
y 1
CP =R+~=R(l+--)=R(y-l+l)
y-1
1
y 1 y-1

Proved.
Degree of Freedom:
QUZZESoffreedom ofa molecule are MORE
The degree NOTES
defined as theONindependent number
ofparameters required, to descnl
the state ofthe molecule completely. The motion ofatom and molecules is generally descnbed in tenn ofdegn
offreedom they possess. When a gaseous molecule is heated then the energy supplied to it may bring abo1
three kind ofmotion in it (translational motion, rotational motion anq vtbrational motion).
It is the number ofindependent variables coordinates required to specify a molecule in the space or
is the kind ofenergies sustained by a molecule or by knowing the value ofwhich we can predict about the kn
ofmotion (viz translational, rotational and Vibrational) executed by a molecule.
Both kind ofmolecules (linear and non linear) show three kinds oftranslational motion (ie alongx,y & z ax~
• ' i
y
,}
om
\ ,/'
----x I
.c
z
C
Non-linear molecules (like N02, S02, S03, CH4 etc) show three kings o~rtationalmotion
B
Linear molecules (like CO, CO2, N2, 0 2 et~) show two kinds ofrotatfonal motion.
Ifthere are N number ofatoms, then the total degree offreedom (DOF) iri the space= 3N i.e. Translation
yA
degree of freedom+ rotational degree offreedom+ vibrational degree of freedom= 3N
tr
!Ntr + Nrot + NVib;:::;: 3NI ... (1)

is
For both kind ofmolecules (ie linear and non-linear)

NIr =3
m
from equation (1)

he
... (2)
.C
For Linear Molecule: Nrot 2

w
from equation (2)

w
=> Nv.b= 3N (3 + 2) ·
w
=> !Nvw = 3N-51

For non-linear molecular: N rot= 3
fromequation(2)
=> NVib 3N - (3 + 3)
=> INVib = 3N -61

Problem-I. Calculate NVib for the following moleculers N02, S02, CO, S03, CO2, CH4, N2, 0 2,
Soln. Here linear molcules are CO, CO2, N2, 02'
Forthesemolecule !Nv;b =3N-51 HAND MADE NOTES www.ChemistryABC.com

(i)CO
.HereN= 2 QUZZES MORE NOTES ON www.ChemistryABC.com
(ii) CO2
HereN= 3
(iii) N2
Here N= 2 :. NVib =3:X 2-5 = l
(iv)02
HereN=2 :. NVib = 3x2-5=1 i
And non-linear moleculers are "
N02, S02, S03, CH4,
om
For these non-linear molecl).lers..
NVib is given as NVib= 3N 6
.c
(i) N02
C
HereN=3
(ii) S02
B
yA
HereN= 3 :. NVib = 3x3-6 = 3
tr
(iii) S03
is
HereN=4 :. NVib =3x4-6=6

(iv) CH4 Here N = 5 :. NVib = 3 x 5 - 6 = 9
m
Relation between Cv and Degree of Freedom(/):

he
Ifthe degree offreedom ofone molecule isJ, then the degree offreedom ofN molecules will be rV,
Now from the equipartition theorem ofenergy, Average internal energy per degree offeedom per molecule is
.C
given as, E = ~ kT (Where, k= Boltzman constant)

w
k= R 8.314 J /Kmol _ x
w
-23 -1
N 6.023xl023moi-1 => k- 1.38 lO JK
w
Averageintemalenergyofallkindofmolecules (i.e ofN molecules)

l I
NfkT => E= R/T
2 2
Now we know that
Cv = -2I
Rjxl::::> Cv=- HAND MADE NOTES www.ChemistryABC.com
. ~ ·2
According to the principle of equipartition of energy
Degree of freedom Symbol . Contribution to Cv
R
Translational N1r
2
R
Rotational Nrot
2
R, at high temperature
Vibrational 0.2R, at room temperature
{
0, at very jow temperature
i.e. at room temperature the contribution to Cv from N~ib is only20% !hen we get
om
20 .
Contributio1;1 to Cv from ij\/ib Rx 100 = 0.2 R .
At high temperature,
.c
C
(Cv)tota/ = N,r(¾) + Nrot(¾) + NVib(R)
At room temperature,
B
yA
(Cv )total = Ntr ( ¾) +Nrot ( ¾) + NVib (0.20R)
tr
Problem-2. Estimate y at room temperature for (a) He, (b) CH4

is
Assuming these gases as ideal gases. What will be the value ofy at higher temperature?
m
Soln : Case-I: At room temperature

he
(a) He Helium consists ofone atom only and atoms have onlytranslatiori motion.
Ntr= 3, Nrot 0, NVib = 0
.C
(Cv\ota! Ntr (:) + Nrot (:) + Nvib (0.20R)

w
R 3R
(C)
V total
=3x-+O+O
2 ~C=-
v 2
w
3 5
And cp= cv + R 2 R + R = 2R
w
~ (5/2)R 5::: 1.
66
y Cv (3/2)R 3
(b) CH4 • it is a non linear molecule
Here N1r== 3, Nrot = 3 and N 5
Nvib= 3N - 6 ~Nvib= 3 X .5-6 ~ Nvib 9

r:a.,r,""""
...,,._ ,,- ... • ~· - - - -
J
=> Cv=4.80 x 8.314
QUZZES Kmol =39. 9072 J/K.mol
Cp= CV+ R = (39.90 + 8.3_140) = 48.22 J/K
Cp 48.22
=> y= Cv = 39.90 = l. 20
Or . CP= Cv + R = 4.80 ~ + R 5. 80 R
Cp 5.80R ·
y = Cv = 4.80R = l.20Ans
. \)
Case-II: At high temperature:
(a) . He: SameValue . ;//
om
· Helium consists ofatom only and atoms have onlytranslation m6t~ti .1. r
Ntr=3,Nrot=Oand,Nvib=O . i 1 • •.
.c
(CV)totat NtrT:) + NrotT:) + Nvib (R) fl
C
R · 3R .. /fl
C=3x
v 2
+0+0:::>C=-
V 2 t, , 3-
·· 3 5
B
yA
CV =C
· V
+R= 2 R+R= -R
2
= ~ = (5/2)R 5 =1. 66
tr
y Cv (3/2)R 3
is
(b) CH4 It is a non-linear molecule

Here N~ 3, Nrot 3, N 5
m
Nv,b. = 3N 6 = 3 x 5 6 => Nv,b. = 9

he
(CV)total Ntr (:) + Nrot {~) +·Nvib (R)

.C
3R 3R
CV= -+-+9
2 2 => RC V=3R+9R => CV= 12R
w
Cp= CV+ R 12 R + R 13 R
w
Cp 13R
y= C = 12 R = 1.o3
w
On the basis of this question, we can say that on increasing temp Cp, Cv increases, but r decreases.

( PRACTICE SET ] '
I. Adiabatic reversible expansion of a monoatomic gas (M) arid a diatomic g~ (D) at an initial temperature T
has been carried out independently from initial volume V1 to final vol~e Vr The final temperature (TM fc
monoatomic and TD for diatomic) attained will be · - [GATE 2003]
(a) TM =TD >T1 (b) TM <TD <TL (c) TM >TD>½ · (d) TM =TD=½
2. Foronemoleofanideaigas (:H::H:\ = [GATE2004]
(a)-1 (c)+l
om
3. One moleofan ideal gas (Cv = 1.5 R) ata temperature 500Kiybmptessed from 1.0 atm to 2.0 atmb!
a reversiblecisothern;ral path. Subs~quently, it is expanded back to 1.0 a'.tm by a reversible adiabatic path. ·
The volume ofthe final state in litre is: 'l [GATE 2004]
.c
(a) 15.6 (b).20.5 (c) 31.1 (d)41.0
C
)
,
B
4. For an ideal gas (:), =
yA
(c) 0 (d) 1
tr
5. The vanderwaal equation is given by
( 7na) (v-nb)=nRT
2
is
P+
m
a
The ratio b has a dimension of
he
(a) atrn/litre (b) litre atm'mol~ (c) litre/mole (d) litreatm mo1e-2•
.C
6. For a reverrisible adiabatic compression which realationship is true:
Ptr Ttr = Ptr Ttr

w
(a)
w
( c) p2i-r T-r -
2 -
pI t-r T, r
I
w
7. For reversible adiabatic expansion of an ideal gas

T = constant x px
The value of 'x' is:
1 I
(a) y (b) 1-- (c) -1 (d) 1-y
y y
8. Among W(work), Q (heat), U(intemalenergy) and S(entropy)

(a) Wand U are path functions but Q and S are state functions
(b) Wand S are path functions but Q and U are state functions.
(c) S and U are path functions but Q and W are state functions.
(d) Wand Q are path functions but U and S are state functions ..
Work (w) involved in isothermal reversible expansion from Vi to Vr ofil moles ofan ideal gas is
(a) w =..:.nRTln(Vr /'½) (b) w = nRTln(Yr I'½) [NET Dec. 2013]
(c) w=-nRT{Vrl'½) (d) w =-nRTlog(Vr /VJ

_;
The heat capacity of 10 mo l ofan ideal gas at a .certain temperature is 300 JK-1 at constant pressure. The heat
capacity ofthe same gas at the same temperature and at constant volume would be [NET Dec. 2013]
(a) 383 JK- 1 (b)217 JK-1 (c) 134 JK-1 (d) 466 JK- 1
;
At 300 K 2 moles of an ideal gas expand reversibly and isothermally from lL to 1OL. What is the entropy
' '
change for the process? (R = 2 cal K- 1mor 1 and in 10=2.303) i

'.\
(a) 9.2 cal K-1 mor1 (b) 4.6 eal K- 1 mo1-1 ,
om
(c) 2.76 cal K- 1 mor' (d)O ..
.c
· Consider the following statement :
C
, a
1. The Boyle temperature for a vander waal gas is T8 = -
· .· 2Rb
B
yA
. . . . . b . .., 2a
2. Themvers1ontemperature1sg1ven yequatton .tr= Rb
3. The Joule~Thomson.coefficient for an ideal gas is zero.

tr
4. H2 and He show cooling at ordinary temperature ofthese.

is
The correct statement is: ·

(a)l,4 (b)2,3 (c)l,2,3 (d) 2, 3, 4
m
Work done in a free expansion process is

he
(a) zero (b)minimum

(c)maximum (d) positive
.C
A gas is expanded adiabatically from 750 K to its twice volume than the'final volume is if ( CP = ½R) , final
temperature is:
w
(a) 100 (b) 300 (c) 475 (d) 750

w
An ideal gas expands by following an equation PVa = constant. In which case does one expect heating?
w
(a) 3 >a>2 (b)2>a> 1 (c)O<a<-1 (d)-1 <a< 0 [NET June 2011]
The ~S for 3 mole of an ideal gas (diatomic) which is heated and compressed from 298K and 1 bar to 398K
, and 5.bar (.Cp = 7 R )

2
(a)-14.88 J/K (b) 14.88 J/K
(c)-4.62 J/K (d) 4.62 J/K
The Cp /Cv ratio ofa gas mixture consisting of8 g ofhelium and 16 g ofoxygen is:
(a) 1.33 (b) 1.47 HAND

(c)MADE
1.59 NOTES (d)2www.ChemistryABC.com
Lerot.h c:x t-lrs"t, Law or I nermoaynam1c6·!
18. Considerthefollowing statement

Cp 8(
(1) The value of CP + Cy for helium is 3 Y = 3
5) '. . Cp ~ Cy
(2) The value of CP +Cy for H is
1(
Y=
7)
2 6 5
Which ofthe following statement is correct.
(a) 1 only (1?}2 only (c) both 1 and 2 (d) neither 1 nor 2
19. If an ideal gas having internal energy U = ½PV is expanded reversible and adiabatically from PI' V1, T 1 to P2,
V2,.T2 then
(a) TV-½= constant (b) TV½ ~ constant ·.) '
·(c) PV-½ = constant (d) PV½ = constanti
om
.· './
t
20. Onemoleofanmonoatomic gas (y,= 1.66) is mixed withonemqJe o.fdiatomic gas ( y = 1.40) .The,valueof
.c
Y for the mixture is:
C
(a) 1.40 (b) 1.50 (c) 1.53 (d) 3.07
/'i'
rj'
B
yA
ANSWER KEY
tr
Questions 1 2 3 4 5
(b) (dl (cl (bl (b)
is
Option
Questions 6 7 8 9 10
m
Option (bl (bl (d) (a) (b)

Questions 11 12 13 14 15
he
Option (bl (b) (al (c) - (cl

Questions 16 17 18 19 20
-
.C
Option (al (cl (bl (a) (bl

w
w
w

[ Chapter 3]
oecond Law ofThermodY;namics

'
,)
Introduction: · ./
The second1awhelps us to determine the direction in .which energy C8._IJ/be transferred. It also helps us to
om
predict whether a given process ora:chemical reaction.can occur,spo~¢eotMy. It introduces a new concept
of entropy. It also helps us to know the,.equihbrium conditions. t /
.c
Statement of the Second Law of Thermodynamics :
"Heat cannot spontaneously pass from a colder to a warmer body." (R.J.E calusius).
C
''In an adiabatic process the entropy either increases or remains unchangedr Ir]
'
B
' . ' .
AS~O
yA
where the sign(>) referes to the .irreversible, the "equal" refers to the reversible case. (P.S. Epstein).
"Work can always be completely converted into heat but heat can't be completely converted into
work without -..
leaving a permanent cha(lge in the system or surroundings. Only a fraction of heat
tr
' ' . •: '
can be converted into useful work & the rest remains unavailable & unconverted. [CSIR-NET]
is
"In any irreversible process the total entropy ofall bodies concerned is increased." (GN. Lewis)
m
"The entropy function ofa system ofbodies tends to increase in all physical and chemical processes occurring
in nature, ifwe include in the system all such bodies which are affected by the change." (Saha)
he
Carnot cycle or Carnot reversible heat engine:

.C
The carnot cycle is a hypotheticai ideal & reversible cyclic process, devised by a.french engineer Sadi Carnot
(1824), to demonstrate the maximumconvertibilityofheat into work.An ideal machine to demonstrate this
w
cycle is called carnot heat engine.

w
The work done in a carnot cycle is maximum because the various steps involved are reversible & the
reversible work is the maximum possible work which a system can do.
w
Carnot heat engine is ideal and regarded as standard (ideal) by which the performance ofan actual reat engine
is compared.
Schematic plan:
It consist of
(i) Source of heat (A), at high temperature T2•
(ii) Sink (B), at lower temperature T1
(iii) Working substance (C)
(iv) Perfect heat insulating stand (D)

u www.ChemistryABC.com
Working
substance
(C)
Source(T} · Non coq.ducting · Sink (T)

(A) stanq (E>) (B)
[It is conducting & heat [It is conducting & heat

capacity very large] . . ., , capacity very large]
om
1
The idealcamot'scycle is the cycle in which working substanc; an ifealgas. if
Operations inv~lved in c;rnot cycle: f ·
.c
This cycle involves four operations. These are as follows. .
C
B
Heat source
(I)
yA
w 'fi>T1
i
tr
,p
is
m
he
Indicator Diagram of Carnot Cycle

.C
Step 1: Isothermal and reversible expansion of working substance:

Cyclin:der is placed on the source, load on the piston is decreased, working substance expands, external wo
w
happens. Fall in temperature, but it is in the contact with the source, Thus maintains the temperatl.re quantity
heat absorbed is q1•
w
T2 =constant. AU =0
w
. v2
F1rst law, q1 = -w 1 = RTit'n V ... (1)
I
Step 2: Adiabatic reversible expansion of working substance:
Cylinder is removed from the source put on the perfectly insulating pad. Load is further decreased 1o cau
expansion ofworking substance adiabatically. ·
q =0 (Fromfirstlaw)
w 2 =AU=Cv(T1 T2 ) ••• (2)
Step 3: Isothermal and reversible compression of substance:

Cylinder is again transfer to the sink at temperature (T1), load on the piston is slowly increased Gas is compreSSI
Heat developed passes into the sink. Temperature remain constant at T1•
V HAND MADE NOTES www.ChemistryABC.com
q2 =-W3 =RT,£nv: ... (3)
Step 4: Adiabatic reversible compression of substance: ' . .
Cylinder is again transfer to the insulating pad. Load
QUZZES MOREon the piston~
NOTES ON againsli.ghtlyincreased, so that substance
undergo slow adiabatic compression. Temperature rises from T1 -4 T2 ..
w 4 =Cv(T2 -T1) ii(4)

..
;
Thus net work during one complete cycle.

WTotal = WI +w2 +W3 +w 4
,}'
... (5)
om
\ /'
Now for adiabatic-eurve BC and DA
TV7.:..1 =Constant
i
.c
for curve BC, T2V/-1 = T1V/-1
C
.!i_ = ( v2 Jr-I \,
B
... {6),
T2 V3
yA
for curve DA 1
.n., '[14
V r-1 -
- T
2V.
,r-
;: =(~:r
tr
;.. (7)''
is
. from equation (6) and (7)
(~r =(if
m
he
I~=~: I ..
.C
(&}
Putting value ofequation (8) in equation (5)
w
V2 V V2 V
WTota1 =-RT2.t'n- -R'Tifn-1 =-RTifn- +RTiRn-2
w
' ~ ¼ ~ ~
w
) ... (9)
Efficiency (11) of carnot heat engine:

The ratio ofwork done by the system to the amount ofheat transferred to the system at the higher temperature.

~
.• ,.(10)
L-2J
Discussion of above equation (10):
(i) TJ does not depend on the working substance used.
(ii) TJ depends only on the ratio ofsink to source temperature
(iii) TJ is always less than one.
.· T • .
1
(iv) TJ is one only when T is zero. i.e. when T1 =0K(-'273°C) or T2 =Cf>. Both these temp. are impossi
2 .,
to obtain. Therefore, it is impossible to construct a he~t engine whose efficiency is unity or I 00%. Tha
heat can't be transformed completely into work. . , , · .\
(v) For all reversible cycle operating between the same initial and :finaJtemperature,
,/
the efficiency
'
is the sar
.
om
1
Refrigerator: "/
Energy will not flow spontaneou§lY from a low temperature to a high/
temperature without any work. Refrigerator does this work. P '
.c
Refrigerator is a machine operates in a manner opposite to that of a
heat engine. (Working substance take heat from the low temp. source
C
then same work is performed on it & :finally it rejects heat to a high_
temperature source. Space to the refrigerated is the low temp. source\
B
& surrounding atmosphere is high temperature source.
yA
In carnot engine heat is input work is output. In refrigerator
heat is output & work is input.
tr
, Comparison of efficiencies of reversible and irreversible cyclic processes.

The efficiency of a reversible Carnot cycle is the theoretically possible maximum value which
is
engine can have. Since the various processes of this type of engine are to be carried out reversib
m
therefore, such type of an engine does not have any realistic basis because reversible procesi
are idealized concepts which can never be realized. A real heat engine, which is irreversible
he
nature, will have efficiency smaller than the reversible heat engine.
Let us have two cycles, one operating reversibly and the other irreversibly. Let both
.C
them operate between the same two temperature T1 and T2 and involve ideal gas as the worki
substance. These two cycles along with q values, are shown in the figure below.
w
(A) Isothermal expansion form volume V1 to V2 • The expressions for the work involved are
w
• V:
-w2 (rev)= q2 (rev)= RT; In-1.
w
. v;
And -w2 (irr) q 2 (irr) =Popp (Vi -v;)
T2
------
qz(rev)
w(rev) w(irr)
qi(rev) qi(irr)
HAND MADE NOTES ------ www.ChemistryABC.com
/ Ti
Since we known that jw2 ( rev )I> lw2 ( irr)I, therefore,
q2 (rev)> q2 (irr)
(B) Isothermal compression from volume V3 to V4 • The expressions for the work involved are
V ;
-w1 (rev)= q1 (r~w) = RT; In; 3
-w1 (irr)=qi(irr)=p' 0PP(V,.-VJ .. . . , . , . . ...

Now since in the irreversible process, inore work is done as compared to that in the reversible
process, we have
w1 ( irr) > w1 (rev)
It follows that
om
1% (irr )I> !%(rev )I
or Jq1 (rev,)j <Jq1 (irr )I
Now the efficiencies of the two cycl~s are . ll
TJ(rev)= q2 (rev)+q,(rev) = q2 (revf-jq1 (rev)] =l_Jq,(revJj
.c
q2 (rev) q2 (rev) q2 (rev)
C
'.
1J ( irr) = q2 ( irr) + %(irr) = q2 .(irr )-d<i1,(irr)j = 1 _ jq1 (irr)j \
q2 (irr) q2 (irr) ·
B
q2 (irr)
yA
Now since q2 (rev) > .q2 ( irr) and jq1 ( rev )I< lq1 ( irr )I, therefore, it follows that
tr
lq1 ( rev)I < 1% (irr)I ·{l

is
q2 (rev)· q2 (irr) or
m
1.e. 1J (rev)> 1J (irr)

he
Basi~ Conclusion from Efficiency of a Carnot Cycle :

F~r a reversible Carnot cycle operating between two temperatures T2 and T1, the efficiency is
.C
given as
rJ = q2 + q, = r; - i;
w
q2 r;
w
where q2 and q1.are the heats exchanged with the thermal reservoirs at temperatures T2 and T 1,
respectively. Rewriting the above expression, we have .
w
Or,
%+~=0
,Or, r; i;
--o
' that is, the sum of the ratios of the heat involved and the corresponding temperature is zero for a
Carnot cycle. T2
:·, · ·T-c-8 diagram of camot cycle:
Total heat taken by system dQ =TdS -- Q.
Q =T2 ( S2 - S1) ·
Heat energy converted into work
_ _ _._.,___,, __.. .... _.,. . . . . _ . · · . · - ~ · · -1 111
-w = dQ = (T2 - Tl) (S2 -SI)

Tl= -w = (T2 -T1)(S 2 -S1) =T2 -I;

. Q . T2 ( S2 -Si) T2
~
~
H vs S or TVs S ~r U vs S diagram for camot's engine.
Qh
A
.----+-+-----, B
t
H AB::: reversible is,othennai expansion
or
om
, ·. '.j .
T BC = reversiblrdiahatic expansion
or
u D C
CD = reversi9.le ~sothennal compression
DA = reversible adiabatic compression
.c
s~
C
U vs V or T vs V or H vs V diagram for camot's engine:
Qh
B
yA
t
u AB= reversible isothennal expansion
tr
or BC = reversible adiabatic expansion

is
T
or CD = reversible isothennal compression
m
H
DA = reversible adiabatic compression
he
v~
.C
Coefficient of performance (P):

The ratio of heat transferred from a lower temperature to a higher temperature to the work done on
w
~ = jqcw I
w
machine to cause this removaL i.e.

w
The less the work done the more efficient the operation and greater the coefficient ofperformance.
wirr > wrev' ~rev> ~irr
At T1 ~O kelvin ~=0
w = JqcJ = (qc) =oo
~ 0
Thus as the temperature of a system is lowered the amount ofwork required to lower the temperature :fur1
increases rapidly and approaches infinity as the zero kelvin temperature is attained.
Efficiency in the form of expansion ration (p)/Adiabatic expansion ratio:
V
p=-3 =-4
V
v2 v1
Tr
Efficiency(ri)= l-T
(V,)
A www.ChemistryABC.com
.2
For adiabatic curve TV'-1

t ~
p
T2V.2(r-1) = TV (r-1)·
r
I 3
T 1 1 .
=> r: = ( ~ = (pr ,y~
,)
(rt)
om
Relation between and· ( ~) :
.c
C
B
yA
tr
A certain engine which operates in a Carnot cycle absorbs 4 kJ at 527°C how much work is done on the engine
is
per cycle and how muchhealis evolved at 127°C in each cyc1e?

m
. The efficiency ofthe Carnot cycle is given by

he
.C
w
ThJS,
w
Thus, the heat evolved in the present case is

w
400K)
q1 =- - . (4k!) -2k.J
( 800K
and the work done on the engine is
w= -(q 2 + q1) =-4+2 =-2kl

Toe negative sign indicates that the work is actually done by the engine.
What% ofT 1 ofT2 for a 10% efficiency?
-1- T, ~=1- T,
1• TJ - T2 => 100 T2
T1 =1-~= 90 =0.9
T2 100 100
~ IT1%=90%T2 I
3. Calculate the maximum efficiency ofan engine operating between l 1;0°C and 25°C. ·
Soln. Maximum efficiency ofan engine working betweentemperatures T2 and T1 is given by
T/ = (T; -T; )/ T2 =(383K 298K) I 383K = 0.22? =22.2%

4. Heat supplied to a Carnot engine is 1897.8 kl How much ~eful work can be done by the engine which wor:
between 0°C and 100°C?
Soln. T2 = 100+ 273 = 373K; T; O+ 273 = 273K; q2 1897.8 kl,

w= q2 (T2 -T;)! T; = 1897.8 klx(373K -273K)/373K = 5f)S3ikJ,
om
Therefore,-work_done by the engjne 508. 7 kJ i/ •/
An important generalisation
.c
C
... (i)
B
yA
Equation (i) may a1so be written, in the general form, as
qrev = constant
tr
T
is
where qrev is the quantity ofheat exchanged in a process carried out rever~ibly at a,temperature t. This is:
q; ,as well shall see presently, represents a definite quantity

m
important generalisation since the quantity

he
state function, the entropy change ofthe system.

5. A Carnot cycle operates on a temperature difference of200K. One-third ofthe heat absorbed from the sour,
.C
at T2 is discharged as waste heat to the sink at T1• The cycle does 400J ofwork. Calculate q1, q2, T 1 and 1
[GATE199
w
Soln. Given, Qh =Q (for source)

.
w
Q
3
w
QC (for sink)
w =-400J and t1.T = 200K
r, I-QC 1-R ~ r,=I

Qh 3Q 3
T,, - T,, 2 200
T/ = T,, or 3 = Th or Th 300
t1. T T,, - T,,

200 =300 I::
I::= 100
~1,1'-~;;
~.--.
w 2 (-400) ....
Since, 11 = -QUZZES
+Qh => 3 = MORE NOTES ON
Qh . => Qh =600. www.ChemistryABC.com
Q = 600 =200
C 3 .1
A Carnot engine operates between 600Kand ~OOKand absorbs 2,000 calories from the source per cycle.
The wbrkdone (in Cal) per cycle is ·· , [GATE 1998]
(a) 1,000 (b) ~,00 · ·(c) 666 (d) 2, 000
T;, -Tc
Since, 11 =
T;,
800- 600 200 I -w
17=---=-=- => rJ=-
800 800 4 +Qh
om
11 = -w or w = 500 cal I
4 2000. ,. . __- ,----'·
I
.c
The coefficient ofperformance ofa perfect refrigerator working reveraiblybetweenthe temperaturesTc and Th
is given by . [GATE 2004]
C
T. -T . 1,T. ./'IJ
B
(a) Tc ;T;, (b) h T c (c) T. -T . (d) T., ~T -
·c h h
yA
C C C
The ratio of heat transferred from a lower temperature to a highertemperatureto the work done on the
~ = JqcJ
tr
machine to cause this remova~ i.e... w·

is
The less the work done the more efficient the operationandgreater Jhe,co.e.fllliient.ofperfor,mance.
~rev> ~irr
m
lqcl = Tc
~=
he
lqhJ-lqcl Th Tc
.C
The plot that describes a Carnot cycle is: [GATE2006]

w
w
p
s V
D D C> C>
s
w
(a) (b) (c) (d)
T
T V T
. , 'Correct option is (a)
In a camot engine 200 cal heat is given to the sink by a reservoir at 27°C. Iftemperature ofthe rource is 57°C,
then find:
(i) How much heat flows from the source.
(ii) Efficiency ofthe engine
(iii) Work done by the reservoir.. HAND MADE NOTES www.ChemistryABC.com
Soln: Here,given, Q2 = 200 cai Q1 =?
QUZZES T2 = 27°C = (27 + 273)MORE
300K
NOTES ON . www.ChemistryABC.com
T1 = 57°C = (57 + 273) = 330K; 11 = f and W ~?
. 01 02
(1)Wehave, - = -
¼ T2
. Q1 200cal Q1 200
330K = 300K ~ ll = 10 cal
02 200 10 1
(ii) 11= 1 -QI ~ri= 1- 220 ~ 11= 1- -1} ~
.
11=, -i-
11
(iii) W= Q1 - Q2= (220-200)cal= 20cal
OR
(ii}11= 1- T2 = 1- 300 1-~=_!_
om
¼ 330 11 11
. 02 '1 200 •' '1 200 10

'
20op,.;·
.
(1) 11= 1- - ~ -= 1-- ~ 1--= - ~-= - ~ Q=220cal
.c
QI · 11 QI 11 Ql 11 QI I .
(iii) W= Q1 Q2~ (220 - 200)cal = 20cal
C
10. Temperature ofthe sink ofa carnot engine is 7°C and efficiency is 50¾, th6ft calculate change in temperatu
B
ofthe source to increase the efficiency ofthe same engine 70%.
yA
. Tsink
Soln. We know that' ri= 1- T ... (i)
source
tr
· 50 1
Case-I: ri= 50% = =- ~ Tsmk
. = 7°C = (7 + 273) K = 280:K
100
..•
, '2.
is
' '
1 280K
from equation (i), - = 1 - T
m
2 source
280K l 1
he
-- 1- - ~ Tsource= 560K
T,owce 2 2
70 · 7
.C
Case-II: n'I = 70% 100 -- 10 ~ . = 7°C = (7 + 273) K = 280K

Tsmk
from equation (i)
w
7 280K 280K 7 3
-=1- ~--=l--=-
w
10 T.ource T.~wce 10 10
w
T
source
= 28
0Kxto
3x l
~ Tsource= 933.33K
Change in temperature ofthe source= (933.33 560)K = 373.33K
11. The efficiency ofa carnot's cycle is 1/6. If on reducing the temperature ofthe sink by 75K, e:fficency becorr
½, then find initial and final temperature between which the cycle is working.
Soln: We have, ri
[Using equation (Q]
l l 75 75 HAND1MADE NOTES
75 www.ChemistryABC.com
-= -
3 6
+ - ~ - = - -- ~
T1 T1 3 6
= -~T=450k
T1 6
' K= 5 x 75 K= 375 K
(ii) =:> T2= 5QUZZES
x 450 MORE NOTES ON www.ChemistryABC.com
6 i'
A Carnot refrigerator talces heat from water 0°C and discard it to a room at 27°C. 1 kg ofwater at ooc is to
be changed into ice at 0°C. How ~y calories ofheat is discarded.to the roopi? What is the work dore by the
refrigerator? In this process what 1s the value of cop? ·•
(Hot)
om
\
.c
C
SinkT2 l 1_1
\
B
(Cold)
yA
Here T I= 27°C = 300K, T 2 0°C = 273K
1kg 0°C Water -t 0°C ice
tr
Cal .
Q2= 1 x 103 g x 80 - g =:> Q2 8 x 104 Cal
is
m
Qi Ti . Q 8 104 1 300
We have, Qz = T2 =:> Q1= Q2 x T2 =:> i= x ca x 273,
he
Work done Q1 - Qz== 8 x 10. 4 x 300 .

Cal-,- 8 x 104 Cal= 8 x 104 x [300 -1 Cal J
273 273
.C
= 8 x 104 x 27 JCal= 72 x IO .· Cal= 0. 79 x .10

4 4
Cal
w
[ 273 91
w
8xl04 Cal 273

and Cop (/3) - ~ =---,,=----~-
4 300 -1Jca1 27
w
Q2 8xl0 [
273
A Carnot engine operates at 55 % efficiency. If the temperature ofreject steam is 105°C, then the absolute
temperature ofinput steam i s _ _ _ _ [GATE 2014]
1. We know that,
Efficiency ofcarnot engine,
T/ 1--
r;
... (1)
~
Given, · T/ = 55% = 0.55
And T2 =105 + 273 =378K
From equation ( 1), HAND MADE NOTES www.ChemistryABC.com
378 378
0.55 =1- K =:> 0.55-1 =- - K
'T' 'T'
-0.45 =_378 => I'i = 378 840 K .
I'i 0.45
Concept of Entropy :
g
Since we know that if dZ =O, then~ is an exact differential and Z is a state function. Similarly,
we can conclude t~at q(rev) IT represents a state fun,.ction because the value of its differnetia]
d(q(rev)!T) over a cyclic process is zero. This stat~ function is given the name entropy of the
1
system and is represented by the symbol S. Thus ,
dr('·_
0 -
d(q(rev)) and S q(rev)
=--'-~
.
T T . ~
om
The entropy ~ction is a ~nction ~f the independen.t variares which are used to de~ne !he s~tt
of a system. It 1s an extensive funct10n. The change ni the yalue' of the entropy function m gomi
from one state to another, is independent of the path, antl tb'.e cyclic integral of dS for cyclic a
.c
change of state is always zero.
C
Unit of Entropy: 11
B
The unit of entropy function is energy-qnit divided by temperature unit.
For example : JK-1, cal K-1.A unit of 1 cal K.-1 ( non-SI) is known as an entropy unit and is representec
yA
by the symbol eu.
tr
The value of dq (irr) for an irreversible cyclic process :

is
T
If there is any irreversibility at any stage of a cycle, the net work obtained lwnetl in the cycle isles;
m
than the maximum work obtainable from the reversible cycle operating between the same tw<
he
temperatures. Consequently, the efficiency of an irreversible cycle is always less than the efficienc~
of the corresponding reversible cycle. It follows that
.C
lwnetl ½-I'i
- - < - - or
q2 + qi ½-I'i
<
q2 ½ q2 ½
w
Or,
!h._<_I'i q2+!h_<O
w
'7' or '7' T.
q2 1 2 12 I •
In the limiting case of infinitesimal heat transfer, this inequality becomes

w
qdq~rr) <O
The fact that the sum of i

2
and i, is zero for a reversible Carnot cycle and is negative for a cycl
I
involving irreversible isothermal expansion or/and irreversible isothermal compression may b
rationalized as follows.
Since, in the isothermal reversible process of a Carnot cycle, the heat transferred is directl
. l tothe temperature, we have T
proport1ona q2 = constant an d T.
qi = - constant ( smce
' ; negative,
q1 IS . '
2 , I
q2 q,HAND
. MADE NOTES www.ChemistryABC.com
It follows that the sum of T and T. IS zero.
2 I
'.iID an irreversible cycle operating between the same two temperatures T 1 andT2, we have
· (i) q2 (zrr}<
q2 (rev) and hence q2 (irr )IT; www.ChemistryABC.com
will be less positive as compared t~d:he·value of
. . . . : '· .
q2 (rev)n;
(ii) ql (irr) < qi (rev) or jqirrl > 'qrevl and hence ql (irr )11; will be mdre negative as compared to the
value of q1 (rev) n;.
Thus, in a cycle involving either of these two irreversible steps or involving both of these
steps, the sum . ·
q2 (irr) q1 (irr)
------'---'- +---'-----'-
I; 7; .,
will always be negative, because q1 ( irr) I 7; is negative anc:l- its magnjthde is greater than q2 ( irr) I I; .
' \ II '
om
' :}
A General Proof for reversible process 4dq(rev)/1,~ q ,nd irreversible proc~ss

·4dq (irr) / T < 0 :
.c
· We have seen that for a Carnot cycle
C
Ji
B
-w=4'iq and cfdq~ev} =0 . 1
yA
· Now consider any other cycle, howsoever complicated (containing any number of temperature
reservoirs and any working substance): For this engine, according to the first law, we have
tr
-w =9dq
1 1
••• (ii)
is
and let for this engine.

m
4d;' > 0 ... (iii)

he
The two engines are coupled together to make a composite cyclic engine, represented by the
symbol, 'eye'. The work involved in the composite cyclic engine is
.C
weye =w+w 1
which, by equation (i) and (ii), are equal to
w
-wq<: =4dq +qdq'= qdq(cyc) ... (iv)

w
If equation, (i) and (iii) are added, we get

w
A(dq(rev) dq') dq(cyc)

'j · + - > 0 or 4 - - - > 0
T T T
If it is adjuste~ that the composite cyclic engine does not produce any work, then wo/e = 0 . With
this condition, equation (iv) becomes
qdq(cyc) = 0 ... (v)
Since each of the cyclic integral can be considered as a sum of terms, we may write equation (v)
and(iv) as
... (vi)
!l!..+ <12 +11+ ...... > 0 ... (vii)

i; I; r; HAND MADE NOTES www.ChemistryABC.com
The expression on the left hand side of equation (vi) includes a number of terms, some positive
and some negative. The positive ones just balance the negative ones and the sum is zero ..
In equation (vii) the net sum can be positive only if the positive terms are divided by sm
numbers
QUZZESand negative terms are divided by large
MORE NOTES ON numbers. This amountsto the fact that we i
associating positive values of q with low temperature and negative values with high temperatur
This is against the second law of thermodynamics. Thus, our approximation that
qd;' >0
must be wrong. It follows that for any engine, we must have .
qd;' ~o.
Now we will show that for a reversible cycle qdq ,'/ T =0 whereas qd1 'IT < 0, for an irreversil
cycle. .)
om
Case-I : Reversible cycle. Let us assume for this cycle\
1i <O' t
.c
We can reverse this engine, which changes all the signs but not the magnitudes of the qs. Thus,·
have,
C
cio/ >0
B
yA
which obviously is wrong as deduced earlier. This leads to the conclusion that for any system.
qdq~ev) 0
tr
for all reversible cycles.

is
m
Case-II : Irreversible cycle. Since the heat and work associated with an irreversible cycle :
different from those associated with the reversible cycles, therefore, the only possibility left
he
qdq' < 0
T .
.C
for all irreversible cycles.

w
Entropy and unavailable energy:

In the carnot cycle working between two temperatures TiCsource) and T1(sink:). Ifq2 is heat absorbed fi:
w
source at T2 and qi be the heat rejected to the sink at TI' then work dorie is given by
w
-w = q 2 - q 1 =energy available for doing useful work

qi =unavailable portion ofheat i.e. unavailable energy
TJ 1-.9:!.=l- Ti
q2 T2
... (1) .

a
Thus for given sink (T 1), unavailable energy;(q1) is determined by (q/T2). Hen~e entropy IS unavailable ·
energy ofsystem, ie. law ofincrease ofentropy forMORE
QUZZES irreversible
NOTES ONprocesses is connected with the hw ofdissipation
ofenergy. Therefore, the second law ofthennodynarnics says that~ntropy is a ~ure ofunavailable energy.
Spontaneous (irreversible) process and non-spontaneous (irreversible) process:
The process which proceed by themselves and bring the system clo~er to the ~uihbrium are called s)'.X)ntaneous
process. All natural process are spontaneous. .:
The process which can't proceed by themselves are called non-spontaneous process. These require
external force and brings the system away from equilibrium.,. ·
In spontaneous processes entropy ofuniverse is completely increasing ie. u~vailable energy goes on
increasing and the workable energy is continuously de<ireasing. A spontaneous process comes to an
end at equilibrium, and since at equilibrium the entropy, isJnaximum, it can·be remarked that nature is
trying to attain the state ofmaximum entropy. ' \
Entropy is a state function
dS = dq = an exact differential
om
\
T
=> Entropy is_an extensive property.
=> , . Entropyofreaction:
.c
Product Reactant
~s(reaction) = L nS L ms
C
.
..
S = absolute entropies ofrespective substances. \
B
n and m = moles ofproducts and reactants respectively
yA
For a reaction, m,A + m2 B--)>n 1C +n 2D
=.[ J
tr
_ AS(reaction) n,S(C) +n2S(D) ] - [ m1S(A) + m2S(B)

is
Entropy· changes in· reversible process:

Let us consider an isothennal reversible process. In this the system absorbs heat q from surroundings at
m
temperature T. This results in an increase in the entropy ofsystem given by.

he
+q
ASsystem = T
.C
-q
ASStrrroundings = T
w
Total entropy ofthe process ASsystem + ASsurroundings
q (-q)
w
ASprocess =T+ T
w
I ASpro=s = 0 I ... (1)
Entropy changes in an irreversible processes:

Let q amount ofheat is transferred from system which is at higher temperature (T2) to the surroundings which
is at lower temperature (T1)
Decrease in entropy of system =

. T2
Incease in entropy of surroundings = +q
T1
Net entropy change ofprocess =AS system + AS surrounding
. .
QUZZES
dS,_.=;>; =qU-;Jf;l;~l] MORE NOTES ON www.ChemistryABC.com
T2 .;;,_ T1- ~ +ve
1~sprocess =+ve I or 1 ~s > o 1 ... (2)

Combining equation (1) and (2)
I ~sprocess -~ o 1
. Adiabatic processes are isoentropic:

dS= dq
T
For adiabatic process dq = 0
om
I dS=O I or S= constant i
!l
.c
Relation between ( expansion coefficient (a) and isothermal compressibility (P):
C
/'I:
f'
B
yA
tr
Problem: For C6 H 6 (.e), a= lxl0-3 K- 1 and~= 9xl0--4MPa-1 at300 Kand 10L325 kPa pressure.:
the change in molar volume which will be required to produce an entropy ch~nge_ of 2JK ~1~01- 1 at 27
is
Assume ~ and ~ to be constants.

m
l a
he
Soln. We know that, (.: =

~
.C
~s a
Or, - =-
~v ~
w
w
w
Entropy as a function of temperature and pressure :

Since, we know that dependence of entropy on temperature and pressure as
S=f(T,p)
On differentiating, we get
88
ds=(as) dT+( ) dP ... (i)
ar p aP r
In order to write the equation,
I p
dS=-dU+-dV ... (ii)
T T
from equation (i), we consider the following relation. -., -·.:j\';
QUZZES
U=H-PV MORE NOTES ON www.ChemistryABC.com
~ dU=dH-PdV-VdP
\
Taking H ~ f(T,P), we get
Ju7{(:;), n+(:!), dP} PdV-VdP ... (iii)
Substituting equation (3) in equation (2), we get
dS l{( 0 H) dT +(
8H) dP} P v/ P
-dV-,-dP+-dV
r ar p aP r T T T
'
',i
Or,
om
Comparing th.is with equation (1 ), we get
- '
.c
.... (iv) .
C
_(as) =.!.[(aH) -v]
B
And . [jp T T [jp T
yA
Entrc,py change in an isobaric variation in.temperature :
tr
Equation (iv) gives the dependence of entropy on temperature at const.a,;it presSl,lre. Since both CP and T are
is
positive, therefore, s increases withincrease in temperature ofthe system when the pressure is hell constant.
For a finite change oftemperature at constant pressure, we have
m
M = r.J2 nCp dT
he
. T
1i
For Cr:
.C
Case-I : If CP is considered independent oftemperature in the range T1 to T2, then

w
M nCP 1n I;
Ti
w
Case-II: If temperature dependence of CP is available in the form ofth'? analytic expression.

