01 Thermodynamic Process Theory1
01 Thermodynamic Process Theory1
01 Thermodynamic Process Theory1
88 Thermodynamic Processes
13.1 Introduction.
(1) Thermodynamics : It is a branch of science which deals with exchange of heat energy
between bodies and conversion of the heat energy into mechanical energy and vice-versa.
(2) Thermodynamic system : A collection of an extremely large number of atoms or
molecules confined with in certain boundaries such that it has a certain value of pressure,
volume and temperature is called a thermodynamic system. Anything outside the
thermodynamic system to which energy or matter is exchanged is called its surroundings.
Example : Gas enclosed in a cylinder fitted with a piston forms the thermodynamic system
but the atmospheric air around the cylinder, movable piston, burner etc. are all the surroundings.
Thermodynamic system may be of three types
(i) Open system : It exchange both energy and matter with the surrounding.
(ii) Closed system : It exchange only energy (not matter) with the surroundings.
(iii) Isolated system : It exchange neither energy nor matter with the surrounding.
(3) Thermodynamic variables and equation of state : A thermodynamic system can be
described by specifying its pressure, volume, temperature, internal energy and the number of
moles. These parameters are called thermodynamic variables. The relation between the
thermodynamic variables (P, V, T) of the system is called equation of state.
For moles of an ideal gas, equation of state is PV = RT and for 1 mole of an it ideal gas is
PV = RT
a 2
For moles of a real gas, equation of state is P 2 (V b) RT and for 1 mole of a real gas it is
V
a
(V b) RT
V2
(iv) Isochoric
Important points
(i) Heat is a form of energy so it is a scalar quantity with dimension [ML2T 2 ] .
(ii) Unit : Joule (S.I.), Calorie (practical unit) and 1 calorie = 4.2 Joule
(iii) Heat is a path dependent quantity e.g. Heat required to change the temperature of a given
gas at a constant pressure is different from that required to change the temperature of same gas
through same amount at constant volume.
(iv) For solids and liquids : Q = mL
temperature]
(Q)V C V T
and
(Q)P C P T
[For constant
pressure]
(2) Work (W) : Work can be defined as the energy that is transferred from one body to the
other owing to a force that acts between them
If P be the pressure of the gas in the cylinder, then force
exerted by the gas on the piston of the cylinder F = PA
In a small displacement of piston through dx, work done by
the gas
dW F .dx PA dx P dV
dW
Vf
Vi
P dV P (Vf Vi )
dx
F=PA
genius
Thermodynamic Processes 90
Important points
(i) Like heat, work is also a path dependent, scalar physical quantity with dimension
[ML T 2 ]
2
Positiv
e
work
Expansio
n
Compressio
n
A
Negativ
e
work
V
(iv) In a cyclic process work done is equal to the area under the cycle.
It is positive if the cycle is clockwise.
It is negative if the cycle is anticlockwise.
P
P2
P1
(v) W
Vf
Vi
P
B
C
Positive
work
P2
P1
D
Negative
work
A
V1
V2 V
V1
V2 V
Clockwise cyclic
Anticlockwise cyclic
process
process
P dV
From this equation it seems as if work done can be calculated only when P-V equation is
known and limits Vi and Vf are known to us. But it is not so. We can calculate work done if we
know the limits of temperature.
For example, the temperature of n moles of an ideal gas is increased from T0 to 2T0
through a process P
and we are interested in finding the work done by the gas. Then
T
PV = nRT
and
..(ii)
..(i)
Vf
Vi
P dV
2nRT
nRT2
dT
or dV
2nRT
dT 2nRT0
T
2T0
T0
So we have found the work done without putting the limits of volume.
(vi) If mass less piston is attached to a spring of force constant K and a mass m is placed
over the piston. If the external pressure is P0 and due to expansion of gas the piston moves up
through a distance x then
Total work done by the gas W W1 W2 W3
M
M
1
Kx 2 mgx
2
(vii) If the gas expands in such a way that other side of the piston is vacuum then work done by
the gas will be zero
As W PV 0
[Here P = 0]
Gas
Vacuum
(3) Internal energy (U) : Internal energy of a system is the energy possessed by the
system due to molecular motion and molecular configuration.