w
· · Cp =a+bT+cT 2 + .....
Then we have,
2
2
r.J (a+bT+cT +..... )
M = n--'---------'-- dT
. T
1i
. . . . .
Problem : Calculate M for 2 mol of nitrogen heated at constant pressure from 300 K to 400 K. Given
the temperature variation of CP of nitrogen as
1
Y'llC mor 1
~20+4(;}-2x10-,(;J
Solo. Since, we know that,
(aras) =cp
P T
Substituting the given expression ofCp, we get
(as) mor =( )+A

oT p
/JK- 1 1 20
T· . . ·
·(.I_)
I K-2x10_3
. K2
dS .·.· =(20 +~- 2x 10'"3

JK- 1mor 1 T K
(I_)dr)
K 2
,. ' ;
On integrating this we have,

4
= ooK(20 J
4
-+--2xl0-
300K T
3 ( T ).
2
K
dT )·
K
om
4 400
' = [20 In ( T j] 00K +[4 (!_)]400K -[2 }·0- x 3
,( r:· ~]
K) 300K K 300K . K ) 300
.c
= 20x 2.303x log(:)+ 4(400-'- 300)-2 xrn-'. [ (4~)' -(300)']
C
\. -
B
3
= 46.06x0.124+ 4xl00-2xl0- ( 400-300)( 400+300)
yA
= 5.70+400-2x100x700xl0-3 ·
· =405.70-140=265.70JK-1
tr
Combination of first and second law:

is
For a reversible process

m
(i)According to first Jaw, dq = dE + PdV

he
(ii) According to second Jaw, dS = dq

T
dE+PdV
.C
From these two, dS =- - -

T
w
I TdS=dE+PdV I ... (1)

w
Entropy change for ideal gas in different cases:

w
.. dS dq = du+PdV
T T
dS = nCvdT + PdV
... (A)
T
Or, dS=nC dT +PdV

VT T
dT dV
Or, dS=nCv-+nR-
T V
Integrating it without limit and taking limit.
dT J dV
JdS= JnCvT+ nRV HAND MADE NOTES www.ChemistryABC.com
Or, IS =nCvinT+nRinV
QUZZES
+S0 I MORE NOTES ON www.ChemistryABC.com
For with limit,
S2 T2 dT V2 dV
~
f dS=JncvT+nR
~
fv
~
)
T V
Or, S2 - S1 = nCv In ___1._ + nR In ___1._
Ti// J1i
. T V
Or, AS= nCv In ___1._ + nR In ___1._ (1)
Ti J1i .',,) .
Equation (1) is reversible, ideal but isothermal.
... (2)
om
\
I
?zJ!; nRT2 / ... (3)

Fromequation(2) and (3), we get (
.c
v; . -
PiVi r; =>-.
-=-
- r;- ·Pi-
C
Pi J1i Ti Vi Ti Pi
,,
'·'
B
.
From(2),
r; r;
yA
P.
M == nCv In-+nRin-+nRin-1 ·
7; Ti Pi
tr
T, P.
M n(Cv+R)In-L+nRin-1
Ti Pi
is
m
T, P.
M=nCPln-2 +nRin-1
Ti Pi
he
PV ==nRT
.C
PdV +VdP=nRdT
PdV =nRdT VdP
w
From(A),
nCvdT + nRdT -VdP
w
dS
T
w
n(Cv +R)dT-VdP
Or, dS=
T
dS = nCpdT VdP
Or,
T
dT VdP
Or, dS =nCp
T T
dT nRTdP
Or, dS=nCp- .•
n
T p p
dT dP HAND MADE NOTES www.ChemistryABC.com
Or, dS nC - - n R -
PT p
5econd Law ot l hermoa:yna1
For widxmt limit,

JdS= JnCP 7dTJ

- nR dP
p
Or S =nCP InT-nRinP+S0
'
For limit,
(i) Isothermal Process:

· In isothennal Process T =C011$tant
om
J
T 1 =T2 =T => 1n(~:J=In'.(l)=O /
fromequation(l)
.c
_
6Sisolhtnnal - nR lnl(VYi)l
2
C
(ii) lsochoric Process!
In isochoric process
B
yA
V=Constant =>· V I =V2 =V.
..
~. . => . (~') . (V)
tr
l
ASisochoric= n CV In 1:) + n R In V
is
m
he
.C
w
w
=>
w
. Now, ideal gas equation is

PV=nRT
=> P1 V 1 nRT1 ... (ii)
=> P2 V2 nRT2 ... (m)
From equation (iii) and (ii), we get
T2 -
Ti P1V1
Now making substitution in equation (i), we get
AS= n CV In (lp2 .v21) + nRln (lv21)

P1 vi Yi
~:~ _
..... .. --
_
--
j ··:::::>
C ln (P2 )+n(Cv +R)ln(V2)
L\S= nQUZZES MORE NOTES ON www.ChemistryABC.com
. V P. V.
I !
1n(p2)+nC 1n(VV .
AS= nCVP. p
2
) (·· -C
CP V
= R)
l I .
This is the general express for the change in entropy for an ideal gas when pressure and vohnric(P&V) vary.
(iii) .. Isobaric process : ,. · .. . - ·
,In isobaric process '.. · · ·
P= constant => PI= P2 = ·P
from equation (iv)
· => (P) (v)

. L\Sisobaric= n CV ln p + n cp 1n l ~ => L\Sisobaric= n'Cv ln _1 + n Slln. l ~)
. ' /(v')
/
om
\ .
L\S isobanc 0"+ n Cp ln vi
(v2)
l-
.c
[ SOLVED PROBLEMS)
C
B
Ten moles of a gas is allowed to expand from a state, A, at 10 atmand 300K;to a state B, at 100 aw.and. .
yA
600K. If the value ofCP is 6. 99 5 Cal deg- 1• Calculate the entropy change ofthis process. [GATE 19981
. /lS = 2.303 nCP ln i,,.. + 2.303 nR log-

1
r ·R
Smce, p
tr
.LI 2
is
600 100
M = 2.303x10x6.955log-+2.303x10x8.314log-
300 10
m
2
M = 2.303x10x6.955 log-+2.303x10x8.314log10
he
3
M = 2.303x10x6.55[log2 log3]+ 2.303x 10x8.314xl
.C
bS =160.1(0.301)+ 191.47
M = 239.66 J!K-mole
w
For an irreversible adiabatic expansion ~fa perfect gas from volume V. to V1 the change in entropy the ga;. of
w
is [GATE 2001]
w
(b) zero (b) less than zero (d) greater than zero
For irreversible adiabatic expansion entropy is greater than zero {dS) irr ~ 0
The change in entropy when one mole of an ideal gas is compressed to one-fourth ofits intial·vol~ and ·
'simultaneously heated to twice its initial temperature is [GATE 20041 .
(a) ( Cv R) ln 4 (b) ( Cv - 2R) ln 2 (c) ( Cv - 2R) ln 4 (d) ( Cv + 2R) ln 2
Since, we know that, M = Cv ln ½+ R ln V2

Ti Vi
V HAND MADE NOTES www.ChemistryABC.com
Given, ~ =V, fl;=-, I;= T and T= 2T
4
----··--- ----·· --· .. -· ···--v··-.••1
QUZZES M=Cvln ( -2T) +Rln-.

T
V NOTES
MORE -.· M=C
ON
4V ~ · • v
.ln2+Rln
· (1 ).
~.
4
==> !1S = Cv 1n 2 - R ln 4
M = Cv ln2-Rln(2)2 ==> M = ln2-2Rln2; ==> M = ln2( Cv -2R)
==> L\S =(Cv - 2R) 1n 2 .

. . - .
Correct option is (b) . . .

4. A box of volume V contains one mole of an ideal gas. The probability that all N parti2tes will. be fo
occupying one halfof the volume leaving the other halfe:r,npty is [GATE 2008]
(a) -
1
2
2
(b)-
N
(c)Gr (d) (!f
Solo. Occupation ofn molecule out ofN molecule in the v~lu~e V out of~otal V.
.· \ I'/ ,
om
N! (V)n(V-v)N-n · I
Total volumer Vi .
/
n!(N-n)! V v ··
.c
V
· v=- andN=n
C
2
. m(Z~nil2; r(v-: 12
r B
yA
tr
o{ff (:V )"

is
(H
m
0
{ff(H (-.- 0!=1 and =I)
he
lx(ff xi ~{ff
.C
Correct option is (c).

w
5. The change in entropy when two moles ofArgon gas are heated at constant volume from 300 K to 500
w
E: [GATE 2010]
w
(a) -12.74 JK- 1mor 1 (b) -6.37 JK- 1mo1- 1 (c) 6.37 JK-1mo1- 1 (d) 12.74 JK- 1mo1- 1
Solo. Given, for Argon Cv = l.SR, n = 2
T., = 300K and T2 = SOOK
½ 500
M = 2.303 nCv log-= 2.303x2xl.5x 8.314x log-
T., 300
= 57.44x 0.22 = 12.74 JK-'mole
= 57.44[log500-log300]
= ( 57.44)x(-0.22) = -12.74 JK-'mole

Correct option is (a) HAND MADE NOTES www.ChemistryABC.com
...,,.
;:,.
..--
..
.-..-
For the process [GATE201l]

1Ar( 300K, 1 bar)~ l Ar( 200K, 10 bar)
assuming ideal gas behavior, the change in molar entropy is:
(a)-27.57 J K-1 mo1- 1 (b) +27.57 J K:-- 1 moI- 1
(c)-24.20 J K- moI-
1 1
(d) +24.20 J K- 1:. moI- 1
1• Argon is monoatomic gas, for monoatomic
Cv ==I.SR and Cp == 2.5R .,
T. R
M == 2.303 Cp logi+ 2.303Rlog-1
Ti Pi
: )
200K J ' J f I bdr
M == 2.303x2.5x8.3I4 log K +2.303x8.3I4 /log
K-mole 300 K-mq,e , IO bar
\/
om
J .· J ... ·
M = 47.86 , [log200-J.og300]+ 19.15-- ,.[Io~Y-loglO]
K-mole ·· .· K,mole
.c
·==47.86 J (-0.176)+19.15 J (-1)
C
K-mole K-mole
l.
\,
B
1 1
=-8.42 J -19.15 J · ==..:..27.57 JK- mole-
K-mole K-mole
yA
The maximumnon-PV work that a ,system can performatconstantP is [GATE2014]
tr
Wm OOW ~M 00M
is
1. From first law ofthermodynamics

m
AE=q+W
the work done, W may be expansion or non-explansion so
he
A£ = q + ~xpansion + w;,on-expansion
Or, q = /lE + p!:.V + Woon-expansion

.C
q = Ml wnon-expan.sions (':!:ill+ pliV =m)

w
Since, we know that,

w
w
TM == m - Woon-expansion
Or, Woon-expansion
,
=m - Tt:.S (': t:.G = m-TM)
Woon-expansion = t:.G
The non-expansion work is also termed as non-pv work or useful work.
The !iG shows the maximum non-pv work that a system can perform at constantp. ·
The van der waals constant a and b ofCO2 are 3.64 L2 bar moI-2 and 0.04 L mo1-1, respectively. The value of
R is 0.083 bar dm3 mot- 1K:- 1• If one mole of CO2 is confined to a volume of0.15L at 300K, then the pressure
(in bar)exerted by the gas, is_--"-- HAND MADE NOTES www.ChemistryABC.com
· [GATE ·2014]
· ·�
Soln. Given,
QUZZES a= 3 .64L2 bar mor2MORE NOTES ON www.ChemistryABC.com
b=0.04L mor 1
.
R = 0.083 bar dm 3 mor1 K-1

T=300K andP=?
(p+ ;2 )(v -b) = RT

Van der Waals equation for I mole of a gas can be writt�n as
(p+ ;, )(V-b)=RT
On substituting values
om
3 64
[p+ · 2 ]{6.15-0.04)=0.083x300 v
(0.15 )
.c
3 64
(p + · )(0 .1 1) = 24.9
C
0.0225
B
p+l6 1.78 = 226.36 or p = 226.30 =161.78 = 64.58 bar
yA
Criteria for Reversible and Irreversible processes :
Since, we know that
tr
dS = dq,ev/T = ( dU + PdV)!T
is
For irreversible process,

m
TdS> dq,ev
We also write, TdS = dU + PdV (for reversible process) ... (1)
he
TdS>dU+PdV (for irreversible process) ... (?)

Combining (1) and(2), we have
.C
TdS�dU+PdV '. .. (3 )
(i) Criterion in terms ofchange ofentropy: IfU and V (internal energy and volume) remain constant, t
w
for an isothermal process (i.e., Tis constant)
I( I
w
dS)u ,v �(j
w
(ii) Criterion in terms of change ofintemal energy : IfS and V (entropy and volume) are kept const
then, for an isothermal process (i. e. Tis constant)
I (dU)s,v �o I
(iii) Criterion in terms ofchange ofenthalpy: IfSandPare kept constant, the expression T dS � dU + f
may be written as
dU+PdV�O
But dU+PdV=dH
I( dH)s,P� 0 j
(iv) Criterion in terms of change of work function. Combining TdS
HAND MADE NOTES
� dU + PdV with dA = dU - j
for a.ti. mfinitesimally small change, we have

dA~-PdV
At constant volume, therefore, ( dA) v,r
(v) Criterion in terms of change of free energy St1bstituting for dG from
dG=dU+PdV +VdP-TdS-SdT, TdS ~dU +PdV, weru,ive,
dG5VdP-SdT
Therefore, at constant temperature and pressusre, I(dG) P,r :s; 0 \:"
Remark:
MH,P?. 0 fiPH,S C. 0
MG,T?. 0 AVu,s 50
om
AVA,T :s; 0 ATA,V 50
!).TG,P s0
.c
(PROBLEMS)
C
or
The criterion for irreversibility for a process involying no work pressure-Jo1~l work is, [GATE 1995]
B
yA
(a) ( dS)v,u < 0 (b) ( dS)v,u > 0 (c) ( dS)T,P < 0 (d) ( dS)r,v > 0
1. Correct option is (b)
tr
Choose the correct criterion of spontaneity in terms ofthe properties ofthe system alone: IGATE 2001]
is
(a) ( dS)u,v > 0 (b) ( dS)T,P > 0 .(c) .(dS)H,P < 0 (d) (dGJr;v <0
m
t1.
·The criterion for spontaneous change in terms ofthe state functions is: [GATE 2004]
he
(a) dUs,v?. 0 (b) dAT,V?. 0 \C} dSu5/~0

0
(d) dGT,v 5-0
n. Correct option is (c)
.C
w
Entropy - change in reversible phase transformation :

w
Melting:
liquid
w
tllir .,,
(a) Solid ATr
Since, AG== AH - TAS .. ; (1)

For equilibrium, !).G == 0, so :from equation (1)
O==Afl-T!).S ~ Ml TAS ~AS== AH

T
Aflfusion
Tfusion
(b) liquid ~r .,, solid
ASfusion = HAND MADE NOTES www.ChemistryABC.com

2. Vapori~ation:
(a) QUZZES
liquid ~· ' vapour MORE NOTES ON www.ChemistryABC.com
where, Tb is boiling point.
(b) vapour .6 Hv ' liquid

Tb
3.
1~v=-~I
Sublimation: \
(a) solid 'vapour
om
AS = AH sub
-sub T'
sub
.c
(b) liquid~ vapour
C
\
B
AS = - AHsub
sub T
yA
sub
Aflv
tr
Remark: svapour - sliquid = T, Or,

b
is
4. Allotropic transformation:
m
Solid( a )~Solid(P)
he
a & P are two allotropicformofasolid.
AS = AHtrans .
.C
Trans T
trans
w
AStans = entropy change for phase transformation process.

w
AHtrans = .enthalpy change for.phase transition at constant pressure & constant temperature.
Trans phase transition temperature
w
Entropy-:-change in irreversible phase transformation:
substance ( S) ----t substance ( V)

(Ti) (T2)
(i) substance (S )----t substance ( S)

(Ti) (Tr)
Tf
In this condition, phase same, temperature different, AS= CP,s ln T
l
(ii) substance (S)----tsubstance ( S)

(Tr) (Tr) HAND MADE NOTES www.ChemistryABC.com
F"· Jn this condition phase differet and temperature same~
QUZZES
· 1 ~S=~~: I
(iii) substance( t)-----4-subtance( t)

(Tr) (Tb)
In this condition phase same,.temperature different.
'E
AS = Cp ,., In ..JL
T
f
~,
(iy) substance (f) ~ substance ( v)
om
(~)
In this condition, phase,same, tempe,rattrre same, ,/
.c
C
B
(v) substance(v)~substance(v)
yA
(Tb) {Tr)
In this condition, phase same, temperature different.

tr
is
m
Note:
he
(a) In adiabatic reversible process"!
ASsystem =.1S
· surround'mg =AS umverse
. =.0
.C
(b) In adiabatic irreversible process:

w
ASsystem z 0;. AS universe z O and.ASsurrounding =0

w
[---so_r;_Y_ED_P_R_O_BLE-M--.S)
w
Calculate the entropy change for the transformation H2 0(liq,l atm) 4 H20(g, 0.1 atm) at 100°c.
AHvap for H20. is 40670 J. [GATE 1996}
ln. (i) H 20(e,latm)--tH2 0(g,latm)
(IOO"C) (lOO"C)
H20( f, 1 attn)--tH 20(g, 1atm)

(373K) (373K)
AS= AHvap = 406701 = 109.03 J/K

Tb 373K
(ii) H2 0(g, 1 atm)--tH 20(g, 0.1 atm) HAND MADE NOTES www.ChemistryABC.com
(373K) (373K)
. In~ J 1
QUZZES
AS= nR MORE NOTES ON log-
- =>AS= 2.303 x 8.314--x www.ChemistryABC.com
(fo~imole)
P2 K-mole ·0.1
J
AS = 2.303 x 8.314- log 10
K
Therefore, total entropy, AS= (109-19.14)J!K: =128.l4J/K
1
2. Calculate the entropy change in the melting ofl kg ofice ~t 0°C. Heat of fusion ofice = 334. 72 Jg-
M =Ml!= 334.72J g-1 =l.226JK-l -I =1226JX-lk -I

Soln. 1 Tf 273K g · g
3. Calculate the entroy increase in the evaporation of one mole qfwater ~t I00°C/Heat ohaporisation ofwi
at 100°c 2259.4 J g-1• ··
1 1
om
M = Mlv =2259.4 Jg~ X 18g mor =109.03 JK-lmol-1'
Soln. v T,, 373K . .. . /
4. Calculat~ the entropy change accompanying the freezing of one mole ofwater at 25°Cto ice at- I 0
.c
given that the heat of fusion ofice at its fusion point (0°C) is 6.00 kJ mo1- 1, the heat capacity ofic
36.82 J K- 1 mol- 1 and heat capacityofliquid water is 75.31 J K-1moI-1.
C
Soln. (0 H20(R) ~ H20(R)
B
(2s•c) (O°c}
yA
H 2 0(f)~H2 0(R)
{298K) (273K)
273K
tr
For this AS= CP, I n - -

' ,, 298K
is
273
=> AS= 75.31 JK- 1mor 1 In
m
298K
=> AS 75.31JK- 1mor 1 In(0.916)
he
AS= -6.60JK-1mor 1
.C
(iI) H 20(f)~H 20(s)"

(273K) (273K)
w
1 1
AS=_ AHf = 6.00kJmor 6000Jmor =-21.97 JK-'morl
T 273K 273K.
w
H 2 0(s)~H 20(s)
(in)
w
{273K) (263K)
263 263
AS= C ln (36.82 JK-1mol-1 ) In
P,S 273 273
AS =-1.50 JK- 1mor 1
Totalchange of entropy AS = -6.60JK- 1mor 1 + ( -21.97) JK- 1 ( mor 1 ) + ( -1.50JK- 1mol- 1 )
= -29.96JK-1mor1
5. The criterion for the spontaneity of a process is [GATE 2000]
(a) ASsys >0 (b) ASSWT>O (c) ASsys+ASsurr >0 (d) ASsys-ASsurr >0
Soln. Correct option is (c)
b.S0univ. for the following reaction, at-298 K is: [GATE2006]
(a) -197 J K-1 (b) 0 J K-1 (c) -308 J K-1 (d) 111 JK- 1
.
,,
. -1
I. Given, ASsys =-197JK and Misys =-91.Rk:J
Since, we know that
ASuni = uAS sys +ASsurr =- 197 JK -1 +91.8

- kJ
238 K
- I
Sll!"f
AHsysJ
( ·: AS/ ;i = - --
T
=-197JK-1 + 918 00J = -197JK-1 +308JK~t= 111 ;K'"1 /

29.8K .
om
1
Correct option is (d) _. \ / _
The enthalpy of fusion ofjce at 273K i~,,6.01 kJ mot .and the enthalpiofyaporizatio_ n ofwater.at 273K ~
1
44.83 kJ moI-1• The enthalpyofsub.limation(mkJ mot1}ofice.at27.1K.,is .. [GATE2014]
.c
1. Given, H20{ s)~H 20(£) MI1 =6.0lkJmor1
C
(273K) (273K)
l
B
.\
H 20 ( £)----+ H 20 ( g) .AH 2 = 44.83 k:Jmol-1
yA
(273K) (273K)
Therefore, enthalpy ofsublimation ofice,

. . '
1 1
H20 ( s )----+ H 2 0 ( s) => MI = MI 1 + Lill2
tr
=>MI= 6.0 l k:Jmoi- =:>AH= 50.84 kJmor

(273K) (273K) . '
is
Calculation of dS when vander waal gas heated and expanded simultaneously:

m
Consider n-moles ofa vander waal gas, the temperature ofgas is changed from T1 to T2 and at the same time
he
the volume ofgas changes from V1 to Vr This process is done at constant external pressure and in reversible
manner. Then by first and second law:
TdS=dqrev =dE+PdV ... (1)
.C
Let E=f{T,V)
w
8 0
dE=(aE) .dT+( E) .dV [where, Cv = ( E) ]
w
=> ofv OVT. ar v

w
dE CvdT + ( :~ tdV ... (2)

" from equations (1) and (2)
·
TdS =Cv.dT + (BE)
av .dV + PdVT
dS=Cv( ~)+ ~[(!l }v +P ... (3)
now for n-moles ofvander waal gas.
(P<: }v nb)=nRT
- - - - , ---,--···' ----· •..•... -.,·- -·••.11
2
p~ n a
QUZZES
= nRT MORE NOTES ON www.ChemistryABC.com
V
2
(V-nb) ./. (4)
Now (
'8VT
oE) = internal pressure
·
(n = n a Jfor real g.as.
V2
2
.
therefore, fromequation(3), we get
dS=C
v
(d1)+l[nva
T T
2
2
+P]dv ... (5)
fromequations(4) and(5)
dS =.Cv (dT)+![ nRT ]dV

T T (V-nb)
om
... (6)
.c
Taldng Cv as independentofT, integrating equation (6) between suitable limits, we get
C
L\S = nCv [RnT]~~ + nR [ in (V -nb) I:
B
yA
L\S = nCvln T2 +nRln[(V2 nb)] .... (7)
T1 (V1 -nb)
Cases - I: When it expands isothermally, Tis.constant.
tr
L\T = O, then from equation (7), we get

is
L\S = nR£n(V2 -nbJ

m
:::} T V1 nb
he
Case - II: When V is constant; ( V1 - nb) =(V2 - nb)

.C
So, from equation (7), we get

w
w
w
Problem : One mole of a van der Waals gas undergoes a change from 298K and 1 dm3 to 373K and 10
dm3• What is the change in its entropy? Given, b = 0.06 dm3 mo1- 1, Cv = 29.0 J K- 1 mo1-1•
Soln. Since, we know that L\S =Cv In T2 + R In V2 b

Ti ~ -b
373 6
We get, L\S=(29.0J K- 1 mor1 )1n +(8.314J K- 1mor 1 )1n(l0-0.0 )
298 · 1-0.06
=(6.51+I9.6I)JK- 1mor 1 =26.12J K- 1mor 1

~-··--- -:---
Entropy Change when two solids at different temperatures are brought together:·
Consider two solid bodies ofthe same material each containing one mole ofa suJ:>stance. Let one be at higher
te~erature Tb and the qther at lower temperature Tc. Let both of thetµ be broughttogether. Heat will flow
from the hotter body to the colder till thet temperatures ofboth of them arH the same. Let the equilibrium
temeprature be T, the value ofwhich can;be caleulated aj;.follows: : ·
Heat lost by hot body= Heat gained by cold bgdy · ·· ·
mhCP (T,,-" T) = mcCP(r-·y;)
Or,
Consequently, entropy changes oftwo bodies are
T T
CJ.Sh Cp In- ~ llSC = Cp ln-
T,, I;
om
\
The total change in entropy is
M=M11 +Mc _,
.c
T T
· M=C ln-+C ln-
P T. p T
C
h C
B
'I' Tj·· .
M =Cp ( ln-+ln-
yA
T,, i:
PROBLEMS
tr
A 50 g mass of Cu at a temperature of393 K is placed in a contact with a 1OOg mass of copper at a

is
temperature of 303K inathermallyim.llated container. Calculateq and ~total for the reversible process. Use
1 1
a value of0.418 4 J g- K- for the specific heat capacity of Cu.
m
1. We have 50 g copper at 393 Kand 1OOg copper at 303 K. The temperature ofteh two bodies when they mve
come to thermal equilibrium can be calculated as follows.
he
Heat gain= heat loss

m1Cp ( ll~ )= m2Cp (Ill;) i.e. m1 1l'I';_
.C
1'11i_!l~
Therefore,
w
(100g)(T 303K)=(50g)(393K-T)
T=333K
w
T
/.1Sh nCpln-
w
Thus,
T,,
50 1 1 1 333
=( g _1 J {( 0.4184 J g- K- )( 63 g moZ- )} x 2.303 log( K)
63 g mol . 393K
-3.466 JK- 1
T
Mc =nCpln-
i:
=(· 63 lOOg
g mol
1 1 333
_ J{(0.4184Jg- K- )(63g mor )}x2.303Iog( K)
1
303K
1
3.951 JK-1 HAND MADE NOTES www.ChemistryABC.com

--- .. ~ -
2. 5g ice at 200K is added to 30g water at 300K in thermally insulated conciiner. What is the final terrperatun
Given : Mi fusion ofice at 200K = 300 J g-1 and average specific hc,at capacity of liquid water= 4 J K-1g-
(a) 239 K (b) 380 K (c) 429 K (d) 129K . ·
· Soln. Heat required to convert 5g of ice at 200K to 5 g water at 200K
= (5g )(3001 g- 1 }= 1500 J
Final temperature after rrilidng.
Heat gained by ice= heat lost by hot water
-(5g )(300 Jg- 1 )+(5g )( 4JK-1g- 1 )(T-200K) (~og )(4J:K- 1 g- )booK ...::r)
1
'
=> 15001 +(201)(:-200) =(1201)(300-:)
om
15001 +20 TJK- 1 -40001 36000J-120JK-I t
.c
33500
140TJK-1 = 335001 => T = J => T=239K
140 K- 11
C
B
Entropy of Mixture ofldeal Gases :
For one mole of an ideal gas,
yA
dS= dT +R dV ... (i)
T V
tr
Integrating equation (i), assuming that Cv remains constant for an ideal gas, we have
is
S = Cv In T + R In V + S0 ... (ii) .
where S0 is the integration constant.
m
RT
We know that, CP Cv = R and V p , we get
he
S CPlnT-RlnP+RlnR+S0 . ... (m)

.C
CplnT-RlnP+S0 1 ... (iv)

where, S0 '( = R In R + S0 ) is another constant.
w
If n1, 71i, n3 , etc. are the numbers ofmoles ofvarious gases present in the mixture and p 1, Pi, p 3 , etc. are
w
their partial pressures, then, the entropy ofthe mixture is given by

w
S = n1 ( CP lnT-Rln p 1 + S0 ')+ n2 ( CP lnT R lnPi +S0 ')+n3 ( Cp lnT + Rln p + S0 ) •••••

=l'.n(CplnT-Rlnp+S0 ') ••• (v)
The partial pressure (p) ofan ideal gas is gi~en by the expression,
p =xP ... (vi)
where, xis the mole fraction of that particular gas in the mixture and Pis the total pressure..
Substituting p by xp in equation (v), we have
S=rn(CPlnT-RlnP:-Rlnx+S0 ') •• .-(vii)
Equation (vii) is known as the entropy of a mixture ofideal gas.
Entropy of Mixing: Entropy ofmixing is defined as the difference between the entropy ofthe mixing of
gases and the sum of the entropies ofHAND MADE NOTES
the separate www.ChemistryABC.com
gases, each at a pressure P, Thus, ·
nc;.4 """"4".., - · ..... __ . ·----·.,·--···----
Mmix = r.ri( Cp lnT R lnP-R lnx + fio1)-r.n(Cp lnT ..:::;Rlli.\'.P'*~if) ., ·

. '·. -,",
Mm1x =-R Lnlnx=-R(n, lnx1 +n2 lm.x2 +~ lnx3 + ....'.. )

Mmix = -R.. En; lnx; }
where, n; = number ofmoles ofitllgas.

and x;=numberofmolefractionoffgas
n.
x
1
--'
- r.n;
.)
om
I
.c
Or,
C
B
{SOLVED P~OBLEMS)
yA
Assuming that there is no chemical reaction, the change in entropywhen2moles o~,3 moles of1½ and 2
moles ofHN3 are mixed at constant temperature is · [GATE 2002]
(a)-62.79 JK-1 (b) 62.79 JK-1 (c) 125.58 JK-1 (d) -:-:l25..58JK-1
tr
Given, n,, =2, n2 =3 and n3 = 2

is
The mol~ fraction are

m
2 2
---=
2+3+2 7
he
3 3
X - ----
2 - n1 +n2 +n3 _- 2+3+2 7
.C
n3 2 2
and X3 = =---- .
w
n1 +n2 +n3 2+3+2 7

Therefore, entropy ofmixing,
w
Mmix = -2.303R[ n1 logx1 + n2 logx2 + n3 logx3 ]

w
2
= -2.303x8.314[2log +3log!+ 2log~]
7 7 7
= -19.147x[-2x0.54-l.102-2x 0.54]
. =-19.147x[-L08-l.102 1.08] =-19.147x-3262 63 JK- 1
,correct option is (b)
The entropy ofmixing of 10 moles ofhelium and 10 moles of oxygen at constant temperature and pressure,
assuming both to be ideal gases, is · · ·[GATE 2007]
1 1 1 1
(a) 115.3 JK- (b) 5.8 JK- . (c) 382.9 JK- (d) 230.6 JK-
Given, n,, = 10 moles and n2 = 10 moles
The mole fractions are,
10 1
QUZZES
Xi =-n_+_n_ = 10 +10 - 2 MORE NOTES ON www.ChemistryABC.com
1 2
10 1
And = -
10+10 2
=> t:.SmL, = -2.303R [rn, logxj]
=> t:.Smlt: = -2:303R[n1 logx1 + n2 logx2 ]
[ 1
=-2.303x8.314 10Iog + 10Iog
1] ,.
2 2
=-2.303x8.314x20x(-0.3010) 1)5.3 JK~ 1
om
3. 1 molof CO 2 , 111101 of N 2 and·2 molof 0 2 were mixed at 3p0K. The entropy ofmixing is
.c
8R£n2
(a) 6R£n2 (b) 8R£n2 (c) (d) 16R£n2 [NETDec.2011
300
C
Soln. Given, 1Zi = 1, n2 = 1, n3 2
B
Mole fractions are,
yA
1 1
X1 = =
ni +n2 +n3 1+1+2 ·4
tr
1 1
Xz = =
is
fli+n2+n3 1+1+2 4
2 2 1
m
And X3 = = ---
n1 +n2 +n3 1+1+2 4 2
he
!::Smix = R [n, ln xi + n2 1n Xz + n3 ln X3 ]
.C
= - R [1 x 1n .!_+I x In.!_ + 2 x 1n .!_]

4 4 2
w
=-R[21n½+2In½+2ln~]=-6R[ 1n½] 6Rln2

w

w
Clapeyron and Clausius-Clapeyron Equations

The Clapeyron Equation:
In a one-component phase diagram the lines separating the phases (for example, solid from liquid, }quid fo
vapor, and so on) are not arbitrary, they are determined bythe principles ofthermodynamics and equilibrit
Thermodynamics provides an equation for each line which gives pressure, p, as a function oftemperature 1
other words an equation ofthe form p = p(T). It is now our task to find the form ofthe function p(T) .1
diagram below is a portion ofa hypothetical phase diagram and the phase line on this diagram separates a i
f3 . In other words, the phases a and f3 are in equilibrium with each other at any and all points on the lir
p /
HAND MADE NOTES . www.ChemistryABC.com
T
Phase transitions occur at constant temperature and pressure at the points on this line. Consider the phase
ransition. QUZZES MORE NOTES ON www.ChemistryABC.com
a---t/3 ... (1)

Since the a and ·/3 phases are in equilibrium with each other at any
.
point on the
' line we know. that the line
we know that the change in the Gibbs free energy for the transition given by equation ( 1) is zero everywhere
on the a f3 phase line. That is,
jj.G = O ... (2)
We already know that for any process at constant temperature it must be true that ·
jj.G !::J:l7'Tt:.S ... (3) ..
From equation (2) andequation (3), we get
\
!::J:l
b.S=- ... (4) i:/
I
T
For a phase transition,
om
i .· \ / / /
The various quantities appearing in these equations are defined ~s u~tial by ..
jj.G=Gp-Ga ... (5) f
.c
AS= Sp -Sa ... (6)
C
!::J:l=Hp-Ha •. . .. (7)
B
We will also define jj.V the same W.:!,Y,
AV=V13 -Va ... (8)
yA
As we have already said, jj.G =0 alog the a - /3 phase line. That is, AG is contant along the line and it is
being held constant at the particular value ofzero. We want the slope ofthe line. Therefore, the derivative we
~)
tr
(ar
. . ·.
is
want is, t:.G. We can write this derivative in tenns Dfquantities we know using the cyclic rule (Euler's
m
chain relation) as,
aP) (a!:),
he
(ar = -(aAG) ... (9)

.C
t:,G
8P r
w
We know that from dG = -SdT + Vdp, or dAG = -MdT + AVdp, that

w
(.8AG)
ar p =-AS ••• (IO)
w
And (8AG) AV ... (11)

i)T T
Using equation (10) and equation (9), we get
... (12)
Equation(12) is the Clapeyron equation. .

The Clapeyron equation is thermodynamically exact. It contains no approximations. There is another version of
the Clapeyron equation which we obtain by inserting equation (4) for AS into equation (12).
(~; lG =-:r-A-V ... (13)

~ ~ ...... ~...,..,.,I , ..... '"""'""Y'I' ""'"I I I ,...,, • I l""""""J I 'Cl"
Equation (13) is useful when we want to integrate dP to' find Pas i;i. function of T. The integratio
QUZZES
equation (12) or (13) is easiest ifweMORE
can NOTES ON
make the www.ChemistryABC.com
approximation that either AS or Mi is reason
-constant over the temperature range. However, Mi 'usually varies 'more slowly with temperature·
does AS, so that the approximation is better when we integrate equtaion ( 13). Prepare equation ( 13
integration. ' ·
Mi
dP = - dT
TtiV ... (14)
To integrate equation (14) we must know how Mi atid tiV depend on temperature. We have aln
said that Mi can be regarded as approximately constant, but what about ti V ? ti V is certainly not 1
near constant if one of the components ofthe phas~ transition is a gas, but if the transition is solid~:
or solid ~ liquid we can approximate tiV as a cosntant,also. Tb.en . .-
equation (14fr
1integrates to,
',i
Pi P.t = Mil Ti
tiV nT. ... '(15.y),.:t
.
om
-
l / .'
If we let T2 and f\ range o ve~Jhe phase line and call them T and~ respectively, we can write pressun
function of temperature for a solid ~ solid or solid ~ liquid phase transition line as follows.
.c
Mi T
Pz =Pi+ tiV ln '.I'i ... (16)
C
The Clausius-Clapeyron Equation: ; "
B
Equation (12) and (13) are exact and are valid for all types ofphase transitions, solid~ liquid, solid~
yA
liquid ~ gas, and so on. However, for solid ~ gas and liquid ~ gas phase transitions we can make
approximations which will give us a very useful equation called the Clausius-Clapeyron equation. The
approximation comes from noting that the volume ofa given amount ofgas is much larger that the vo lune
tr
equivalent amount ofsolid or liquid. This being true, we can approximate ti V by the volume ofthe gas.
is
JS,
ti V = Vgas - ~iquid ~ Vgas ... (17)

m
Or, ti V = Vgas 'V.01u1 ~ Vgas ... (18)

he
In the second approximation we replace Vgas by the volume the gas_ would have ifit were ideal. That is, v.
nRT
.C
Vgas = p. With these two approximation equation (13) becomes

w
(!:L =-T-:-r-=-ei;:--; ... (19)

w
p .
w
where the bar over the enthalpy, H means that-it is the enthalpy per mole. Equation (19) is known a
Clausine-Claperyron equation.
We should emphasize that equation (19) is not exact (because it contains two approximately) and it
applies when one ofthe phases in the phase· transition is a gas. Equation (19) can be prepared for ntegrat
follows.
dP Mi dT
-=-- ... (20)
2
P R T
Equation (20) can be integrated if we know how Mf changes with temperature or, as we will now do,
regard Mf as approximately constant. With the Mf regarded as constant equation (20) integrates to
... (21)
Equation (21) is the integrated form ofthe Clausius-Clapeyron equation.
·.,.!""
Ifwe want pressure as a function oftemperature we can let the point (Pi, T;,) range1*~ the phase line and call
it simply ( P, T) and then take the antilog ofboth sides.
... ;(22)
Notice that ifwe know two points on the curve'~e can solve for Afl, or ifwe know one point on the curve
and Afl we can solve for any 9ther point or even get the entire curve:Equation (22) is ·very useful for estinat-
ing vapor pressures at various temperatures when Ml and the v~~r pressure at one temperature are known.
Application of Clausius-Clapeyron Equation:

\
(1) Effect of pressure on boiling point : For liquid~ vapotfr '
Since we know that,
om
dP Aflv
dT= T(J/,,"-Ve)
.c
dP
IfV >>V: then-=(+)ve
. V e, dT
C
When pressure increases then dP is positive and ifdT is positive so boiling poil).t inciJase.
. ~
B
dP
If Ye >> V,,. So, dT = (-) ve
yA
When pressure decreases then dP is. negative and ifdT is negative. So, boiling point is decreases.
(2) Effect of pressure on melting point:
tr
For fusion, solid~ liquid

is
Since, we know that

m
dP Afl1
dT= (Ve-V:)
he
1r
1
If V: >> Ve (e.g.: ice, gallium, bismuthese)
.C
dP =(-)ve
dT
w
When pressure increases then dP is negative. So, melting point decreases.

And if Ve>> V: (e.g. wax,sulphuretc.).
w
· -dP- ( +) ve
w
dT
When dP is positive and dT is positive then boiling point increases.
( SOLVED PROBLEMS]
/
The specific volume ofliquid·water is 1.001 mLg-1 and that ofice is 1.0907 mL g-1 at 0°C. Ifthe heat offusion
1
'ofice at this temperature is 333.88 J g- 1, the rate ofchange ofmelting point ofice with pressure ofdeg atm-
will be [GATE 2007]
(a) 0.0075 (b) 0.0075 (c) 0.075 (d)--0.075
• Given, Ve= 1.001 mLg-1, V: =1.0907 mLg-1
dT
Therefore, the rate ofchange ofmelting point ofice with pressure
HAND MADE NOTES
is dP www.ChemistryABC.com
Since we know that,

dP till . Ml
dT = TtlV = T(Ve-V.) www.ChemistryABC.com
dT T(Ve-V.)
Or, -=~---'-
dP till
dT 273K(l.00I mLg- 1 -1.0907 mLg- 1)
Or,
ctr = 333.88 J g- 1
. . . . -I ..
dT <:::: -24. 7881 mLg K = _ . ,
Or, 0 0733
dP 333.88 Jg-I ,
dT I
Or, dP = -0.073 degatm-
om
Correct option is (d) ' 1/
. . - ·. dP 1 ; -
2. For the-liquid ~ vapour etJ_uilibrium of a substance dt at r'bar and 400 K is 8 x 1o-3 b~r K- 1 •
.c
molar volume inthevapourformis 200Lmor 1 and the molar volume in the liquidformisnegligibl
C
molarenthalpyofvapourisation is (l.ObarL = lOOJ) /[NET Dec. 2011)
B
" 'F
(a) 640kJmor' (b) I00kJmor 1 (c) 80kJmor1 (d) 64kJmor'
yA
Soln. Given, :;, = 8 X 10-J bar k-l' T =. 400K, V.; = 200 L mor 1' Ve= 0, tillv =?
tr
dP tillv
is
dT = T(Vv -Ve)
m
8 X l 0-3 bar = tillv

he
k 400k (200 _!:_ - 0 \)

mol
.C
6.H ""8x10-3 bar x400kx200J::_

" · k mol
w
l'lH = 640 bar L

w
" mole.
J
w
6H 1 =640xl00-- .
' mol
6.Hl' = 64000 J mor'
=> llH,, ;,;: 64 kJ mor'
So, correct option is (d)
3. The vapour pressure ofdecane is 10 torr at 55.7°C and 40 torr at 150.6°C. The 6.Svap at 100°C is
(a) 12 J K- 1 (b) 24 J K- 1 (c) 1.2 J K- 1 (d) 2.4 J K= 1
Soln. Given, Pi~ 10 torr, r; 273 + 55.7 = 328.7 K
~ 400 torr, T2 = 273 + 150.6 = 423.6 K
Since we know that, HAND MADE NOTES www.ChemistryABC.com
P,_ _· AHV -
1og--
QUZZES
[T2-T;J MORE NOTES ON www.ChemistryABC.com
Pi 2.303R · T;I;_
log 400 = Allv [423.6-328.7]

10 2:303x8.3I4 423.6x328.7 ,;
lo 40 = Allv [94.9] _ . ..
g 2.303x 139237.32 -=;, Aflv -4504 JOUie
::
/iS 4504} _. . -I
And vap = T = 373K -12.075 JK
. . ,\ {
Calculate change in temp ofboiling water when the pressure is,,increased bi 27. l 2mm of Hg. The normal
boiling point of water at.latm is I00°C. .· • . // , ·
om
Given, Latent heat ofsteam =537caVgand specific volume ofstea~,/4~7/C.C
t _,;-;,~ ' ' ,. - '
(a) lK (b).2K .(c) 3K . ,(a}4K
.c
.n : From clausius-clapeyron EquatiQn,we have
C
,1,r,_.·.~
.
\ r.
B
yA
dT= (V2 - v,) X T X dP ... (i)
MI
tr
Here, dP= 27. l 2mm of Hg

is
27.12
m
dP= 760 atm [ ·: latm = 760 mm of Hg]

he
27.12
;;::;, dP = 760 x 1:01 x lO'N/ml
.C
dP:;,a 0.036041 x 105N/m2 ... (ii)

w
cal
LiH= lg x 537 g = 537cal = 537 x 4.18J
w
L\H= 2244.66J ... (iii)

w
V, = 1.c.c 10-6 m3 => V, = 1674c.c = 1674 x 10- 0 111

3
6.V=V2 -V1 = 1673 x 10=6m3 ... (iv)
100°C = 373K .,. (v)
And dT=?
Now making substitutionfromtheequation(ii), (iii), (iv) and (v) in(l), we get
1673 xlO,,.{i m3 N
dT= . . · · x 373K
2244.66J ·
x 0.036041 x IO'-,
m·
· _ 2249.0629 (N - m) K _
=> QUZZES
dT- 2244.66 J =>
MORE
dT-NOTES
lK ON www.ChemistryABC.com

5. When lead is melt at atmospheric pressure (The melting point is 600K), the density decreases from 11.01
10.65g/C.C and the latent heat offusion is 24.5J/g. What is the melting point at a pressure ofl 00 atmosphe1
(a) 800K (b) 100 K (c) 200K/ · (d) 600K
Soln ·: From clausius-clapeyron equation, we have
om
(V2 -V)
=> dT= l XTX dP ... {i)
AH
i
Here, dP= (100 - 1)atm = 99atm = 99 x 1.01 x 105 N/m2
.c
:::::> dP= 99.99 X 105N/m2 ••• (iJ.)
C
24.SJ
B
AH= lg x - g = 24.5J. ... (m)
yA
· m1 lg c.c l · --o · ·
V= -
di
=- -=
11.0lg
-xlO m
11.01
3
=> V1= 0.090 x 1Q-6m3
tr
I
is
ffi2 C. C l -6 3 -6 3
V2= ~= 10 _65g _ xl0 m
10 65
=> V2 =0.09389x10 m
m
AV= v2 - v1 = (0.09389 - o.09082) x 10-6m3

he
AV= 0.00307 x 10·6m3 ... (iv)

.C
T=600K ... (v)

3
w
And dT= O.OOJ07 xl0--om x 600K x 99 99 x 105 __!i_2

24.SJ . m
w
dT= 18.419 (N-m) K

=>
w
24.5 J
=> dT=0.75K
Therefore, at melting point,
100 atm = (600 + 0. 75)K = 600.75 K
' 6. At what pressure will ice melt at- 1°C assuming thatAH1 is independent of pressure andAHr 6.0(
KJ/ mol. [Given, dH,o = 0.9998 g/c.c and dice= 0.9170 g/c.c].
(a) 175 atm (b) 160 atm (c) 135 atm (d) 200 atm
Soln: From clausius clapeyrom equation
log Pi
AH! ( 1
= 2.303R T1 -
1
T2
J HAND MADE NOTES www.ChemistryABC.com
... (i)
. AH (½1
p2 1QUZZES
atm= - AV1 x 2.303 logIO T
1
l-J MORE NOTES ON ... (ii) www.ChemistryABC.com
Given, Lilly 6.0095 x 103J/mol ... (pi)

The molar mass ofwater= 18 g/mol
= 18 x 10-3 Kg/mol
3
18 X Hr kg'.mor1 18 X 10-{j 3 z-1
V = - = 0.9170xl0-3kg
mice = 0.9170 m mo
I dice 10-6 m2
18.10-3 kg.moZ-1 18xl0-{j 3 ,-1·

mH 2o _3 - - - m mo . .,
v2 = d= 0.9998x10 kg 0.9998
om
H20 10-tim3
V2= 18.00 X 10-ti,m3mor 1
.c
!lV= V 2-V 1=(18.00- 19.62) x 10-ti m3mol- 1
!lV= - 1.629 x 10-{j m3mol-.1
C
... (iv)
1, /1
B
\
T=0°C
1
273K
yA
T=
,2
1°C 272K
)=
tr
2 272
log 10 (T log 10 ( K) =log 10 (0.996)
T1 273K
is
i. )=
m
log10 ( -0.0015 ... (v)

he
Nowmakingsubstitutionfromthe.equation(iii),(iv)and(v) in(ii}weget
6.0095 x 103Jmol- 1 x2.303x[-0.0015]
.C
p2 1 atm- -1.629 m3mor 1 xlO-{j

w
0.022xl09 J
w
p2 1 atm= 1.629 mJ.

w
Nm
.'.. latm = l.OxlO5 - x- 2
' 9 m m
0.013xl0 ___ 5
p2 latm= m3 x l.OlxlOs atmm
2
latm.m 3 =l.0lxl0 Nm.
=> latm.m 3 1.01x105 J
, - => P2 -1 atm= 0.01343 x 104 atm

=> P2= latm +0.01343 x 104 atm; P2= 1 atm + 134.3 atm
=> P2= 135.3 atm

( PRACTICE SET )
1. The M accompanying the evaporation of one mole of water at I00°C, assuming the MI W)P ofwater i
cal g- 1 is
(a) 540 calg-1 (b) 25 calk- 1 mo1- 1 (c) 20 cal lc1 mot-1 (d) 26.0 calk:-1 mo1- 1
2. A carnot engine converts one-sixth of heat input into work. When the temperature ofthe sink is reduc
62°C, its efficie_ncy is doubled. The temperature ofthe source and the sink respectively
(a) 372K, 310K (b) 372K, 410K (c),4\0K, 372K (d) 410K, 310K
3. A carno t engine operates between a source at temperature 300K and a sink at temperature 200 K. Le·
the efficiencyofthe Carnot engine when the temperatures ofboth the so~fce and tb,.e sink are decrea~
the same amount. Then, ' / ' '
(a) 11<½ (b) 11> ½ (c) 11= .· ½ )/ :'
(d) rt=½
om
4. The change in entropy when two mole ofice are heated frof-:1 (fc to 10°C
cp,ice=37.7J/k-;mole t J
.c
CP' l = 75.3 J/k-mole (Heat capacity is independent of tempert1
kJ
C
LiHr = 6.01--
mole
B
(a) 52.26 J/K (b) 44.03 J/K (c) 46.84 J/K (d) 49.45 J/K
yA
S. The LiS for 3 mole ofan ideal gas. (diatomic) which is heated and compressed from 298Kand 1 bar to
tr
S
and bar ( Cp =iR)
is
(a)-14.88 J/K (b) 14.88 J/K (c)-4.62 J/K (d) 4.62 J/K
m
6. · Consider the statement

he
( 1) liU J,,,,ian and have same vlaue.

t:Ji_r,,sion
.C
(2) The change in enthalpy when 50 litre ethanole initially at 1 bar and 300K is heated to 9 bar nd
is SOOOJ/mole. (IfCp methanol is 80 J/k-mole)
av) (BP) (av\ v
w
BP t for 1 mole ofideal gas is-?£

2
(3) The value of (BT BT P v\
w
Correct statement is
w
(a)l,2 (b)2,3 (c) l, 3 (d) 1 only
7. What is /iA for the process
2He(g, 27°C, 1atm)-t 2He(g, 27°C, 5 atm)

assuming that the process is reversible
(a) 8.029 kJ (b) 802.9 kJ (c) 80.29 kJ (d) 0.8029kJ
8. /iH~98 for the reaction,
C2H 40(g) =·>CH 4 (g)+CO(g)

is -16.0 kJ. From the given d8ta, evaluate the temperature at which ~H will be zero.
Substance : C2H40(g)HAND MADE NOTESCH/g) . www.ChemistryABC.com
CO(g)
Cp(J/K mole) : 50 36 . 30
(a) 1298K (b) 1000K fo) 1298°C (d) 1100°r,
Choose the corn~ct criterian of spontaneity in term of the properties of the system alone
(a) (dS)T,P <0 (b) (dS)H,P <0 (c) (dG)T,v <0, (d)(d~Ju:~ >0
One mole ofHelium is mixed with 2 mole ofNeonboth at the same temperature and at same pressure ifthe
total volume remains constant than LiSHe and LiSNe respectively are ;
(a) 6.743 J/k, 9.136 J/k (b) 9.136 J/k, ~.743 JI~
(c) 13.486 J/k, 18.272 J/k (d) 6.743, 13.486 J/k
I00 gm nitrogen at 298 K~e held by a piston under 30 atmpress-gre. The pressure is suddentlyreleased to 10
atm and the expands adiabatically if (Cv = 20 .8 J I k - mole) v~lue
.
of LiSsystem
.
, LiS surrounding , LiSuniverse is:
(a) All are zero because process is adiabatic (b) 2.932 J/k, 0, 2.932 J/k
(c) 10.47 J/k, 0, 10.47 J/k (d) 3.571 J/k, 0, 3.5:71 J/k /
The molar heat capacity at constant pressureofH2O(g) at .
1.00 ~tm. is·. give,rrty
/ '
om
. ....-\C-p-=-30-.5-4_+_10_.3_x_10__2-T~\. ' '/
The change in entropy wlien 2.0 mole ofwater at 1.00 atm. The terqkrature of8i0(g) is increased from
.c
300K to 1000 K is:
(a)87.96J/k (b)8.796J/k (c)879.6J/k (d)0.8796J/k.
C
·Which ofthe following are the correct criteria for naturallyoccuring (spontiQ.eouS,processes?
B
(E =internal energy, S = Entropy, G = Gibb's free energy, A= Helmholtz free energy)
yA
(a) (LiG)T,V so (b) (LiA)T.I' $0 (G)(LiS\;,P ~o (d) (LiE)s,v so
For mixing oftwo ideal gases at 25°C and I atm, which one of the following is incorrect?
tr
(a) LiTmix =0 (b) LiHmix. = 0 (c) LiVmix.:;:: 0 (d) LiSmix. = 0

is
The plot that describes a Carnot cycle is:

m
p
he
s s V
D D C (>
.C
..
(a) (b) (c) (d)

w
T V T T
w
dP.
w
The correct order ofratio of the dT for three phases transfomiation is: .
(a) (dP) > (~p) .> (~~) .(b) (dP)

dt
·(dP) > (dP)
> -dt
L-g dt S-g S-L
dt S=L dt . L=gdt S-g
, (c) ldtdP) > (dP)

f
S-g
(dP)
-dt- > dt S-L L=g
(d) (dP)
dt S=L
> (-~~ \J
. dt S-g
> (dP)
dt L-g
Ifthe enthalpy and entropy changes for the following process, C6Hif)-+ C6H6 (g) are respectively, 33.90
KJmo1= 1 and 96.4 Jmol- 1K= 1, the temperature at which the vapour pressure of C6H6 (g) is 35 Torr_will be
HAND MADE NOTES (d)www.ChemistryABC.com
273 K
(a) 352 K (b) 507 K (c) 278 K
18. Consider· the fullowing statement
I. QUZZES MORE NOTES ON
Entropy ofuniverse is continually increasing. . www.ChemistryABC.com
II. Total entropy change for a reverrisible isothermal cycle is zero.
ill. AS mix =0 for ideal gas rv. Entropy is unava,ilable work
Correct statement is:
(a) l, II, III, IV (b) I, II, III (c) I, II, IV (d) I, IV
19. For heating inJouly Thomson coefficient which ofthe following option is true·
(a) (oH) < 0 means T(

~T
8
V) < V
8T,
8 8
(b) (, ~) > 0 means T( V.·) < V
WT 8T,
(8H)
(c) -
8Py
< 0 means T -
8Tp
>0 (8V) · ·( ) >.' Om.<;ans..., T (8V)
(d) " ,
T, 1
-
.•
> ,V· ·
8Tp
om
20. A camot_ engine works between the tempera~re 300Kand'£oJ(.and takes 6kJ energy. Then what is
work done ' ,.. t
.c
(a) 1 kJ (b) 1.5 kJ (c) 2.5 kJ (d) 2.0 kJ
C
B
yA
tr
ANSWER KEY
is
Questions 1 2 3 4 5
Option (d) (a) (b) (a) (a)
m
Option (a) (a) (a) (d) (a}
he
Questions 11 12 13 14 15
Option - (c) (a) (d) (d) (a)
.C
Questions 16 17 18 19 20
Option (a) (c) (c) (b) (b)
w
w
w

[ Chapter 4]
,, '
iuilibrium Oriteria free E;nergy Function
Introduction: , ) . ,
This chapter begins with a discussion ofmathematical prop~rties ofthetotal djffer~ntial ofa dependent variable.
om
Three extensive state functions with dimensions ofenergy are introducegv.'en;palpy, Hehnholtz energy, and
Gibbs' energy. These functions together with internal energy, ate ca.1Ieafhe1'19otlynainics potentials. Some
formal mathematical manipu1ation ofthe four thermodynamicpotentialsmre,descnbeq that lead to exp:essio'ns
.c
for heat capacities surface work and criteria for spontaneity in closed system
Free Energy Functions . · ... · .
C
In thermodynamics, two new functions have been introduced to explain various N:iysicll and chemical processes
B
occuring at constant temperature and at constant/pressure. These functions are called free energies or free
yA
energy functions. These are
(a) Helmholtz free energy or work function
(b) Gibb's free energy or Gibb's potential.
tr
Helmholtz's free energy (A) : A = U -TS :

is
Helmholtz free energy functional (A) is some time called work/unction or maximum work content ofthe
m
system At a constant temperature, the maximum work obtainable from a system is at the expanse of decrease
in the Helmholtz free energy ofthe sytsem .
he
dq
We know that, dS ~ T
.C
And dq::::; dU + PdV

dU+PdV
w
dS~ T ~ TdS~dU+PdV
w
dU+PdV-TdS:::;;O
Ifvolume is constant i.e. dV = 0
w
( dU - T dS )v : :; 0
Iftemperature ofthe system is also constant
(8U-TS)V1T:::;; 0
The quantity ( U - TS) is referred to as Helmholtz free energy and is represented byA
IA=U-TS]
Therefore, for a given process
(BA)vr :5: 0
A is an extensive property and state function. Hence, dA is an exact differential for isotehrrnal process
dA=dU-TdS HAND MADE NOTES www.ChemistryABC.com
Or, I M=!iU-T!iS J
Significane of Helmholtz free energy:
SinceQUZZES
we know that, MORE NOTES ON www.ChemistryABC.com
· M=!),JJ-TM
Or, M = !:J.U -T q:;: . [: !1Srev = q;v]
Or, M =!:J.U qrev

Butfrom:firstlawofthethermodynamics, !:J.U = q+w /
wrev ~ !:J.U qrev
. In case ofwork is done by the sytem, then sign is neg~t~te.
-wrev !:J.U -qrev
-wrev =M
om
Or, J -M = Wrev I·
where wr~ represents rnaximumuseful work done during the refersible process~ Hence, decreases in Helmh<
.c
free energy gives the maximum Work obtained from the system.
Variation ofAwith temperature and volume:
C
A U:_TS
B
... (1)
yA
From first law
dq=dU+PdV
From second law
tr
dq =TdS
is
Therefore, equation ( 1) becomes

I dA=-PdV-SdT I
m
At constant volume, dV=O
(:a
he
=-S
.C
=> At constant temperature, dT=O
(!~1
w
=-P
w
Note: (M).ideal gas >(tiA) Vander waal gas

w
Gibb's free energy functions (G): G = H- TS :

It is that thermodynamics quantity ofa system the decrease in which value during a process is equalto the us
work done other than work of expansions.
' dq
Since, we know that, dS 2 T
dS:?:: dU +PdV
T
dU + PdV - TdS::; 0 ... (1)
If P & Tare constant, then equation (1) becomes
8(U + PV TS)P,T::; 0 HAND MADE NOTES www.ChemistryABC.com... {2)
The quantity (U + PV -TS) is referred to as the Gibb's free energy & is represe,µted byG Hence G is
· defined as.
- ~IG_=_E_+_P_V_--TS--,! ... (3)
Since H == E + PV )
1 G=H-TS 1 ... (4)

Therefore, for a given proces~.
( 8G)P,T ~ 0
G is an extensive process and it is a state function because it depends' on H, T & S which are all state functions.
If dG is exact differentia~ so '
[iG=dH-TdS I
\
om
or, [ L\G =Llli-TL\S I .
1
Significance of: Gibb''S free energy function !
.c
The change i,n Gibb's free energy is given by
L\G = Lili -TL\S
C
Since, we know that
B
Lili = L\U + PL\V
yA
L\G =Lili+ PL\G TL\S but M = L\[J -TL\S
!lG = M + PL\V =-wrev + PL\V
tr
-L\G=wrev -PL\V
Hence, the decrease in free energy functions gives the maximum work that can be obtained from a system at
is
constant temperature and pressure.

m
At constant volume PL\V == 0

Therefore,
he
or, -wrev =L\G

.C
or,
w
Variation of (G) with temperature and pressure:

w
·· G :;;.H-TS
dG = dH-TdS-SdT
w
But H=U+PV
dH dU +PdV + VdP
dG = dU + PdV + VdP-TdS-SdT
from first and second law ofthermodynamics
TdS dU+PdV
,
' . [<lG"=VdP-SdT I ... (1)
=> When pressure is constant, dP 0, ( dG) P = -( SdT) P
dG) -S
( dT p
When temperature is constant, dT 0 HAND MADE NOTES www.ChemistryABC.com
(dG)T = (VdP)T
; t:.qunu,num l,M"t:erla t"ree t:.nergy rune
QUZZES dG). =V MORE NOTES ON www.ChemistryABC.com

( dT T ' ... (2)
Integrating equation (2)
=>
But for idealgas?iPV = RT
02 P2 dP
=>
01
J dG=RT
P1
JP
'I
,)
p
=> G2 -GI =RTfo_l.
pl
,)
om
'\
p I
1
~G =RT.eni /
' P,
I
.c
For 'n' mole ofideal gas.
C
~G=nRTln Pi
B
Fi
yA
For an isothermal process,
tr
is
V
~G = 2.303 nRT log-1
v2
m
he
Importance of Gibb's free energy or efficiency of energy Conversion:

Since we know that
~G =Llli-T~S => MI =~G+ T~S
.C
. . Net useful work obtainable ~G MI T~S

Efficiency of energy conversion = - - - - - - - - - - = = -
w
Total energy ~H ~H
w
.
Effiiciency . = 1- -·
of energy conservation T~S
-
MI
w
Case-I: For spontaneous process, the entropy increases i.e. ~S = +ve

Therefore, efficiency ofenergy conversion ~ 100%
Case-II: When the change is at equihbrium, entropy is maximum, then ~S = 0
Therefore, efficiency of energy conversion= 100%
Calculation of AG :
Since we know that
~G = MI - T~S
Product · reactant
MI{reaction) = L ni~Hf_ - L n jMig

product reactant
~s(reaction) = L niS- L njS

)i
Exoergonic and Endoergonic reaction: .
The chemical reaction in which free energy decreaseg('{ AG = -~~) are called exo-ergonic and the chemical
reactions accompanied by increase of free energy i.e. 8G. is positive are kq.own as endo-ergonic.
Standard free energy change.
product. reactant
AG =
0
L niAG~ - L njAG~
Coupled reaction:
Those biological reaction which are non-spontaneous ( AG =~ve), becom~s sponqtneous ";hen coupled
. if '
i ·,) -;:'-
with a reaction whose AG. is highly negative. So that change offree energypfboth the readions is negative.
1
These pair ofreactions are called as coupled. reactions. ·
om
Biological reactions are generally coupled reaction. \ /? / '
Example : Conversion ofGlucose into ylucose -6-phosphate ,
(i) Glucose +HlO4 4 Glucose .. 6 - phosphate + 1¾0 f iG 0 =+ 12.5 kJ / mole
.c
(ii) ATP +H20 4 ADP+ H3P04..
AG 0-=-34.1 kJ / mole
C
Adding both the reaction, we get coupled reaction is
B
Glucose+ A1P 4 Glucose - 6 - phosphate+ ADP
AG 0 =-34.1 + 12.5
yA
AG 0 =-12.6kJ'/mole
tr
Thus the conversion of Glucose into Glucose - 6 - phosphate becomes spontaneous.

Enthalpy, Entropy and Free energy changes and the nature of reaction:
is
AH AS AG=AH-TAS Conclusion
m
- + Spontaneous
he
- - [-atlowTJ fsponraneous J
+at high T non-spontaneous
.C
+ + [ +at low T J [Non-spontaneous J

w
-at hihg T spontaneous

+ - + Non - spontaneous
w
w
Gibb's - Helmholtz's equation: (Temperature dependence of free energy).

First form of Gibb's - Helmholtz equation:
The change in the free energy with temperature at constant pressure as given by
(!~)P =-S ... (1)
From the definition of G(= H..., TS), we have
-S= G-H ... (2)