The energy due to molecular motion is called internal kinetic energy UK and that due to
molecular configuration is called internal potential energy UP.
i.e. Total internal energy U U K U P
(i) For an ideal gas, as there is no molecular attraction U p 0
i.e. internal energy of an ideal gas is totally kinetic and is given by U U K
and change in internal energy U
3
RT
2
3
R T
2
T
RT C V T
1
( 1)
1
2
(P f Vf Pi Vi )
(iii) Change in internal energy does not depends on the path of the process. So it is called a
point function i.e. it depends only on the initial and final states of the system, i.e. U U f U i
(iv) Change in internal energy in a cyclic process is always zero as for cyclic process U f U i
U U f U i 0
So
(a) 20 J
200 Kp
100 Kp
P
100 cc
R
Q
300 cc
genius
Thermodynamic Processes 92
(b) 20 J
(c) 400 J
(d) 374 J
Solution : (b) Work done by the system = Area of shaded portion on P-V diagram
(300 100)106 (200 10) 103 20 J
Problem 2. An ideal gas is taken around ABCA as shown in the above P-V diagram. The work done during
a cycle is
[KCET (Engg./Med.) 2001]
P
(3P, 3V)
(a) 2PV
(b) PV
C
(P, 3V)
A
(P,V)
(c) 1/2PV
(d) Zero
Solution : (a) Work done = Area enclosed by triangle ABC
1
1
AC BC (3V V) (3P P ) 2 PV
2
2
Problem 3. The P-V diagram shows seven curved paths (connected by vertical paths) that can be
followed by a gas. Which two of them should be parts of a closed cycle if the net work done
by the gas is to be at its maximum value
[AMU (Engg.) 2000]
a
b
c
(a) ac
(b) cg
(c) af
(d) cd
Solution : (c) Area enclosed between a and f is maximum. So work done in closed cycles follows a and f is
maximum.
Problem 4. If Cv 4.96cal/ moleK , then increase in internal energy when temperature of 2 moles of
this gas is increased from 340 K to 342 K
[RPET 1997]
Solution : (b) Increase in internal energy U .C v .T 2 4.96 (342 340) 2 4.96 2 19.84cal
Problem 5. An ideal gas of mass m in a state A goes to another state B via three different processes as
shown in figure. If Q1, Q2 and Q3 denote the heat absorbed by the gas along the three
paths, then
[MP PET 1992]
(a) Q1 Q2 Q3
(b) Q1 Q2 Q3
(c) Q1 Q2 Q3
B
V
(d) Q1 Q2 Q3
Solution : (a) Area enclosed by curve 1 < Area enclosed by curve 2 < Area enclosed by curve 3
Q1 Q2 Q3
Problem 6. The relation between the internal energy U and adiabatic constant is
(a) U
PV
1
(b) U
PV
1
(c) U
PV
(d) U
PV
U cvT T cv
cv
PV
R
PV
R
1 1
[As
PV RT
PV
R
and
R
]
1
W Q or
W
J
Q
Important points
(1) From W = JQ if Q = 1 then J = W. Hence the amount of work done necessary to produce
unit amount of heat is defined as the mechanical equivalent of heat.
(2) J is neither a constant, nor a physical quantity rather it is a conversion factor which used
to convert Joule or erg into calorie or kilo calories vice-versa.
(3) Value of J 4.2
J
erg
J
4.2 107
4.2 103
.
calorie
calorie
kilocalori
e
(4) When water in a stream falls from height h, then its potential energy is converted into
heat and temperature of water rises slightly.
From
W = JQ
mgh = J ms t [where m = Mass, s = Specific heat of water]
Rise in temperature t
gh
C
Js
(5) The kinetic energy of a bullet fired from a gun gets converted into heat on striking the
target. By this heat the temperature of bullet increases by t.
From
W = JQ
1
mv2 J ms t [where m = Mass, v = Velocity of the bullet, s = Specific
2
v2
C
2Js
If the temperature of bullet rises upto the melting point of the bullet and bullet melts then.
From
W = JQ
1
mv2 J [mst mL]
2
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Thermodynamic Processes 94
Rise in temperature
v2
t
L
2J
s C
(6) If ice-block falls down through some height and melts partially then
gets converted into heat of melting.
From
W = JQ
mgh J m' L
melts]
m'
So
mgh
kg
JL
JL
meter
g
(a) 42C
(b) 49C
(c) 0.49C
(d) 4.9C
Solution : (c) Loss in potential energy of water = Increment in thermal energy of water
mgh J mst 9.8 210 4.3 1000t t 0.49C
Problem 8. A block of mass 100 gm slides on a rough horizontal surface. If the speed of the block
decreases from 10 m/s to 5 m/s, the thermal energy developed in the process is
[UPSEAT 2002]
(a) 3.75 J
Solution : (a) Thermal
(b) 37.5 J
energy
developed
(c) 0.375 J
=
(d) 0.75 J
Loss
in
kinetic
energy
1
1
m(v22 v12) 0.1 (102 52) 3.75 J
2
2
Problem 9. The weight of a person is 60 kg. If he gets 105 calories heat through food and the efficiency
of his body is 28%, then upto how much height he can climb (approximately)
[AFMC 1997]
(a) 100 m
(b) 200 m
(c) 400 m
(d) 1000 m
28
28
mgh J
h 200m
Problem 10. Hailstone at 0C falls from a height of 1 km on an insulating surface converting whole of its
kinetic energy into heat. What part of it will melt (g 10m / s2 )
[MP PMT 1994]
(a)
1
33
(b)
1
8
(c)
1
10 4
33
calorie/kg]
m'
1
m 33
J mL mgh
[As L = 80 103
Problem 11. A bullet moving with a uniform velocity v, stops suddenly after hitting the target and the
whole mass melts be m, specific heat S, initial temperature 25C, melting point 475C and
the latent heat L. Then
(a) mL ms(475 25)
mv2
2J
mv2
2J
mv2
(d)
J
mv2
ms(475 25) mL
2J
Solution : (b) K.E. of bullet = Heat required to raise the temperature of bullet from 25C to 475C + heat
required to melt the bullet
(c) ms(475 25) mL
1
mv2
.
mv2 J [ms(475 25) mL] ms(475 25) mL
2
2J
Problem 12. A lead bullet at 27C just melts when stopped by an obstacle. Assuming that 25% of heat
is absorbed by the obstacle, then the velocity of the bullet at the time of striking (M.P. of
lead = 327C, specific heat of lead = 0.03 cal/gmC, latent heat of fusion of lead = 6
cal/gm and J = 4.2 J/cal)
[IIT-JEE 1981]
(a) 410 m/sec
1
2
Solution : (a) Using expression obtained in problem (11) we get 75% mv J [ms(327 27 mL]
2
Problem 13. A drilling machine of power P watts is used to drill a hole in Cu block of mass m kg. If the
specific heat of Cu is 5 J kg1C 1 and 40% of power lost due to heating of machine the
rise in temperature of the block in T sec (will be in C)
(a)
0.6 PT
ms
(b)
0.6 P
msT
(c)
0.4 PT
ms
(d)
0.4 P
msT
Work(W)
W=PT
Time(T )
As 40% energy is lost due to heating of machine so only 60% energy will increase the
temperature of the block
60% of W = m s t t
0.6W
0.6PT
.
ms
ms
Important points
(1) It makes no distinction between work and heat as according to it the internal energy (and
hence temperature) of a system may be increased either by adding heat to it or doing work on it
or both.
genius
Thermodynamic Processes 96
(2) Q and W are the path functions but U is the point function.
(3) In the above equation all three quantities Q, U and W must be expressed either in
Joule or in calorie.
(4) Just as zeroth law of thermodynamics introduces the concept of temperature, the first
law introduces the concept of internal energy.
(5) Sign conventions
Positive
Negative
Positive
Negative
Positive
Negative
(a) 260 J
(b) 150 J
(c) 110 J
(d) 40 J
Problem 15. 110 J of heat is added to a gaseous system, whose internal energy change is 40 J, then the
amount of external work done is
[CBSE PMT 1993; AFMC 1999; JIPMER 2000]
(a) 150 J
(b) 70 J
(c) 110 J
(d) 40 J
Problem 16. When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy
supplied which increases the internal energy of the gas, is
[IIT-JEE 1990; RPET 2000]
(a)
Solution : (d)
2
5
(b)
3
5
R /( 1)
U C v T
1
1
5
Q C P T R /( 1)
7/5 7
(c)
3
7
(d)
5
7
Problem 17. An electric fan is switched on in a closed room. The air in the room is
[MP PET 1996]
(a) Cooled
(b) Heated
(c) Maintains its temperature
(d) Heated or cooled depending on the atmospheric pressure
Solution : (b) When an electric fan is switched on in a closed room conventional current of air flows. Hence
due to viscous force mechanical energy is converted into heat and some heat is also
produced due to thermal effect of electric current in motor of fan.
volume of 4 m3 . Energy of 100 J is then added to the gas by heating. Its internal energy is
[MNR 1994]
Q U dV
100 U 300
U 400J
Problem 19. A thermodynamic process is shown in the figure. The pressures and volumes corresponding
to
some
points
in
the
figure
are
and
VA 2 103 m3 , VD 5 103 m3
In process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added
to the system. The change in internal energy of the system in process AC would be
[CBSE PMT 1992]
P
(a) 560 J
(b) 800 J
(c) 600 J
O
(d) 640 J
Solution : (a) By adjoining graph WAB 0 and WBC 8 104 [5 2] 103 240J
WAC WAB WBC 0 240 240J
Now, Q AC Q AB Q BC 600 200 800J
From
first
law
of
thermodynamics
Q AC U AC WAC
800 U AC 240
U AC 560J .