T
Substituting this in equation (2), we get
(aa)
8T
=a-n,
T P or
G=H.+r(aa).
.
HAND MADE NOTES
8T P
... (3) .
'S··
in actual practice, one commonly employes the relations which give the .variation in llG with tempera
Such expressions can be conveneniently
QUZZES MOREwritten
NOTESstarting
ON ~om equation (J) with the replacement ofGwith
~~~~M. . ./ .·
We have,
8(!:J.G)J =M= llG-Mf
. ( oT
... (4)
P T
~G=Mf +r(o(llG)J
Or,
. oT P
This equation is called Gibbs-Helmholtz equation.
. .
2. Second form of Gibb's Helmholtz's equation:, )

Variation ofHelmholtz's free energy with respept to temp(?I'atur~;{t CQnstant volume is called secondfm
om
Gibb's-Helmholtz's equation. 1
\ /'' ·
Since, we know that

A= u _:TS => M = llU - Tas 1 ' ... (I) ,
.c
At constant temperature, we get
-[o(f~!J
C
(8M) =(BU)
B
of ofy V of 'V
0
yA
=(:1Dv -T( :f-~s(:l
tr
Since we know that,
(-8UJ Cv· and ,T (8/lS)· .=Cv .

is
-
ar v ar v
m
So, putting these values in above equation, we get

he
(8MJ
ar
=Cv-Cv M
v
.C
w
Fromequation(l), we get
w
M=~u-r[(-~JJ
w
IM=W+r(~ll
Temperature dependence of AG/ T:
Since, we know that at a constant temperature, !:J.G = M{ - TM, so that
llG = Af!-TllS = !lH -llS

T T T
Differentiating llTG with respect to temperature at constant pressure, we get
, ···- ... _ ..... _. •• __...· - ... - - - · .... "OJ • - . · - - · - · . . ·l!!!!!!!!!!!!!B
_:
... (1)
Since we know that,
a( Mf)J . (a(AS)J
·( . .ar p =C
. p and T. ar . ._=C
. p
p
om
Putting this values in equation (I), we get
o(AGIT)} =~ Ml +Acp·'~ ACP
.c
{ ar p T2 T T
C
B ... (2)
yA
tr
o(AGIT) }
Or, {-(Ilr 2 )ar = Mf
is
p
m
Since, d ( ~) =·- ) 2 dT. So, from above equation, we get .

he
Ir)}
.C
Afl = {o(AG
o(II T) p
w
TAG versus TI will be a straight. line with slope equal to iVl.

w
If AH is independent oftemperature; the plot of

w
I
If Afl depends on temperature, then the slope of the plot at T gives Ml at temperature T.
( SOLVED PROBLEMS )
The free energy change ( AG) accompanying a given process is-85.77 kJ at 25°C and-83.68 kJ at 35°C.
Calculate the change in enthalpy (Aff) for the process at 30°C.
In. AG at 25°C = -85.77 kJ and at 35°C = -83.68 kJ
o(AG)J = -83.68kl -(-85.77kl) 0. 209 kl K_ 1

( ar p 308K _,_ 298K HAND MADE NOTES www.ChemistryABC.com
t:..GQUZZES
at 30°C may be taken as the average ofthe values at 25°C andwww.ChemistryABC.com
MORE NOTES ON
35~C:
· (85.77kJ +83.68kl) .. 84 :7·25kl

t:..G at 30°C = - .. 2 . =- · .
-84.725kl = Ml +(303K)~0.209kl K-1 )
Hence, Ml =-148.05kl
2. For the reaction,
'
~ N2 (g)+ ~ H2 (g)~ NH3 (g), t:..G~98 K = -16.45 kJ mor ~d t:..H~98 = -4p.h kJ mor 1 • Estir
1
f
/
the value of t:..G ~K if m 0 is considered independent oftenw7ture,.
om
Soln. We know that ·
' f
.c
t:..Go=t:..Ho+_!_(t:..Go -t:..Ho)
T 'f. T0
C
0
We get,
B
yA
5
t:..G~ = -46.11 kJ mor 1 +( 00K)(-16.45 kJ mor1 +46.11 kJ mor 1 )
·298K
tr
= 3.66 kJ mor 1
is
3. · What is the t:..G 0 for the reaction

m
he
from the following data

.C
(i) cq (g) + 2NH3 (g)-+ H20(g) + CO( NH2 )i; t:..G 0 =1.91 kl mor1
w
1
2fJ (g );
w
(ii) H20(g )-+ H 2 (g )+ 2 t:..Gi::;: 227.44 kl mor1

w
(iii) C(graphite) + 0 2 (g )-+ cq (g); t:..G: =-394.38 kl moZ- 1

(iv) N2 (g)+3H2 (g)-+2NH3 (g); t:..G1 = -32.43 kl mor 1
(a) 197.46 kJ moI-1 (b)-197.46 kJ mo1- 1 (c) 157.46 kJ mo1- 1 ( d)-157.46 kJ mo1- 1
Soln. The given reaction is,
C(graphite )+ 2H2 (g )+ N2 (g )+..!:.02 (g )-+ CO(NH2 ) 2

. 2
The above reaction can be obtained by adding the given reactions. Thus, we have
I
t:..G = t:..G1°+t:..G~ +t:..G: +t:..G1

0 1
HAND MADE NOTES kl moZ-
=-197.46 www.ChemistryABC.com

0
4. AG at 298K for the following reaction is -3.202 MJ mor1 ,
QUZZES 1 MORE NOTES ON www.ChemistryABC.com

C6H 6 (1)+7 0 2 (g)~ 6C02 (g)+3H2 tp(£)
2
0
What is M at 298K.
(a)-3.198 MJ mo1-1 (b) 3.198 MJ moJ.-- 1 '
(c) 2.188 MJ mo1- 1 (d)-2.188 MJ mo1-1
So1n. For this reaction,
, 1 1
Av =6-7-=-l-
s 2 .2
0 0 0
Now, AG =M +A(PV)=M +(Avg)RT/
,\
0 0
Or, M = AG -(Avg )RT /
;:
Substituting the given values, we get · .· . ., )· ,
om
- M~ =-3.202x!9 J moZ-
6 1
-{(-1.5)( 8314 k;"'1Jr 1)(298K)}
. ' 1
=-3.202x 106 J mor1 +3716.36 J mor1
.c
=-3':198xl06 J mor1 ::::: -3.198 MJ moZ-1
C
Correct option is (a) \ ,,,
5.
B
For an ideal gas undergoing isotheµnalreversible expansion, AG is equal to
yA
(a) AU (b) ·;M' (c) M (d) AT
SoJn. The :function G and A are given by
tr
G=H-TS and A=U-TS

The :functionH is given by
is
H=U+PV
Substituting this in the function G, we get
m
G=(U +PV)-TS =(U-TS)+PV

he
Or, G=A+PV
Thus, for a process
.C
AG=M+A(PV)
Now for anisotbermalexpansionofanidealgas, we have
w
A(PV)=A(nRT)=O as AT=O
w
So, that AG = M
w

QUZZES
·-
(PRACTICE SET) ,
I. The free enetgychange ( LiG) fora given process is-79.8

. .
kJ at20°C ~nd-77.4kJ at 30°C. What is the
change in enthalpy of the process at 25°C
. (a)-150.12 kJ (b) 150.12 kJ (c) 101.12 kJ (d) 101.12 kJ
2. Which ofthe following properties are characteristics ofan ideal solution? [GATE2~15]
(i) ( LimaxG)T,P is negative (ii) ( LimaxS)T,P is positive
(iii) ( Lirnaxv)r p is positive (iv) (t:..n;.'.axH)T,P isnegative

' '1 ·\
: ,)
(a) (i) and (iv) (b) (i) and (ii) (c) (i), anµ (iii) ,;··· (d) (iii) and (iv) ·
'I
/
om
3. What is the ( Li G) for the conversion of3 mol ofliquid benzene aji8b0 Q(nonnal boiling point) to vapour at
same temperature,and a press~!! of0.66 bar. Consider the var,our a:s an ideal gas. ·
(a)-3.66 kJ mo1-1 (b) 3.66 kJ mo1- 1 (c) 1.66 kJ mo1- 1 (d) .66 kJ mol-1
.c
4. What is the difference between LiG and Mat 20°C for the reaction,
C
. . . 1
H2 (g, 1 atm)+ 0 2 (g, 1 atm) = H 2 0( R)
B
2
yA
(a)-3.716kJmol- 1 (b) 3.716kJmol- 1 (c)-2.3I6kJmol-t (d)2.316kJmol- 1
5. For a certain process, LiG = -50.208 kJ and LiH = -73.22 kJ at400K. Fortheprocessatthis temperat
tr
is
8( LiG)]
LiS=-57.47 JK-1 . Whatisthevalueof [ of P
m
(a) 48.57 JK-l (b) 54.47 JK-1 (c)' 66.47 JK-1 (d) 77.47 JK-1
he
ANSWER KEY -
.C
Questions 1 2 3 4· 5
Option (a) (b) (a) (a) (b)
w
w
w

[ Oh~pter s J·
fhird Law of Thermodyn~l11ics I
! '
Introduction: · Jc
The first and second law of thennodynamics have led to' new conceRtS of~nergy content and entropy. The
om
third law, however, does not lead to any new concept. It only places a lifuitation on the value of tre entropy
of a crystalline solid. The· third law ofthennodyna~cs is state that feg~rding the properties ofsystem in
equilibrium at absolute zero temperature. The entropy of perfect crystal, at absolute zeroth law is exactly
.c
equal to zero. ·
C
\
B
Nernst - Heat Theorem:
yA
W-Nemst in1906, studid thevariatien ofenthalpy (Ml) and free energy (AG) with decrease oftemperature
(in figure). ·
tr
is
m
Temperature
he
According to Gibbs-Hehnholtz,equation
AG=Mf +T(oAG)
~r i .,. (i)
.C
u p i
at absolute zero, i.e. when T= oj AG= Ml. However, Nernst observed that as the temperature is lowered
w
. / a(AG) . . h
towards absolute zero, the value of oT decrases and thenapproach zero asymptotically. This means t at
w
AG and Ml are not only equal at the absolute zero but the value approach each o_ther asymptotically in the
w
vicinity ofthe temperature. The result is,knownas Nemst,heat.theorem

d(AG) d(Afi,J
lim--'-_-a_= lim--'-----'- = 0 ... (ii)
T-+0 dt T-+0 dT
AG has been showas greater than M{ temperature away from the absolute zero.
,' . .. d(AG)
However, the reverse is also possible because __:._,.;_ can becomes positive or negative.
· dT
But we know that ·
( d~G)J~-M ... (m)
And d(Aff)J=-AC. ... (iv)

( dT P
1 n1ra 1,..c;:1w V1 1 ru::::rn1vayn;:,irr
Combining
QUZZES equation (iii) and (iv), we get
lim M =O and lim8CP =0

T--j,O T--j,O . .
Since gases do not exist at the absolute zero, this means that the heat theorem_is not applicable toga
Similarly, it has been found to be inapplicable to liquid also. :
Problem: The free energy change accompanying a given process-85. 77 and-83.68 kJ at 25°C and 3~
respectively. The change in enthalpy for the process at 30~C is
(a)-148.05 kJ · (b) 148.05 kJ (c) 180.05 kJ (c)-180.05kJ
. -85.77 +(-83.68) . · .
Soln. 8G at30°C= -------'----"-=-84.725k.J
. 2
':\
d(8G)) = -83.68(85.77) =0. 209 kl
( dT 308-298
om
P \
.;G=Ml +T('d(8G))
.c
dT P
C
-84.72=Afl +303x0.209
B
MI =-148.05 kl
yA
. The Third law of thermodynamics:
tr
The third law of thermodynamics concerns the entropy perfectly-ordered crystal at zero kelvins.
When a chemical reactions or phase transition is studied at low temperatures, and all substanc<
is
pure crystals presumed to be perfectly ordered, the entropy change is found to be approach as the tempera
approaches zero kelvins.
m
limM
T-;.0
=0 (")
••• l
he
(Pure, perfectly-ordered crystals)

Equations (i) is the mathematical statement of the Nernst heat theorem or third law ofthermodynamics.
true in general only ifeach reactant and product is a pure crystal with identical unit cells arranged in pe1
.C
spatial order.
w
The zero ofEntropy : "Ifthe entropy ofeach element in some crystalline state be taken as zero at the abs<
zero of the temperature : every substance has a finite positive entropy may becomes zero and does so beco
w
in the case ofperfect crystalline substances."

w
According to this principle, every substance (element or compound) in a pure, perfectly-orderd er

at OK, at any pressure has a molar entropy of zero ·
Sm (OK) = 0 ... (ii)
(Pure, perfectly-ordered crystal)
This conventionestablieshes a scale ofabsolute entropies at temperature above zero kelvins called third
entropies.
The temperature dependence of Entropy:
We have seen calculations for the entropy change for processes. However, .it is also possible to calculat<
dS
absolute entropy. We can begin with the definities ;ev .The heat transferred during a process at com
;rev = CvdT . Thus, the entropy change at constant volume is
vo lurne is dS
dS =S(T)-S(O)= Icv(T)dT
QUZZES o T MORE NOTES ON www.ChemistryABC.com
We have kept the equation general by showing Cv ( T) as a fun~tion oftemperature., This calculation ofthe
· entropy is valid only at contant V. At contant P we find an analoguous,expression. Starting with the heat
transferred, dqp,rev = CpdT
. TC (T)dT
We have, M =S(T)-s(o) J
o
p .
T
j.
The temperature dependence of the heat capacity: ~\
The thirdJaw ofthermodynamics shows that ·,)
.)
'/
CP ~Oas T ~OK /
om
It is necessary to treat CP ( T) as a function oftemperatur..efoi;.~r"o~one. The Einstseinand Debye
. -- . ~ ' -
theories ofheat capacity can be used.to determinecthe functional fprm;of the heat capacity of at thes~ low
temperatures. The Debye law states that the heat capacity depends oriT3 neat T =OK. ·
.c
3
At temperatures close to. I T =0, Cv,m =aT · 1
C
The constant a is an empirical constant. For practical calculation ofthe eµtroPjexperimental values can be
B
used and the integrals are evaluated numerically) ' ·
yA
Problem : Show that the entropy of any substance at very1owtemperature (0 < T < 20 K), where Debye's
relation for heat capacities ofcrystals is valid, is one-third ofthe molar heat capacity.
tr
:oln. Since we knpw that, at low temperature {O < T < 20 K)

CP = '
aT3 ... (i)
is
m
dS=(~ )dT_
he
T T ,
0
f dS = f( Cp !T)dT
0 .
.C
T T3 T
Or, ST -S0 = f.9:__dT = aJT 2 dT = a T3

w
0 T I O 3
aT3
w
Or, ST=-
3
w
From equation (i), we get

s = Cp
T 3
Absolute Entropy: .
I '
\ \ C (T)
The absolute entropy can be calculated from Oto any temperature Tusing the integral ofthe function, PT .
If there is a phase transition between Oand temperature T we can also calcjlate the contnbution to the entropy
from the transition. · '
Tfasf cp (T)ir MI fas T..,fp cp (T)dT Mfvap TI cp (T)dT
ST=
( ) +--+ ..-!--'---.:.-+--+
0
T, Tf a s ~HAND MADE T NOTEST.~.~· www.ChemistryABC.com
T
The heat capacity ofeach phase is different and the heat capacities are also a function oftempera1llre.
------------- -- ----:--------...,------··
Residual Entropy: .
QUZZES
Residual entropy is the different MORE
in entropy NOTES ON
between www.ChemistryABC.com
a non-equilibrium state and crystal state of a substar.
close to absolute zero. ·
There are certain solid which do not have perfect crystalline structure at 0°K like as H2, D 2 , CO, N
N20 and ~O. So, they have certain amount ofentropy even at absghite zero. This is called residual entro
According to third law ofthermodynamics for perfect crystalline solid at zero Kelvin the entro
change in zero. But certain solid at zero,kelvin the entropy does not found the zero. So, this is not follow t
rule.
Residual Entropy ex: probability of arrangement of one molecule of the atom.
So, for Boltzmann'~ rule,
• 'i· ;x .
where S is residual entropy, K boltzmann's constant and w is probability ofarrahgmen,t of 1 mole ofthe ate
·' ;j'
!
om
CO, an imperted Crystal: .· · , ./ .·
The molecule CO has a very small dipole moment and there is a ;finite ,thance that CO will crystalline as 0
CO : .CO instead ofCb: CO: CO:· For each CO molecule thereF1"e.tJo possible orientations ofthe mokct
therefore, there are two ways each CO can exist in the lattice. The number ofways per molecule is w= 2
.c
each CO. If we have NCO molecules there are wN ways or 2N ways that all of the CO can be distribut
Therefore, the entropy a.t zero Kelvin is
C
S = k 1n W = k 1n ( wN) = Nk 1n w = R ln w = R 1n 2 \ i;:
Where,R=Nk
B
yA
The entropy at zero Kelvin is known 'as residual entropy. There are number of substances that show sim:
statistical variations in orientation that lead to a residual entropy,
tr
(l>ROBLEl\lIS)
is
m
1. The molar entropy ofcrystalline CO at absolute zero is: [GATE2010]

(a)Zero . (b)-Rln2 (c)Rln2 (d)2Rln2
he
Soln. . For 1 moleofcertainmonoxide, w= 2N

Since we know that, S =k ln w
.C
S = kln2N
w
S = Nkln2
S=Rln2 (·: R = Nk)
w

w

- .... -- . . . ---- - - -----.1 - ·--:--- - - - - -
Remarks:
· TranslationalQUZZES
entropy of the molecule MORE NOTES ON www.ChemistryABC.com
_ (2mnkT)½ v e½
sir -.Rn 2
h NA )
Where, V= Volume ofthe system, NA= Avogadro's number, m= Mass ofthe molecule
h = Planck's constant; k = Boltzmann's constant, Ty= absolute temperature (K).
,'•, . .
Rotational entropy of the molecule ..

:\
•i
.,s
.-/
/
om
Where, Ij = Principal moments of inertia ofthe molecule,
-
cr = Symmetrynumber ofthe molecule.
. - f
.c
h = Planck's constant,k =Boltzmann's constant,
C
T = absolute temperature (K) '· /t
B
yA 't
Vibrational entropy of the .molecule.
··.a{ hViexp(-hVi/2kT) ·J}

tr
[
Svib =L. [ ( )ffn !-exp(-hV;f2kT)
k! 1-exp -hVj /2kT
i=I .
is
Where, n.J = Frequencies ofnormal VIbrations in molecule,

m
k= Boltzmann's constant
he
T = Absolute temperature (K)

.C
h = Planck's constant
w
w
w

( PRACTICE SET)
1. Standard entropy ofcrystalline carbon monoxide (in kJtmol) at OK is atound
(a)0.03 (b)2.50 (c)zero : (d)5.76
2. Suppose there are four ways the molecule CFCJBrl can be oriented in a crystal lattice at zero Kelvin W
the molar residual entropy?
(a)4Rln(2) (b)2Rln(4) (c)Rln(2) (d)Rln(4)
3. Which phrase best summarizes residual entropy?
(a) The entropy of imperfect crystals (b) ']Jle entropy ofthe lowest energy state
(C) The entropy ofrotation (d) J'he entropy ofresidues
4. Ifthere is no residual entropy
.
then at T =. 0 K we can.state that
' ,.
. ,., " ,)
(a)S>OandCP>O (b)S=OandCP>O (c).S>OandCp;·O (d)S=OandCP=O

. /
5. Consider the statement CP -t O as T -t O. Which consequen, is valid.
om
·
(a) It is harder to heat matter near absolute zero, thatis it requts ~6re energy per degree oftemperatur1
(b) It is easier to heat matter near absolute zero, that is it reqbires less en~rgy per degree of temperature
.c
(c) It gets easier to reach absolute zero because it is easier coolmatter at lower temperature
(d) Both (b) and (c) are correct.
C
B
yA
tr
ANSWER KEY , !
Questions 1 2 3 4 5
is
Option (d) (d) (a) . (d) (b)

m
he
.C
w
w
w

[Ch~ptef 6]
:,artial Molar Quantity
')
'.()
Expression of Partial Molar Quantity ..

. ,
.
1
l
/
.
i
Thennodynamic relations derived earlier are applicable !ci the closed systt;:Jill bt1t ifwe discuss a system where.
om
not only the work and heat but matter are being exchanged, wen~ patifal n;tolar quantity.
For example : Gibb's free energy ofa ~tern is a functton not only ofte~rafure and pressure, but also ofthe
amount ofeach substance in the system, such that · ' ' '
.c
• I
G = f (T,P,n 1,n 2 ,n3....nk)
C
where n1, n2, •••••••••• nk represent the amount of each of the k components in the system. For simplicity, let a
system contain only two components. The total duferential of G is \ /'/!
t,,,,,a.,,, +( :L,,,:,, '"~ {: L,,,.,, '• +(:, L,,,,,,, -•• B

yA
dG = (::
tr
In this equation (.: ) and ( !;'. ) are known as partial molar free energy ofcomponent I and 2, respectively.
is
m
In general, the partial derivatives ofa thermodynamic function ( y) with respect to the amount ofcomponent (i)
he
of a mixture, when tem.perature, pressure and other constituents are kept constant is known as partial molar
quantity ofith component and is represented as ¥;,pm (or simply as r; ). Thus
.C
. (av)
Yipm=
'
-
GI\
w
T,p,np
j.. i
The various partialmolal quantityare3

w
=(: )
w
(I) Partial molal internal energy E,

, 1 P,T,ni,n 2 ......... .
(2) Partial molal enthalpy energy Hi - =(8H)

on.
1 P,T,n 1,n2 ..•.
, (3) Partial molar Gibb's free G1 (:;) ( chemical potentiaO

1 P,T,nt,n 2.........
(4) Partial molal work function i\ =(::)

1 P,T,n 1,n 2
(5) Partial mo1al volume

- (av)
v; = Olli P,T,n1,n2····
The most useful partial molal quantity is partial molal free ene~. It is known as chemical potental (µ.i) .1
name chemical potential is not given to the partial Gibb's free energies but also all partial thermodynar
potential in respect oftheir natural variables. · ·
Toe various chemical potential are as:
(oG)
(l) Olli T,P,n .ni-........
1
(
2 (BAJ
) Olli V,T,n ,n
1 2
..........
( J'
8H
(J) Olli S,P,n ,Di-..... .
1
(
4
l
l ( :~ v,,,.,, ............
Definition of partial Molar quantity:

The partial molar quantity may be defined in either ofthefo llowing two ways: . _
(1) It is the change in Y when 1 mole of component i is added tCl a systetin which,is so large that 1
addition has a negligible effect on the composi~ioq of the syst,r.n. '
(2) Let dY be the change in value ofY when anin:finitestimal amount dn. of component i is added tc
om
It is change in 'y' when 1 mole of component 'i' is ad<led fhe §yste~ of definite composition. B~
infinitesimal amount dni we mean that .its additi911 does not,i.Cause any appreciable change in
!.l
' ,,.. . f . ·• '
.c
composition ofthe system Ifwe divide dY by cit\, we get the partial mo Jar quantity ( Thus,
C
partial molar quantity of componenti may be defined as the change in :X: per mo le ofcomponent i whe1
infinitesimal amount ofthis compone~t is added to a system of de'tinite kmposition. ·
ChemicalPotential(µ,):
B
yA
The most useful partial molar quantity is the partial molar free energy Gi,pm • It is so useful that it is given
name of chemical potential and a separate symbo11,1i to emphasis its generality.
tr
8GJ ·
is
µ- ( -1
i - 8r!1 ;T,p,njs ••• (1)
m
j-F-i
Chemical potential is an intensive property because it is a molar quantity.

he
The thermodynamics properties likes E, H, S, A and Gare extensive property. In case ofopen system, exten
property live G is a function not only of temperature and pressure but also the number ofmoles of vari
components present in the system
.C
G = f(T,P,nl'n 2 , n3 , ••••••••••• n;)
:L.,, , ~ +. . . . +(:~L.,.,dn,
w
dG =(:~\.,,, dT+(:a,,,, dP+(

w
w
OOJ .
The quantity (an, T,p,n2 ...••• nj
is the partial molar free energy and is represented by G; or µi known as chen
potential ofthe component i.
G, =A=(:a; ni T,p,n1,n2
Hence, chemical potential of a system substance may be defined as the change in free energy of the sys
when 1mole ofthat substance is added at constant temperature and pressure.
Thus, equation for change in free energy ofthe systesm may be written as
aa = (~~)
P,n1s
dT +(~0 )
1P T,n 1s
dP+ µ1dn 1 µ 2dn2 + +. . +Adn;
Physical Significance:
(1) It represents
QUZZES the partial change in free energy
MOREwith respect
NOTES ON to its number ofmoles at constant temperature
.and pressure. : '
(BG)r,P =Adfii. +µ2dn2+..... +µin;
on integrating we get,
GT,p,n = µ.nl + µ2n2 + ....... ,+ µjni
(2) It is an intensive property and may be regarded as the driving force that complex a system to attain an
equilibriwn /
(3) At equilibrium the chemical potential ofa substance has the #ame value through the system ie. L µ;dni =0
(4) For 1 mole ofpure substance, the chemical potentials is identical to the. free energyµ = G
(5) It has great importance to drive Gibbs Duhem equation. '
Expressions of ~U, dH, dA and d.G for multicomP,one~t ~pen.~y7s_t fu:.
om
Ifa system contams a total ofk components, then · . .A
~. G = f(T,P,nr,n 2 ,.,,.-:.,nk) t
.c
Its differential is given by
=(BGJ +(BG] +±(BG] .dni . \
C
BG dT
BT P,n ,S Bp Tns Bni T,P,n1s
i=l /'{/
B
J , ,J j,t.i ~
Substituting equation ( 1) in the above equation, we get.

yA
BG=(BGJ
BT p' n.s
dT+(B?J
BP Tn·S
dp+ ±Adn,
i=I
tr
) , J
k
is
dG=-SdT+VdP+ LAdn; .;;(2)'

m
Now, since A= G- PV, therefore,

dA = dG-PdV -VdP
he
Substituting for dG from equation (2), we get
d4 = (-sdI' +V,JP + tµ,dn, )- PdV - VdP

.C
k
w
= -SdT PdV + L Adn; ... (3)

i=l .
w
Similarly, for enthalpy

w
H=G+TS
dH=dG+TdS+SdT
= (-sdI'+VdP+ tµ,dn, )+ Td\' +Sal'

k
=TdS+VdP+ LAdn, ... (4)
i=I
and for energy U = H PV
dU =dH-PdV-VdP
=(TdS+VdP+ tµ,dn;)-PdV-VdP
k
TdS-PdV + LAdn, ... (5)
Ifthe amounts ofall components are held constant, then equation () to ( ) are redµced to
QUZZES dG=-SdT+VdP MORE NOTES ON www.ChemistryABC.com ... (6)
dA = -SdT - PdP ... (7)
dH =TdS+VdP ... (8)
dU =TdS-PdV ... (9)
Equations (3) to (9) are applicable to closed systems;
Fromequation(2 ), it follows that ·
A=" (8AJ
8n; r,v,,, 1
s
fr.i
... (10)
Similarly from equation (4) and (5), we have · J -,i

i
µ.= -
[aH·J ... (11)
om
I 8ni S,P,n1s
j'F-i
. (auJ
µ.= -
.c
1
8ni S,V,111s ... (12)
j#
C
B
Hence, ... (13)
yA
From equation ( 13), it is obvious that the chemical potential plays a different role in each equation (1) to
even through it is the same quantity in all cases.
tr
Variation of chemical potential with temperature and pressure:

is
m
he
(dGt =(dG)
dT
dT+(dG)
dP T11-s
dP
.C
P11s
' I ,- J
dsG =VdP-SdT
w
by equating the coefficient of dT and dP in the above two equations, we get

w
(dG) (dG)
w
-S d -V
dT P,11 an dP T,n -
Which are applicable to a single-component system become
(dµ;)
ar pns
' J
-S. and (dµ;)
I 8P T n ,S
' j
= V-
·
I
for a multicomponent open system. These equations may be proved as follows:

[since the order of differential does not matter; as G :is~avstate function]
a
= {-(-S) }· =-S.
. ani T,P,njs I .
j'#i
Similarly, [
aµ.)·
al' ,,.,( aP an,
·{ a(aGJ ,Pj"!;:}
T T,, =
{a(aa) .} .
an, ap' i.,. r,,,;:,
J
a }
= {-(V) =~
an, T,P,n1s
j=i
om
\
(LaP
8A) , =V: r,n.s , ... (l)
J
.c
Since, we know that PV =RT => V = :
C
8µ;) RT ',.
\
B
Fromequation(l), [aP =P
yA
:::::,
tr
is
m
0 p
µ.1 =µ.I +RTRn-
he
pO
where, µ? = standard chemical potential.

.C
·µi is the standard chemical potential when the pressure ofthe gas is 1°bar.It may be noted that µi depend on
w
both T and P but µi 0 depend only on T.

Gibb's -Duhem equation:
w

w
GT,P,n = An1 + µ2n2 + .... + µA

On differentially,
dG = Adn1 + n1dµ 1 + ~dµ 2 + ...... + Adni + n;dA
dG =[Adn1 + µ2dn2+ ..... + µidni] + ( n1dµ 1 + ~dµ 2+ .... + n;dA)
, But we know that,
( dG)r ' P =(Adn 1 + ~dn2+ ..... + µ 1dn 1 )
.
Therefore,
Or, I In;dA =0 I '

This reaction is general form ofGibbs Duhem equation. For a binary solution.
n1dµ 1 + n2dµ 2 = 0
"1
· · Chemical Potential of a gas in mixture of ideal gases:

The expression for the chemical potential of an ideal gas in a mixture of ideal gases can be obtainf(:
replacing P in equation ( µ = µ' + RT In ( ; ) ) by the partial p17ssure ofthe gas in the gaseous mixt
Thus.
µ(. .) =P:+RTln(;)
where, p 0 = l bar. Since E; =x;P, therefore,
A(mix) =l +RT1n(; )+RTlnx;,
. .
=A +RTlnx;
.
.··· .· \ /
.
, ... (1)
om
where, µ; represe,nts the chemical potential ofthe pure ideal gafat t9rilperature T and pressure P and is g:
as
.c
µi• = A0 + RT 1n ( pO . p)
C
. ' 1¥ .
Since, < l, the term RTlnx, in equation ( 1) has a negative value. \It, therefore, follows that
B
Xj
•
yA
. A-cm1x) < A
that is, the chemical potential ofan ideal gas in a mixture ofideal gases with a total pressure Pis less thar
corresponding value for the pure ideal gas at the smae total pressureP.,
tr
Graph between chemical potential Jl<T,P> and pressure (P) :

is
µ = µ 0 +RTlnP
m
Repulsion takes place

.,,. ... - - For real gas
he
/ "'
I'
I
I
I For ideal gas
.C
I
t I
w
µ I
I
I
w
I
I'
I'
__ .,,,,,, ,, "' "'

w
Attraction takes place
P-¼
( 1) In moderate range ofpressure attraction between the molecule is considerable in real gas.
Pressure for real gas in terms of chemical potential is less than the pressure ofideal gas. ·
(2) For high pressure region the molecule for real gas start repalled to each other so P for real gas in tern
chemical potential becomes higher than the ideal gas.

..._,uu.........p1. UI. .I.' Uj;A\.,11,,Y • . , .
Making use of the free energy function, G, Lewis introduced the concept offugacity forrepresenting the
actual behaviour ofreal gases which is distinctly different
QUZZES fromON
MORE NOTES the ~haviour ofideaf gases.
Variation offree energy with pressure at constant temperature is given by ·
(8G).
oP
=V.
T
,:
... {1)
This equation is applicable to all gases whether i~al or non-ideal ' ·· · ·
Ifone mole ofa gas is under consideration, then Vrefers to molar volume. For an ideal gas, the above equation
may be written as
(dG)T =RT~ ' ... (2)

and furn moles as,
. . dP •·
(dG)r =nRTF=nRT(lnP) ... (3)
om
futegration ofequation, (3) yields
G=G·+nRTlnP .,... f. ... (4) '
whe~e G* , the integration constant, is the free energy ofn
moles ofthe ideal gas at temperature T when the
.c
pressure P is unity.
Equation (4) evidently, gives the .free energy of an ideal gas at temperature T and pressure P..
C
Integration of equation (3) between pressures P1 and P 2, at constant temperat\}re T(fields, ·
B
Pi dP P,,
LiG = jnRT- =
yA
p nRT In -1:.
P. · ... (5)
J\ . 1.
The corresponding equation for 1 mole ofthe gas would be
tr
LiG =RTln~ ... (6)

is
RT
m
Equation (4) and (6) are not valid for real gases since Vis not exactly equal to p
he
In order to make these simple equations applicable to real gases, Lewis introduced a new function/, called
Jugacity Junction. It takes the place of Pin equation (3) which, for real gases, may be expressed as
.C
(dG)r=nRTd(Jnf) ... (7)

and equation (4) may be represented as
w
G = G* + nRT Inf ... (8)

w
where, G• is the free energy of n moles of a real gas when its fugacity happens to be 1.
Thus, fugacity is a sort of 'frictions pressure' which is used in order to retain for real gases simple forms of
w
equations which are applicable to ideal gases only. .

Equation (8), evidently, gives the free energy of a real gas at temperature T and pressure Pat which its
fugacity.can be taken asf .
Equation (7), on integration between fugacitiesf. andJ;,, at constant temperature T, yields
!!.G=G, G, =nRTln(J.J ... (9)

The corresponding equation for 1 mole ofthe gas would be
l!.G G, G, Rr1n(J. J ... (10)
Or, 11!.µ = µ, ~ µ, = RT m_i] HAND MADE NOTES www.ChemistryABC.com

.•. (11)
The Fugacity Coefficient:
, InQUZZES
general, the fugacity of a real gas MORE NOTES
is related to ON
its pryssure byanwww.ChemistryABC.com
eq_1Jation
' .
where y is known as the fugacity coefficient and is a measure of deviations of a real gas from the idt
behaviour. Since all gases approach'ideality in the limit ofzero pressure, it is obvious that
'iim f = lim y = 1 .... ( 13)

p~ 0 p p~ 0
The fugacity is expected in the same units as pressur~'and thus the fugacity ~oefficient is a pure rumbe1
:,)
Evaluation of Fugacity: , ,
Equation dG = VdP is applicable for all substanc~s. ExpressingJi' in terms ofmolar quantities and ther
- '
om
.' '/.
equation(lO) and(ll) separately, we get ' /'

p. /
' J2 ., P..jf
Vm , ideal dP = µ P.2, ideat - µP.1, idea1 =RT 1n ....1.
P. (where, Vmis molar value)
.c
Pi - I
C
Pi f
I 2 /¥
Vm, real dP= µ P2 , real - ,µPi, real =RTln-
J; f
B
Pi I
yA
Talcing the difference, we have
p. . .
2
tr
J(Vm
Pi
,r
eat-Vmideai)dP=RTlnJ;-RTln?i
• Ji P,
I I
is
p. .
(/2Ji J(. PiPi J=1(Vm, real - Vm, ideal ) dP

2
m
Or, RT ln
he
Let P 1 be allowed to approach zero value, then from equation ( 13 ), we get

J; .
lim _l =1
.C
Pi~O Pi
With this, the previous equation becomes
w
J= PiJ(vm,reai -Vm,idea1)dP
w
RTln ( f~·
w
Now fora real gas,
Vm,real =ZRT
p =ZV.
m, ideal
where Z is the compression factor. Introducing this in the previous expression, we get
Fugacity at low pressure:
The ratio ; , where Pis the actual HAND MADE

pressure, NOTES
approaches www.ChemistryABC.com
unity when P approaches zero since in that case
gas approximate to ideal behaviour. The fugacity function, therefore, may be defined as
ltm..::'..-=l
. P~O p '..,j i .... (12) .
at low pressures, fugacity is equal to
. Evidently,QUZZES pressure.
MORE NOTESThe
ON two. terms differ mafenally only athigh pressures.
· Fugacity forvander Waals gas: . ·
Since, we know that
,I
Or,
i.e. Z <1
Hence,/ less than Pis due to the constant 'a'. At high pressure, th~ vander,:Waals eguation takes the fonn
:i' . :
,)'
( P) (Vm - b) = RT /
,' , l)' :
om
\ ,.1··
/ .·j ·
Or, ~ z =PVm =I+b_!_ '
/
RT RT 0
••• p'
.c
i.e. Z >1
Hence,!greaterthan·Pis duetoconstant b. Thus, theneglectof'f:hetermlYin vander Waals equationj makes
C
f less than P and that of a makes it greater than P. ,
!' \ ,,,
( PRbBLEMS)
B
yA
l. At any temperature T, the fugacity coefficient ( y) is given by [GATE2010]
tr
In y = PJZ-1 dP'
P'
is
, 0
where Z is the compressiblity factor. The fugacity coefficient of a real gas governed by equations of state,
m
P(V-b) = RT with 'b' a constant is given by

RT RT bP bP
he
(a) bP (b) ehP (c) RT .(d} eRT

.C
,oln. Inr = .brp zP-I dP ... (1)

w
Since, we know that the compressibility factor, Z =pV

w
• RT
w
P(V b) RT; PV-Pb=RT; PV=RT+Pb

PV RT Pb
Both side divided by RT, RT = RT + RT
Pb
Z =1+ RT Put this value in equation ( 1)
lny =
.b
f.'P
I+ Pb -I
R: dP => Inr r Pb
RT dP= Pb
.P · RT
hP.
:::> r =eRT
Correct option is(d) HAND MADE NOTES www.ChemistryABC.com
. . -- . ' - .- . -. . . -.- ' . .
2. What is the free energy change accompanying the compressi6n ofl mole ofa gas at 57°C from 25 to 200
The fugacities ofthe gas at 57°C mayMORE
QUZZES be taken as ON
NOTES 23 and 91 atrn respectively at pressure of25 and 200
(a) 3730.0 J · (b) 3630.0 J (c) 3530.1 J / (d) 3330.2 J
SoJn. We know that,
2
AG= nRTln Pz = 1 mo/(8.314 JK-' mar' )(330K)In 00 atm = 5702.SJ
Pi ; 25atm
Now,
9
AG=nRTlnh =lmol(8.314JK-1 mor1,)(330K)In latm =3730.0J
· i I: Tiatm . .
Correct optio~is (a)
3. Showthat, .
om
·Aff.
=--'
T2
.c
C
B
yA
1 · 1 Ml. ·
=~T2(Ml; TAS;)+T(-M;):=- Ti'
4. Show that the change
tr
2 mol of an ideal gas (2 bar, 273K) ~ 2 mol of gas (1 bar, 273 K)

is
carried out irreversibly against an external pressure of 1 bar is spontaneous. ·

SoJn. AG for the process can be calculated using the expression
m
p,.
AG=nRTin-2
he
Pi
Substituting the given values, we get
.C
1
. AG= (2.0 mo/)(8.314 JK- 1mor')(273K)(2.303)log( bar)
w
. 2bar
w
= -3.147.06 J.
Since, there occurs a decrease in the value of AG, it must therefore, be a spontaneous change.
w
5. Show that ifequation of state for a gas is

· P(Vm-b)=RT
where b is a constant, the fugacity ofthe gas is given by
Inf= hp
p RT
~ f p
Also show that R·T is small (applicable at low pressure) this equation reduces to - = - -
p P;deal
where P;deat is the pressure the gas would exert ifit were to be ideal but with the same molar volume as th
gas at pressure P. HAND MADE NOTES www.ChemistryABC.com
SoJn. We know that, P(Vm -b) =RT

. PVm· Pb
Or, Z=-=1+-
QUZZES
RT RT MORE NOTES ON www.ChemistryABC.com
Substituting this in the relation,
In/ =rZ-ldP
p O p
and integrating the resultant expression, we get
/
In / = bP or In = exp ( bP )
P RT .P RT
bP
If RT is small, we may write
om
f =l + bP = RT+ bP
P RT RT
. From the given equation ofstate, we have . (
.c
RT+bP=PVm
C
.
Hence, p
I =P( RTVm )-__!_
- P.
'..
.\
B
ideal
yA
2
· Or, P =I ~deal
that is, pressure ofthe gas is the geometric mean of the ideal pressure and fugacity.
tr
For the reaction,

is
.!..N2 (g )+~H2 (g) ~ NH3 (g), AG~98K = -16.45 kl mar' and m~98K =-46.11 kl mar 1•
2 2
m
What is the value of AGiooK if m 0 is considered independent oftemperature..

he
(a) 5.89 kJ mo1- 1 (b)-5.89 kJ mot-1 ( c) 3.66kJ mot' (d) ~3:6'6'kJmol-1

In. We know that,
.C
AGT0 =m0 +!_(AG

L
0
To
-m 0 )
w
We get,
.
w
AG~ =-46.11 kl mar 1

+( 5
00K)(-16.45 kl mar 1 +46.11 kl mar1 )
w
298K
3.66kl mar1
, The fugacity coefficient qi is given by In qi Az; 1

z
) dp where is the compressibility factor, and p the
pressure. The fugacityof a gas governed by the gas law p(Vm -b) = RT is [GATE2004]
(a) pln(Vm/RT) (b) pi/RT (c) pe-bplRT (d) pip/RT
In. In q> = rz; l dP HAND MADE NOTES www.ChemistryABC.com

... (1)
Par.t:.ial Molar .Guattt
= ~ ,'
Since, we know that the compressibilityfactor, Z
P(Vm -b)=RT
PVm -Pb=RT
PVm =RT+Pb
Both side divided by RT
PVm RT Pb
-=-+-
RT RT RT
Pb
Z =1+ -RT Put this value in equation ( 1)
om
·
Jnq,:::: r +RT-
l Pb ,
1
Pb ·.
,··
dP => Jnq>= rp RT dP= Pb '
.c
P ~ P RT
C
bP bP
=> q) =eRT => q) =p eRT
B
yA
[PRACTICE SET )
tr
is
1. The internal pressure ( BU/BV )T of a real gas is related to the compressibility factor Z = pV/RT
J
m
[Vis the molar volume [NET Dec. 201

he
(a) ( BU/BV)T == RT(BZ/BV)T (b) (BU/BV)T =RT/(v z)

.C
2 2
(c) (8U/8V)r ==(RT IV)(BZ/8V)v (d) (BU/BV)T =(V /RT )(8Z/8T)v
w
2. The fugacity ofa gas depends on pressure and the compressibility factor Z ( == p V/RT) through the rela
w
(V is the molar volume) •

w
For most gases at temperature T and.up to moderate pressure, this equation shows that [NET Dec. 201
(a) f < p, if T-+ 0 (b) f < p, if T--+ oo
(c) f > p, if T--+ 0 (d) f =p, if T--+ O
3. The chemical potential(µ) ofthe i'11 component is defined as [NET Dec. 201
(c) µi =(:)
. I T,P
4. Given the following two relations, x 1df.1i + x 2dJLi = 0 {A) [NET Dec. 201

and x1dV. + x2df'i =0, (B)
for a binary liquid mixture at constant temperature and pressure, the true statement is that,
(a) Both the relations are correct
QUZZES
(b) RelationAis correct, butB is not MORE NOTES ON www.ChemistryABC.com
(c)~elation Bis correct, but A is not
(d) Both the relations are incorrect, except for veiy dilute solutions.
For a vander Waals gas
Whatisthefugacityat 100 barand298K.[Given: a= 0.2416dm6 bar mot-2 and b -:f= 0.02661 dm 3 moZ-1 ]
j •
(a) 106.9 bar (b) 101.9 bar (c) 110.9 bar (d)l,20.9 bar ·1
• .'i
What is the free energy change accompanyingthe compression ofl ,mole 9'a gas at 25°C from20 to 200 atm
om
The fugacities ofthe gas be taken as 18 and 120 atm afm, respectiv~lyftt p~essure of 18 and 200 a1In.
(a) 4700 J .. (\:>) 4000 J
'
(c) 4100 J
.
. (d),3900 J
' ' ,
·
) '
What is the fugacity ofH:z(g) at l00°C and 300 atm, assuming that it is a vander Waals gas, with a= 0.244 dm6
.c
atmmol-2,b 0.0266dm3 mot1 and Vm O.ll9dm3 mol- 1•
C
(a) 485.5 atm (b) 386.6 atm (c) 286.6 atm (d\586.~atm
.,
B
The expression which represents the chemical potential ofthe i'1 species(µi) in a mixture ( i -:t j) is: ·
yA
(a) (BE/ oni)s,v,nj (b) ( 8H / oni)s,v,nj · ( C) ( 8~ / Olli )s,v,nj ( d) ( 8G / Olli t,v,nj [GATE 2007]
tr
is
ANSWER KEY
m
Questions 1 2 3 4 s
he
Option {c) {a) {d) (a) (a)

"
Questions 6 7 8
Option (a) (b) (a)
.C
w
w
w

[ Chapter~
Thermoc·hemistry
Introduction:
The chemistry dealing with heat changes in chemical reactio~ islmown as thennochemistry.
om
Heat of formation: · .· · / · ·
The enthalpy change associateu· with the formation of i moI6 ofoompound from its element under stan
.c
condition
e.g. C+2H2 ~CH 4 AHr = -18.50 Kcal
C
c+o 2 ~co 2 Aflf = -96.96 Kcal
.Heat of combustion:
B
yA