Problem 20. If R = universal gas constant, the amount of heat needed to raise the temperature of 2 mole
of an ideal monoatomic gas from 273 K to 373 K when no work is done
[MP PET 1990]
(a) 100 R
(b) 150 R
(c) 300 R
(d) 500 R
genius
Thermodynamic Processes 98
Solution : (c)
gas
Q U
R
R
T 2
Cv T
[373 273]
= 300 R
5
1
1
3
5
]
3
Conductin
g walls
Gas
Since these two conditions are not fully realised in practice, therefore, no process is
perfectly isothermal.
(2) Equation of state : From ideal gas equation PV = RT
If temperature remains constant then PV= constant i.e. in all isothermal process Boyles law
is obeyed.
Hence equation of state is PV = constant.
(3) Example of isothermal process
(i) Melting process [Ice melts at constant temperature 0C]
(ii) Boiling process [water boils at constant temperature 100C].
(4) Indicator diagram
P
T1<T2< T3
T3
T2
T1
Work
V
(i) Curves obtained on PV graph are called isotherms and they are hyperbolic in nature.
(ii) Slope of isothermal curve : By differentiating PV = constant. We get
P dV V dP 0
tan
P dV V dP
dP
P
dV
V
dP
P
dV
V
(iii) Area between the isotherm and volume axis represents the work done in isothermal
process.
Q
Q
mT m 0
[As T = 0]
E P
dP
Stress
E
dV / V Strain
Vf
Vi
P dV
Vf
Vi
RT
dV
V
[As PV = RT]
Vf
W RT loge
Vf
2.303 RT log10
or
Vi
Vi
Pi
2.303RT log10 Pi
P
P
f
f
W RT loge
but
U T
[As T = 0]
U 0
external surrounding.
or if the work is done on the system than equal amount of heat energy will be liberated by
the system.
(b) 1728 J
(c) 1728 J
(d) 1572.5 J
Vf
W RT loge
11.2
8.3 273 ( 0.69) 1572J
22.4
Vi
Problem 22. An ideal gas A and a real gas B have their volumes increased from V to 2V under isothermal
conditions. The increase in internal energy
[CBSE PMT 1993; JIPMER 2001, 2002]
Solution : (c) In real gases an additional work is also done in expansion due to intermolecular attraction.
genius
Thermodynamic Processes 100
Problem 23. Which of the following graphs correctly represents the variation of
(a)
(b)
(c)
(d)
1 dV
1
V dP
P
1
So,
graph will be rectangular hyperbola.
P
[SCRA 1994]
1
V
(Boyle's law)
Problem 25. How much energy is absorbed by 10 kg molecule of an ideal gas if it expands from an initial
pressure of 8 atm to 4 atm at a constant temperature of 27C
[Roorkee 1992]
Pi
(10 103 ) 8.3 300 loge 8 104 8.3 300 0.693 1.728 107 J
Pf
4
W RT loge
Problem 26. 5 moles of an ideal gas undergoes an isothermal process at 500K in which its volume is
doubled. The work done by the gas system is
(a) 3500 J
(b) 14400 J
Vf
(c) 17800 J
(d) 5200 J
Vi
2V
5 8.3 500 0.69 14400J .
V
Problem 27. Work done by a system under isothermal change from a volume V1 to V2 for a gas which
n2
nRT
V
V2 n
V1 n
V1 V2
V1V2
V2
V1
n2
V2 n
V V2
n2 1
VV
V
1
1
2
V1 n
V2 n
V2
V1
V1 V2
V1V2
n2
V1V2
V
1
2
n2
nRT
n2
2
V n
V
PdV nRT
V2
V1
dV
n2
V n
V2
V2
dV
V1
V2
nRT loge
V1
V V2
V2 n
n2 1
V1 n
V1V2
Gas
.. (i)
genius
Thermodynamic Processes 102
.. (ii)
.. (iii)
(P dV V dP)
P dV 0
( 1)
or
P dV V dP 0
asC v
( 1)
dV dP
0
V
P
i.e.,
(P dV V dP)
P dV 0
R
loge V loge P C ,
log(PV ) C
i.e.,
PV constant
or
.. (iv)
Equation (iv) is called equation of state for adiabatic change and can also be re-written as
and
TV 1 constant
[as P = (RT/V)]
T
constant
P 1
asV
.. (iv)
RT
P
.. (vi)
By differentiating, we get
1
dP
PV
dV
V
P
V
Cp
Slopeof adiabaticcurve (P / V)
1
Slopeof isothermal
curve (P / V)
Cv
(6) Specific heat : Specific heat of a gas during adiabatic change is zero
As
Q
0
0
mT mT
[As Q = 0]
dP
Stress
E
dV / V Strain
E P
i.e. adiabatic elasticity is times that of pressure but we know isothermal elasticity E P
E
E
Adiabaticelasticity P
Isothermal
elasticity P
i.e. the ratio of two elasticities of gases is equal to the ratio of two specific heats.
(8) Work done in adiabatic process
W
or
or
or
So
Vf
Vi
P dV
Vf
Vi
As P
V
K
dV
V
1 K
K
[1 ] Vf 1 Vi 1
[P f Vf Pi Vi ]
R
[Tf Ti ]
(1 )
( 1)
V 1
( 1)
[As K PV P f Vf Pi Vi ]
(1 )
[Pi Vi P f Vf ]
-
As V dV
(9) Free expansion : Free expansion is adiabatic process in which no work is performed on
or by the system. Consider two vessels placed in a system which is enclosed with thermal
insulation (asbestos-covered). One vessel contains a gas and the other is evacuated. The two
vessels are connected by a stopcock. When suddenly the stopcock is opened, the gas rushes into
the evacuated vessel and expands freely. The process is adiabatic as the vessels are placed in
thermal insulating system (dQ = 0) moreover, the walls of the vessel are rigid and hence no
external work is performed (dW = 0).
Now according to the first law of thermodynamics dU = 0
If U i and U f be the initial and final internal energies of the gas
then
U f Ui 0
[As U f U i ]
Thus the final and initial energies are equal in free expansion.
(10) Special cases of adiabatic process
PV constant P
T
1
,
V
PT
constant
PT
and
TV 1 constant
1
V 1
Type of gas
Monoatomic = 5/3
Diatomic = 7/5
Polyatomic = 4/3
P
P
P
P
1
V
1
5/ 3
7/ 5
1
1
V4/3
PT
P T 5/ 2
P T7/ 2
P T4
1
V 1
1
V 2/ 3
1
V 2/ 5
1
V1/ 3
genius
Thermodynamic Processes 104
Final pressure
Final temperature
1
2
Vf
Vi
(ii) Expansion : If a gas expands isothermally and adiabatically from volume Vi to Vf then
from the slope of the graph it is clear that graph 1 represent
P
isothermal process, graph 2 represent adiabatic process.
Work done
Final pressure
Final temperature
1
2
Vi
Vf
Problem 28. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of
its absolute temperature. The ratio C p / Cv for the gas is
[AIEEE 2003]
(a)
3
2
(b)
4
3
(c) 2
PT
1 .
(d)
5
3
Cp 3
3
3
1
Cv
2
2
So,
8
5
of its original volume. If , then
27
3
(a) 450 K
(b) 375 K
T1 V2
T2 V1
(d) 405 K
constant
V
T2 T1 1
V2
(c) 225 K
27
300
5
1
3
27
300
2/ 3
675K
Problem 30. If = 2.5 and volume is equal to 1/8 times to the initial volume then pressure P is equal to
(initial pressure = P)
[RPET 2003]
(a)
P P
(b)
P 2P
(c) P P (2)15 / 2
P2 V1
P1 V2
(d) P 7P
P
85 / 2 P P (2)15/ 2
P
(b)
(d)
Its
thermal
energy
Problem 32. P-V plots for two gases during adiabatic process are shown in the figure. Plots 1 and 2 should
correspond respectively to
[IIT-JEE (Screening) 2001]
(a) He and O 2
(b) O 2 and He
2
V
(c) He and Ar
(d) O 2 and N2
Solution : (b) Slope of adiabatic curve
1
. So is inversely proportional to
Atomicity
of thegas
Problem 33. A monoatomic ideal gas, initially at temperature T1 , is enclosed in a cylinder fitted with a
frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by
releasing the piston suddenly. If L1 and L2 are the lengths of the gas column before and
after expansion respectively, then T1 / T2 is given by
[IIT-JEE (Screening) 2000]
L
(a) 1
L2
2/ 3
L1
(b)
L2
L2
(c)
L1
T1 V2
T2 V1
L
(d) 2
L1
1
L A
2
L1A
5 / 31
2/ 3
L
2
L1
2/ 3
Problem 34. Four curves A, B, C and D are drawn in the adjoining figure for a given amount of gas. The
curves which represent adiabatic and isothermal changes are
[CPMT 1986; UPSEAT 1999]
P
B
D
V
Solution : (c) As we know that slope of isothermal and adiabatic curves are always negative and slope of
adiabatic curve is always greater than that of isothermal curve so is the given graph curve A
and curve B represents adiabatic and isothermal changes respectively.
genius
Thermodynamic Processes 106
Problem 35. A thermally insulated container is divided into two parts by a screen. In one part the
pressure and temperature are P and T for an ideal gas filled. In the second part it is vacuum.