The enthalpy change associated with the complete combustion of 1 mole ofsubstance.
1
e.g. ~ H20(g) MI =-68.40 Kcal
tr
H2(g) +202(g)
is
s(g) +0 2(g) ~so2(g) l'.lH =-71 Kcal

m
Measuring the enthalpy of Combustion (Bomb Calorimeter) :

The calorimeter used for determining enthalpies of combustion, known as the bomb calorimeter, is sho,
he
figure below (1) and (2).

The iru)er vessel or the 'bomb' (figure-2) and its cover are made ofstrong steel coated inside with
.C
or platinum or some other non-oxidisable material. The cover can be fitted tightly to the vessel bymearu
metal lid screwed down on a lead washer. A weighed amount ofthe substance is taken in a platinum c
w
which is supported on a rod R.

w
S\irrer
w
Figure-(1 ): Bomb calorimeter Figure-(2): The Bomb

.. A thin platinum wire Wis connected betweenthe rods Rand G; as shown. This serves to initiate the combus-
tion when heated
QUZZESelectrically. The bomb is tightly
MOREclosed andONoxygen introduc~
NOTES through the inlet tube T until a
pressure of about 20-25 atmospheres is attained. The bombi:is then lowered in water placed in a double
jacketed and polished metallic calorimeter so as to minimise .error due to radiaton. A mechanical stirrer is
_ provided as shown (figure (1)). When the temperature ofthe water has be9otne steady, the substance idgnited
by passing electric current through the platinum wire W The rise oftemperature ofthe water in the calorimeter
is noted after every minute by means of a Beckmann thermometer graduated to read up to a hundredth ofa
degree. The final temperatµre when corrected for the radiation error in the usual way, minus the ini:ial tempera-
ture, gives the rise oftemperature. '·
The heat capacity ofthe calorimeter system is obtairted by burning a known weight ofa substance of
known enthalpy ofcombustion. For this purpose, usually benzoic acid ofhigh grade putityis taken. Its enthalpy
ofcombustion which has been very carefully measured is taken as .:-3226. 7ld mot1:l . ) '
Suppose the thermal capacity ofthe calorimeter system including~ater is z and 0' is the rise in tem-
Jar
om
perature produced by burning a quantity 'm' ofthe given subst.ance of:tno mass M. Then, evidently, the
enthalpy ofcombustion ofthe substaq~ is given by, q v = z x 0x
1 I
, I
.c
Heat of neutralisation:
The enthalpy change associated with complete neutralisation ofl equivalent ofacid reacts with 1 equilivalent of
C
~~ 1 ~
B
'i.,
AH{ neutralisation) = AH(ionisation) + AH (H+ +0 H-)

yA
=> For strong acid-base pair AH = -13. 7 Kcal
=> For weak acid - base pair .AfI = -11.4 Kcal
tr
Heat of solution:
is
It is defined as the enthalpy changed when 1 mole of solute is dissolved completely in excess ofsolvent.
m
Heat of hydration:
Enthalpy change associated with the formation ofl mole ofa specified hydrate from 1 mole ofan anhydrous
he
substance on combustion with the required number ofmoles ofwater.

Hess's law of constant heat summation:
.C
At constant pressure the total enthalpy change ( AH) accompanying a chemical reaction is the same whether
the reaction takes place in one or more steps.
w
e.g. for a reaction

w
A--),D AH=q
w
& for A~ q,
B--C~qJ
D
Thus, accordingto this law
I q ::::: q, + q2 + q3 I
Bond energy:
It is defined as the average amount of energy required to break 1 mole of such bonds present in di:flerent
compounds, the original molecules and the dissociated products being in the gaseous 1state. Bond energy is
endothermic process. ·
The energy of formation a bond is numerically·equal to its corresponding bond energy but with a
negative sign(ie. exothermic process). ·
Relation between enthalpy of reaction at constant-volume-and at constant pressure.

HAND MADE NOTES
·: AH=AU+PAV ... www.ChemistryABC.com
{l)·
·Butweknow
MI Qp
and AU qv
Therefore, equation(!) becomes
!Qp =qv +PAV I ... (2)
for 'n' moles of an ideal gas
PV=nRT
.Let n1 and 11i are the number of moles of gaseous reactant and gaseous products respectively.
PAV=AngRT
'
Mg=(number ofmole of gaseous products]-'-[number of moles ~fgas~ous reacr1nt]
IQp =:c:qy+MgRT I ... (3)

. -., j '
om
Adiabatic flame temperature: · '/ ,
The temperature wpich the system attains ifthe changes in the s~eiµ are carried out under adiabati: conditi
It takes at either constant P or at~onstant volume for exothermic reaction. '
.c
At constant pressure:
C
For exothermic reaction under adiabatic condition, heat evolved would not be es9ape and hence increa
temperature ofsystem. \ /1
B
The increase in temperature can be takes in two steps.
yA
(!)Reactant (T0, P) ~ product (T0, P) MIT 0
(2) Product (T0, P) ~ product (Tr,P) MI 2

tr
Theheatevolvedin2ndstepisgivenas MI2 = f CP(product)dT

is
For overall reaction, Reactant ( T0 , P) ---+ Product (Tr, P)

m
Net heat change

he
MI = AHTo + MI2
Since adiabatic process are isoenthalpic
.C
MI= 0 ~ 0 = AHTo + MI 2
8H2 -~HT0
w
w
w
Note: Ifthe compound is burnt in air number ofmole ofnitrogen also changes to final temperature and he
included in the CP{product)
I mole of02 = 4 mole ofN2 associated.

At constant volume:
Ifreaction takes place closed vessel under constant volume condition
AU
Tf =- r To + To
CV(product)
( SOLVED PROBLEM~) ,
1. 0.50 g ofbenzoic acid was subjected to combustion in a bdtnb calo~nwihen the temperature of the
. calorimeter system (including water) was found to to rise by 0.55°C. What is the enthalpy ofcombustion of
benzoicacidatconstantvolurne; · .1 '· · ·
1 1 1
(a)3200k:Jmol- (b)-3200k:Jmot (c) 1600kJmor (d)-1600kJmol-~
Som. · Enthalpy ofcombustion at constant volume
qv =zx0xM/m
=23.85 kl mar 1 x 0.55K x 122 g mar1 I 0.5o' g/= 3200.7 kl mar1 = -3200.7 kl mar'
[Note: The entha]pyofcornbustion has alwaysanegative;sfgn]
:,/
/
om
2. In a bomb calorimeter, the combustion of0.-5 g ofcompoundk(~r,mass = 50 g mot 1) increased the
temperature by 4K. If the heat capacity of the calorimeter along with th~t of the material is 2.5 k:JK- 1, the
1
molarintemafenergyofcornbustionrink:J, is : ,- [NET June Z014)
.c
(a) 1000 (b)-1000 (c) 20 (d)-20
S0 m Molarinternalenergyofcombution= zx0x M = 50x 4 x 2· 5 5 50 5

0xl0 = .xlOxlO = ooo = 1000 k:J=
C
• ~ 0. 5 0.5 l,' /10.5 5
B
-1000 kJ
Note : Energy ofcornbution has always a negative sign.
yA
Correct answer is (b)
tr
3. What is the relationship between qp and q v Haber synthesis ofammonia assuming that the gaseous reactants
and products are ideal.
is
(a)qp=qv-2RT (b)qp=qv+2RT (c)qp=qv-RT (d)qp=qv+RT

m
Som. N 2(g) + 3H2(g) -.:-- 2NH3(g) ' Ang = 2-( 2 + 3) = -2

he
qP = qv + AngRT=:> qP =qv -2RT

.C
4. One mole ofnaphthalene was burnt in oxygen gas at constant volume to give carbon dioxide gas and liquid
w
waer at 25°C. Heat evolved was found to be 5138:8 kl The enthalpy ofreaction at constant pressure is
[Given: R = 8.314 J/K-mol]
w
(a) 5143.8 kJ (b)-514~.8 kJ (c) 21430.01 kJ (d)-2143.01 kJ

w
Som. CIOH8(s) +1202(g) ~10C02(g) +4H20(l)

Ang =10-12=-2
qP =qv +AngRT =5138.8-2x8.314xl0-3 kJ /molex298
1 qp =-5143.sk:J I
5. The enthalpy of combustion of glucose C6H,20 6(s) is-2816 kJ/mol at 25°C. Calculate AH~ ( C6H 12 0 6)
values for CO2, 0 2(&) & H 20(e) are-393.5 and 285.9 kJ.mot 1, respectively.
· (a) I360.5kJ/mole (b)-1360.5kJ/mole (c)-I260.4k:J/mol (d) I260.4k:J/mol
11101·111~11c:;;1r110,:;.c
0
Soln. · C 6 HQUZZES MORE NOTES ON AH =-2816www.ChemistryABC.com
12 0(s) +60 2(g)~6C0 2(g) +6H 2 0(t)
kJ. ,
Since' AH= LAHtrodue1$) - LAH~(reacunt)
But AH~(0 2 )=0; ·m~(C6 H 12 0 6 )=-1260.4kJ/mole

6. The heats ofcombustion o ~ and 1¾ are 9.06 and 68.9 kcal respectively. The heat offurmation ofammor
B ;
(a) 94.29 kcal (b)-94.29 kcal (c) 91.30 kcal . ( d) -91;20 kcal,,
",,,' ;:,
Soln. We have to find,
1 3 .. 0
om
(1) 2N2+2H2~NH3 Mic=?
3 · 1 3,,··
(2) NH3 + 0 2 ~ N 2 + H 20 AH= -9.06 kcal ,·
4 2 2
.c
10 . AH = -68.9 kcal
(3) H 2 + ~ H 20
C
2 2
/'f
t
B
3 /
multiplying equation(3) by and then substractingequation(2) from equation (3)
2
yA
3 3 . 3 3. 1 3 ·
-H +-0 -NH --0 ~ - H 0--N --H 0
2 2 4 2' 3 4 2 2 2 2 2 2 2
tr
l 3 3
N 2 + H 2 ~ NH 3 AH = x(-68.9) (-9.06) =-94.29 kcal
is
2 2 2
m

he
7. The molar heats of combustion of C 2H 2(g), C(graphite) and H 2(g) are 31.62 kcal, 94.05 kcal and 68.32 k<
respectively. The standard heat of formation of C 2H2(g) is
.C
(a) 54.20 kcal (b) 64.20 kcal (c) 74.20 kcal (d) 84.20 kcal
Soln. Given,
w
(0 C+0 2 ~CO 2 AH=-94.05kcal

w
AH= -68.32 kcal

w
5
(iii) C 2H 2 + 02 ~ 2C0 2 +H 20 AH =-310.62 kcal
2
We have to find,
J
[ 2 x equation ( i) + equation (ii)- equation (iii)
1 5
2C+20 2 +H 2 +-0 2 C 2H 2 -0 2 ~2C0 2 +H 20-2C0 2 -H20
2 2
2C+H 2 ~C 2H 2 ; AH=2x(-94.05)+(-68.32)-(-310.62) 54.20kcal
What is the heat offonnation of benzene from the following data, assuming no resonance bond energies.
C-C =83 kcal; C = C = 140 kcalMORE NOTES ON
QUZZES www.ChemistryABC.com
Heat of atomisation ofC = 170.9 kcal
Heat of atomisation ofH = 52.1 kcal
(a) 75 kcal (b) 85 kcal (c) 100 kcal (ft)120 kcal
oln. We have to find AH offollowing reaction
6C<s> + 3H 2(g) ~ C 6 H~g)
I
C /
"'-c~~c
I
c~C/c"-
I
/
I
om
For reactant:
Heat ofatomisation of moles ofC = 6,x 170.9 kcal · i
.c
Heat ofatomisation ofmoles ofH =6x 52.1 kcal
For products:
C
Heat of fonnation of 6 moles of C-H bonds= -6 x 99 kcal
B
Heat of fonnation of3 moles ofe--C bonds= ;..3 x 83kca1
yA
Heat offormationof3 molesofC=C bonds;= -3x140 kcal
On adding, we get heat offormation of C 6 H 6 i.e.
tr
MI= 6xl 70.9 +6x52.l-6x99-3x83-3x140 = 75.0 kcal

is
What is the resonance energy in C}\COOH from the following data if the observed heat of formation of
m
CH3COOH is -439.7 kJ
Bond energy (kJ) Heat of atomisation (kJ)
he
C-H= 143 C 716.7

C--C =::: 348 H=218.0
C=O 732 0 = 249.1
.C
C-0 351
0-H 463
w
(a) 200.3 k:J/mole (b)-203.3 k:J/mole (c) 100.3 k:J/mole (d)-100.3 k:J/mole
w
foln. Calculate Af-If ( CH3COOH)

2C<s) + 2H 2(g) +0 2(g) ~ CH 3COOH H 0
w
For reactant: .
Heatofatomisationof'2' moles ofC=.2x 716.7 = 1433.4k:J
I I
H-C-C-O-H
Heat ofatomisation of'4'moles of H = 4x 218 = 872 kJ I
H
Heat of atomisation of'2' moles ofO 2 x 249 .1 = 498.2 kJ
For products:
Heatofformatinof3 molesofC-H= -(3x413) =-1239 kJ
Heat offormationofl molesofC-C = -lx348 =-348 kJ
Heat of formation ofl moles ofC=C = -732 kJ

Heatofformationofl molesofC-0=-351 kJ
Heat of formation of 1 moles of 0-·H =-468 kJ
Let, resonance energy in C~COOH = x kJ
Adding these, we get AHr of CH3COOH
-~39.4+x = -439.7(given)
Ix=-200.3kJ/mole [
Correct option is ,(b)
(P~CTICE ,SJ:T]
1.
v1aue of 6U 0
(a)-1, 364 kJ (b) l, 364 k:J · (c) 2, 364 kJ ; , (d)-2, 364 kJ .
om
2. The enthalpy offormation of ethane at 25°C and at consant p;Jsure;is-84.68 kJ. The enthalpy of foam
1
at constant volume at this tempmture is , .· f
(a) 79.70 (b)-79.70 (c) 89.70 (d)-89.70
.c
3. The enthalpy ofcombustion ofC2H4 at l 7°C and at constant volume is-1389 .9 kJ. The enthalpy ofcombu:
C
at constant pressure is · /t
(a) 1394.7kJ (b) 1294.7kJ (c)-1394.7kJ (d)-1294.7kJ
4.
B
The enthalpy ofneutralization ofNH40H with HCl is-51.46 kJ mot 1• What is the enthalpy ofioniatic
yA
NH40H. Assume that theenthalpyqfneutralization of a strong acid with a strong base is-57.35 kJ mol
(a) 5.89 kJ mo1- 1 (b)-5.89 kJ mo1-1 (c) 4.89 kJ mot-1 (d)-4.89 kJ mol- 1
tr
5. 6H~eutr forCH 3COOHwithNaOHis-51.63 kJmot1• What is the AHionize ofCH3COOH.Assumetha

is
enthalpy ofneutralization ~fa strong acid with a strong base is-57.3 5 kJ mo t 1•

(a) 4.72 kJ moI- 1 (b)-4.72 kJ mo1- 1 . (c)-5.72 kJ mot-1 (d) 5.72 kJ mol- 1
m
6. 2 gram of C6Hig) was burnt in excess ofoxygen in a bomb calorimeter according to the reaction
he
C6 H 6 (g)+ 7.50 2 (g)~6C0 2 (g)+3H 2 0( £)

Ifthe temperature rise is 42.1 °C and the heat capacity ofthe system is 2.0 kJ K- 1, the enthalpy ofcombm
.C
ofbenzene at constant pressure at 25°C is

(a)-39.20 kJ moI- 1 (b) 39.20 kJ mol- 1 (c) -49.20 kJ mo1- 1 (d) 49.20 kJ mot 1
w
7. For a gaseous reaction at 300K, AH - 6U = -4.98 kJ, assuming that R = 8.3JK- 1mor 1, 6ng is
w
001 ~2 ~3 ~4
w
8. The heat change at constant pressure, Clp, is equal to

(a) 6U (b) AH (c) 6G (d) RT
1
9. AH~ ofS02 is-297 .5 kJ moI- • The energy required for the decomposition of32g ofS02 is
(a) -148.75 kJ (b) 148.75 kJ (c)-297.5 kJ (d) 297.5 kJ
10. For the combustion of one mole ofCH3COOH(l) at 25°C, 6n is
(a)-1 (b)l (c)O (d)-1/2
ANSWER KEY
Questions 1 2 3 4 5
Option (a) (b) (c) (a} (d}
HAND MADE NOTES (b) www.ChemistryABC.com
Option (a} (b} (b} (c)
[ Chapter a]
ttatistical 1hermodynat,riics
Statistical Thermodynamics i
:;'•
.' 'j
om
Chemical thennodynarnics is a macroscopic view ofdescribingthe beJJiivio~ oflarge number of molecules
in tenns of pressure vohun~, compositi~J.J, and ·exchange ofheat:and wor~. Whereas Quantum mechanics
provides a microscopic description for the structure and interactions offueleeules. '
.c
Statistical thermodynamics provides the bridge between microscqpic mechanics (quantum) and
macroscopic thermodynamics (classical).
C
Types of Statistics: \ /1'
B
Different physical situations encountered in nature are described by three types of statistics, the Maxwell-
yA
Botzmann (or M-B) statistics, the Bose-Einstein(orB ..E}statistics~the-F.~·,.Oirac (or F-D) statistics.
The M-B statistics, developed long before the·advent of quantum mechanics, is also called classical statistics
whereas the Bose-Einstein statistics and the Fermi-Dirac statistics are oollectivelyealled quantum statistics.
tr
The characteristics ofthree types ofstatistics are summed up as follows:

is
Maxwell-Botlzmann Statistics:
m
The particles are assumed to be distinguishable and any number ofparticles may occupy the same energy level.
Particles obeying M-B statistics are called boltzons or maxwellons.
he
Consider a system ofN distinguishable particles occupying energy levels s0 , s 1, s 2 · etc. The total number of
arrangements for placing~ particles in the ground state energylevcl .s0 .,Jl1 .particles.in the. first excited state
.C
energy level n 2 particles in the second excited state energy level , and so on, is known as the
Si,
w
thermodynamic probability, (or number ofmicrostates) W, of the given macrostate. It is, in general, a
very large number. Thermodynamics probability Wis given by
w
w
N! N!
w -------
nl !n2 !n3 !... ....nj ! IIni !
where, N =En;, Nis the total number and summation is over all the energy levels.
The thermodynamic probability for the system ofN particles is given by
n.
W = N!ll g;' x constant
; n!
. l
Where, gi degeneracy(or multiplicity) ofthe energy level &; •

It is well known that the entropy S and probability W of a given state ofsystem_m:e related by the Boltzmann
equation, the most famous equation in statistical mechanics, · ·· ,,
I S=klnW I
2. Bose-Einstein Statistics:
The QUZZES MORE NOTES ON
particles are indistinguishable and any www.ChemistryABC.com
number ofparticles may occupy a given energy level 1b.B statist
is obeyed by the particles having integral spin, such as hydrogen (H:i) deuterium (D2), nitrogen (N2), heliun
(4He) and photons. Particles obeying B-E statistics are called bosons.
In Bose-Einstein statistics, we deal with the distribution ~findistinguishable particle among the energy le,
with no limit on the number ofparticles,.in any ofthe energy state:
The numbe:r ofways ofplacing N indistinguishable/particles in g distinguishable energy states witb
restriction on the nmber per energy state is given by
W=(g+N-1)!
(g-l)!N!
The thermodynamic probability of a given distributi~n ofparticle~;~ver the energy levels is the produc·
om
'! '.
number ofarrangements of all energy levels, i.e. ·· ' /
'
W=TI. (g.l +N.I -1)'• f
.c
I (gj-l)!N,!.
C
3. Fermi-Dirac Statistics: .
. O(_
..'
The particles are indistinguishable but only one particles may occupy a given energy level. This statistic
B
obeyed by particles having half-integral spin, e.g. the protons, electrons, helium-3 t'He) and nitric oxide (N
yA
Particles obeying F-D statistics are called fermions.
In Fermi-Dirac statistics, we deal with the distnbution of indistinguishable particles among the ene
levels with only particle in any ofthe energy state.
tr
Let N indistinguishable identical particles be distnbuted over g distinguishable degeneracy energy st.
is
with the restriction of only one particle in any ofenergy state. The later condition requires that N ::; g .
m
For distinguishable particles, we will have

g choices for the 1st particle
he
g-1 choices for the 2nd particle

.C
g - N + 1 choice for the Nth particle.

The total choices or arrangements will be
w
(g)(g-1) ...... (g-N+l)=( '

. )
w
. g-N !
Since the particles are indistinguishable, the above arrangement has to be divided by N! permutations ofth
w
particles. Hence, the number ofways ofplacing N indistinguishable particles in g distinguishable regene1
energy states with a limit ofno more than one particle per energy state is
W= g'. .•
(g-N)!N!
The thermodynamics probability ofa given distribution ofparticles over energy levels is the product ofnurr
ofarrangements ofall energy levels, i.e.
. The expression for the most probableHAND MADE NOTES

distribution www.ChemistryABC.com
ofN particles among the energy levels according to
Fermi-Dirac statistics.
QUZZES n. = gi MORE NOTES ON www.ChemistryABC.com
' exp(a+,lh';)+l
Summary for the three types of Statistics:
Statistics Thermodynamic probability. Number e>f particles in the ith state Entropy
I . n. •.
W =N!IT g;' . gi
Maxwell-Boltzmann f n.=-- S = klnWM8
MB •
' n;. ' ea+ps,
W -IT(N1+g1-l)! g
Bose-Einstein n.= i $, = klnW8 E
BE- i N i.'( gi -1)'. I . ea+Pe,/...:. t
, . I
..
:i' i
Fermi-Dirac W =II g;! ·n,= .:':g; S = klnWFD
om
FD i N;(g/ -N;)! ,, -/+P\+ 1
..
- /
. t
The probability must be a maximum for an equilibrium state so that at equilibrium ··
.c
., S=klnWmax]
C
l ,,, ...
\
B
Stirling's Theorem-: ·Stirlintstheoremor appi;oximationgives
Rn n ! =n Rn - n
yA
(Provided n is very large.)
( SOLVED PROBLEMS) .
tr
"
What is the number ofways ofdistributing 20 identical objects with the arrangements 1, 0, 3, 5, 10, 1.
is
(a) 9.3lxl0-8 (b) 9.3lxI08 (c) 93.lx·I0-8 (d) 93.lxl08

m
201
·
oln. Smewekonwthat, w-- l!0! 9·31 x 108
3!5!10!l!
he

.C
What is the weight ofthe configuration in which 20-objects are distnbuted in the arrangement 0, 1, 5, 0, 3, 2,0,
1, 0, 8.
w
(a) 4.19x10 10 (b) 4.19xl0 12 (c) 4.19xl014 (d) 4.19xl0 16

w
· · W= 2
· o! = 4 I9xl010
oln. Smewekonwthat,
w
0!1! 5 !0! 3!2 IOll!0! 8 ! .

~orrect option is (a)
(a) In how many ways (W) can six distinguishable molecules be placed in three different energy level; with 3
molecules in the first levei two in the second and one in the third.
(b) Write down the different ways ofdistribution of three balls a, b, c between two boxes
' W _ 6I _ 720 _
:01n. (a) - 3!2!1! - 12 -
60
abc ab ac · be O c b a
(b) The eight different distnbutions are - , - , -b , - , '--b , -b ' - ' -b
0 c a aca ac.c
The number ofconfigurations in the most probable state, according to Boltzmann formula, is
(b) e-S/kB (d) e-LiG/kBT [NET June 2014]
Soln. Sinceweknow that, S = k 8 lnW
S /k8 =lnW
W =eStkB.
5. Using Boltzmann distnbution, the probability ofan oscillator occupying the first three levels (n = 0, 1 ar
found to be p0 = 0.633, p1 = 0.233 and p2 0.086.
;',
The probability offinding an oscillator in energy levels in n 2 3 is

(a)0.032 . (b)0.048 · (c)0;952 (d) 1.000
Soln. Total probability= PT= 1
Probability of occupying first three levels PO + P1 + P2 , _
p T=po+ p I + p 2 + probability offinding an OSCil.JatOf in energ}' l~vels in n 2 3
om
1 = 0.952 + Pn~ = 1-0.952 = 0.048 ' //
, t
6 .. The number ofways in which four molecules can be distnbuted in two different energy levels is
.c
(a) 6 (b) 3 (c) 16 · (d) 8
=
C
Soln. Number ofenergy levels (g) 2
Number of molecules (N) =4
Thenumberofwaysofdistribution= ~ri =24 =16
B
yA
(since, it is not given either particles are distinguisable or indistinguishable, therefore, we willconsid
distinguishable)
tr

is
Canonical Ensemble Partition Function:

m
If g.l be the weight factor for the ith level, then

ni gi e-a. .e-~ei => ni = e-a. .:E ( gi. e-~q )
he
Where a and ~ are constant and ~ = 1/k:T .

.C
Il• gj e-ej/kT
According to Boltzmann distnbution law, --;:- :Egi e-Ej/kT
w
w
the factor in denominator• :Eg.I e-ei/kT is called partition function .denoted by q.

w
n. e-ej/kT e-Ej/kT
Thus, . -'=~-- => ni =ngi
n q q
Thus partition function is the sum ofthe probability factor for different energy levels or 'the nu:n:i>er of
in which energy ofa system is distributed among the molecules constituting the system.'
Thus the partition function can be expressed
The energy Ei ofa system containing non-interacting particles a, b, c ............. is given by

Ei =E 1 {a)+E2 (b)+ (c)+ ............+E(N)
The partition function '1c is defined as (for ~ = 1)
4c = Le(-Ei/kT)
i
Since, a, b, c et-c· are identical therefore, each particle will have same set of energies andaswellas partition
function, ::::::> qa = q b = qc = ........ ·· = qN . :•
Therefore, for a system ofN independent molecules the canoQical ensemble partition_funaien is giwn by
I qc = qN ( for distinguishable entities) 1 . -1 .- ·,
The relation is applicable for the systems ofi perfect crystal

In the case ofa gas the_molecules are indistinguishable.J'herefore the energy states.that differ by the
two particles are not distinguishable and should be counted qnly once:
E1 =E1 (a)+ (b)+ .............states
,\
E2 =E 2 (b )+ E (a)+ ..............states
Therefore the partition function for an ideal gas is:
om
\
Iqc =qN /N! I Forindist~ishablemolecules
.c
Physical significance of partition function (q) :
(i) It indicates how the particles. are distnbuted among the various energyfevels.
C
(ii) It reflects the diversity ofenergy states ofthe molecules ofa system , /-t
B
(iii) It determines the relative population ofany :two quantumstatespJUld4:by.tliexatio.n/n.
We can derive the ratio as ' P q
yA
tr
is
Since partition function for a molecule is a function of temperature and since the exponents in e-ElkT are
m
inversely proportional to the temperature, the terms in partition function approach zero as T ~ O•except for
one which is g0e-eotkT for the ground state because E0 =0 so that this term is equal to g0 • Thus as T ~ O q
he
approaches the degeneracy ofthe ground state.

.C
As T ~ oo , kT ~ 0 , so that e-q tkT approaches unity and q approaches the sum ofdegenercies ofthe levels
w
w
Problem: When T ~ oo, value of the single-particle partition function will be (given: degeneracy of
w
level} g.) J
[CSIRDEC. 2014]
(a) 1
oln. Correct option is (c)

Translational Partition Function:
The particle can execute translational motion along x, y and z directions. The translational partiti:m function for
a molecules along x, y and z axis is given by
-Et(x)lkT ( ) -Et(y)lkT ( ) -Et(z)/kT
qt.( x ) = i:gte ; qt y = I:gte ; qt z = I:gte
where Et is the translational energy, k is Boltzmann constant and~ is statistical
HAND MADE NOTES
weight factor.
Hence the net translational partition function ofa molecule will be
QUZZES
(z) NOTES ON
qt =qt (x)qt (y)qtMORE www.ChemistryABC.com
Since ~ = 1 for each level
. qt (x) = Le-EtlkT and SO Oll.
Now in a box of sides l,X ly and l.Z
{-n2h2/8mli kT)
:Ee
Or, q = ( 2mnkT)½ and so cin , I/
om
~----.C-
~x) h .
1
Hence the net tr~lational partition function of molecule wiJI be;
.c
(2mnkT)½
q,= · ·3 {(r,.t'y,fz)
C
h .
B
But volume of box(V) =lx,l Y' lz
yA
2
(2mnkT)3'
Hence, qt = h3 xv
tr
Since <1 for a molecule is proportional to the voluem Vin which it moves. Therefore, Cl: is an extensive va
is
· 1/2 .
The quantity ( h 2 / 2mnkT) is called thennal wavelength (A). Thus
m
Iq, ~;, I
he
The condition fur the applicability ofBoltzmann statistics is that thennal wavelength must be small corrq:
.C
mean distance between molecules.

Rotational Partition Function:
w
Rotational energy (Er) for a diatomic molecule in~ rotational energy level is given by
w
j(j+ i) h 2 / 81t 2I
w
where j is rotational quantum number and I is moment of inertia.

The degenercy ofjth rotational level will be 2j+ 1 and is equal to rotational statistical weight factor(&.).
The rotational partition function for a diatomic molecule will be
qr :Egr e-ErlkT
2 2
• ) [-. j(j+l)h /81t IkTJ
Thus qr =L ( 2J+l e
Which on solving we obtain
81t2IkT T
qr= =-
er
h2 HAND MADE NOTES www.ChemistryABC.com
h1
Where erQUZZES
= ~Ik is rotational temperature.MORE NOTES ON www.ChemistryABC.com
81t
This equation is for heteronuclear diatomic molecules like NO, HCl etc.
For homonuclear diatomic molecules ·
)
Ingenerai
Where, cr is symmetry number and cr is 2 for homonuclear diatomic mole~tiles.

Fornonlinearpolyatomicmolecules , ; f
. /
om
'/
.c
Where IA, I8 , le are moment of.inertia aboutthree principa.laxis.
C
Quantum Mechanical explanation of symmetry number
The total wave function ofa molecule is given by
'I' total ='I' t 'I' r 'I' v 'I'e 'I' n
B
yA
where the subscripts t, r, v, e and n stand for transitional rotational, VIl>rational, electronic and nuclear, respec-
tively. ·
tr
On the basis ofthe nuclear spin quantum number, the atom are two types:
(i) Boson: The nuclei ofthese atom have integral spin quantum number.
is
e.g. 2H (I = 1), 14N (T = 1) and 1°B(I =3)

m
For bosons, the total wave function is symmetric.

he
(ii) Fermions : The nuclei of these atom have half. integral spin quantum number.
e.g.
1
H(I=½} N(I=½} 0(1=%)
15 17
and
35
c{1 ¾)
.C
For fermions, the total wave function is antisymmetric.

w
• The rotational wave functions have the same symmetry characteristics as t!iose ofthe angular functions of
w
the hydrogen atom

w
'I' r are symmetric for even values of rotational quantum number J.

'I' r are antisymmetric for odd values of rotational quantum number J.
• A nucleus with spin quantum number I has a total of2I + 1 spin states. These states are represented by the
magnetic spin quantum number which have values of+I, +(I-1 ), ....... (I-1 ), -I.
For example, for a proton ( I = ½) the two states are,

m1 +½ ,known as a. spin. Its wave function is represented by the symbol a..
m1 -½, known as ~ spin. Its wave function is represented by the symbol-~. ·
A diatomic molecule involving identical nuclei has:atotal of(2I + 1)2 nuclear wave
with spin quantum number I, www.ChemistryABC.com
HAND MADE NOTES
functions. Out ofthese wave functions,
(2I_ + 1)2 (2I)/2 i.e. I(2I+ 1) wave functions are antisymmetric.
and (2I+ 1)2- I(2I+ 1), i.e. (I + 1) (2I+
QUZZES 1) wave
MORE NOTESfunctions
ON are symmetric:
Example: In hydrogen molecule, we have three symmetric nuclear spin wave functions. These are ac
One antisymmetric nuclear spin wave function. This'1s ( }i)(ap - Pa) .

Factor affecting symmetry of total wave function:
(1) For bosons, the total wave function has to be ~yfi1metric. This is possible whenl(2I+ 1) antisyn:n
nuclear spin functions couple with antisymmetric rotational w,sive functions fqr '*hich rotation qu:
number J has odd values. (I + 1) (2I + 1) symmetric nucle~r spin functions couple with symr
om
rotational wave functions for which rotational quantump:?1mber J has even values.
(2) For fermions, the total wave function has to be antisyfnm~iric. This is possible only when I(
antisymmetric nuclear spin functions couple with syrrnnetric rotational wave functions for whicl
.c
tional quantum number J has even values. (I + 1) (2I + 1) symmetric nuclear spin functions c
with antisymmetri9 rotational wave functions for which rotational quantum number Jhas odd v
C
These are also applicable to linear polyatomic molecules spch a,;C0 2 and C 2H2•
\, J-
Expressions of partition functions:

B
yA
For Bosons, we have
2 8112
qr,n =(1+1)(21+1) L (21 +1) e-J(J+I)h / IkT +1(21+1) L (21 +1) e-J(J+I)h 218112 IkT
tr
] ] ... (i)
(even) (odd)
is
For fermions, we have

m
qr,n =(1+1)(21+1) L (21+l)e-J(J+I)h 2/Sn2IkT +1(21+1) L (21+1)e-J(J+l)h 2/8112 IkT

... (ii)
he
J J
(odd) (even)
-
.C
The heat capacity of hydrogen gas

Rotational partition function:
w
For hydrogen atom, I=½, hence from equation (ii), we obtain that,
w
qr,n =(I) L (21+1)e-J(J+I)h2/8112IkT +(3) L (21+1)e-J(J+l)h2/8112IkT

w
] ]
(even) (odd)
Ortho and para hydrogens:

The hydrogen molecule with opposite nuclear spin (i.e. antisymmetric nuclear spin function) is calle
hydrogen. It is associated with rotational wave functions having only even values of rotational qu:
numbers.
The hydrogen molecule with parallel nuclear spin (i.e. symmetric nuclear spin function) is called
hydrogen. It is associated with rotational wave functions having only odd values ofrotational qu:
numbers.
The ratio of molecules of ortho and para dihydrogen is given by

(3) t (2J +1) e-J(J+1)h2 /8n2IkT
Northo=QUZZES
__ oo_dJ_ _ _ _ _ __
Npara L
(1) (2J + 1) e-J(J+I)h2/8irlkT
evenJ
1
This ratio depends on the temperature ofthe system. The percentage Ofpara dihydrogen is 100 at OK. It
decreases with increase in temperature and·attains a value 25 perce~t at high temperatures.
Effect of nuclear spin.
The effect of nuclear spin is important in case ofhoqmµuclear di-atomic molecules such as the two
isotopes ofhydrogen atom~ and D2). In case ofheteronucl~·diatomic molecules, the effect ofnuclear spin
is not so important. . ' . ·; .,
Let us consider a homonuclear diatomic molecule where each ofits cgbstitutent atomJia,s spin quantum '
number i. Hence the resultant spin quantum number I ofthe molecule can ~ume the values ranging from Oto
,a
Of an even integer, and the relative
om
(2 i) in steps?funity. F_or a halfinteger spin i, total nudear spin I can be
nuclear part1t10n funct10ns for even and odd values ofl are . /
f ~ • ,.
•z, i Z' ~ i+l
.c
even . 2i + 1 odd - 2i +1 .
For an integer spin i, total nuclear spin lean be an odd or an even integer, and the relative nuclear partition
C
functions for even and odd values ofl are \ /1 ,. · · -
Z' i+l Z' _ _i_
B
yA
Zi +i
even · odd ~ 2i +-i
Taking into account the nuclear spin effect, the rotational partition function for such molecules ran be modified
tr
to
is
m
Where, zeven = I (2J +1)exp[-crrJ(J +1)]

he
J=0,2,4
I
.C
zodd = (2J +1)exp[-~J(J +1)]

. J""l,3,5
w
At high temperatures, Zev!!n = =21

Zodd Zrat = ( kT I 2B). It shows that at high temperature, the free energy,
w
entropy, internal energy and specific heat ofa diatomic molecules does not depend on the nature ofthe atoms
w
in the molecule. At low temperatures, the situation is different and the gas may be regarded as a combination of
two components corresponding to even and odd values oftotal spin quantum number.
For hydrogen isotopes, the molecules with greater nuclear statistical weight are termed as ortho- .
hydrogen molecules, whereas those with smaller weight are known as para-hydrogen molecules. Thus, for 8i
and D2, the nuclear partition functions are as follows:
ortho Z'even =½
Z'
para odd J1
The relative abundances ofthe two specfes in equilibrium are determined by the relative magnitude ofthe two
parts ofthe partition function Zrot. Thus, the ortho-para ratio is given by
QUZZES
N( ortho-Hz) _ Z'ot1a Zot1aMORE
_ 3 Z dd
NOTES ON
0
N (para-H2) - Z even 2 even 1
- · 2 even
As t -> 0, all the f!lOlecules tend to be in the lowest rotational state with J = 0, and thus the f\ gas is onlyp
=
whereas the D2 gas is only ortho. At high temperatures~ Zeven Zodd , the abundance ratio for~ is three;
as
for D2 is two. Hence, the temperature varies from ze~6 to infinite, the ratio v~es from zero to three fo1
and from zero to two for D2• However, for all practical purp~ses, we use the rjtio 3 an4 2, respectively.·
densities ofthe molecules can be expressed as , . · 1
3 I
om
N H2 =4 Northo + 4 Npara \
.. 2 ·· I .x'
f
NDi =3 Northo + 3N para
.c
Accordingly, the specific heat is
C
3 I
)H2 =-cortho
B
(crot +-cpara
4 4
yA
2 I
-cortho +-cpara .
3 · 3
tr
Vibrational Partition Function:

is
The partition function for vibrational energy ofa diatomic molecule is q v = :Egve-~v tkT
m
where gv :::: statistical weight fuctor for vibration, vibrational energy ofa diatomic molecule is Ev= (v + ~) l
he
vis vibrational quantum number with values 0, 1, 2, ........... .oo • u O is fundamental :frequency ofvibration. S
.C
gv is unity for each energy level, we may express

= L e-EylkT
w
qv
Thus, VIbrational partition function can be expressed as
.
w
-(v+.!.)huolkT
w
qv = Le
OJ
2
V=O
On solving above expression, we get
e 2kT I
or qv =I -e-huotkT or q v
2sinh ( hvo)
2kT
If ground state energy is zero ( + v ~ v) then

. qv = I e1 ;~ or qv = [ I - eHAND
hvo MADE NOTES
kT
J-l www.ChemistryABC.com
For most diatomic molecules at ordinary temperature tl;ie value ofCl is nearly unitY, because huo is appreciably
QUZZES
greater than kT. MORE NOTES ON www.ChemistryABC.com
Electronic Partition Function:

The electronic partition function can be expressed as
qe = :l:ge e-EelkT
where, Ee = electronic energy of molecule.
ge = electronic statistical wt factor.
In general ge = 2j + l '
, .
·,.
Where j is total angular momentum and it~ values are denoted by symbols SJ12, P112, P312 where the
subscript equals J. Thus , . ' J:' . ' ,
/
·.. qe=L(2j+l)e-Ee/k'.f / \ .. / '
om
The value of 'le way exceed unity because ofdegeneracy. For examp1{02 Jfus two unpaired electrons in the
ground state and is triply'degenerate ((= 3). Hence.,· f ·
.c
qe = :l:ge e -ee/kT = :l:J.e-OfkT =.3 . (:. eo 1)
C
For atoms and molecules with one unpaired electron spin degeneracy is 2 in ground state.
1, 11
B
Hence, qe = :l:2. e-O/icT = 2 ·· "
yA
Generally the electronic energy acquired l,)y the molecule at ordinary temperatured is extremely low so its
contnbution to the net partition function is immaterial.
tr
Remark: The energy term E represents the total energy ofthe molecule such as translationaL (Et ) ,
is
rotational (er) , vibrational (Ev) ;·electronic (Ee) and zero ·pointenergy{€0}is

m
E=Et + Er + Ev +Ee + Eo + Enuc.

Also the statistical wt factor is defined as
he
g = gt X tr X gv X ge X go X gnuc
Thuspartition function becomes
.C
-[Et+E +E +E +EQ+E ]/kT

q = L ( gt X gr X gv X ge X go X gnuc. ) X e r v e nuc
w
Or, q = Lg t e-EtlkT X :l:gr e-ErlkT X :l:g v e-£v/kT X :l:ge e-e,fkT X :l:gOe-eo/kT X :l:g nuc e-e.uc/kT
w
Or, I q =qt xqr xqv xqe xqo xqnuc. I

w
Thus net partition function is the product of the separate partition functions with respect to different
types ofenergies associated with the molecules.
Relation between Partition function and thermodynamic functions:

Internal Energy: IfE is internal energy of the system at temperature T and Q is the partition function of
the particles in the system. Then the expression for their relationship is
E~kr'(ai;f t
For N molecules ofthe system. . . .;;,,"
·t ,;.,
For 1 mole of the system HAND MADE NOTES www.ChemistryABC.com
2. Heat capacity:
·: E=Rr'(8;'t).
... (i)
Since, we know that
(:t =Cv
om
So, from equation (i)
.c
... (ii)
C
Cp-Cv =R
B
Cp R+Cv
yA
From equation (ii), we get
tr
is
3. Entropy and partition function:

m
For a system containing n molecules at temperature T entropy is related with partition function as
he
.C
And for 1 mole ofan ideal gas,

w
S=+n(q)+r(~). +1]
w
w
In another form, the relation, between entropy (for 1 mo le) and partition function is:
S=kfu[fr)+kr(~).
4. Work function (A) and partition function:
SinceA=E TS .
puttng the values ofE and S for 1 mole of an ideal gas, we obtain
A=Rr'(~).-Rren(!)-Rr'(~)-RT
(SOLVED PROBLEMS)
What is the vibrational contribution to tl:$Jolar heat capacityo~Nz(g};1t 1000,K. The vibration temperature
ofN/g) is 3374 K. ·
1
(a) 4.48JK- mot- 1
(b) 3.48JK- mot\ (c) 5.38 JK- mo1-1
1
-~
1
(dJ 6.25JK-1mo1:.._1
Jn. Sinceweknowthat,
C .
V(Vlb}
=R (0T ) _e_ _/T_)2
vib•'
2
-evib
(1-e,..evib/T
8.314(3374)2 e-3.374 2 =3.48 JK-1 morl I
1000 ( l-e-3.374) .' ; .
om
CorLectoption is (b)
/.
What isthe frac;ion of~2(g) molecule;inthe v= Oind v = I vibrati~nafstates at 300K.
.c
0vib ofN2 is3374K.
C
(a) l.31xl05 (b) l.3lxl04 (c) l.3lxl03 (d), 1.3 J:'}< 102
B
"
-(0 · IT) '