If now a small hole is created in the screen, then the temperature of the gas will
[RPET 1999]
(a) Decrease
(b) Increase
Solution : (c) In second part there is a vacuum i.e. P = 0. So work done in expansion = PV = 0
Problem 36. Two samples A and B of a gas initially at the same pressure and temperature are
compressed from volume V to V/2 (A isothermally and B adiabatically). The final pressure of
A is
[MP PET 1996, 99; MP PMT 1997, 99]
(a) Greater than the final pressure of B
V
P2' 2P1
2
..(i)
P2 2 P1
..(ii)
1
times the
32
Problem 37. A gas has pressure P and volume V. It is now compressed adiabatically to
original volume. If (32)1.4 128, the final pressure is
(a) 32 P
(b) 128 P
(c)
P
128
V
P
2 1
P1 V2
(d)
V / 32
1.4
P
32
(32)1.4 128
or
P
P1
V1 V2
constant
T1 T2
P2
B
II
dP
0
dV
f
1 R
2
P
0
V / V
Vf
Vi
[As P = 0]
P dV P
Vf
Vi
dV P[Vf Vi ]
[As
= constant]
W P(Vf Vi ) R[Tf Ti ] R T
Q U W
R
T
( 1)
W R T
and
R T
T R T R T
1 R T
( 1)
1
1
Q C P T
U P mL P[Vf Vi ]
or
[As Q = mL]
[ U K 0
as
there
is
no
change
in
temperature]
U P mL
negligible]
Note :
in the form of heat energy, while in isobaric expansion, temperature increases and
heat flows into the system.
Isobaric expansion of the volume of a gas is given by Vt V0 (1 V t)
where V
1
perC coefficient of volume expansion.
273
1cm3 of water at its boiling point absorbs 540 calories of heat to become steam with a
volume of 1671cm3 . If the atmospheric pressure is 1.013 105 N / m2 and the mechanical
genius
Thermodynamic Processes 108
equivalent of heat = 4.19 J/calorie, the energy spent in this process in overcoming
intermolecular forces is
[MP PET 1999, 2001; Orissa JEE 2002]
(a) 540 calorie
(b) 40 calorie
(d) Zero
Q P(V2 V1 ) 540
(b) 250 J
(c) 250 W
(d) 250 N
Problem 40. 5 mole of hydrogen gas is heated from 30C to 60C at constant pressure. Heat given to
the gas is (given R = 2 cal/mole degree)
[MP PET 2002]
R T
Solution : (c) (Q) p C p T
1
(Q)p 5
5
5 2
1
5
for H2]
Problem 41. The latent heat of vaporisation of water is 2240 J/gm. If the work done in the process of
expansion of 1g is 168 J, then increase in internal energy is
[CPMT 2000]
(a) 2408 J
Solution : (c)
(b) 2240 J
(c) 2072 J
(d) 1904 J
Problem 42. When an ideal gas ( = 5/3) is heated under constant pressure, then what percentage of
given heat energy will be utilised in doing external work
[RPET 1999]
(a) 40%
(b) 30%
W Q U C p C v
1
1 1
Solution : (a) Q
Q
Cp
(c) 60%
1
1
5
3
2
5
(d) 20%
2
100= 40%.
5
pressure is 1.671m3 . The latent heat of steam is 2.3 106 J / kg and the normal pressure
is 105 N / m2 . If 5 kg of water at 100C is converted into steam, the increase in the internal
energy of water in this process will be
(a) 8.35 105 J
(d) Zero
Solution : (b) Heat required to convert 5 kg of water into steam Q mL 5 2.3 106 11.5 106 J
Work done in expanding volume, W PV 5 105 [1.671 103 ] 0.835 106 J
by
first
law
of
U Q W
thermodynamics
P1 P2
constant
T1 T2
P
A
V1
dP
dV
V2
f
R
2
P
P
V / V
0
II
[As V = 0]
W 0
Note :
[As W = 0]
P f Vf Pi Vi
1
perC coefficient of pressure expansion.
273
where P
genius
Thermodynamic Processes 110
Solution : (d) Process CD is isochoric as volume is constant, Process DA is isothermal as temperature
constant and Process AB is isobaric as pressure is constant.
Problem 45. Molar specific heat of oxygen at constant pressure C p 7.2 cal/ mol/ C and R = 8.3
J/mol/K. At constant volume, 5 mol of oxygen is heated from 10C to 20C, the quantity of
heat required is approximately
[MP PMT 1987]
(a) 25 cal
(b) 50 cal
[As U = 0]
(3) For cyclic process P-V graph is a closed curve and area enclosed by the closed path
represents the work done.
If the cycle is clockwise work done is positive and if the cycle is anticlockwise work done is
negative.
P
P
A
B
Positive
work
Negative
work
B
C
V
(4) Work done in non cyclic process depends upon the path chosen or the series of changes
involved and can be calculated by the area covered between the curve and volume axis on PV
diagram.
P
P
A
WABC= + Shaded area
B
V
genius
Thermodynamic Processes 112
(a) 34 J
(b) 70 J
(c) 84 J
(d) 134 J
Solution : (d) Heat given Q 20cal 20 4.2 84 J .
Work done W = 50 J
is anticlockwise]
By first law of thermodynamics U Q W 84 ( 50) 134J
[As process
Problem 47. An ideal gas is taken through the cycle A B C A, as shown in the figure. If the net
heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C A is
[IIT-JEE (Screening) 2002]
V(m3)
(a) 5 J
(b) 10 J
1
(c) 15 J
A
10
(d) 20 J
P(N/m2
)
Solution : (a) For a cyclic process. Total work done WAB WBC WCA
along BC]
1
1.0 10 10 (2 1) 0 WCA
2
5J 10 J WCA WCA 5 J
Problem 48. In the following indicator diagram, the net amount of work done will be
P
(a) Positive
(b) Negative
(c) Zero
V
(d) Infinity
Solution : (b) Work done during process 1 is positive while during process 2 it is negative. Because process
1 is clockwise while process 2 is anticlockwise. But area enclosed by P-V graph (i.e. work
done) in process 1 is smaller so, net work done will be negative.
Problem 49. A cyclic process for 1 mole of an ideal gas is shown in figure in the V-T, diagram. The work
done in AB, BC and CA respectively
V1
, R (T1 T2)
V2
(a) 0, RT2 ln
V
V2
V1
V2
V1
V2
, R (T1 T2)
V1
(c) 0, RT2 ln
A
T1
B
T2
V2
, R (T2 T1)
V
1
(d) 0, RT2 ln
WAB P V 0
V2
V1
WBC RT2. ln
Problem 50. A cyclic process ABCD is shown in the figure P-V diagram. Which of the following curves
represent the same process
P
C
D
V
(a)
(b)
(c)
(d)
A
C
D
Solution : (a)
C
D
D
T
C
T
Isochoric
Isoba
rIsotherm
Adiabati
13.13 Heat Engine.
c
O
Adiabati
c
Isochoric
Isoba
r
Isotherm
Isoba
r
Isochoric
Adiabati
c
T
Isotherm
V
Heat engine is a device which converts heat into work continuously through a cyclic process.
The essential parts of a heat engine are
Source : It is a reservoir of heat at high temperature and infinite thermal capacity. Any
amount of heat can be extracted from it.
Working substance : Steam, petrol etc.
Sink : It is a reservoir of heat at low temperature and infinite thermal capacity. Any amount
of heat can be given to the sink.
The working substance absorbs heat Q1 from the source, does
an amount of work W, returns the remaining amount of heat to the
sink and comes back to its original state and there occurs no
change in its internal energy.
By repeating the same cycle over and over again, work is
continuously obtained.
The performance of heat engine is expressed by means of
Source
(T1)
Q1
Heat
Engin
e
W = Q 1 Q2
Q2
Sink
(T2)
Workdone W
Heatinput Q1
Q = W
so
genius
Thermodynamic Processes 114
Q1 Q2
Q
1 2
Q1
Q1
A perfect heat engine is one which converts all heat into work i.e. W Q1 so that Q2 0 and
hence 1 .
But practically efficiency of an engine is always less than 1.
Q2
Heatextracted Q2
workdone
W Q1 Q2
Q2
Q1 Q2
(T2)
A perfect refrigerator is one which transfers heat from cold to hot body without doing work
i.e. W = 0 so that Q1 Q2 and hence
(1) Carnot refrigerator
For Carnot refrigerator
Q1 T1
Q2 T2
So coefficient of performance
Q1 Q2 T1 T2
Q2
T2
or
Q2
T2
Q1 Q2 T1 T2
T2
T1 T2
Q2
Q2 / Q1
or
Q1 Q2
1 Q2 / Q1
.. (i)
Q2
Q2
1
or
Q1
Q1
..(ii)
Ideal
gas
Source
T1 K
Insulatin
g stand
Sink
T2 K
genius
Thermodynamic Processes 116
W = Q1
Q2
W1 Q1
V2
V1
Q2
V
P dV RT1 loge 2 Area
V1
ABGE
(ii) Second stroke (Adiabatic expansion) (curve BC) :
The cylinder is then placed on the non conducting stand and the gas is allowed to expand
adiabatically till the temperature falls from T1 to T2.
W2
V3
V2
P dV =
R
[T1 T2 ] Area BCHG
( 1)
V4
V3
P dV RT2 loge
V4
V
RT2 loge 3 AreaCDFH
V3
V4
(iv) Fourth stroke (adiabatic compression) (curve DA) : Finally the cylinder is again placed on
non-conducting stand and the compression is continued so that gas returns to its initial stage.