,Jn. Sincewe:firstcalculate, e vib =e-33141300 =e-11.2s =l.3lx10-s
yA
Therefore,
tr
is

m
The characteristic rotational temperature er for H 2( g) is 87.5 K. What is the value ofthe molecular partition .
function for rotation at 3000K?
he
(a) 17.14 (b) 15.26 (c) 18.21 (d) 19.20

1>ln. Sinceweknowthat, -
.C
T 3000K
qr= cr0r = 2x87.5K=17.14
w

w
The translational partition function ofa H atom is 7.8x 103° at 3000K in a volume of0.25 m'. What is thermai
w
wavelength?
(a) 3.l75xI0-10 m (b) 3.175xl0-11 m (c) 3.175xl0-12 m (d) 3.175x10-13 m
oln. Since we know that,
Thennalwavelength, A= (4tVJ½ = (7.80.25

x
)½ =3.175x10-
3o
11
m
10
What is the vibrational partition function for Br2 at 300 Kifthe vibrational frequency is 151.2 cm-1•
(a) 0.725 (b) 7.25 (c) 1.94 (d) 19.4
1
lpln. qv. l-e-huo/kT ... (i)HAND MADE NOTES www.ChemistryABC.com
QUZZES
34 10
hu 0 _hero'- 6.62x10- x3xI0 MORE NOTES ON
x151.2 www.ChemistryABC.com
kT - kT - 23
1.38xl0- x300 =0 ·725
From equation (i), we get
1
qv = - 0.725 = 1.94
1 -e
Correct option is (c) . ,' .-'
6. (a) What is the electronic partition function ofCl atomjf it is assumed that energies of ist and higher e,
states are very large.
004 ~3 W6 00~
(b) Ifenergy ofthe first excited state ofCI atom is o. II eV, what is tlie
contnbution to the partition funct
this term at 1000 K.
om
(a) 5.56_ (b) 4.56 (c) 7.32 (d) 8.56
2
Solo. (a) The ground state spectros~pic term for Cl is I).; whek su'bscript is the value ofJ.
.c
The degeneracyofany states is (2J+ 1). Thus
C
=2J + 1 = 2 x-3 + 1 =4
B
g0
2
yA
qe = g0 ( elec) = 4
tr
2
(b) The J values for first excited state is ( P ½) is 1/2,
is
m
1
So, g1 = 2J + 1 = 2 x + 1=2
2
he
J
-l.8xl0-20 J
_ + r 2
(-e/kT) _ + \ i.38xto-=- ~1000 =4.56
.C
4e -go gt e - 4 2e
w
7. What is the rotational partition function of11i molecule at 300K. Moment ofinertia o~ molecule is 4 .5~
w
kg m2, symmetry number cr = 2 .

47
(a) 19.20 (b) 18.20 (c) 17.20 (d) 15.20

w
8
1t2Ik:T = 8x ( 3.14 ) 2 x4.59x10-47 xl.38x10-23 x300
Solo. qr= crh2 2x(6.626xI0-34r =17.20

8. What is the molar residual entropy ofa crystal in which the molecules can adopt 6 orientations ofequal e
at OK.
(a) 14.89 JK- 1moi- 1 (b) 13.89 JK- 1mo1-1 (c) 16.89 JK- 1mo1- 1 (d) 12.89 JK- 1mol- 1
Solo. The residual entropy is given by S =NklnW
W == number of orientations, N = number ofparticles= NA= 1 mol
NAk = R = Gas constant. HAND MADE NOTES www.ChemistryABC.com
= 8.314 JK:1 mo1- 1•
r .... - - ..-,-r-1 ,-1
Assuming I\ and HD molecules having equal bond length. What, is the ratio ofthe rotational partition functions
ofthese molecules at temperatur~ above 1000 K? ·.
,,
5 2 8 : 3
(a) - (b) - (c) - (c) -
3 8 3 8
81t 2 IkT I µr 2 µ
)In.. qr=---=> qr «:.-or - or qr oc-
crfz2 cr cr cr
lxl 1
qH 2
-=
1+ 1 1
--X-
qH2 2
1 3 1 3
=> -=-X-=-X-=-
qHD lx 2 2 qHD ~ 2 4 2 8 i
/
1+2 3 . '/ ,
om
-
\ f , I
/.
). What is the translational partitionfun~tion ofa~ ·moleculeconfin&iiii:1J;.f00 ml flask at 1298 K . [Given :
.c
Molecular wt. of}\ =2.016.J
(a) 2.8xl0 26 (b) 2,8~10-26 (c) 28x 10 26 (d) 28x.10- 16
C
[)In.. Since, we know that, \. /f
B
'
(21tmkT)½ . ·
yA
qt= h3 xV =.JM 3T 3 xVxL89xl0 26
tr
= .J2.0I63 x 2983 x lOOxl0-4 x l.89x 1026 = 2.8x 1026

is

m
2 2
l. The first excited state of Cl atom P½ lies at 0.11 eV above the ground state P½. What is the electronic
he
partitionfunctionofClatomat 1000K
(a) 5.26 (b) 4.56 (c) 6.i'8 (d) 7.18
.C
oln.. For the ground state g 0 =2j+l= 2xi+1=4

2
w
For the first excited state. g1 = 2 ( ~) + 1 = 2

w
w
19
_ /kT _ /kT O. llxl.6xl0- ( )
Hence at 1OOOK, Cle= goe Eo + gle e, = 4eo + 2el.38xl0-23xl000 =4 + 2 0.28 = 4.56
2. For H 2gas at 3000 K, what is the vibrational temperature 0vib and the vibrational partition function qvib,
respectively, given that the fundamental vibrational frequency of8i molecule is 4405.3 cm-1.
(a) 6338K, 1.137 (b) 7456K, 1.137
(c) 6338K, 2.234 (d) 7456K,2.234
34 8 2
. = hu = hcu = 6.626xl0- x3xl0 x4.4053x10 = 6338 .3K
oln. qv,b K K · l.38xl0-23
1 1 1 1 =-1-=1.137
qvib = 1- e-0/T = 1- e-6338.3/3000 = 1- e-2.1128
HAND1-0.1210
MADE NOTES 0.879 www.ChemistryABC.com

13. Give the symmetry number for each ofthe following molecules
(a) CO (b) 0 2 (c) CO2 (d) H2S. (e) 0 3 (g) Sill
(h) CHC~ (i) SF6 (k)A1iC16 ; .
Soln. For homonuclear diatomic molecule or a symmetrical linear molecule symmetry number cr = 2 and for h
nuclear diatomic molecule cr = 1 .
(a) CO -+ cr= 1
(b) 02 -+ cr=2
(c) CO 2 .
-+ cr=l
(d) H2S -+ cr=2
(e) 03 -+ cr=2
(f) S0 3 -+
om
cr=6
(g) SiH4 ..,+ cr= 12,, ..
(h) CHC~ f
-+ cr=3
.c
(i) SF 6 -+ cr=24
(k) A1iCl6
C
-+ cr=4
B
14. The energies of first three energy levels offluorine atom are as follows
yA
Energy level Energy (cm-1)
0.0
tr
404.0
is
2
D½ 102406.5
m
Calculate:
he
(i) Electronic partition function and

(ii) The fraction of fluorine atoms in the three energy levels at 1000 K
Soln. g0 =2x½+1=4; g1 =2x½+l=2; g2 =2x½+1=6
.C
q e -- g0e-e,i/kT + g le-e1/kT + g2e-e2/kT = 4 .xe0 + 2e-404/!c/kT + 6e-I02406.51!c/kT

w
= 4 + 2e-05813 +6e- 147 A = 5.118

w
(ii) From Maxwell Boltzmann distnbution.

w
Ng.e-e,1kT
ni=--'--
qe1
n g e-eolkT 4x 1
- 0= 0 =--=0.72
N ge1 5.118
n ge-e1/kT 2xe-0.5813
_I= I =---=0.218
N qe1 5.118

• ;~r I - .' .. ·--·.
. _-'. -' -::··-:.,...:
Nnon-interactingmoleculesaredistnbuted amongthreenon-degen~energj'tevels s 0 ,=:~,,t1 = l.38x 10-21 J

and s 2 -= 2.76x 10-21 J, at 100K Ifthe averagetotalenergyofthe ~ystemat this t6mperatureis l.38x 10-1s J,
the number ofmolecules in the system is: ·[GATE 2007]
(a) 1000 (b) 1503 (c) 2354 (d}2-987
_:i =e-{Ei:-E'o]lkT
1. Since,
no
n ;... -[!.38x!0-2l-l.38xl0-21 ]tkT

. -1- e
no
n
=eo =1
om
-.
i \
no
.c
Or, ... (1)
C
[ 2.76xl0-21 ;'..l.38xl0-21 J
B
For n2 ::::: e 23
-l·.38xto- .x!OGK
n1
yA
tr
.. ::(2)
is
From equation (1) and (2)

m
he
E = noEo + n1E1 + n2E2

.C
no +11i+ni
w
Or,
w
Or, [-: (no+ 11i + n2)E = 1.38 X 1-0-ts J J

w
Or, l.38xl0- 18 J .= ~ [ 1.38xl0-21 + 1.38xl0-21 + 2.76x 10-21 J

Or, l.38xl0- 18 J = n0 x[ 2.76x 10-21 +1.0lxl0-21 J
Or, 1.38 x 10-is J = n0 3. 77 x 10-21
n0 = 0.366 x 103
~=366
From equation ( 1), we get
1
· 366 . ~ .
/Jr· rn = 366+366+ 2.71 = 867 ~ 1000 (approximate)
16. The molecular partition function for a system in which the energy levels are equispaced by£, is
1 1 1
(a) 1+ (c) 1+ (d) l-e,_~6 [GATE2014
' ' '
Soln. The molecular partition function, ifenergy level o~a ~ystem are equispaced by £, is gives as
1 ,• 1
Molecular partition function= _ e-Ps
1
w~ere, /3 ·= kT /
Correct option is (d) . i ,
om
17. The temperature dependence ofpartition functions are as f~/ws:r [GATE2013
,
qtranslational oc T
312
qvibration OC f O
.c
qrotation oc T(linear molecule) qrotation oc T3/2( non-linear molecule)
C
According to the conventional transition state theory (CTST), the t,e~ture dependence factor fora n
B
ofthe type given below is · " .
yA
Linear molecule+ linear molecule ;:=::=:: non-linear transition state ~ products
(a)11 (b)T° (c)T 1 (d)T2
tr
Soln. A+ B ~ A B # ~ Product
is
m
he
.C
k =(RT)l!.ng [.iJJL] =(RT)-t [T T]

312
• oc r-1
w
p p qA qB p T3/2. T '
w
Rate"" AB#= kp [ Al[ B].

w
AB# = Rate~ frequency
A=_!_ AocT- 1
r'
r(S,r':,k,T r,
18. The molecular partition function ofa system is given by [GATE20U
q (T) ~ ( k;; where the symbols have their usual meaniD
The heat capacity at constant volume for this system is

(a) 3R (b) 6R (c) 9R/2 (d) 3R/2
312 312
)-(·k8 T) (81C mk 8 T)
2
·
.. ( QUZZES
Given· q T - 2
ID· · he h
£nq(T) = £n [ ( S1C h":B2)3/2] +fnT

•. 2
3
For I mole ofthe system internal energy is given by ,\
U=RT 2 (B£nq) =RT 2 xi=3RT

ar ·
om
.V
T . \
The molar heat capacity a,t contant vo 13!me {Cv) is given by
=(au)
.c
Cv =3R
ar v
C
B
For an ideal gas with molar mass M, the molar translational entropy at a given temperature is proportional to
yA
(a) M312 (b) M112 (c)eM {d)ln(M) [GATE2015]
iln. Since, we know that
tr
S=klnW
For molar mass M, the molar translational entropy is
is
Sm= klnM Or, Sm oc lnM

m

). The vtbrational partition function for a molecule which can be descnbed as a simple harmonic oscilhtor with
he
ir
fundamental frequency u is given by {GATE 2005]
(a)O'Ph:; l (b)fi-exph:T
.C
w
(c) exp[-k:Tl[i-exp[-k:;)T' (d) ex{ 2!:TJ[i-ex{ k:;Jr

w
w
1bt. For harmonic oscillator, its vibrational energy is
&V (v +_!.)2 hv · v - 0 I 2
0, - , , ,•••••
where, vO, the classical frequency of oscillation

, Since, the vtbrational energy levels are non-degenerate, the expression ofvtbrational partition fim:~tion is given
by

.· ~
hvo - 3hvo 5hvo

QUZZES 2kT MORE NOTES ON www.ChemistryABC.com
- 2kT - 2kT
q=e +e +e + ..... ~ ..
_hvo 1 .
=
2
e k~
-
= exp(-~)[1-exp(..:.~)]- :
2k 8 T k8 T
1-e kT .
1
q=
where, 2sinh( hvo)
2kT
om
21. Rotational partition function for~ molecule
Soln. H2 is a half integer (spin i = ½)., l
Rotational partition function for such molecule
.c
Zrot =(Z'even Zeven +Z'odd Zor1d) ... (i)
C
CalcuJate
Z' =-i-=_1_½_=.!_
B
yA
even 2i+l · 1 4
2x +1
2
tr
_ i+l _ ½+1 -
'odd____ _-3
Z
is
2i+l 2 xl+l 4
2
m
zeven = I (21 +1) e-PBhc.!(J+I)

he
J=0,2,4
zodd= I (21+1)e"'"'PBhcJ(J+I)
.C
J=l,3,5
w
Put the vlaue in equation first
=: L !L
w
zrot (21 +l) e~PBhcJ(J+I) + (21 + 1) e-PBhc.l(J+I).

J=0,2,4 J=l,3,5
w
Zror=:[ L (21+l)e-PBhcJ(J+I)+3 L (21+l)e-PBhcJ(J+I)]

J=0,2,4 J=I,3,5
22. Calculate the rotational partition function for D2 molecule.

D2 is spin one integer (i = 1) rotational partition function for such molecule
2 rot = ( z 1even zeven + z 1odd zodd) ... (i)
Calculate
Z' i+I l+l 2
even = 2i + 1 = 2 + 1 =3
~
. i -1
Z'ood =2i+l =3
Zeven = L (2J +I)e-PBhcJ(J+l) .· 1 ..·'

[Where, /J' =-·-{ ]t'/•
J=0,2,4... : · ·· k8 T
Zodtl = L (21 + I)e-PBhcJ(J+l)

. J=l,3,4... ,
Put the value in equation (i)

Zro, =(Z'even Zeven +Z'oc1d Zoc1d)
.=!
\
L (2J+l)e-PBhc.l{J+t)+½ L (21+1)/PBhcJ(J+i) /
I
om
J=o.2,4,.... J=l,3,s . / ,
= ½[2 L (2J ~l) e-fiBhcJ(J+l} + L (1J + l )e-PBhc.l(J·li)]
.c
J=0,2,4,.... J==l,3,5
C
,,
The rotational partition function of~ is: \ [NET Dec. 2012)
(a) L (2J+l)e-phcBJ{J+t)
B
yA
J=O, 1,2.....
L (21 + 1) e-l}hcB.J{J+l)
tr
(b)
J=l,3,5,......
is
L, (ZJ+l)e-PhcBJ(J+l)
m
(c) .
J=0,2,4 ..... .
he
(d) ¼[,);. . . (2J +I) .;-ll'd'l(J+I) + 3,.,f.r.. . YJ +I) e -j!hcl!J( J+t)]

.C
w
w
w
if.

Statis;.tical Thermodyt
QUZZES ( PRACTICE
MORE NOTES ON SET) www.ChemistryABC.com
1. Which ofthe following is true
(a) qlr > qvib > qel > qrot (b) qtr > qrot > qvib > qel
J
(C) qtr >> qrot >> qvib >> qel (d) qel > qvib > qrot > qtr
2. In terms of the me>lar partition function q, the internal energy ofa molecule is given by
(a) U =nRT (o£nq/ av)T u ;= nRT ( o£nq t av )r

2
(b)
1
(c) U=nRT 2 (o£nq/aT)v (d) U.=nRT(o£nq/ aT)v:
3. A system has 100 degenerate energy levels and 100 bosons are k~pt in it. Find the ehtropyofthe S)
equilibrium · · • / '
om
r •• ,.'/
(a) 10-2 kB (b) 102 kB (c) 460.6 k.B/ (d) 4.606 kB

__.,, ...
4. Calculate the total number ofmicrostatesfor 6 identical pafticlbs with their occupation numbers {1,
.c
three states is:
(a) 6 · (b) 12 (c) 60 (d) 720
C
1 /~ ;
. The energy levels ofthe harmonic osc9Iator (neglecting zero poini energy) are ev .= nhv for n =0, 1, :
B
5.
yA
Assuming hv = kB T , the partition function is:
1
tr
1 1
(a)e (b) - (c) 1-- (d) 1-!
e e e
is
6. A system consisting of4 identical and distinguishable particle, each possessing three ava,ilable states,
m
and 3 units, has 10 units of energy. The number of ways, W, in which these conditions are satisfied is
he
(a) 2 (b) 4 (c) 6 (d)l 0

7. The number of ways in which four molecule~ can be distnbuted in two different energy levels is
.C
(a) 6 (b) 3 (c) 16 (d) 8

w
8. The translational partition function ofa hydrogen molecule confined in a 100 mL flask at 298 K (Mo
hydrogen =2.016) is:
w
(a) 2.8xI020 (b) 2.8xl025 (c) 2.8xI026 (d) 2.8xI027 •

w
9. The translationaL rotational and vibrational partition functions for a molecule are
ftranslation :;: : 1010 m-\' f rotation z. fvibration :;: : 1, ( ks T / h) z. 1013 at room temperature, NA z. 6 X 1023
Using the approximate data given above, the frequency factor (A) for a reaction ofthe type:
atom+ diatomic molecule ~ non-linear transtion state ~ product, according to the conventional tr,
state theory is
(a) 2xl03 (b) 6xI07 (c) 2xl012 (d) 6x1013

A ,1 ' [CSIR DEC. 2014]
I
,).
I ;,~
Identify the speed distnbution functions ofNe, Ar, and Kr ;with the curves nlthe figure above
(a) Ne-A, Ar-B, Kr~C . (b) Ne-B, Ar-C, Kt:--A (!' [CSIRDEC. 2014]
om
(c) Ne-C, Ar-B, Kr-A >.{d).N~-C, Ar-A~Kr~~ J . ..
- . . /
' .F' f ;
(0 = ; )
.c
The low and high temperature limits ofvibrational partition function are [CSIRDEC. 2014]
C
B
yA
( c e-0/2T and T e-0/T
. )
0
tr
The probability offinding the harmonic oscillator in the energy leveln = 1 is (neglect zero point energy and
assume hv = kBT) [CSIR DEC : 2014]
is
(a)e (b)e2 (c)I-e-2 (d)e-2 (e-1)

m
I. If er denotes the characteristic tempreture of rotation then the magnitude of

he
[ er (H er (D 2) 2) J/[er (HD )J (assume the bond lengths to be the same for all the molecules) is
.C
wm ~m ~w ~m
w
i. The rotational partition function ofa diatomic molecule with energy levels corresponding to J = 0, 1, is
(where, E is a constant) [GATE 2015]
w
(a) 1+2e-2 e (b) 1+3e-2e' (c) l+e-3e (d) 1+3e-3e

w
5. A molecule has a ground state and two excited electronic energy levels, all ofwhich are nondegenerate with
the energies: E0 = 0, E1 = 1x10-20 J , and E2 = 3 x 10-20 J . If P0, P1 and P3 are fractions of molecules
occupied in ground, first and second excited states, respectively, at 298K, then, P0 : P1 : P2 =?
(a) 0.919: 0.081 : 0.001 (b) 0.900: 0.098: 0.002
(C) 0.666 : 0.333 : 0.111 (d) 0.880 : 0.088 : 0.022
ANSWER KEY
Questions 1 2 3 4 5
Option (c) (c) (b) (c) (d)
Option (d) (c) (c) (b) (c)
Questions 11 13 ·
12HAND MADE NOTES 14 15
Option (b) (d) (c) (b) (a)
['Chapter 9
Kinetic Theory of Gases
Kinetic Theory of Gases:

Kinetic theory of gases which provided sound theoretical basis for tli~ various gas la~s, mainly Boye 's h
Charle's law, Graham's law andAvogadio's law. Tpese laws were eqwirical generalization based on exp.riner
om
observation. There was no theoretical background to justify then( / . .
. . '! /
It is based on following postulates f
.c
(i) Every gas consists oflarge number oftiny particles called gas molecules.
(ii) Gas molecules can moye rapidly in every possible direction and colloid with one another and with walli
C
container. ,_ ,
(iii) Their collisions are perfectly elastic i.e. there is no loss ofmomedtum J.d energy. ·
B
(iv) The volume occupied by the gas molecule is negligible as compared to the total volume of gas.
yA
(v) There is no attraction or repulsion between gas molecules or gas and waals ofvessel.
(vi) The pressure of the gas is due to bombardment of gas molecules on the waals of vessel.
(vii) Their is no effect of gravity on the distnbution of gas in the container.
tr
(viii) The average kinetic energy of gas molecule is directly proportional to the absolute temperature.
IK.E.oc TI
is
m
Derivation of Kinetic gas equation :

Let us consider that gas is kept in a cubical box oflength 'a' & move with velocity 'v'. The mass of one !
he
particle be 'm'. The resolution ofvelocity 'v' along x, y, z axes.

y
.C
V
V
... (1) ----x

w
v,
z
w
If a gas particle moves in a cubical box along +ve direction ofx-axis with vx velocity between two faces ',
and 'B'.
w
momentum ofthe particle before collision to face 'B' = mv

. X
momentumaftercollision=-mv. X
Therefor~, Changeinmomentumofparticle= -mvx -mvx =-2mvx

y
---a---
-
m v,
X
z
Therefore, Momentum transferred to the face B, Ap =2mv.x .
On colliding with the same race 'B' the molecule travel2a distance with v velodty
2a / X '
6t==-
vx )
Change in momentum per unit time at Brace,
' 2
6p 2mvx mvx
-==--.v = - -
6t 2a ~, a
Change in momentum between A and B races
According to Newton's law, .-/

/
Change in momentum per unit time= Force
om
·:2mv2
F=----x
a
.c
. . .Lip . 2mv/ . 2mv 2 • .
Similarly, along y and z axes ~ are - - - and z respectively.
C
. ut a a , /"f.
\ f '
B
· 2mv2· ,~mv 2 ·2mv 2 ··. ··2m( 2 2. 2)
Totalchangeinmomentum= x + Y + z = - vx +vy +vz
yA
a a a a ·
2
I F.. __ 2mav I (from equation(l ), v2 =v 2x+ v2y:+ v 2)z
tr
is
Area of cube= 6x(a)2

m
· F 2mv2
Pressure, P = - =>
he
A a.6a 2
.C
1 mv2
p = --- (·: a3 = volume of cube v)
3 a3
w
v2 = Root mean.square velocityJ

2
P = 1V
mv [
w
3
w
For I mole, . [ PV = ½mv' [
For 'n' mole, I PV =½mnv' I ... (2)
Equation (2) is known as kinetic gas equation.
Kinetic Energy of one mole of gas :
From Kinetic gas equation.

. 1 2
PV = mn v (m.n =M total mass of gas)
3
1 2
PV=-Mv
QUZZES MORE
(multiply and divide byNOTES ON
2 on RH.S 1)
3
PV =
2
3 2
(! Mv
2
) :::> PV =
2
3
KE :::> (KE = _!_ mv
2
2
) :::>
•
3
KE=-PV
21
From ideal gas equation PV RT 'for 1 mole'
. Id
IKE=~RTI ... (3)
Where, R = universal gas constant.
I KEocT J
Thus kinetic energy for 1 mo le ofgas is directly proportional to' the teWl)erature
/
om
Kinetic energy for 1 molecule:
~etic energy for I mole·;, !

RT.
.c
3( R J
C
KEformolecule= T NA (NA= Avogadro\ umber).
2 1
EB B
yA
k=Boltzmannconstant(l.38xto-23 J /K) ... (4)
tr
JKEocTI
is
Thus kinetic energy for one molecule also depends on temperature.

m
Deduction of various gas laws from kinetic gas equation:

(1) Boyle's law:
he
Pv = 1 m.nv
-2
=> PV = 1 mv
-2
(m.n=Mtotalmassofgas)
3 3
.C
2
PV=
KE
3
w
Fromequation(3), KE oc T
w
PV=),T
3
w
If temperature of gas kept constant

PV = constant
K
PV=K => P=-
V
Or IP oc ~ IThis is Boyle's Jaw.
(2) Charle's law:
1 -2
Pv =-m.nv
1 -2
=> Pv =-mv (m.n=M)
3 3
PV ~(~ Mv 2
) => PV=3_KE
3
2 V 2
KEocT => PV =-T HAND MADE=>NOTES
-=_;_P www.ChemistryABC.com
3 T 3
At constant pressure
V . .
QUZZES
T =k ::} V =kT I I
MORE NOTES ON
::} V oc T i.e. Charle's la"f·
(3) Avogadro's law: . 1

One mo le ofany substance contains the Avogadro constant(Q.022x 1023) ofmolecules. It was Avogadro who
discovered the law ofnature for gases. This law is known as Avogadro's law, which states:
The volumes ofthe same number ofmoles ofall gases measured at constant temperature and
pressure are the same. That ~, at the same temperature.:'and pressure equal volumes of all gases
contain equal number of moles or molecules. , ·
. Let us consider that two gases having masses II1i , ,:rn/and number ofmoles .°i and 11z are at the same
temperature pressure and Volumes. ' -· ·
.. From kinetic gas equation
PV=3m1n1V1
1 2
... (i//
om
. 1 .· ' 2 /
PV =-m2n2 v2 ••• (lli) )
3
.c
On comparing equation(i) an~(ii), we get
C
1 ·2 1 2
Jffi1Il1V1 = JmznzVz \.
m1n1vf =m2n2v~
B ... (m)
yA
Since both gases are at the same temperature hence they will have same KE.
1 2 1 2
tr
i.e. 2m1V1 =2m2v2 ... (iv)

is
On dividing equation (iii) by equation{iv)

m
he
::} n1 11z, this isknownasAvo..gadro's law.

.C
(4) Graham's law of diffusion:

"The rate ofdiffusion of a gas is inversely proportional to the square root ofits density at constant pressure."
From kinetic gas equation.
w
1 2
w
PV = -mnv [mn = M total mass of gas]

3
w
PV=-Mv
1 2
~ v2 = 3p where, d = density of gas
3 d
v=!f
IfP = constant.
1
voc-
Jct
This equation is known as Graham's law.
Maxwell's distribution of molecular velocities:
QUZZESto assumption ofkinetic theory
According MOREofNOTES
gases,ON
molecules of gaswww.ChemistryABC.com
colloids with one another and with t
walls ofvessei thus their velocities keep on changing which results in different speeds for different fraction
molecules at a gain temperature.
If N is the total number ofmolecules in a gas out ofwhich AN molecules-have the same speed, ti
LiN' dN
fraction ofmolecules having same speed is given by N or N .
dN · . .•·
The quantity N gives the fraction ofmolecules with ~peeds between c and c+dc. It can also given by
1
p(c)dc.
Maxwell plotted a graph between different values of the spee~ and the corresp<;mding molecul~
fractions which is known as 'Maxwell's distribution curve'. ' i " , ' -.1
,J
.
,/
om
.. dN
N
.c
Or
Fraction of
C
molecules
B
r exp~~~)cttc
Velocity~
yA
2
~=4n(2;
tr
The important features of curve are given as:

(1) The fraction ofmolecules with too low or too high velocities are very small.
is
(2) The fraction ofmolecules possessing higher and higher speeds keeps on increasing till it reaches a cert:
maximum value and then starts decreasing as shown in the curve.
m
(3) There is a certain velocity for which the fraction ofmolecules is maximum This is called the rrost proba
velocity. Its value at a given temperature depends upon the volume ofthe gas.
he
Effect of temperature on distribution of molecular velocities:

.C
212
In molecular speeds at x-axes the factor e-( mc kT) called Boltzmann factor increases with the rise in ternperat
w
and thus molecular speeds increases. But the fraction ofmolecules having most probable velocity decrea
with the rise in temperature.
w
· Thus thickness ofcurye increases along with the decrease in height and curve shifts towards right
with the rise oftemperature
w
',
i '' \
\
\
\
Fraction of I \
I \
molecules I
''
I
,, I
''
., ., ,, ' ' ,, .
.,"
Molecular Speeds

Maxwell's distribution of molecules kinetic energies: '
It is a curve between the fr~ction ofmolecules having kinetic e.nergies in the $}ge of s and s + ds i.e. ~8
versus kinetic energies s .
,1 .
om
e(KE) '£· /
.· .· /
!',
The maximum in the pr~bability.functi~n corresponds to the most pf~bable kinetic energy.
.c
Most probable kinetic energy
1
C
permole= -RT l.
2 \
B
I
[k =Boltzmann constant]
yA
per molecule= -kT
2
Types of Molecular velocities:

tr
Molecular motions are of three types:

is
(1) Most probable speeds (vm/ · ·. . ··

The most probable speed is defined as the velocity possessed by maximum number ofmolecules ofa gas at
m
a given temperature. It is given by

he
vmp = J2~T . . (l)

.C
M = molecularmass ofgas.
(2) Average Speed (va):
w
It is the arithmatic mean ofdifferent velocities possessed by the molecules ofthe gas at a given temperature.
w
Ifc" c2, c3 : ••••• en are the in1ividual velocities ofthe gas.molecules and 'n' is their total i:mmber, then,
average velocity is given by.
w
v) =Ci+ C2 + C3 .. ,..cn
( .... (2)
n
If cl' c2, c3 ..... are the velocities possessed by groups of n" ~, ~ ..... molecules of the gas respectively,
average velocity is given by. ·
(c)= n1C1 +n2c2 +n3c3 + .... .
n1 +n2 +n3 + ....... ..
on solving Boltzmann - distnbution ofmolecular velocities, average speed comes out to be
(v)=J8RT ... (3)

nM
(3) Root mean Square velocity:
It is the square root ofthe means ofthe squares oftheNOTES
HAND MADE speeds ofthe different molecules in a gas at a given
temperature.
Ifcl' c2, c3 •••••• are the velocities possessed by groups' of~, 1½,1½ .7 .'. molecules respectively..·
BysolvingBoltzmannequationvrmsisgivenby vnns
.,·
" lRT
= M =J3PV
M ... (4)
•.1
Relation between different types of Speeds:

(i) Average speed= 0.9213 x Root mean square spe~d
(ii) Most probable speed= 0.816xRoot mean square speed.
:. Vnns :vav :vmp =l.224:1.128:1= 1: 0.92: 0.82
(iii) > Vav > V mp
om
V nns .·· / \ ;;/ .
Thus, at a given temperature root mean square velocity ofagas is maximum and probable speed is minim
, ~
,. ! .
.c
i
C
fraction
B
of molecule .
yA
tr
velocity__,,.
is
Collision diameter:
The distance between the centers of the two molecules at the point oftheir closest approach is know:
m
collision diameter.
he
.C
w
For two molecules to colloid they must have closer approach than collision diameter.
w
Collision number (z1): •

The number ofcollisions suffered by a single molecule per unit volume ofthe gas. It is given by
w
z1 = .Ji :rcd2 (v)p ... (1)

where, p number ofmolecules per unit volume ofthe gas
(v) =Average velocity
d =collision diameter.
The total number ofmolecules colliding per unit time per unit volume ofthe gas
=.Ji :rcd 2 (v)p2 (2) ••.
Since each collisions involves two molecules, the number ofcollisions oflik:e molecules per unit tine per\
volume ofgas is given by ·
zn =}_.Ji
2
:rcd2 ( v) r:>2
... (3)
QUZZES
Collision frequency:_The MORE
number ofcollisions NOTES ONby ~oleculesper
experienced www.ChemistryABC.com
.o.c.ofa gas per second is known
as collision frequency. ·
·· Totalno.~fmoleculescollidingperccpersec.is z=-1i.'rc(v)N 2d 2 •. · .
. :: .
z
Collision frequency n = [N number ·ofgas molecules]
2
1t(v)d2N~, /
n= Ji ... (4)
·Mean free path: . ,

The mean free path is defined as the average distance travelled by. a molecule petween fwo succe~ivecollisions. \
.' )
1 = . average distanace travelled per sec,. = .· (v) )/ , . => = 1
om
1
number of collisions byamoleculepersee. ./i1t{v(<i Nj
2
. -Ji:n:d2N
Assuming that the collision diameter, d:·is independeirt oftemperawre/tbehmperature and pressure depenoence
.c
of the mean free path is given by ideal gas law as
C
B
yA
... 1 oc. 1 .·. at constant T
'[ p .
A oc T at constant. •p1
tr
is
Degr,ee ofJreedom: .
The degrees of freedom of a molecule are defined as the independent number of parameters required to
m
describe the state of the molecule completely. Each molecule has 3N degree of freedom. Where N is total
number ofatoms in 1 mole of molecular formula.
he
It is of tlrree types:
(i) Translational degree of freedom: .
.C
Each molecule has tlrree translational degree of freedom along tlrree axes x, y and z.
Monoatomic gases, He, Ne, Ar etc. posses only3 translational degree of freedom. They haveuo.
w
rotational or vtbrational degree offreedom.

w
(ii) Rotational degrees of freedom:

•An linear molecule, such as CO2, C28i, HCl has 2 rotational degrees of freedom.
w
• Non-linear molecule, such as 8i0, NH3, CH4,."2S etc.has 3 rotational degrees of freedom.
(iii) Vibrational degree's of freedom:

• for linear molecule (CO2, C2H2) =3N 5
• for non-linear molecule (H20, NH3, CH4) = 3N-6
Note:
(i) Rotational degrees offreedom present only in liquid and gases.
(ii) In solid state molecule possess only vibrational degrees offreedom.. Translational and rotational motion in
solids converted into vtbrational motion.. ·
Principle of Equipartition of Energy:

According to the law ofequipartition ofenergy,
HANDthe total
MADE energy ofa molecule
NOTES is divided equally mnongst the
various degrees offreedom of molecule.
Nnev1c; .1 neory or \.:7a~
. . 3
QUZZES
Since MORE NOTES
average kinetic energy ofamo lecule kT ON = www.ChemistryABC.com
2
Therefore according to equipartition principle, energy divided to each degree offreedom
1 3 1 1
=-x-kT =-kT or -RT per mole.
3 2 2 2
. ·1 · 1
Thus kinetic energy for: each degrees of freedom is kT per molecule or
RT per Illolecule.
2 2
In Vibrational motion two atoms oscillate against,each other, and possess both potential & kinetii
energy. Thus energy ofvibration invo Ives two degree$ offreedom
1
Energy ofvibration per molecule= 2 x kT =:= kT Of RT P~vmole.
2
ornf
\ ,')
om
. ,• . ' : . .
There!im;, . (i), Energy oftra,r,slation fur each degree p>,L ~ k:T = ~ RT

1
.c
· . . 1
(it) Energy ofrotation for each degree offreedom = kT = RT
2 2
C
(iii) Energy ofvibration for each degree offreedomf kLf RT
,,
B
Thus, if1\,1½ and °:3 are the translational, rotational and vibrational degrees offreedom, the total energy
yA
l 1
= n1.-kT +n 2 .-kT +n 3kT
2 2
tr
Root mean square velocity from kinetic gas equation :

is
m
From kinetic gas equation,

he
1 -2
Pv =-mnv ... (1)
3
Where, p = Pressure ofgas, v . volume ofgas, m = mass ofl gas molecule
.C
n number ofmoles of gas, v Root mean square velocity.

1 2
w
PV = Mv ; (m.n = M total mass ofgas/mo1. mass)

3
w
.L. 2 3PV
·3PV=Mv 2 ; => V =--
w
Real gases: Van der Waals equation:

To display the deviation ofa gas from ideal behaviour clearly, the ratio ofthe observed molar volmne, v, tc
.
ideal molar volume
(
Yid
ITT) is plotted against pressure at constant temperature. This ratio is .callee
=p
.compressibility factor, Z,
~c Theory ot c.::,aees ·.. ~
2
QUZZES MORE NOTES ON .· Realgas
l
V _ RT] z
· [ ideal - p
ideal gas
Foran ideal gas= Z = 1, Fora real gas z ¢ 1 ·

/ 0 '---------'---
p
(i) If Z > 1, the gas is less compressible compared to an ideal,gas.
(ii) If Z < 1, the gas is more compressible compared to an ideal;~as.
The gas which do not obey ideal gas laws is known as real gas.'
~ l ,; ; - .
A gas tinder low P and high Tobey ideal gas equation but at low temperature ;iiid high pressure it deviates from ·
ideargas laws. · · .; . .. ,/ .! . .. • · . ·
om
Van der ~~ modified ideal gas equation PV= nRT to malc.e it appticalie ~or real gas at all temperature
and pressure, which JS as · · , · /
(p<: }v-n~)=nRTfo;'~'toola 1
.c
. ,
C
na
2
.
\ /1 .
B
1.
Where - 2 pressure correctionfuctorandmeasure ofintermolecular force ofattraction.
V
yA
nb = volume correction andcalledexcluded :volume.
Unit ofa= atm. lit2 mole-2 and b= lit. mole-1.
where, 'b' is four times the actual volumeofgaseous molecules.
tr
is
4 .
b = 4 x Avogadro constant x rrr 3 where, r = radius of gas molecules.
m
3
he
Boyle's Temperature (Tb):
The temperature of a gas at which it behaves ideally even at high pressure, is called 'Boyle temperature'(Tb).
.C
w
[ SOLVED PROBLEMS ]
w
.
Estimate CV (molar heat capacity at constant volume) for
w
(i) HCl (ii) He (iii) CO2 (iv) NH3

n. (0 HCl - linear diatomic molecul~.
3
Translationaldegreeoffreedom 3 ET =-RT
' 2
1
Rotationaldegreeoffreedom 2(linear), :. ER =2x RT=RT
2
Vibrational degree offreedom 3x2 5 = l; Ev= RT
3 7 .
Totalenergy= RT+RT+RT => U = RT
2 2
, -._ ,---~(!~t = ~R(:a HAND MADE NOTES www.ChemistryABC.com
.·~ f,,.IMen;;.tC I neory or C

(iI) ForHe:
Only 3 translational degree offreedom

1- 3
U =3x-RT =-RT
2 2
. (!i). =~Rr:a
(fu) For CO2 - linear:
'
Translational degree of freedom= 3 .. ET =~RT

:. . 2
Rotational degree of freedom= 2 / :. ER = ~T-i
om
Vibrationaldegreeoffreedom= 3x3-5, Ev·=4R'f =f
. , ,."' 3 13 . t' .
.c
Totalenergy(U) =4RT+ RT+ RT=
2 2 RT
C
C =~R(dT)
v 2 . . dT
B
. V
yA
(iv) NH3 - Bent:
3 .
tr
Translational degree offreedom= 3; ET= RT

2
is
3
Rotational degree offreedom= 3; ER =
2RT
m
Vibrational degree of freedom= 3x4-6 = 6; Ev= 6RT

he
3 3
Total energy (U) =-RT+ RT+6RT=9RT
2 2 . .
.C
( dU)
dT V
=(-.!9RT)
dT
~1 CV =9R I
w
V . .
w
w
Non- linear molecule.

3
Translationaldegreeoffreedom=3; ET =
2RT
3
Rotationaldegreeoffreedom=3; ER =
2RT
Vibrational degree offreedom = 3 x 5-6 = 9 Ev = 9RT
3 3
U =-RT+-RT+9RT
2 2
U=I2RT

( dU)
dT V
=(_!I2RT)
dT V
(, ·-~
· QUZZES
N()te:_At room tempeature the contribution ofMORE NOTES ON
vibrational www.ChemistryABC.com
enerwtowards tota,J'~nergyi,is 20% (or 0.20) fur
each Vibrational degree offreedom . i' \ .. ..
Find Y ( ~: Jfur (i) He and (ii) CH at room temp.
4
· , i·
1• (i) He only Translational degree of freedofu = 3

. R.T. 3 3R
C v=-. (
U=ET=- =:,.
2 2
3 5
=> Cp=Cv+R=-R+R=-R
. 2 2
'
.)
om
.c
C
3
Translationaldegreeoffreedom=3; ET =
2RT
/"'ii
\ y
B
. ! 3
yA
Rotational degree offreedom= 3; ER= RT
2
Vibrational degree of freedom = 3 x 5-6 = 9 ; :. Ev =0.20 x 91(£' ~1.8RT
tr
3 3
is
Totalenergy = RT+ RT+L8RT

2 2
m
U = 4.8RT, Cy = 4.8R
=c;P = 5.8R = l 2
he
Cp =4.8R+R 5.8R, Y Cv .SR 4 ·
Calculate the temperature at which the average velocity ofoxygen equals that:ofhydrogen at 20 K.
.C
n.
w
(v)=~3~T
Jli;
w
~ ( v) oc [M mo!. mass ofgas]

w
(v) oxygen _ T02 .x2

( v) hydrogen - 32x20
, Given that TH 2 = 20K, MH 2 = 2 and M 02 32
I=~ [·:(v)0 , =(v)H,]

4. QUZZES the root mean square velocity
Calculate MOREof
NOTES ON at j27°C and 70
nitrogeQ. cm pressure. Density of Hg = 13
Soln. P1 =70cm P2 =76cm }

3
atSTP
Vi=? V2 =.0224cm
TI = 300 K ; T2 = 273 K
. 3'
Pi V1 = P2.:V2 _ Vi = P2V2T1 = 76 x .0224 m ·.x 300 _ 3
T1 T2 _.., . pT 70x27J _.., Vi =0026725m
1 2
70 2
Pressure in MKS unit (P) = x101325 Nm-
· 76
v =
M
lPV = 28 1o-'3 k 1 ~I' J
3
3x(70 / 76)x 101325 x0.026725 m _1)16le- = . _,
2
51 7 ms
.
om
rms
. x g.mo e , // . i
..
5. The maxwell Boltzmann distnbution for molecular speeds"is sho'1 in,the following fi~:In the figure, H i
height of the Peak, Lis the location at the maxima and W islt:he \vidth at half height. As the temperatu
.c
decreased.
C
B
yA
f(v)
tr
is
v--+
m
(a) H increases, L decreases and W increases. (b) H increases, L decreases and W decreases.
(c) H decreases, L decreases and W increases (d) H decreases, L decreases and W decreases.
he
Soln. Correct answer is (b)

.C
6. For the distribution ofmolecules velocities ofgases, identify the correct order from following.
(a) Vnns > Yav > Vmp (b) Vmp > Vnns > Vav ( C) Vav > Vrms > Vmp ( d) V mp > Vav > V rms
w
Soln. Correct answer is (a)

w
7. Justify the considering co; as an ideal gas, equipartition theorem predicts its total energy as 6.5 KT.
w
Soln. CO2 is linear molecule.

Translation DOF 3; Rotational DOF= 2; Vibrational DOF = 3 x 3- 5 = 4
Totalenergy=~RT+RT+4RT 6.5RT permole

2
For 1 molecule:
ETotal TR) = 6.5kT

= 6.5 ( NA [where k !!:._,NA is a Avogadro number]
NA

~""""' .....-~-.., - ·'"'.·.~
QUZZES
. [ PRACTICE SET
MORE NOTES ON
J, www.ChemistryABC.com
'\if
Ifthe temperature ofa gas is doubled, therms s~d ofmolegules.

(a) increases bya factor of2. (b) dt:}creases bya factor of 2.
(c) increases by a factor of .Ji (d) decreases by a fac~ot of Ji .
The ratio ofthe temperatures ·at which the average velocity ofethane (T1) is equal to the root mean square
.velocity ofbenzene (T2) is:/· ·
(a) 0.151 (b)0.48 (c) 1.209 (d) 0.38~
Given that the mean speed ofH2 is 1.78 kms- 1, the mean speed ofD2 will be 1
(a) 1.26kms-1 (b) 2.52kms-1 (c) 5.04kms-'1 (d)/3.I7kms-1•
' ... ' : • .# .;, ./
The vnns of a gas at 300 Kis 30 RJ( Themolar mass ofthe gas, inkgy(otl, is
om
(a) LO (b) l.Oxlo-1 (c) l.Oxl0'-2 ' /(d)}I.Oxl0-3
. . . l
·The molecular weight ofan idealgas naving a~!}Sity ofl .5 gL-1 at~ ooiic and 600 Torr is
.c
• (a)45.9g/mol (b)
.
4.59.g/mol
. . . . (c}5.-82-g/mol (d) 58.2 g/mol
C
The virial expansion for a real gas can be written in either ofthe following ~rms:,)
..
B
. . . \
PV
-·_. = 1+ ·BpP·+ Cp·p2 + ............
. .
yA
RT
=l+Bv V +Cv V 2 + .. ~ ........ .
tr
If Bv aBp, the value of a would be

is
(a) PV/RT {b}RT/PV ·(c}PV (d)RT

m
For an assembly of molecules (molar mass= M) at temperature T, the standard deviation of Max.weHer 's
speed is approximately
he
··~ ·~ ~ ~
(a) 0.7VM (b) l.4~M (c) 0.7 Vii (d) 1.4 Vii
.C
The temperature of a sample ofa gas is raised from l 27°C to 527°C. The average kinetic energy of the gas.
w
(a) does not change (b) is doubled (c) is halved (d)cannot be calculated.
· The kinetic energy ofN molecules of 0 2 is x Joules at-123°C. Another sample of02 at 27°C has kinetic
w
energyof2x Joules. The latter sample contains.

w
(a) N molecules of 0 2 (b) 2N molecules of 0 2

N .
(c) moleculesof0 2 (d) Noneofthese.
2
0. Toe kinetic energy of any gas molecules at 0°C is.
(a) 5.66xl0-21 J (b) 3408 J (c) 2 cal (d)O
1. The ratio between the vnns velocity ofH2 at 50K& that of02 at 800Kis
(a)4 (b) 1 (c) i (d) 1/4
l2. The Vnns speed ofN2 piolecules ina gas sample can be changed by-
(a)Anincrease in volume ofsample (b) MixingwithArsample ·
(c) An increase in pressure on the gas HAND(MADE
d) an NOTES www.ChemistryABC.com
increase in temperature
Kinetic Theoeyof Ga
13. Among
QUZZES
the following statements about tl}~ a.vet:age velocity; <iP> ofa
MORE NOTES ON
gaseous molecule the one that is
NOT true according to kinetic theory ofgases is :
(a) <iP> increases a:s the square root oftemperah:tte increases
(b) <v0 > decreases as the square root ofmass decreases.
(c) <v0 > increases as pressure decre~es..
(d) <iP> does not change with the volume of container at a given temperature.
14. Therms velocity o:fN2 molecules in a gas at a given temperature is 500 m/se.c. Therms velocity ofHe a1
temperature is (rounded to the nearest m/sec).
(a) 1750m/sec (b)1322m/sec (c)6125m/sec (d)661 m/sec
• : a ' ,: ,} ,) •
15. Given that therms velocityQfN2 molecu\es in a gas is 500 m/sec at;27°C, therms value at 320°C is
(a) lOOOm/sec (b)2000m/sec (c)707m/sec(dtl 1740m/sec
om
' '/
.c
'
ANSWER KEY
Questions l 2 ' 3 4 5
C
Option (c) (b) {b) 't. /J {d) {d)
B
Questions 6 ,7 8 9 10
Option (d) (a) (b) (a) (a)
yA
Questions 11 12 13 14 15
Option (c) (d) (b) (b} (c)
tr
is
m
he
.C
w
w
w

fcflapter 1 ~
_;
Chemical Equilibria
Introduction: . ,:;
In a chemical reaction, chemical equilibrium is tqe state in which both reactant arid products are present in
om
concentration which have no further tendency to change with time. Usu~this state result, when tre forward
-reactant proceeds at the ~e rate as the reverse r~ction. Th~foacti6n ratbs of the forward and backward
reactions are generally ~ot zero but equal. Thus, there are not net ¢hang~ in the concentration of reactant
.c
and products.
C
Le-Chatelier's Principle:
6i
In order to predict qualitativeJy, the effect ofvari~on in pressure ortempeniture a systeu'i inequilibrium, we
B
use Le Chatelier's principle.
yA
The principl~ states,that, ·~ene¥er stress is placed on any system in a state ofequilibrium, then the
system will always react in a direction which will tend to counteract the applied stress".
tr
(a) Effect of pressure or Volume change on equilibrium:

The general rule of thumb;
is
''When the pressure is increased (i.e. volume is decreased without any addition or removal ofgas to form the
m
system) then composition ofa gas phase equilibrium adjusts so as to reduce the number ofmolecules in the gas
phase."
he
e.g. N2 (g)+3H2 (g).._- ·2NH3 (gJ

.C
Condition (i):
Ifthe above reaction is at equilibrium and suddenly pressure is increased, then it will move in f01ward direction,
so that it will reduce the nurnberofgaseous molecules.
w
Condition (ii):
w
If pressure is decreased, then reaction will move in backward direction such that, it will increase the overall
number ofmolecules in gas
w
e.g. H1 (g)+I1 (g)~2HI(g)

Here number ofgaseous molecules on reactant side is same as number ofgaseous molecules on product side,
So this reaction is no way (by either going forward or backward) can reduce or increase number ofmolecules.
So, above reaction will have no effect ofchange in pressure.
Problem-I. If the volume of the container in which the reaction H 2 (g) +/2 (s) ~ 2HI(g), is at
equilibrium, is suddenly increased, so predict the direction ofnet reaction thereafter.
loin. As volume has been increased, this implies that pressure has been decreased, so reaction will move in forward
direction so that it can increase its number gaseous molecules.
Note: Pressure or Volume change affects the equilibrium only when there is difference in number ofmoles of
gaseous molecules on the reactant side andHAND
product side.
MADE The numbeiofmoles
NOTES of solid or liquid reactant and
product are not counted.
(b) Effect of temperature: , .
WhenQUZZES MORE NOTES
the reaction is enQ.5)thennic, i.e., absorbs heat,ON www.ChemistryABC.com
addit~on ofheat will favour it and reaction mould proc
" to a greater ex~~nt athiglier temperature ·
The opposite is observed, if the reaction is exothennic.
e.g. N 2 (g)+3H 2 (g)~2NH 3 (g) Mf0 =,....92kJ :
As the forward reaction is exothennic, so increase oftemperature will led to reaction in backward directiori
dissociation ofmore N1¾ molecules.
0
e.g. N 2 (g)+02 (g)~2NO(g), Mf =+180kJ
As the forward reaction is endotherrrdc so higher temperature will favour the fonnation of NO molecules
. . ·- r ,' . . , ..
(c) Effect of addition of inert gas to the equilibrium react;on mixture: \ ·

Case I: Addition of inert gas at constant pressure l}as the same effect of decrease,in pressure ofgase
molecules at equilibrium · ,/ ' .
·
om
. . . ~
e.g. N 2 +3H2 (g)~2NH 3 (g) / . \/ · J ·

Ifinert gas 'Ar' is added to abovereaction at {)q_uilibrium,then jt has i~e effect as decreasing the,pressur
system only. ·
.c
· Reaction will move in backward direction.
Case TI: Addition ofinerl gas at constant volume has no effect on equilibrium
C
B
Problem-2 ..For the equation reaction:
yA
Na 2 HP04 .2H20(s) ~ N a2 HP04 (s)+2H2 0(g);. +ve
Predict the favorable condition for the decomposition of above dihydrate salt in terms ofpressure chm
temperature change and inert gas addition (ifrequired). ·
tr
Soln. (i) Pressure should be decreased.

is
(ii) Temperature should be increased

(iii) Inert gas should be added.
m
Mathematicalchemistry of reaction which proceed at equilibrium by using free energy concept ofthermc
namics. We use Gibbs Helmholtz equation and concept of feasibility or !lG value to determine that
he
reaction is in a equilibrium or not llG = llH - T AS

.C
1. Calculate llG for the Reaction:

C6 H120 6 + 60 2 --+ 6C0 2 + 6H 2 0
w
The standard free energy of glucose, CO2 and H20 are -910.5, - 394.4 , -237 .2 respectively, show th
w
is feasible or not.
w
AG = AG product - AG reactant
Glucose=-910.5
CO 2 = 6x-394.4
H20= 6x-237.2
AG =[6x-394.4]+[6x-237.2]-910.5 =-2879.1 kJ
Therefore, reaction is feasible
AH =AHproduct - AHreactant
AS = ASproduct - ASreactant

· Lllir = -46.19 J / Mol
QUZZES
The standard entropy MORE NOTES ON www.ChemistryABC.com
For, N2 =191,H 2 =130,NH 3 =192

1 . 3 ] · , '· ..
~S = ~Spro<luct -~Sreactant = 192- [ -x 191 +-x 130 = 192-[95.5 +195] =-98.5 J /K-mole
. 2 2 · ·
T~s = 298 x (-98.5) = -29353 J / mole
~G=~Hr-TAS=-46.19 + (-29353)=-46.19 + 293~3 ·
= 29306.8 l J / mol = 29306 kJ mol- 1
Chemical Equilibrium:
The rate at which a substance reacts is p~oportional to its active mass ~rid the rate of ch~mical reaction is
om
directly proportional to product of active mass ofreacting substance./;
aA + bB ~ mM
' .
+. nN
. ,>''~
· .. :. . f ,'.· / ·
rr oc [At [Btand rb {Mr [Nt . .
.c
oc
C
rr = kr [A]a[ BJb andrb
·
=kb [M]m [N]n
At equilibrium,
B
yA
tr
kr [At[B]b =kb [Mt [Nf

is
~_[Mt[Nr.
m
kb - [At [Bt
he
~=K
k eq
b
.C
w
=>
w
'
The equilibrium candition can be written in terms of activity, partial pressure, molar concentration or mo le
w
fraction of the species involved in the reaction and Keq will have different numerical value and different repre-
sentation. The representation is Ka, ~t'
K , KC and KX
Thermodynamic Treatment of Law of mass action

~G=-RTlnKP
Consider the reaction, aA + bB ~ mM + nN
We know that chemical potential ofa reaction consisting of'a' mole ofA and 'b' mole ofB is given by
dG .
-=µ·
dll I
Greactant =axµ A +bxµ 8

Gproduct =mxµ M +n x11
rN
Change in Gibbs free energy
AG=Gproduct Greacblnt =(mµM +nµN )-(a µA tbµB)
The chemical potentialof r11 species
µi =µt+ RT In Pi ... (i)
Putting the value
µM =[µt/+-RTlrt PM], µN =[µt + RT lrt PN J µA=[µ!+ RTlrt PA]

µB=[µi+RTlrtPB] ,
AG= m[µt + RT.In PM]+ n[µt + RT lrt PN]-~[µ! + RT lrt Pi}·b[µt + R'I; }d PB
- /
1
=0 /:1.
om
At equilibriwn, AG _ I
0 0 - . 0 ; --_ _ -0 -
0 = m µM·+ mRT 1rt PM.+ nµN +nRT lrt PN -aµA ,-a l,lTlil PA -bµB -bRTln:PB
.c
0= mµ~ + RTlrtP~ +nµt +RTlrtP~ -aµ1-RTlnP!-bµ~ -RTmP:
C
-RTlrtP~ -RTlnP~ +RTirtP! +RTmP: =mµ~ +n~ -a~~ -bµ~'
p~ p~] B
yA
-RT[m = AG 0
pa p,b
_ A B
tr
is
AG 0 =-RT[m P~P~]
papb ... (i)
A B
m
he
... (ii)
.C
pm pn
where ' kP = _M__!!
paA pbB = constant
w
-
w
PM' PN' PA and P8 be the pa~ial pressure ofcomponent M, N, Aand R

w
The chemical potential in term ofmolar fraction (xi)
µ . =µ?+RTlnx.
! I I
From this the following expression analogous to equation (ii) is obtained.
and in term of molar concentrations ( C;)
µ. = µ? + RTlrt C.
l I l
From which we obtain that

.·~
Ifthe reactants ahd the product are not ideal gases, then the equilibrium cosntants, l(!h,
(a ma n).
K=K'h MN
(a~ a!)
where ais are the activities. For non-ideal, i.e., real gases
K =K = (J;; J;)
. I (J; fff)
where fis are the fugacities. For a mixture ofr~lgases;
J; . . . r ; .
'i =_; or J; =r; Pt, where ris are the activity coefficients.,
Pt
om
rZ. rZ P;; P;
Hence, KJ -- -=--"-'-X-=--"-'-
a b · na P,b
Y"A YB r.4. B {
.c
C
B
yA
r Van't Hoff reaction isotherm:
tr
It readily follows
From equation (i) and (ii) that
is
001
KP =e-A RT orj h.G=-RTlnKP I
m
Problem-3. Calculate KP for the reaction, co(g) + H20(s) ~CO2+ H2(g)

he
If h.G forthereacti9nis 16.3 kJ/mol

.C
3
,In. 8G = -RT 1n K lo K = -l 6.3xl0 .
P => g P 8.3I4x298x2.303
w
log KP = - 28.5 => KP = anti log ( - 2.85);

w
KP= lAlxl0-3
w
Relation between
·
Kp , K and K : C X
aA +bB ~ mM+nN
m n
K =PM xpN
p, paxpb ... (1)
, A B
We know that , PM= ~P,

PA xAP,
I
Putting value in equation (1)

. -~
=>QUZZESI KP =Kx pAn :I MORE NOTES ON www.ChemistryABC.com
~=(m+n)-(a+b)
At atmospheric pressure j KP = Kx I
Relation between
..
Kp and KC : .
PV =nRT => p = ~ RT => : = CRT => ,.P-i,

..... =-C-iR_T_,
Pm =CMRT; Pn =CNRT; PA =CART; f'8 =C8 RT;
om
=>
.c
Problem-4. Calculate K and K X for the reaction, N 20 4 ~ 2NOi(g) for which K p = 0.157 atm.
C
C .. . .
. . . \ ?
Soln. KP= Kc [RT]rui
B
yA
=> K = KP = 0.157 =6.37x10-3
c [RT]rui 0.0821 x 300
tr
is
Homogeneous Equilibria
m
There are two categories specified in the case of homogeneous equilibria.

(i) Number ofmole ofproduct are equal to the number ofmoles ofreactant
he
(ii) Number ofproduct of mole doesn't equal to the number ofreactant mole.
.C
In Case-I:
l\n g = 0 ; KP =Kc = Kx
w
In this case the addition ofinert gas to the reaction doesn't change the relative amount of the species.
w
A+B~C+D
w
t =0 a b C -d
t=t a-x b-x x x
Total number ofmoles =a+ b
. _ number of mole x p
Partial pressure - total of moles
a-x)
Partial pressure of A = ( - - P
a+b
Partial pressure of ·B = -b-x)

- p
( a+b
. ·-· ... -·· --~
' ---- . ---
QUZZESof C = (-x-) P
Partial press~e MORE NOTES ON www.ChemistryABC.com
a+b
Partial pressure of D = ( x ) P
. .·~ . .
a+b
. .-- .
X)··(X)
= (a+b
- - Px - · - P
K a+b x2
=> KP = - - - - -
P (a:-x)p(b-x)p (a-x)(b:...x) i
a+b a+b ~
In Case-II:
om
Consider a case in which An .ispostive_ if ~n g "" O
, KP,:/; Kc,:/; Kx
.c
A ~ B+C
t=O 0 ff
C
a
t=t {a-x) X X
B
yA
Total number of moles= a +x
a-x)
Partial pressure of A;PA =( - - P
tr
a+x
is
Partial pressure of B, P8 = (-x-) p

m
· - a+x
he
Partialpressure ofC, Pc =(~1p a+x)

.C
)p( )p
w
X X
K = ( a+x a+x x2
w
P (a-x)p =>!(.p=( a2-x 2)p

a+x
w
In form of degree of dissociation:

Ifa=l;x=a
2N0 2 • Find out Kp

-:~:;
~oln. N204 2N02

t =0 l 0
t =t (1-a) 2a
Total number ofmoles =l - a + 2a =MORE
QUZZES
1+ a NOTES ON www.ChemistryABC.com
Problem-6. The dissociation ofPC~ was studied at 29°C at total pressure of I atm. The value ofKP is fou
0.46 calculate the d~gree ofdissociation ofPCt IfkeepingJhe temperature constant the pressure ofsyste1
raised to IO atm. what will be the degree ofdissociation. .
Soln. PC15 ~ PC13 + Cl2
t O I O 0
t=t (1-a) a a
Totalnumber ofmoles = 1 - a + 2a I+ a
/,/
om
\ /
Partial pressure of PC15 = ( ~'":) P ~
.c
Partialpressure·of PC\·= ( :a)p
C
1
B
yA
Partial pressure of C1i = ( : a) P
1
tr
a 2P
K = - -2-
is
p (1-a )
m
Given that, KP= 0.46, P 1, a=?

he
K,=t I2 2
x 2)=>(I-a)'=~=> a =0.46 (1-a. )
2 2
.C
1 a. 0.46
~ a. 2 = 0 .46 - 0.46a. 2 ~ a. 2 [1 + 0.46] = 0.46
w
a.2 = 0.46 ~ a= Jo.46 ~

w
a= 056
1.46 'f 1.46
w
Kp = 0.46, P = 10, a = ?
2
Pxa. 2 = 0.46(1-~1.2)~ a. 2 = 0.46-0.46a. ~ 0.46-0.46 a.2 = I0a.2
10 .
a. = 0.043 ~ a= 0.02
2
We see that ifwe raise the pressure ofthe system lower the extent of dissociation.
Problem-7. The extent ofdissociation ofPC~ at a certain temp is 20% at 1 atm pressure calculate the pres:
at which the substance is halfdissociated at the same temperature.

a, 2 p _ (0.2)2 xi
KP = (1 __ 2\ - /1 tr. '"l\ 2 \
= 0.04
KQUZZES _!_ =0:04 MORE NOTES ON www.ChemistryABC.com
... (i) /
P 0.96 24
a 2P . (o.5)2
K = --=--=---'---P 2
P
2
(1-a ) 1-(0.5)2
K
p
=( 1-0.25
0.025 )P" 2
KP=( 0.5 ) P2 =0.33P2 .. ju)

· 0.75 I
•J
om
Comparing equations (i) and (ii) we get
0.04 = _Q.33 p 2 /
f
Pi= 0·04 = 0.12 atm
.c
0.33
C
Temperature Dependence of Equilibrium constant (Van't Hq~ eqyation} ,
B
' '
LiG = ;_RT In KP
yA
In K = - LiGO
P RT
tr
0
is
~-lnK = _ 1 d LiG
dT P R dT T
m
he
.. ~{i)
.C
From Gibb'~Helmholtz equation,

w
... (ii)
w
From equation (i) and (ii), we get

w
dlnKP] Mio
[ dT RT2 Proved.
Where, MI0 is the standard enthalpy ofthe reaction.

Integrated form of this equation:
JdlnKP = TMI0 ... (i)

. dT RT2
Integrate the Van't Hoffequation and detennine the constant of integration
LiH 0
JdlnKP = JRT 2 dT
Chemical Eq,uilU
QUZZES
J-
MORE NOTES ON
1 '- .T' ..
1]
-2+1,
.Jt.
1
ln K = AH
P R T2
dT
[ T2J
- = dT => . 2 d1' = -·."_'_.
-- . -.
-2+1 ·- T
. j
where, I is the constant of integration.

In standard condition it is consider that enthalpy doesri'tdepend on the temper~ture
-AH0
lnK = - - + I ... (ii)
P RT
om
·: lnK ;, _ AGO ... (m)
· P RT t
.c
From equation (ii) and (iii)
C
AG0 - AH 0 ... /1
B
-=--I .... (iv)
RT RT
yA
AH 0 .
=> AG 0 =--xRT-1 xRT= 88°-IxRT
RT
tr
=> IRT=TAS => I= AS

is
R
From equation (iv) and (v), we obtain that,
m
I AG 0
=AH 0 - TAS
0
I
he
Integrated within limit:

.C
Ifwe integrate the equation (1) between two temperature T1 and T2, when the equilibrium constants
w
Kp 1 and Kp 2 respectively.
w
w
Problem-8. The equilibrium constant ofa reaction double on rising the temperature 25 to 35 calculate Ml
Soln. Given that,
Kp =K, KP =2K, T1 =25°C, T2 =35°C
1 2

0 0
In 2K _ AH .[35-25] _ AH [35-25]
QUZZES 2 MORE NOTES ON www.ChemistryABC.com
K - 8.314 35x25 =>In - 8.314 35x25
0 0
AH [308 298] AH [ 10' ]
log 2 = 8.314 308x298 => log 2 = 2.303x8.314 91784 '
Au .30I0x2.303x8.314x91784 : ·.
Llll = lO , = ~2897.7 I= 52.89 lcJ
Problem-9. For the reaction:
CO +S0 3 ~ CO 2 + S0 2
AG = 187 kJ /mole
om
' \
Lili =184 kJJ mole at 25° ,·· ;
. , . >.. /'
Calculate AG at 398 Kand KP at 398 K assw.ningthat AH retmlinsconstantoverthetemperature inter-
.c
val
C
8(AG/T) 8H
oln. - at25°
of T2 .
B
yA
6 2
.=> ·[aG]
T
:
t.G1
[_:;m]r
= . I. .
2
Tt
tr
T1
Enthalpy and entropy is independent oftemperature but AG dependent on temperature
is
AGT2 AGT1 = -[AH_ AH]- AG398 _ AG 298 = AH[398-298]

m
T2 Tl T2 Tl 398 298 398 x 298

he
= AG398 _ 187 = + 184 [ 100 ·]

398 298 398x298
.C
= 298 x AG 398 -187 x 398 = 184 x l 00

= 298x AG398 =184 x 100 + 187 x 398
w
18400+ 74426= 298x AG391 =92826 = 56026

w
;,, 92826 = 3 ll ,;
w
298
AG -311.5
In KP = - => log KP = - - - - - -
RT 8.314x2.303x398
then, KP 0.91
, Dependence of Equilibrium Constant

Dependence of!(, on the pressure:
Since AG depends upon the temperature· alone. KP also depends on the temperature alone and is 'indepen-
dent ofpressure. ·
Hence, a1nKP
BP = 0 => -a,:r [-RAtG] =HAND
o MADE NOTES www.ChemistryABC.com
-l. Pressure dependence of Kc :
KP .[ 1-A.n
KP =Kc [ RT ] An ::> Kc
·
= A ::> Kc = KP RT .
[RT] n
In Kc = In KP - An In RT
d d .. d d
::> dP In Kc = dP In KP - dP L\n In RT=> dP In Kc,=0
i.e. Kc is also independent ofpressure.
3. Pressure dependence of Kx :
. K .· ~ ~
KP= K, [r]"" ~ K, = (P):..; K, = ic. (Pf ~ ln ~1hl f• :..An 1n r
om
;5>··"
/
differentiating both side above the equation, ,·
.c
~lnK =~lnK -~AnlnP
dP xdP Pdp
C
d d d 1
B
-lnK =0--L\nlnP::>·-lnK = -An-
dP X dP dP X p
yA
-An
~lnK = ... (1)
dP X p
tr
For ideal gas:

is
PL\V = L\nRT::> An= L\V

m
P RT
Put value in equation (1)
he
I _c!_lnK
dP
= -AV
RT
I where, L\V = Vproduct - Vreactant
.C
when L\n = 0, Kx is pressure independent. when An * 0, then two case arises.

w
1. Case-I: An > 0
w
There is increase in the number ofmoles means L\V > 0

w
. dlnKX L\V
i.e. dP = - RT
Kx: decrease with increase in pressure.
2. Case-Il: An < 0
1.e. L\V < O
Kx: increases with increase in pressure.

'
Problem-1. 1 mo1e ofPC~ is dissociated at pressure P and temperature T as.fo~ws:
PC15 ~ PCI3 + Cl 2
ifdegree ofdissociation is 50% then,
(a) P = 3KP (b) KP =3P
PC15 ~ PC1 3 + Cl2

,In.
Initially 1 mole. 0 0
at equilibrium 1- 0.5 0.5 0.5
=0.5
.-r
I
Totalmolesofequilibrium= 0.5 +0.5 +0;5 = 1.5
om
, 0.5
Ppc15 =-xP [if total pressure is P] .
1.5
.c
. 0.5
Ppc1 =-xP
l 1.5
C
. _ 0.5 p
P.PCl2 - -1.X
B
5
yA
Since, weknowthat,
tr
0.5)Px(0.5)p
(
P'pe1 x P'ct K 1.5 1.5
is
K, = P\0 , ' -=> ,= ( ~:;)p

m
he
K = 0.5 p = .!_ p
P 1.5 3
.C
Or, I P=3Kp I
Correct answer is (a)
w
Problem-2. One mole of each of PC15, PC½ and C~ are present at equilibrium in a vessel at 1 atm pressure,
w
calculate Kp, given, PCl.5 ~ PCl3 + Cl2 [GATE]

w
(a) P. 3KP (b) P 2Kp (c)P =,Kp (d) none ofthese
)In.
at equilbrium 1 1 1
Total moles= 1 + 1 + 1 =3
, Since, given, P = I atm
P 'PCl5 =3Ip ; P'PCl =!p.

3
P'
3 , ~12 =3Ip

- I-
= P or I P=3KP I
KPQUZZES
3 MORE NOTES ON www.ChemistryABC.com
Problem-3. For the PCl5 ~ PC13 + Cl2 at equilibrium. One thole ofeach present in a closed vesse:
litre volume:
1 . . . .
It is found that equilibrium constant
3.Ifconcentration 6feach component is doubled, then equilibrium
stant will be
1 I 1 '
(a) 9 (b)- (c)- (d) 9] ·
6 ?
Soln. (c) . . .1
om
Because equilib~ constant d~s not depenc! upon col1Cen~£1l- ~£, the equilibrium constant ~ ½(wh
given) .
.c
Problem-4. The equilibrium constant K.p for the reaction,
C
.
H2 (g)+ S(g)~H 2S(g) is 20.2 atm-1 at 945° and9.l atm-1 at I06'J°C. Cal~ulate MI0 •
Soln. According to the integrated van't hoffequation,
B
yA
0
J
ln[KP, 2 == £iH (T2 -T1 J
Kp,t R T1T2
tr
Rln( ::·: ]( T~,~~ J

is
Or, MI' -
m
9 21 (l218K)(1338K)]
)1n( 20.2
he
0
=> MI ( 8.314 JIC 1mor 1 · )
[ l338K-12I8K
.C
=> MI0 = -88126.3 Jmor1

=> MI 0 = -88.126 kJ mor 1
w
w
w

QUZZES
(PRACTICE.SET
MORE NOTES ON
] www.ChemistryABC.com
An example ofa reversible reaction is

(a) Pb (N03 ) 2(aq) + 2Nal (aq) = PbI 2(s) + 2NaN03 ( aq)
(b) AgN03 ( aq) + HCl (aq) AgCl(s) + HN03 { aq)
(c) 2Na(s) + 2H20 (l) = 2NaOH(aq) + H2(g)
(d) KN03(aq) + NaCl( aq) = KCl( aq) + NaN03 ( aq)
For the reaction: CO(g) + H20(g) ~ CO 2 (g) + H2 (g),

at a given temperature, the equilibrium amount of CO2(g) can be increased by
(a) adding a suitable catayst (b) adding a~i(able catalyst".
(b) decreasing the volume ofthe container · (d) increasing the ru,nount,pfC9(g)
om
At constant temperature, the eqtiih"brium constant { KP) for the:faec~4siton reaction, N204 2N02' is
-
2
·'
•
' ,. . '
4
.c
expressed by KP = ( :.:2 ), where p = pressure, x = extent ofdecomposition. Which one of the following
1
C
statements is true? .· .. l, 11
B
.(a) KP increases withmcreases-0fp 0-.\,Kp ·filcreases:with,increases ofX
"\'.~'
yA
(c) KP increases with decreases ofx (d) KP remains constant with change n p and x
Consider the following equilibrium in a closed containerN20 4 (g) ~ 2N02(g).

tr
At a fixed temperature, the volume ofthe reaction container is halved. For this change, which oftre following
is
statements hold true regardingtheequilibriumconstant{KP) aoo-degree ofdissociation (a.)?

(a) Neither KP nor a. changes (b) Both ~Panda. change
m
(c) KP changes but a. does not change (d) KP does not change but a changes
he
Ag+ + NH3 ~ [ Ag(NH3)]\ k1 =3.5 x 10-3; [ Ag(NH3)T + NH3 ~ [Ag(NH3)z]\ K2 1.7 x 10-3
.C
then the formation constant of[ Ag( NH3 )z]+ is

w
(a) 6.08 x 10-{i (b) 6.08 x106 (c) 6.08 x 10-9 (d) None of these
w
A mixture of2 moles each ofhelium a9d an unknown gas (normal boiling point =0°C) is kept in a 22. 4 L flask.
Ifthe flask is cooled to 0.1 °C, the resultant pressure (in atm) inside the flask is
w
(a) LO (b) 2.0 (c) 3.0 (d)4.0
For the presumed equilibrium at 25°C, A ( g) + B ( g) C (g) + D(g) Llli = -50 kJ . The equilibrium
constant ofthis process is:
(a) Shifted to the right by increasing the temperature.
(b) Shifted to the left by increasing pressure
, ' (c) Unaffected by change in pressure
(d) Shifted to the right by increasing pressure
For the pair ofreactions given below
(i) N 2 (g)+3H 2 (g)~2NH 3(g) (il) ~ N 2(g)+%H 2 (g)~NH 3 (g)

Ifat a particular temperature, ~ 1 and ~ 2 are the equilibrium constants for reactions (i) and (ii) respectively
then,
(a) KP. = 2Kl)_ (b) Ku = K\ (c) 2Kp =Kl)_ (d) K: = Kp?
._,r,c:,rrn~,·~'1.Ulllti
9. Forther~action:
QUZZES Br2 (g)+ BF2 (g):-~2BrF 3 (g),theequihbriumco~tant
MORE NOTES ON at2000 Kand 1.0baris 5.
When the pressure is increased by 8-fold, the equihbriumiconstant
(a) Increases bya factor ofl.86 (b) Decreases bya factor of 1.86 .
(c) Remains same (d) Increases by a factor of 8. ·
:
10. In a chemical reaction, A ( s) + B ( g) ~ C ( g)

the total pressure at equilibrium is 6 atm. The value ofequilibrium constant is:
(a) 1/2 (b) 9 (c) 1 ,. (d) 36
11. Consider the reaction :
A+B~C
The unit ofthe thermodynamic equilibrium constant fqr the reaction~.•
(a)molL-1 (b)Lmot1 (c)moFL-2 / . (d)dimensionless
om
12. At a given temperature coI15ider \ // 1
/
Fe 20 3 (s)+3CO(g)~2Fe{s)+3C0 2 (g);k 1 =0.05 1
.c
2C0 2 {g)~2CO(g)+0 2 (g); K 2 = 2x10- 12
C
The equilibrium constant for the reaction /'{
f
B
yA
(a) 1X 10-lJ (b) 2x10-38 . (c) 4x10-15 (d) 2x10-24
tr
13. The relationship between the equilibrium constant ~ for the reaction:
is
CO(g)+.!_02 (g)~C02 (g)

2
m
and the equilibrium constant~ for the reaction:

2CO(g )+02 (g )~2C02 (g) is:
he
.C
14. The hydrolysis constant(~) ofNH4Cl is 5.6xl0-10 • Theconcentrationof~o+ inaO.l M solµtionofN1

at equilibrium is:
w
(c) 5.6x10- 10 . _(d) 2.8 X 10-S_

w
(a) ~5.6x10- 11 (b) ~5.6x10- 10

The equilibrium-constant Kc for the following reaction at 842°C is 7.90xio-3 . What is KP at same tempi
w
15.
ture?
1
2F2(g)~F(g)
(a) 8.64x 10-5 (b) 8.26x 10-4 (c) 7.90x 10-2 (d) 7.56xl0~2
ANSWER KEY
Questions 1 2 3 4 5
Option (d) (d) (d) (d) (d)
Option (d) (c) (b) (c) (c)
Questions 11 12 MADE NOTES 13
HAND www.ChemistryABC.com
14 15
Option (d) (b) (d) .(a) (d)
[ 9'hapter 11]
Phase Rule
'
,)
Introduction .-/
I
, . - _.:ii ·'
om
Pha~~ ~ a ho~ogenous, physically distinct and mech~<:_ally.seP,arapfe Pjlrt of a system and is in dynamic
equilibrrum wtt!t the oth~ parts (phases) through thetrans1tion,:()1'sub~C€JS· .
. , .:>Ji'"' . ,/ ., / '
.c
Explanation:
(1) Mixture ofgases·is always·homogenous anrl'constitute'Single·phase i.e;·P ·=I
C
Aw + Bcg) P =1 1 111
\
B
f
But if van derwaal force of Aw >>Beg)

yA
i. l
liquify liquify
tr
easily hardly
is
P=2
m
(2) Colloidal solutions are heterogeneous

The~fore, P =number ofsolute+ solvent
he
(3) True or aqueous solution is always homogenous;therefore,·P=·1.

(4) Two polar solvents or non-polar sol~nt is always homogeneous, P 1
.C
H20(t) + C2H50H(t) ~ Homogeneous

Polar Polar
w
CC1 4( e) + C2H5 - 0 - C2H 5( i) ~ Homogenous

w
. non-polar non-polru,
w
But, H20(i) + CC1 4(i) ~ not missible, P=2

polar non-polar (Heterogeneous)
(5) A gas in contact ofa liquid generally constitute separate phase
S0 2(g) + H 20(t) ~ P = 2
But as the dipole moment ofgas increases solubility ofgas in 8iO increases, P = 1
(i) NH3(g) + H 20(i) ~ P = 1
(ii) HCl(g) + H20(t) ~ P =1

(6) In saturated solution excess solid is in contact with aqueous solution, P = 2
e.g. Saturated solution ofsugar. HAND MADE NOTES www.ChemistryABC.com
Aqueous Sugar + Excess sugar~ P = 2
(l) (s)
. ;
Phase:F
Below saturation only aqueous sugar P 1

(7) · · In a mixture of solid ; P = number of solids. ,.; .
tg. F¢(s) +'i:<'eO(s) + Fe20 3(s) P =3
(8) In an allotropic mixture
P = number ofallotropes.
(9) An alloy is always homogeneous, P = 1
e.g. Cues> + Z°<•l ~ P = 1
Brass (Homogenous)
(l Q) In a chemicalreactio11 )
;
P =different chemical substance i
;'
CaC03csl CaO<s> + co21g) .· P 3,(two solid+ gas)
\·/·
om
;
' t .
.·.··
.( . '
NH4 Cl<•l, ,.,...NH3(g) + HCl(g) P1.=2(iolid

.• + gas)
.c
Homogenous(P=l)
e.g.
H2(g) + ½(g) 2HI(g) P=l
C
B
Homogenous , ,
(l l) Each hydrate in a solution constitute a phase.
yA
e.g. CuS04 .5H 20(s) +CuS0 4 .3H 2 0(s) +CuS0 4 .H 20(s) +CuS0 4(s) +Ice(s) +vapour(s) ~I
tr
Component (C):
is
The minimum number ofconstituents (i.e. molecular species) sufficient for determining the compositi:m <
the phases of the system. ·
m
Every substance that can be separated from a system and exist outside ofit, is called constituent
he
Number ofcomponent(c) = Constituent - equations relating them ·

(S) (R)
.C
e.g. (i) NH4Cl<s) ~ NH3(g) + HCl(g>

R
w
Constitutent (s) 3 (NH4Cl, NH3, HCl) & Equation(R) 1

w
IC=S-Rf
w
C = 3-l ·= 2
(ii) In Ice r= water r= vapour
R R
S 3 andR=2
Therefore, C = 3-2 = I i.e. one component.
(iii) NH4Cl(s) is heated in a closed vessel.
NH4C1<.i NH3cgJ + HC1<g)
S=3
(a) NH3 and HClare present in equimolar ratio [N'I\] [HCl]
R 2 :. C = 3-2 I
(b) lfin the vessel NH3 or HCl present
[NH 3 ]:t:[HCl] :. R=l :::>C 3-1=2 www.ChemistryABC.com
HAND MADE NOTES
(iv) Mixture ofgases which do not chemically combine (R = 0), component C = number ofgases (Constiti
but Phase ~ I ·
. . . '_. . . ~- ·?,;- •
• T~us number ofcomponent is either

QUZZES
equal to constituent{ifR =l)~,91:~
MORE NOTES ON
it ( If R :;t O)
• Ifions are also present, then in electroneutrality condi~·(positive··~n~~gative ion)
C S-(R+l) -, ... :\;,_:
Degree of Freedom or variance (F) : .- ·{ -1 _
The minimum number ofindependent vari~bles, such as temperature· pressure and concentration which are
utilised to characterize the system completely.
'I F=C-P+2 I"

F =0 it is nonvariant or invariant
(e.g. Triple point, Eutectic point and congruent point)
'
,)
F = 1, monovariant
F =2 Bivariant etc.
om
P will be maximum when F = 0
O:d:C-P+2 => Pmax =C+2'f
.c
J
(a) For one componen_t system C=l
C
Pmax =1+2=3
B
(b) For two - component system C =·2
yA
Pmax=2+2=4
F will be maximum when P = minimum (=1)
tr
Fmax C-1 + 2 [P = 0 means system does not exist]

is
IFmax =C+l \
m
. Thus, C is increases F also increases

But P is increases then F also increases
he
=> If Equilibrium depends on electrical potentialotherthanT.and P.
F = C- P + 2 + y I wher, y number of such variables_

.C
Problem- I. Calculate the number ofphases, component and the number of degrees offreedom in following
w
system.
w
When PNH 3 =PHc1

w
When PNH3 '* PHc1
(iii) I_2 ( sh==I2 (H2 0) + I2 (Benzene) (iv) CaC03 (s)~CaO(s)+C02 (g)

(v) Water at its freezing point. (vi) water at its boiling po int.
(vii) Aqueous solution ofJ\SO4•
(viii) Aqueous solution cont;i.ining J\SO4 and acetic acid.
(ix) Aqueous solution of glucose. (x) Sugar solution contained in an open beaker.
(xi) Na 2S0 4 .lOH 20~Na 2 S04 (s)+IOHp (xii) Abeaker containing liquid water and sand.
· (xiii) Liquid water and water vapour in equilibrium at 1 atm
PhaseR
(xv) Sulphur<JJ ~ SulphuregJ (xvi) Fe(s)+H2 0(g)~FeO(s)+H2 (s)

(xvit) Solid camphor in equilibrium with its vapour (xviii) Binary azeoptropic liquidmixture.
(xix) Asubtance at critical point. (xx) Asubstance above critical point.
Solo. (i) P = 2 (solid'NHFI + gas mixture of~+ HCl)/
C = 3-2 1 :. F = 1-2+2 = 1
(i:t) P=2,C=3-1=2 :. F=2-2+2=2
(fu) P = 3 (I2(s) + H20,1) + Benzene,1)) /
... . _:}: ,'
om
C = 3-0 = 3 (Actually there is no equilibrium (R = O}'befve~n I2 (s ), I2 (H2o) and I2 {Benzene)
I
F = 3 - 3 4- 2 =2 {
.c
(iv) CaC0 3 ( s)~GaO( s) + CO2 (g)
C
P = 3, C = 3 - 1 = 2
F =2-3 +2 = 1
B
yA
(v) At freezing point ice water ~ vapour
P=3, C=3-2=1
tr
F = I - 3 + 2 = 0 (non variant)
is
(vi) At boiling point water~ vapour

m
P = 2, C = 2- I =I F 1
he
(vit) P 1 (Homogenous soln. of H2S04 + H20)

.C
C = 2, F 3
(vfu) Aqueous H2SO 4 + CH 3COOH

w
(not miscible)
w
C = 3, P 2, F = 3 .
w
(ix) P =1, C =2, F =3

(x) Container does not constitute phase
P =1, C =2, F = 3
(xi) P = 3 (Na 2S0 4 .10H 20(s ),Na 2S0 4 (s) and H20(£)).
C=3-1=2
F=l
(:xit) Sand does not constitute phase as there is no exchange of any chemical species between sand an
other phase.
P 2 (water and its vapour)
f'=l · P=1
rr•c::a~·~·-
(xiii) At constant 'P' reduced phase rule F = C - P + 1·used·

P 2 C=l; F=l-2+1=0
(xiv) P = 2 {NH4Cl solid and gaseous phase)
C = 1 (when PNH3 = P8 c1) F = 1
C =2, F=2
(xv) P 2, C = 2-1 = 1 :.F=l
(xvi) P = 3 (FeO(s)' Fe(s) and gas phase)
C=4-1=3 :. F=2
(xvit) P =2, C =I
om
1
.:. F = 1 / .. .· \ / / .
(xvfu) In· ~otropic ~ture compo~ition ofliquid and VaJJ9ttr phase ar)' same, thus composition remains
1
constant and reduced phase fule.is used. / f
.c
P=2 , C=2
F=C-P+l =2-.2+1=1
C
(xix) At critical point water r=== liquid
B
or
yA
. (gas) {R=l)
P=2, C=2 1 =1
tr
· F =C- P + I ··[ At critical point liquid and vapour phase are identical]
F =0 (Invariant)
is
(xx) Above critical point orily gas present and behaves ideally
F C - P+2 = 2
m
P = 1, C =1
Problem-2. Detennine number ofcomponent for following.
he
(i)- NaCl(s)' KCl(s) Na +(aq) K\aq) Cl-(aq)' H 20(,n• H 20(g)

.C
(n) CaCl2 .6H 2 0(s)' Ca(~)' Cl(aq) H2 0{l) H20(g)

w
Soln. (i) (a) In electroneutrality condition

NaCl{s} + KCl(s) + H20(l} ~ K(aq)+Na(aq)+ 2Cl(aq} +H 20(g} .·
w
(b) NaCl Na+ + er

w
(c) KCl K+ + er S = 7, R = 4
(d) H20c1J H 20(g)
Component (c) = 7 -4 = 3
(ii) CaCl2 .6H 2 0(s)'ca++ +2Cl-, H20c1J ~ H20(g)
S = 5 and R= 1
C=S-1=4
Phase Transition:
In a system ofvarious phases, at equilibrium transition ofsubstance between various phases takes place and
Clapeyron equation is applicable ..
dP L\H HAND MADE NOTES www.ChemistryABC.com
dT-T(V2 -Vl)
Phase RI
.LlH = Heat of phase transformation.

v1 =molar volume ofpure substance in phase 1,
v2 = molar volume ofpure substance in phase 2
First order Phase transition :

Clayperon equation is applicable and only one variable temperature or pressure or concentration is used
define a system .
Melting, vaporisation, sublimation etc.
Melting: Llttr = +ve (Heat absorbed for melting)

Solid~ liquid
(V.) (V,)
(a) If Vs > Ve Ve - Vs = -ve
om
dP .LlH
=----,--
at Tr (Ve -\~J f
.c
Thus RHS = negative
C
dP
-=-ve
B
dT
Thus on increases pressure m.p. ofdecreases ice. e.g. ice, Bismuth, Gallium
yA
(b) If v, >vs' vi - vs= +ve
tr
dP
-=+ve
is
dT
Thus on increases pressure m. p. increases etc. e.g. CO2 and majority of substance.
m
Vaporisation:
Liquid~ vapour
he
(V,) (Vg) Llttv = +ve (Enthalpy of vaporisation)

Since Vg >> V/
.C
dP
-=+ve
w
dT
Thus, on increasing pressure boiling point increases
w
C
218 Critical Pressure A
Phase Diagram:
w
e
[A] One component system:
~<1)
Phase diagram ofH20:
4.58 mm
I
~
It consist of three parts. ofHg vapour .g
·i::
(l)Areas ,u
(2) Curves
(3) Points o.001s c
0
374°c
Areas: T° ( C) ~
AOB, AOC and BOC only gas, liquid and solid exists respectively.
P =1, C=l
F = C - P + 2 = 1 - 1 +2 = 2 (Bivari:~.nt)
Thus two variable are needed to define a point in these areas.
Curves:
QUZZES fu,ion MORE NOTES ON www.ChemistryABC.com
OC=solid 'liquid
rn (ice) (v;at,:r)
~rization
(ii) OA=liquid 'vapour
( water) (gas)
sublimation
(fu) OB=solid
· (ice)
'vapour
(gas)
Along these curves P = 2, C=2 1 1

F =1- 2 + 2 1, ( Mono variant)
Thus either temperature or pressure needed to locate the point in these curves.
I '
Point:
om
\
At point 'O' three phases co-:existmequilibrium
• _.!,. _.!,.
ice _-'-- water ,_- vapour,,,..
P = 3, C = 3 2 = 1,- :. F 1-3 + 2 = 0 (non variant)
.c
The point 'O' is called 'triple PQint~ and is non-variant.
C
At triple point P =4.58mmofHg.
T =.0075°C
B
yA
Along ON super ~ooledliquij: wiliemt-in equilibrium withi.ts vapour. fA liquid below its free~ point in liquid
from called super cooled liquid] ·
tr
In presence ofice or'Otherimpurity supercooled liquid changes into so lid ice and curve ON merges into OB.
Along metastable curve F 1 (univariant)
is
Phase diagram of CO 2 :
m
The phase diagram ofCO2 is similar to that of}\O except towards the behaviour of fusion curve.
he
i
.C
p
w
5.11
atm -
w
w
A
-56.6°C T --,..
=---- (MI +ve)
dP
dT = +ve Thus on increasing pressure boiling point increases
ButinH20 V8 > V1,

dP
-=-ve if P increases boiling
HANDpoint
MADEdecreases
NOTES www.ChemistryABC.com
dT
QUZZES
Phase diagram of sulphur: MORE NOTES ON www.ChemistryABC.com \
Su1phur exists in four phases:

(i) Rhombic (SR) (ii) Monoclinic (SM)
(iii) Liquid sulphur(SJ (iv) Sulphurvapour(Sv)
But all the phases does not exist in equilibrium. P 4, C = 1 then F (meaningless)
Thus only three phases exist in equilibrium
F=O
om
Liquid sulphur
I F=2
.c
Rhombic
C
I
F 2
B
H
yA
p
(atm)
Sulphur vapour
tr
I
A I
I F=2
F=O:
is
I
. I
I
m
I
I
I
I
he
I
I
I
I
.C
95.6°C 113°c 119.°C

T(c)
w
The diagram consists of

w
(i) Curves: Six stable curves AB, BC, CD, CE, EG and BE. Four metastable curves, BH, BF, CF and l
(ii) Points: B, C, E and F
w
(iii) Areas: Four Areas, namely ABEG, ABCD, GECD and BCE.
Description:
(i) (a) Stable curves.
Sublimation curve of SR.
Sublimation curve ofSM.
CD, S L ~ Sv Vaporisation curve of SL.
CE, SM~ SL FusioncurveofSM.
EG,sR ~ SL Fusion curve of SR.
BE,SR ~ SM Transition curve of SR.
(b) Metastable Curve: HAND MADE NOTES www.ChemistryABC.com

·BH metastable sublimation curve of SM.( SM ~ Sv)
RC;= ,t a le su · .io en e of S ( ~ .===! ~
CF metastable curve ofsupercooled SL. ( SL ....-= Sv) ,
EFQUZZES SR ....-= S1J
= metastable fusion curve of SR. (MORE NOTES ON www.ChemistryABC.com
Along all these curves P =2, C 1
Therefore, F=C-P +2 =1-2+2 =1 (Monovariant)
.:
(il) Points: . .
Point B, C and E are 3 triple point and F is metastable triple point.
AtpointB, SRFSM F Sv
At point C, SM SL F Sv
At point E, SR F SMF SL
At metastable point, SR SL F Sv
Thus along triple point P = 3, C = 1
F = 0, (Triple points are invariant)
om
(fu.) . ·Along areas only SR,SM' Sv and SL present. .
P = 1, 'c =1, F =t''' (Bivariant)
.c
95 6
Note: SR · ·c 'SM
C
1
Thus 95.6°C is transition temp. ofSR. \ 1
B
(i) At room temp. (25°C)below 95.6°C SR will exist. [P = 1, C = 1, F =2]
yA
(ii) Above 95.6°C SM will exist. [P = 1, C = 1, F = 2]
At95.6°C P=2, C=l, :. F=l
tr
Two Component system:

For two component system the phase rule becomes, F = C-P+2 => F = 4-P
is
The minimum number ofthe phases,P in the system is 1.

If P 1, then F will be maximum
m
Therefore, for two component system.

he
Fmax =2-1+2=3
The maximum number ofthe degree of freedom, Fis 3.
.C
Thus 3 variables temperature; pressure and compositions are used to locate a point.
Thus phase diagram would be 3D or space model This can't be represented on paper. Thus one variable is
w
kept constant.
Generally P = constant and phase diagram is 2D temperature - concentration diagram.
w
Reduced phase Rule:

w
Since, P constant.
F=[C-P+2]-1 (Removing one variable)
I F=C-P+l I
There are 2 cases for 2-component system.
(i) Two components are completely miscible.
(ii) Two components are completely immiscible.
!
Two comp'onent completely miscible
3 types
l
Eutectic Congruent b1congruent
fonnation m.p. m.p.
·····---· ............
Eutectic Systems (Eutectic= Easy melting):

QUZZES
Aliquidmixtureoftwo components, which MOREhasNOTES ONfre~zingpoint (thu.seasilymelt)
lowest www.ChemistryABC.com
compared to allotb
liquid mixtures and on cooling such a mixture, both the components sepatate out as solid phases.
The temperature at eutectic point is called eutectic temperature (TE).
General J>hase Diagram of Eutectic system
B
[Solution of:A~ B] . 961°C
A
., .
. .::-
Solid A+
r solution
r
om
f.p f.p
.c
D D'
,c
F=2 SolidA+ : SolidB
C
I
I (eutectic)
B
100%A L 100%8
yA
Composition(%)
A= freezing point ofpure A

tr
B = freezing point ofpure B

C = Eutectic point
is
L = Eutectic composition,
m
In Eutectic mixture component do not enter into chemical combination and separate crystals can be seen und
he
tn1croscope.
Silver - Lead System:

.C
AC= freezing point curve ofPb. Addition of solution along AC lowers the freezing point ofPb.
BC = freezing point curve ofAg
w
C == Eutectic point
w
B (961°C)
w
F=2
Liquid+ Solid Silver
F=2
(303°C)D1----1"'-c-----------1 D (303°C)
Solidlead:
Eut<tdt I
Solid silver+ Eutectic
I
1
F=2
100% 97.6 % of Pb 100%
Pb HAND MADE NOTES
COMPOSITION www.ChemistryABC.com
Ag
Areas:
{0 TheQUZZES MORE NOTES ON
region above ACB contains,miscible solution. www.ChemistryABC.com
;,, ..
P=l, C=2
F= C - P + 1 = 2 1 + 1 = 2(Bivariant) ,
(iI) Region ADC and BD'C containipw:erl>b andAg in equilibrium with solution
P = 1, C =2, F =2{Bivarianf)
Curves:
AC, solid Pb ~ solution
BC, solid Ag r== solution

P =2, C =2, :. F = C-P + 1
om
\
F= 2 - 2 + 1=1 {univariant)
Eutectie·.Point (C):
.c
Three phases coexist (solidAg, .solid Pb andsolution)
C
P=3, C=2
B
F =C- P+l= 2-3 ++=O(nonvariant~,
yA
Thus eutectic point has fixed temperature and con1pqsition. Above.eutectic (300°C) solutioµ:exist and below
Er pure solids along eutectics exist.
Along with DD'O'Opure substance exist. Itisbivariant.
tr
is
Importance of Pb -Ag system;

It fonns the basis ofpattinson's process for desilverisation oflead.
m
On cooling the mixture, Pb seperates out as pure solid due to low :freezing point (321'C) thus ·remaining
solution be richer in Ag. On further cooling upto eutectic temperature the solution becomes 2.4% ofAg.
he
Bismuth-cadmium system:
.C
A
w
317°C F=2
Solution ____,8
.
w
F=2 F=2 268°C

w
Cd+ solution
1 solid Bi + solid Ag
F=2 : + Euectic
I
100% Bi 100%
40%cd.
Bi Cd

·.·;"-.- , . " -
KI-H20 System:
B do not reach on Y-coordinate becauseMORE
QUZZES the two compon,ent
NOTES ON KI and IIzOare not completely miscible.
A F =2 Soln.
(OOC)
'Terg.p
i
(-22°C) 1---__,:,a,-.;-_---1
I/
F = 2 Solid1KI + Ice
I
100% · SO~ofKI 100%

~O KI·
System with congruent melting point: , ,;

The compound which completely melts at a constant temperature. aqd'the liquid state has compostion, same
om
t~t ofthe solid~ ca~ed congruent melting compound The tempe~.~ this point is called congruent meltn
pomt ordystacttc pomt. · /
t
FeC~-H20-System:
.c
[Soln. Phase]
C
K
B
C
yA G
tr
([
~
r
is
§"'
Temperature
27.4°C
l Temperature
m
Fe,Cl~5H,O Fe,Cf..4H,O
Fe,Cl.-7H,0
he
-55°C
Fe,Cl..12H,O
ice+
.C
Fe,C~.12fii F=2
w
w
Composition of Fe2Cl6
(1) Eutectic points: B, D, F, H, J. (minima)
w
P =3, C =2, F =C- P + 1

= 2- 3 + 1 = 0 ( non variant)
(2) Congruent melting point: Maxima C, E, G and I
P=3, C=2 :. F=O(non variant)
(3) Curves:
Ais freezing point ofwater(D°C).AdditionofFe2Cl6 lowerfreezingpoint to-55°C.
BCD solubility curve ofFe2Cl6• 12:EizO.
DEF solubility curve ofFe2Cl6•7H20
FGH solubiltiycuive ofFe2Cl6.5H20
HIJ solubility curve ofFe2Cl6.4H20.
Along curves P = 2, C = 2, F = 1 (monovariant)
(4) Areas: Area above and below ABCDEFGHIJ
P QUZZES
=1, C 2 = MORE NOTES ON
F =2(bivariant) www.ChemistryABC.com
Vapour Pressure-Composition Diagram of a Binary ,'Liquid Solution

Application of Phase mle:
(a) System consisting of only one phase ,,
(b) System consisting of two phases: Ifthe system has twg phases in equilibrium, namely, liq1:1id and vapour,
then the variance oftlie system is two. This means that the yalues ofonly two variables wiUhave to be
stated in order to describe the system completely. Any tw,b pf the three variables, temperature, pre$ure
and composition can be stated. The third one autoinaticallyAvill have a definite valµe. ·
The straight line of Figure-( 1) and and the curved line Figure- (2) ~epresenJ the two.phases, liquid of
and vapour, in equilibrium with each other. ' · ·· ·
om
Temperature oonstant
., ., .,
., .,
., .,
.c
.,
, .,
'
.,
1 .
l Temperature constant
C
,;
,;
,,.,>"'PA
B
,;
,; p
., .,.
,;
yA
.,
tr
is
m
0 x,- 1 0 0
-x. 0
he
Figure - (1) Figure - (2)

The two states, namely~ liquid and vapour, can be shown in one diagram This is shows in figure-(3)
.C
w
Temperature constant
w
. Liquid
w
Pure Amount fraction, xA - Pure

13 Figure-(3)
A
The upper curve is called the liquidus curve since this represents the line above Which only liquid phase e
A QUZZES MOREthe
point anywhere on this line represents NOTES
two.ONphases, liquid andwww.ChemistryABC.com
va.pour, in equilibrium with each c
Similarly, the lower curve is called the vaporous curve s.ince it represents the curve below which onlyva
exsits. Along this line,also the two phases liquid and vapour, are in equilibrilllil With each other.
A point anywhere in the region between the two curves represents the system in which both liqui1
vapour coexist in equilibrium with eaph other. It is for this re~on, this. region is known as the li:iuid-va
region. For a given composition, the points on lhe liquidus and vaporous curves represents, respectivel
maximum andthe-minnnum pressures within ~hich the tw6 phases can exist in equilibrium with each otre1
above facts can be summarized as follows. · ·
. (i) Ifpsystem > pmax; only liquid ppaSe e:ilits.
(ii) ffpsystem < Pmm; onlyvapourpnaseexists. , ; . . ·'
(iii) For pmax > psystem > pmin; the liquid ~d v,apotlr phas~tare in equilibriµm with each other
. •/
I
om
Tie.Line: As indicated above, t~e r~gion ~~een t?e li~~1ds al)tf~apbr~us cu_rves ofFigure-(3) represc
region where bot,h the phases exist m equilibnum with friCh otl{e~: !)or this region, P 2 and thus F= 2. :
the temperature is fixed, one'other variable (anyone ofp,i'XA and yJ is sufficient to describe the S)
.c
completely. For instance, ifpis chosen, then the amount fractions ofthe component A in the liquil and V<
phases are represented by the intersection points ofa horiz.ontal line, known as tie line, drawn from the J
C
'p' with those ofliquidus and vaporous curves, respectively. IfxA ~chos~Q. as the describing variable, thf
B
intersection ofa vertical line from xA wifh the liquidus curve yields the value ofp and a horizontal line fi
cutting the vaprous curve gives the vapour composition yA.
yA
The Lever Rule: A point within the liquid and vapour curves represents two phases, liquid and vapo
tr
equilibrium with each other. For the point a showninFigure-(4). The amount fraction ofthe consti1uentA
liquid phase is given by the point b (i.e. xA) and the that in the vapour phase is given the point c (i.e. yJ
is
Temperature constant
m
Liquid
he
l
.C
w
w
w
I
I
Vapour 1
I
I
I
I
Pure XA XA Pure
B Amount fraction- A
Figure.; (4)
Thus, ifthe point 'a' is nearer to 'b', the system will consist ofa larger amount ofliquid and a relatively sr
amount ofvapour, but ifit is nearer to 'c', then the amount ofliquid present is relatively small as comparec
the amount ofvapour present. As a moves from 'b' to 'c', more and more of the vapour is formed, th1
HAND MADE NOTES
distance 'ab'represents the amount of vapour formed. Similarly, as www.ChemistryABC.com
a moves from 'c'to 'b', more anim
the liquid is formed and thus the distance 'ac' represents the amount ofliquid formed. Taking thesetogethc
1.,. ___ _
distance ab oc amount ofthe vapour formed
distance ac oc amount ofthe liquid fonned.
Taking the ratio, we have
ab _ amount of the vapour formed
.;. ( 1)
ac amount of the liquid formed
'
Equation (i) is knows as the lever rule.
1
Quantitative derivation of the lever rule: The amount :fractioriXA corresponding to the point a represents
the amount :fraction ofthe component Ain the entire system consisting ofliquid and vapour phases, i.e.
:·
nA(I) +nA(v)
XA =---~-~"---- ... (f)
nA(l) +nA(v) +nB(l) +nB(v)
FromFigure-(4), we have
om
- _ ' nA(l}P' /
( ab ) -XA-xA-X11.---..:...;_- ... (3)
nA(I) +nB(I)
.c
C
... (4)
B
yA
Multiplyingequation{3) bynA(I) + ~(I) andequation(4) bynA(v)+°i3(v)' we have
( nA(I) +¾1)J(ab) =(nA(t) +na(1))x,A :::QA\!). . .. (5)

tr
is
( nA(v) +nB(v) )(ac)-=nA(v) -(nA(v}+nB{vJ)XA ... (6)

Subtracting equation (6) from equation (5) becomes
m
( nA(I) + nB(I} )(ab)-( nA(v) + nB(v) ( ac)) XA ( nA(I) +na(1) + n A( v) + nB(v) )-( n A(l) + n A(v))
he
Which on: making use 9f equation (2) becomes,

.C
( nA(I) +nB(I) )(ab)-{ nA(v) + nB(v) (ac )) = 0

w
ab)= nA(v) +nB(v)

Or,
w
(
ac n A(I) + nB(l)
w
ab) = amount'in ihe vapour phase

i.e. ( ac amount in the liquid phase
Problem-I. At 353 K, the vapour pressures ofpure ethylene bromide and propylene bromide are 22.93 and
16.93 kN m-2, respectively, and these compounds form nearly ideal solution. 3 mol ofethylene bromide and 2
mol ofpropylene bromide are equilibrated at 353 Kand a total pressure of20.4 kN m-2•
(a) What is the composition ofthe liquid phase?
(b) What amount ofeach compound is present in the vapour phase?
:01n. The given that
2 2
P~ = P:thylene = 22.93 kN m- and p~ = P;ropylene = 16.93 kN m-
bromide bromide
2
nA = 3 mol, n8 = 2 mol, p = 20.4 kN m-
Phase
Therefore,
·'
Substituting the given data, we have
2 2
x = 20.4kNm- -16.93kNm- =0.
578
A 22.93 kN m-2 -16.93 kN m-2 ·
XB =1-0.578 = 0.422
.
=PA= XAPA = 0.578x22.93'kNm =0. 65
~ ~
(b)Now, YA . 204.:kN -2 . . / •
om
p p ... m.·.... ' / ,i . .
Let nA and~ be the amounts of vaponzed e(hylene bronude and propylene bronude, respectively, whe
20.4 kN m-2• Hence, we have:(· · f 1 ,
.c
nA ;:: 0.6497
C
nA +nB
"}
B
xA = 3mol-nA . = 3mol-nA = 3mol- =0.5 7
yA
(3 mol-nA)+(2 mol-n 8 ) 5 mol-( nA +n 8 ) 5 mol-nA/0.65
o~ts)
tr
Or, (3 mol-nA) =0.57(5 mol-

is
n = 3 mol-0.58x5 mol =0. 99 mol

Therefore,
m
A 1-0.58/0.65
he
Since, IlA =0.65

nA +na
we, therefore, have
.C
I ) (0.995mol)(0.355)
= 0. 65 nA = nA ( O.65 -1 =
w
Ila (O ) = 0.54 mol

.65
w
Problem-2. For the preceding problem, use the lever rule to obtain the amount of each compound
w
vapour phase.
Soln. Now, we have
Amount in the liquid phase _ YA XA
Amount in the vapour phase XA x A
where XA is the amount offraction ofethylene bromide in the overall system Substituting the values of
and XA, we have
Amount in the liquid phase _ 0.65-(3/5) 0.05 = .
2 25
Amount in the (Iiquid+vapour) phase - (3/5)-0.57 0.02
Adding one on both sides and taking inverse, we get
HAND MADE NOTES 1
Amount in the vapour phase www.ChemistryABC.com
----------"-----=---
Amount in the (liquid+ vapour) phase 1+ 2.26
. . .· l .
AmountQUZZES
inthevapourphase= (l + 2_26 ) (MORE
5 mol)NOTES ON /
5 mol) ( (5 mol) ::
. Therefore, nA =YA ( .
3 26
= 0.65) .
3 26
= 0.99~ mol
5 mol)
n 8 = Ya ( -_.- = ( 0.3.50-" ) (5
-mol)
.-· = 0.0.54 mol
,, 3.26 3.26
( PROBLEMS)' ,i
! ,f,
l. The phase diagram of a pure substance is sketched below; The nu,mber ofdegree qf :freedoD,1at points P1,
P2 andP3 are . . . · ,
om
.c
C
T~
(a) 21,0 (b)J,2,0 ·. (c)2, O, 1 Cd) o:I, 1
B
Soln. AtP 1
yA
P= l C= 1
F=C-P+2 =1-1+2=2 ·
tr
Two phase in equation

is
P==2, C = 1, F = 1
AtP3
m
3 phase exist
P = 3, C = 1, F = 0
he
2. For the following system in equation, CaC03<,; ~ CaO<•> + C02(g)

.C
(a)2,2,2 (b)l,3,0 (c)3,3,2 (d)3,2,l

w
Soln. CaC0 3 ...-CaO+C0 2

(s) (s) (g)
w
Number ofphases= 3
w
Number ofcomponent= 2{because compositionofall the phases can be expressed in term ofCaO and CC\)
F=C P+2=2-3+2
F=l
So, correct answer is (c)
3. The mathematical statement ofGibbs phase rule is
(a)F=C P+2 (b)C=F P+2 (c)C=F-P+l (d) P=C + F+2
Soln. Correct answer is (a)
4. The number ofphase in the system made ofwater and benzene are
(a)l (b)2 (c)3 (d)O
Soln.. F=C-P+2
P = F - C + 2 = 2-2+2
P=2
4. The maximum degrees offreedom in aone component system would be ,
002QUZZES @3. MORE NOTES
~4ON 005
Soln. F = C + 2 - P = 1 + 2 - 1 .
F=2
5. The number ofdegrees offreedom in h9mogeneous liquid region ofa two component system with a eutec
point, at one atmosphere pressure is
ooo @l ~2 W3
Soln. For two component, F =4 - P = 4 - 2 = 2
Correct answer is (c)
s '
om
/
.c
C
B
yA
tr
is
m
he
.C
w
w
w

·.~
( PRACTICE SET ]
At the triplet point ofwater C, P, Fare respectively

{a)l,3,0 (b)l,3,1 (c)l,2,0 (d) 3 3 2 ,.
Ji ' ' '
Which ofthe following equations predicts the lowering the melting point ofice with~ increase ofpressure?
dP (Vi -Vs) dP (Vi+ VJ' dP T(Vi + Vs) dP T(Vi-Vs)
(a) dT = AH ~) dT = dH (c).dT = AH (d) dT = dH
f . . .f . ' f . f
Clausiu-Clapeyron equation is derived from

(a) First low ofthermodynamics (b) Secondfo,wofthermodynamics
(c) Distnbution law (d) None ofthese ·
0
There are many physical or chemical reactions which are accompanied bythe appearance 'or disappearan;e
ofphase. Theseareknownas .· ./ .· \ ./ .'
om
(a) Displacement reactions (b) Decomposition r~tiotjs
(c) Phase reactions .r· ·· (d)'.fhermalre~ctioj /
.c
The lowest temperatures attainable with a freezing mixture is
(a) critical solution temperatur~ (b) Eutectic temperature
C
(c) Triplet point .· (d) Critical temperature ,.
. l 11
B
What is the degree of freedom ofa system haviµg rated in equilibrium wat~r vapour?
003 002 ~l ooo
yA
The minimum number ofphases existing in a system would be
ooo 001 ~2 003
tr
Fortheequilibrium, Fe(s)+ H20(g)~Fe0(s)+H 2 (g)

is
What are the number ofcomponents, phases and degrees offreedom respectively?
m
(a)3,2,3 (b)2,3,3 (c)3,3,2 (d)2,2,3

1. Which ofthe following statements is/are not correct?
he
1. The phase diagram ofa substance shows the regions ofpressure and temperature at which the various
phases are kmetically stable.
.C
2. The freezing point when the pressure is 1 bar is called standard freezing point.
3. The normal and standard freezing points are negligibly different for most ofthe cases.
w
Select the correct answer using the code given below

(a) 1 only (b) 2 and 3 (c) 1 and 3 (d) 2 only.
.
w
.0. What will be the number ofphases, components and degrees of freedom respectively ofthe azeotropic
w
mixture ofwater and ethano 1at 10 atm at equilibrium?

(a)2,2and2 (b)l,2and3 (c)2,2and0 (d)2,3andl
.1. Solid carbon reacts with oxygen in presence of a catalyst to form the Qa'SeOUS oxides CO and CO2• What is the
number ofdegrees of freedom (variance) for the system once equilibrium has been attained?
(a) 0 (b) 1 (c) 2 (d) 3.
l2. Consider the following statements
1. For a one component system, the maximum number ofphases that can exist in equilibrium is three.
2. A system can have negative degrees offreedom.
3. The number ofphases in a system does not depend on the amounts of the various substances present at
. equilibrium.
Which ofthe statements given below aboveHAND
is/are correct?
MADE NOTES www.ChemistryABC.com
(a) 1 and 3 (b) 1 only · (c) 3 only (d) 2 and 3.
13. What are the number ofcomponents and the number ofdegrees of:freedom in
Fe{s) +H 2 0{g) ~ FeO{s) + H 2{s)
respectively?
(a)3,2 (b)4,3 (c}3, 1 (d) 4, 2
14. For a pure susbtance, the triple point ina phase diagram has
(a) One degree offreedom (b)Two degrees offreedom
(C) Three degrees of freedom (d) Zero degree offreedom'.
15. Sulfur can exist in four phases. The possible number oftriple point is
001 ~2 ~3. ~4
om
\ y/ /
ANSWE.RKEY l
I /!:
Questio_ns 1 2 '
3 / 4 5
;<>~
Option (a) (d) (b) t (c) (b) ;
.c
Questions 6 7 8 9. 10
Option (al· (b) (c) (a) (c}
C
Questions 11 12 13 ' J4 15
. (a)
B
\
Option (d) (a) (d) ' (d)
yA
tr
is
m
he
.C
w
w
w

[ Chapter 12]
·'
Solution l§. CoUjgative Pfopertiea
Introduction: ·)
.i .
om
Solution: The Homogeneous mixture oftwo or mo~ than two comp9fieno/ are known as solution. Componen1
present in lar~est quantity is. known as so1vent and~omponept othe(tbat;&olventis known as solute. Solution
. are of various kinds; gaseous solution, liquid solu,tion, solid solution (alloys, amalgam etc.) ,
.c
All the properties which depend upon the number ofparticles of solute irrespective of their nature relative
to the total number of particles present iri the·solution:These properties are known as Colligative Properties.
C
These are
(0 Relative lowering ofvapour pressure
(h) Depression offteezing point
B
yA
(fu) Elevation in boiling point
(iv) Osmotic pressure
tr
(i) Relative lowering of vapour pressure:

is
When a solute is added to the solution its vapour pressure get decreases due to decrease in interaction between
m
solvent molecule this decrease in vapour pressure is known as lowering of vapour pressure and its relative
decrease with initial vapour pressure is known as relative lowering of vapour pressure.
he
pO _p
Relative lowering ofvapour pressure = po where , Po = initial pressure
.C
P = Pressure after adding solute

w
Relative lowering of vapour pressure is equal to more traction of sohrte.

w
According to Raouh's law ''Relaavelowering of vapour pressure is directly proportional to mole froction o1
solute".
w
po -P n2
or, --=
Il1 + n2
where, n1 =numberofmolesofsolvent
11i = n~berofmoles ofsolute
weight HAND MADE NOTES www.ChemistryABC.com
n=------
molecular weight
~0- - mz .
p w·; w
- + -2
m m
For every dilute solJtions ~ 1 >> n2
wz,
pO p n mz Wzm1 .
--=-1.
pO nl
-=-- 11 atm ~ ~60 torr I
·'
I
(i) Forideal solution, Mlmixing =0; SVmixing =0 i
om
eg. Benzene and Toluene, Ethyl Bromide a1;1d Eth;l iodide~ h-~~:/mie ~d n-Heptane
(ii) Relative lowerin~ ofvapour p;ssure can bti measured byOJtw~ld and Walker's dynamics ~ethod.
.c
Problem-1. The vapour pr~sure oftwo liquids P and Qare 80 and 60 torr respe<?tively. TheJoJal vapour o
C
the solution obtained by mixing 3 moles ofP and 2 moles ofQ would b,e
B
';,
3 2
n :::::: 2 XA = - Xa =
Soln. n A = 3'-r3 5' 5
yA
'
PA = 80 torr and PB = 60 torr

O O
3 2
tr
Total vapour Pressure xAP/ +xaPa 0 =sx80+ x60 =48+24 =72 torr
5
is
Problem-2. Benzene and toluene form nearly ideal solution. At 20°C the vapour pressure ofbenzene is 75 tor
and that oftoluene is 22 torr. The partial vapour pressure ofbenzene at 20°C in a solution containing 78 gm o
m
benzene and 46 gm of toluene is

(a)50torr (b)25torr (c)37.5torr (d)53.5torr
he
Soln. Given: P8 0 = 75 torr p 0

22 torr
T
Wa =78gm WT =46gm
.C
ma =78 mT = 92
w
78 46
na = =l llf =-=0.5
..
w
78 92
. na o 1 1
The partial vapour pressure of benzene = xaPa = na + nT Pa = 1+ 0_5 x 75 =- x 75
w
50 torr.
1.5
So, correct option is (a)
Problem-3. At 80°C, the vapour pressure of pure liquid A and Bare 520 and 1000 mm respectively. If
mixture o fAand B boils at 80°C and 1 atm pressure the amount ofA in the mixture is
(a)34%mole (b)48%mole (c)50%mole (d)52%mole
,Soln. Vapour Pressure ofA= 520 mm P/
Vapour Pressure ofB = 1000 mm= P8 °.
Vapour Pressure ofmixture= 760 mm.
760 = xA 520+xal000 ... (1)
Weknowthat xA +xa =l ... (2)
From equation (i) and (ii) we get HAND MADE NOTES www.ChemistryABC.com
760 (l-x 8 )520+ 1000 Xa => 760 =520-520x 8 + lOOOxa

240 = 480
.
QUZZES
.
X B => XB = 2l. => XA = 21 I.e.
. 50% 1
mo e
0
So, correct option is (c).

(ii) Depression of Freezing Point: . .,
On adding solute to the pure solvent the freezing point ofsobition decreases, this decrease in freezing point
is known as depression offreezing point. '·
Depression offreezing-point oc molal concentration (m)
ATr oc m or I ATr = kr,m /
If n is the number ofmoles of solute and W, the-weight in
. .
grains ofthe solvent, .
. m = ; x 1000 (moles of solute per 1OOOgmoftbe splvent) ·
om
n
ATr = Kr x-x 1000
' W .~, , I
~ is the molal depression constant or cryoscopic constant ofthe sbtvent,which is defined as the deir~ssion
.c
in freezing point produced whenl mole ofthe solute is dissolved in 1000.g ofthe solvent. ·
Kr may also be calculated fr,om the following equation,
C
'
RT 2 \
B
K=--
r lOOOCr
yA
where Tis the freezing point ofthe solvent and /r is the latent heat offusion per gram ofsolvent. .
• Boiling point ofa liquid increases with increase in external pressure. Therefore pressure cooker reduces
tr
the cooking time because boiling point ofwater inside a cooker is higher than the normal boiling point.
• Elevation in boiling point can be determined by ··
is
(a) Landberger's method. (b) Cottrell's method.

m
he
R(Tit R(Ti) M1
• K =---=----
b f v 1000 AHvap.1000
.C
(iii) Elevation in boiling point:

w
On adding solute to the pure solvent boiling point of solution gets increases, this increase in ho iling point is
known as elevation in boiling point.
w
Elevation in boiling piont oc molal concentration

w
ATb ocm
¾ is known as the molal elevation constant ofthe solvent or boiling point constant and
K
RT
=---
2
b 1000 Rv
Problem-4. Kr for water is l .86°C m- 1• What is the molality ofa solution which freezes at-0.192°C?
Assuming no change in the solute by raising the temperature, at what temperature will solution boil?
~ for H20:::: 0.515°C)
AT 0.192
Soln. Molality= _ r = - - = 0.103m
Kr 1.86
Again, ATb =Kb.m=0.515x0.103=0.0532°C HAND MADE NOTES www.ChemistryABC.com
Assuming the boiling point ofpure water to be 100°C the boiling point ofthe solution will be 100.0532°C.
. '
Problem-5. Asolutioncontaining2.44 gmofa solute dissolved in 75 gms of~O, boiled at 100.413 °1

Calculate themolari:nassofthe solute. Kb for H20 = 0.52,l(Kg/mole.
Soln. W1 = 75 gms, W2 = 2.44 gm, Kb = 0.52 K Kg/ mole
L\Tb = 100.413-100 = 0.413K
0.4l = 1000x0.52x2.44
3
75xM 2
52 2 4
M2 = 0x .4 = 40.96 => M 2 = 41
· · 75x0.413 ·
'i;"
Problem-6. 5% aquous solution by mass of anon-volatile solute boilsJt 1OO. l 5°C. Calculate the molar tn2
. \/
om
ofthe solute. Kb = 0.52 K kg I mole · ' /
(a) 821.7 gm/mole . (b) 182.7Jmlmole (c)287.3 ghvmoi~ / (d)92.3 gm/mole
f .
.c
Soln. L\Tb =0.15 K Kb = 0.52 K Kg/ mole
5% aquous solution by mass means if the solution is 100 gm then there will be 5 gm of solute and95 gm oJ
C
solvent. \
/¥
f
B
W2 =5gm, W1 =95gm
yA
0.1 5 = 1000x0.52x 5
M2 x95
tr
520x5 27.4 ·
M2 = . =- = 182.66; M 2 =182.7 gm/mole
is
95x0.15 0.15 ·
So, correct. option is (b).
m
Problem-7. An aquous solution ofnon volatile solute boils at I00.l 7°C. At what temperature would it free2
he
SoJn. Kb =0.52 K Kg I mole and Kr = 1.86 K Kg/mole

.C
L\Tb =Kbxm
7
O.I7=0.52m => m= o.1
w
0.52
w
7
L\Tr =Krxm =l.86x 0.1 ·=.608 =0.61
0.52
w
IL\Tr = -0.61 !negativ~ due to depression.

(iv) Osmotic Pressure:
The external pressure which is applied to the solution side to prevent the process of osmosis is known
osmatic process, it can be represented by 1t. Osmotic pressure is directly proportional to concentration
solute.
1t oc C and 1t oc T
rroc cT
. ·n
1t=cRT => 1t=-RT => 1tV =nRT
V
• The best -semipermeable memebrane is that of Gelatinoiis precipitates :,of
) \
cupric ferrocyanide
. / ' ' ,

Cu 2 [ Fe( CN) 6 ]. This is prepared artificially bythe rea9tion between q6.S0
4~d KiFe(CN)6].
• Depression in freezing point can be determined by

(a) Beckmann's method (b) Rast's method.
• Ditillation ofsea water is based on reverse Osmosis.
• When placed in hypotonic solutions, cells, swell and burst(futemolysis). On the other hand, when placed in
hypertonic solutions, cells contract in size (plasmolysis). When excess of fertilizers like urea are applied,
1
plasmolysis takes place. Hence, the plants dry up. ·
• Addition ofH~ to an aqueous solution ofKI results in an incr~ase in w,pour pr~ssure ofsolution because
number ofparticles present in solutiqn decreases. /
2KI + HgI2 ~ ~HgI4 ~ 2K+ + Hgi/.
om
. ·. \ // / .
• Osmotic pressure is generally pr~ferred for the, determination ofmqlecular masses ofproteins.
• · RBC when come in contact with.O. l % NaCtiotutiongrow in'ize'as water from salive water may enter
.c
them and they can ultimately burst.
C
Probl~m-8. Asol~tion containing 8.6 gm per dm3 ofurea was found to pe isa,,Jnic with a 5% solution ofan
orgarucnon-volatile solute.Calculatethemola,rmass oftheJater.
B
t · ·
. . . I .
= 5 gm 100 mL.
yA
Som. 5% sol.
=5 gm per 0.1 lit
tr
is
m
8.6 5
60
he
m2 =· S X ::::C 348.33
60xl mz xO.l => 8.6x0.l
.C
Problem-9. At 37°C the osmotic pressure of blood is 7.65 atm.How much·glucose should be used per litre
for a intravenous injection.
w
Som. M. wt ofglucose= 180 and T = 37 + 273 = 310 K

re= 7.65 atm, R = 0.082 i lit atpi/K-mole
w
w
re= w RT=> ?. 65 ~x 0.0821x310 w = 7.65x180xl 54 .lO gm

m V 180 1 => 0.0821x310
Vant Hoff factor (i):
normal molar mass
i=-------
observed molar mass
Dissociation Association
i-1
i-1 ex.=--
cx.=--
n-1 ! 1
n
BaC12 ~Ba++ + 2Cl- n=3
.
-
Problem-to. Acetic acid associates in benzene.to form double molecules.1.65 gm ofAcOH when dissoh
QUZZES
in 100 MOREby
gm ofbenzene raised the boiling point NOTES ON Qµculate Van't
0.36°C. www.ChemistryABC.com
Hpff factor and degree ofassociati
ofAcOH in benzene.
(Given, Kb = 0.52K - Kg/mole)
Kb = 0.52 K - Kg/ mole, W2 = 1.65 gm,.ATb = 0.36k -.
,"'
W1 = 100 gm M2 =?, i = ?, a=?

ATb = lOOOKb W2 = 1000x2.57xl.65
-w,M 2 M 2 xlOO
036
= I000x2.57xl.65 · M = I0x2.57x1.65 ~ 117 79 :
M 2 xl00 =;, 2 0.36 · '/
.· ' -J
om
i = normal molar mass =;, i = _ 60 , = 0. 51
1
'I
observed molar mass· 117.79
.c
a = : ...: 1 =;, a= 0.51 -1 = -0.49 = 0. 98
--1 . _!__1 -0.5
C
n 2 '\ .
B
. . '
Problem-11. A0.5% aq. solutionofKCl was found to :freezeat-0.24°C. Calculate van'tHofffactor ~

yA
degree ofdissociation of solute at this concentration.
Kr = 1.86 K - Kg/mole
tr
Soln. Kr = 1.86 K - Kg/mole ATr = 0-(-0.24) = 0.24 K

is
W2 =5 gm, W1 =100 gm, M2 =?, i=?, a=?

m
ATr = 1000KrW2 =::} 0. 24 = I000xl.86x5

M2 W1 M2 xlOO
he
M = 1000xl.86x5 = 38 .94
2
.C
lOOx0.24
74 5
i=
w
· = 01.92
38.94
w
1.92-1 •
a=--=0.92 or 92%
2-1
w
1
Problem-12. Freezing point depression ofa molal solution ofN~SO4 in 8i0 was found to be 0.0265
200
Calculate degree of dissociation of salt at this concentration. Kr = 1.86 K - Kg I mole.
- Soln. Kr =1.86 K - kg/ mole ATr = 0.0265 K n=3
1 HAND MADE NOTES www.ChemistryABC.com

I000xl.86x--xl42
0.0265 = 200
M2 xlOOO
"
M2 = QUZZES
l.86x142 = 49.83 MORE NOTES ON www.ChemistryABC.com
200x0.026
142
i= =2.85
49.83
2 85 85
a= · -l 1. = 0.925 = 0.925 or 92.5%
3-1 2 /
Problem-13~ Which ofthe following salt has the samevalue;~fvan't Hoff factor as that ofK 3 [ Fe( CN)6 ].
(a) Al 2 (S0 4 ) 3 (b) NaCl (c) Na2SO~ (d)Al(N03) 3•

. '
Soln. (d)
Problem-14. Van't Hoff factor for AIC~ is:
om
\
(a) 3 (b)4 °-{e{1-l3
Soln. (d) .r·
.c
Problem-15. 75.2 gmofphenolisdissolvedin lKgofsolvent~= 14). Thedepressioninfreezingpoint
is 7K Fmd the percentage ofphenol that dimerlses
C
(a) 75 (b) 25 (c) 65 1,(d) 3,-j:
B
1,
Soln. . L\Tr = 1000 Kr W2 ~ 7= I000x14x75.2

yA
.M2 W1 M 2 x1000
M _ 1000xl4x75.2 tS0.4
tr
2
- 1000x7
is
. 94
l=--=0.625
150.4
m
. 75 %o
a = 0.625 I= 0.375 = O.75 1.e.
he
1 1
-1 -
2 2
.C
So, correct option is (a).

w
Problem-16. What is molarity ofH2SO4 solution that has a density 1:84 gm/cc at 35°C and contain 98% b)
weight.
w
(a)4.18molar (b)8.14M (c)l8.14M (d}l8.40M

w
M 98
d=l.84gm/cc V = - = -
. ' d 1.84
Soln.
W=98gm V=lOOmL
Molarity= 1000W 1000x98 =18 .40

MV J!_xlOO
1.84
So, correct option is (d)
Problem-I 7. Equal volumes ofl o-4 molar CaC½ and 3 x 10-s molar NaF solution are mixed with a ppt oJ
CaF2 occur.
Soln. Ksp =lxl0-11 for CaF2 .
10-4M CaC1 2 3x10-s NaF
,---., ····11!!!!!!!!!!119 -,-.-- -----...,------ - -...--- ---
CaCl2- - t ca++ + 2cr
I0-4M 10-4 M 2x10-4M
NaF--t Na+ + p-
3x10-5 3xl0-5 3xl0-5

CaF2 ~Ca+++ 2F-
.. .
Ionic product= [ca++][ p-J

· 2
= [10-4][2x3x10~
5
2
J ,::)0-4 x36x10-1° =36x10-
F
~
1.4
Ksp > Ionic product so no ppt.

I
Problem-18. 250 mL of0.01 M ~S04 is mixed with250 mL of0.2 ~6Jar KCl solution. What is the conce
tration of K+ ions in the resulting solution.
om
(a)O.lMr (b)0.2Mr · · (c)0.3 Mr }(d)0.4Mr
. I
Soln. 250 mL ;</·-·
.c
K 2S04 --t2K+ + so~~
0.lOM 2x0.10 M OJOM
C
250mL
KCl--tK+ + er
B
yA
0.2 M 0.2M 0.2 M
tr
M1V1 +M2V2 =M3V3

2x0.10x250+0.2x250 = M 3 (250+250) => 250x0.4 = M 3 x500
is
m
0.4 '
M3 =2=0.2
So, correct option is (b)
he
Molarity:
.C
Molarity= number ofmoles ofsolute present in per litre ofsolution

w
wt
Molarity= no. of moles o~solu~e = mol.wt =1000W
w
volumeo'fsolution(ht) _:!__ MV
w
100
1 M'
d=M [ m + 1000
J
where, M ~ Molality, m ~ molarity, M' ~ molecular mass
·Problem-19. The densityof3.60MH2S04 i.e29%ofH2S04 by mass will be

(a) 1.88 (b) 1.22 (c) 1.45 (d) 1.64
Soln. Molarity= 3.6 W = 29 gm V= Mass = 100

density d
-.---,
. :,J,~;_ - :........_
Molarity= 1000W = 1000W = lOOOWd

MV MxMass MxMass
d
= 1000x29xd => d= 3.6x98xl00 . = ·
36
· 98x100 1000x29 => d l.ZZ
. So, Correct _answer is (b)
/
Molality (m): MoJality= number ofmoles ofsolute preseµt per kgofsolvent.

wt
m = numberofmolesofsolute = mo l.wt. wtx,,1000
_ _..______ ,
---'--"'--
mass of solvent • gm M.wtxwt of solvent
1
-
wt.·Of sov
l ent 1n -.- /
/ 1000 \,.,·./
om
.•
'
'.
=>···
Wx1000
m=-~- · [M. /·o1·e uact1on.o .component=
c.. · ·r· · t t,futhberofmoles
· . . of co.mp.onent
, ·]
.c
. · Total number of mole of component
Problem-20. IfthemoJality(m) ofanaq. solution ofcane sugar (C128n011 ) is9.45. What is the mole fraction
C
ofca.ne su.gar in solution. . \ /1
B
(a)0.008 (b) 0.08 (c) 0.016
1
(d) 0.018 ·
yA
Soln. m= 0.45 W1 =? . M=342 W2 =1000.gm
0.45 gmin 1000 gm of solvent.
tr
O _ wt X 1000 .
·45 - 342 xlOOO, wt=0.45x342 =153.9=154
is
153.9
Number ofmoles = - - = 0.45 (for cane sugar)
m
. . 342
1000 .
he
forsolventnumberofmoles = - - =55.55
18
.C
. = 0.45 = 0.45 =0 008

mole :fract10nofsolute 0.4 + . .
5 55 55 56
w
So, correct answer is (a)

w
Problem-21. The mole :fractionQfsolute inonemolalaq. solution is

(a)0.027 (b) 0.036 (c) 0.018 (d) 0.009
w
1 1
X = - - - = - - = 0.0176:::: 0.018
55.55 + 1 56.55
So, correct answer is (c)
Henry law: This law gives a quantitative relation between pressure and solubility ofgas.
It is only applicable ifgas is not largely soluble in the solution, even through saturated, may be very dilute.
It states, at ~ntant temperature the solubility ofgas in a liquid is directly proportional to the partial pressure o1
gas present above surface of solution.
solubility oc p
solubility = K8 P, where, KH =Henry constant
OR
"The concentration is proportional to the mole :fraction ofthe dissolved substance".
So, x oc C
x= KH C
QUZZES
where, x=moleMORE :fraction and C:=. cpncentration.
NOTES ON www.ChemistryABC.com
In tenns ofmole :fraction mole :fraction ofgas is directly proportional to the partial fraction ofgas.
. !· .
i
P=KHx
H igher the value ofKH' lower will be the solubility ofgas .
Problem-22. Henry law constant ofmethyl bromide is 0.159moles/lit atm at 25°C.What is the solubility
,
·moles/lit) ofCH 3Bdn water at 25°Cand partial pressure ot 125 mm Hg .

125
Soln. p:::; 125 mm of Hg= =atm
760
125
s=K P =0.159x =0.026mole / lit
· 760
Non-Ide solution (Deviation from Raoult's. la�)f/
)
om
'
r
I· , I _ ,; - - t
/
Positive deviation Neg�tive deviation
.c
AS=+ve AS::;;: +ve
AV:::;+ve AV=-ve
C
AH=+ve(endotherm.ic) AH=-ve(expthermic)
l
\,
B
Problem-23. The complex compound K 4 [Fe( CN)6 J is 45% dissociated in 0.1M aqueous solutior
A
complex at 27°C.What will be the osmotic pressure ofthe solution?
y
Soln. a=0.45, T=27+ 273=300K, R =0.0821
tr
i-1
0.45=- =>i=2.80
is
5 1
m
O .IM means that IOO mL is 1000 mL

rcV =mRT
he
2.80x0.082I x300x100 _
1t= =6 896atm
1000
.C
Problem-24. Ethylene glycol is a colourless liquid used as automobile antifreeze. Ifthe density at 2D°C 1
4.028m solution of ethylene glycol in11i0 is 1.0241 gm/I4. What is the molarity ofsoln(molar mas:
w
ethylene glycol=62.07gm/mole) .
w
Soln. 4.028m =4.028mole in 1000gm of H 20

w
m M 62.07
d= => v= = = 60.61mL
V d 1.0241
1000 W
M=
MV
· 1 M'
d=[ + JM
m lOO
l 62.07
1.0241=[- -+ JM
4.028 1000
1.024=[0.24+ 0.06]M => I.024=0.3IOM

1.024 3 33
M= = .
fl -:t 1
QUZZES
PRACTICE SET
According to Roults law the relative lowering ofvapour pressure ofa

1. · ofa· solution non-' volatile non-electro]yte
is equal to
(a) molefraction ofsolvent (b)weightpercentofsolute
(c) mole fraction ofsolute (d) we_ight percent ofsolvent
2. . The freezing point of 0.,2molal oolution of non-volatile ooluteJn benzene is 1.02' lower than the freezingpoint
ofpure benzene. The molar depression constant ofbenzene �:
(a)2.04 (b)5.1 ' ·
(c)1.86 (d)0.102 ,
3. Apressure cooker reduces the cooking time because:. <, . .
(a)The heating is more uniform (b) tpe higher p,ress�fendierizes the food.
om
(C) The vapour pressure ofwater is increased (d)'fue bo� pomy6f�fter is elevated
4. The best method for determination ofmolecular \'f�ight ofa ilon-vplatile 'on-electrolyte is:
(a)Hoffinan'smethod (b)VictorMayer"smethod
.c
(c)Cryoscopic method (d)Densitymeasurement
C
5. Ideal solution followsRoult'sLaw at: . . \ . 11
(a)low temperature and high concentration ·(b) pightemperature and law concentration
(c)lowtemperature andJow concentration
B
(d)all temperature and all concentration
yA
6_. For ideal solution:
(a) LiVmix >0,AH>O (b) AVmix <0,LiH mix <0
tr
(c) LiVmix =0,Aff mix =0 (d) .LiVmix < .O�Af:lmix >0

is
7. An ideal solution is formed bymixingtwo pure liquids. For_this solution

m
1. Li H mix = 0 2. LiVmix = 0 3. LiG mix = 0 4. LiSmix = 0

he
(a)Only 1 is true (b)Only l and 2 aretroe

(c) Only 2 and c are true -(d)All are true.
.C
8. The osmotic pressure(in atm)of0.1 M NaCl solution at the room temperature is:
(a)2.5 (b)5.0 (c)7.5 .(d)10
w
9. Ethanol and methanol form ideal.solutions and their vapour pressures in the pure form are 44.5 torr and
w
88.7 torr at 295 K, respectively. Their partial pressures(in torr) in a solution obtained by mixing 50 gms of
w
each, are, respectively

(a) 18.3 and 52.3 (b)52.3 and 18.3 (c)36.5 and 26 . .3 (d) 26.3 and 36.5
10. The osmotic pressure ofblood is 7.4 atm at 27°C. The number ofmoles of glucose(per litre)that should be
used for an intravenous injection so that it has the same osmotic pressure as that qfblood is:
(a)0.1 (b)0.2 (c)0.3 (d)0.4
'
11. The magnitude ofthe colligative property ofa two-component solution depends on the
(a)Number ofsolute molecules (b)Number ofsolvent molecule
(c)Number ofsolute molecules relative to that ofsolvent molecules.
(d)Total number ofsolvent and solute molecules.
12. The vapour pressure ofcrystalline zirconium chloride follows the equation logp(inmm)=-5400/T +
HAND1 MADE NOTES www.ChemistryABC.com
11.766. The heat ofsublimation(in kcal mot ) at 437°C is:
(a)24700 (b)24.7 (c) 103.2 (d)4.7
···a=J 5oh.,tion &Colligative Properties
13. 5.0 gm QUZZES

ofa non-dissociating organic substance dissolves :in 100 g of water
MORE NOTES ON
to generate an osmotic iressure of
7.15 atm at 25°C. The molar mass ofthe substance is:
( R = 0.08206 atm L mor 1 K- 1 )
(a) 17lgmoi-'1 (b)342gmor1 (c) 684gmor1 (d) 35640gmor1
14. The vapour pressure of a solvent decresed by 5 mm of Hg, when a non-volatile solute was added to the
solvent. The mole fraction ofthe solute in the solution is 0.2. 'o/}iat would be the mole fraction ofthe solvent, if
decrease in the vapour pressure is to be 15 mm of Hg? , ;
1
(a) 0.8 (b) 0.6 (c) 0.4 (d) 0.2
1
'
.
. .
15. 1.2 g of acetic acid when dissolved in 100 g ofbenzene raised the boilingofbenkene by0.26°. What is the
value ofvan'tHo:ff's factor of acetic acid in benzene? · 1
• '
(Molal elevation constant ofbenzene is 2.58°)
om
(a)2 (b)l (c),5
.c
ANSWER KEY
C
Questions 1 2 3 4 5
Option (c) (b) (b) l.
\. (c)/1 (d)
B
;
Questions 6 7 /
8 9 10
yA
Option (c) (b) (b) (a) (c)
Questions 11 12 13 14 15
Option (a) (b) (a) (c) (d)
tr
is
m
he
.C
w
w
w

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$)&.

HAND MADE NOTES www.ChemistryABC.com

weightage of total marks. ' /

Thermodynamics. Students preparing for competitive examinations are advised to go through

Mohammad Firoz for formatting, designing and type-setting the book.

HAND MADE NOTES www.ChemistryABC.com

Chapter 1: Introduction to Thermodynamics •- .l

Chapter 2: Zeroth & First Law of Thermodynall)icS 26-60

Chapter 3: Second Law of Thermodynamics! :' 61-104

Chapter 6: Partial Molar Quantity 121-133

Chapter 9: Kinetic Theory of Gases 164-178

Chapter 10: Chemical Equilibria 179-194

Chapter 11: Phase Rule 195-214

Chapter 12: Solution & ColligativeProperties 215-226

HAND MADE NOTES www.ChemistryABC.com

HAND MADE NOTES www.ChemistryABC.com

(v) J!dx = f!ne x + c _(vi) Jsinxdx =-cos�+c

cvm) 1u cx)g(x>; dx :f(x> Jg(x)dx _{ ! J

( vii) log10 8 := 0.903 0 (viii) log 10 -9 = 0.954 . (ix) log!O 10 =1

Basic Units Name S~bol ·

Amount ofsubstance mole rool

Vohnne cubic metre mJ

Density kilogram per kg/rrr

HAND MADE NOTES www.ChemistryABC.com

Conversion of Selected Non-system Units to SI units:

Quantity Unit . Com·ersion factor to SI

Mass gram · lxl0-31$/

electron volt( eV)

Energy, work, amount ofheat kilocalorie (kcal) 4186.8J

Values of Selected Fundamental Physical Constants

Quantity · Symbol _ Value

Gas constant R = 8.314x 107 ergdeg-1 mole-1

= 8.314JK- 1mor1 (SI)

=22.414xl0-3 m3mole-1 (SI)

HAND MADE NOTES www.ChemistryABC.com

g(x) = ax2 +ac.

dependence ofyuponxmeansthatyis a function ofx. This functional relationship is oftendenotedy= f(x),

Similarly, dG=VdP-SdT (where, negative sigti because T at head ofarrow)

IfV is constant so that dV: is zero, then,

(au) =T ... (1)

If S constant /so that dS is zero,

respect to S keeeping V constant, we get ·

And ... (4)

It follows from equation (3) and( 4) that

(:n =-(:1 ... (5)

(:a=-(:a, HAND MADE NOTES

Equation (5) and (8) are known as Maxwell's relationships: ....

(:l =-(!;), ... (9)

(!~), =-(: ), ... (10)

(!~t =-(!~t ... (fl)

dU;; TdS- PdV

Since, (:ar(:;), [From Maxwell's relation]

tions, derive the two thermodynamics equations of state.

( au) =T(as) 1-P

This is the second thermodynamics equation of state.

The correct option is (c)

The triple product rule for such interrelated variables x, y, and z

Cyclic Rule in the form of P, V & T:

(:a (dl')P +(!;), (dV)P =0