W4
V1
V4
P dV
(T2 T1 )
(2) Efficiency of Carnot cycle : The efficiency of engine is defined as the ratio of work
workdone W
[As W2 W4 ]
W
Q
W W1 W3 Q1 Q2
1 3 1 2
Q1
W1
Q1
W1
Q1
or
T1V2 1
T2 V3 1
V
T
or 1 3
T2 V2
..(i)
V3 V4
V3
V
2
or
V2 V1
V4
V1
T1 V4
T2 V1
..(ii)
3
loge 2
loge
V
4
V1
T2
T1
(i) Efficiency of a heat engine depends only on temperatures of source and sink and is
independent of all other factors.
(ii) All reversible heat engines working between same temperatures are equally efficient and
no heat engine can be more efficient than Carnot engine (as it is ideal).
(iii) As on Kelvin scale, temperature can never be negative (as 0 K is defined as the lowest
possible temperature) and Tl and T2 are finite, efficiency of a heat engine is always lesser than
unity, i.e., whole of heat can never be converted into work which is in accordance with second
law.
Note :
The efficiency of an actual engine is much lesser than that of an ideal engine.
Actually the practical efficiency of a steam engine is about (8-15)% while that of a
petrol engine is 40%. The efficiency of a diesel engine is maximum and is about (5055)%.
(3) Carnot theorem : The efficiency of Carnots heat engine depends only on the
T2
temperature of source (T1) and temperature of sink (T2), i.e., 1
.
T1
Carnot stated that no heat engine working between two given temperatures of source and
sink can be more efficient than a perfectly reversible engine (Carnot engine) working between
the same two temperatures. Carnot's reversible engine working between two given temperatures
is considered to be the most efficient engine.
Diesel engine
genius
Thermodynamic Processes 118
Solution : (b) In a refrigerator, the working substance takes heat Q2 from the sink at lower temperature
T2 and gives out a larger amount of heat Q1 to a hot body at higher temperature T1 .
Therefore the room gets heated if the door of a refrigerator is kept open.
Problem 52. The coefficient of performance of a Carnot refrigerator working between 30oC and 0oC is
[UPSEAT 2002]
(a) 10
(b) 1
(c) 9
(d) 0
Solution : (c) Coefficient of performance of a Carnot refrigerator working between 30C and 0C is
T2
273C
273C
9
T1 T2 303C 273C
30C
Problem 53. A Carnot engine working between 300 K and 600 K has work output of 800 J per cycle. What
is amount of heat energy supplied to the engine from source per cycle
[Pb. PMT 2002]
Workdone
Heatinput
T1 T2 600 300 1
T1
600
2
Heat input
Work done
800
1600J .
1/ 2
Problem 54. Carnot cycle (reversible) of a gas represented by a Pressure-Volume curve is shown in the
diagram
A
B
C
V
(a) I only
(b) II only
Solution : (c) Work done by the gas (as cyclic process is clockwise) W = Area ABCD
So from the first law of thermodynamics Q (net heat absorbed) = W = Area ABCD
As change in internal energy in cycle U = 0.
Problem 55. A Carnot engine takes 103 kcal of heat from a reservoir at 627C and exhausts it to a sink at
27C. The efficiency of the engine will be
(a) 22.2%
(b) 33.3%
(c) 44.4%
(d) 66.6%
T1 T2 900 300 6
or 66.6%.
T1
900
9
13.18 Entropy.
Entropy is a measure of disorder of molecular motion of a system. Greater is the disorder,
greater is the entropy.
The change in entropy i.e. dS
Heatabsorbedby system
dQ
or dS
Absolutetemperatu
re
T
Important points
(1) For solids and liquids
dQ
mL
T
T
where positive sign refers to heat absorption and negative sign to heat evolution.
(ii) When heat given to a substance raises its temperature from T1 to T2, then change in
entropy
dS
dQ
T2
T1
mc
T
dT
mcloge 2
T
T1
T2
.
T1
S 2.303mcloge
(2) For a perfect gas : Perfect gas equation for n moles is PV = nRT
dQ
nCV dT P dV
T
nCV dT
T
[As dQ = dU + dW}
nRT
T2 dT
V2 dV
dV
nCV
nR
V
T
V
T
[As PV = nRT]
T2
V
nRloge 2
T1
V1
S nCV loge
T
P
S nCP loge 2 nRloge 2
T1
P1
P2
V
nCP loge 2
P1
V1
S nCV loge
(a) Heated
(b) Cooled
(d)
Solution : (b) For an adiabatic expansion PV constantand for the given process PV 2 = constant
It is also an adiabatic expansion and during adiabatic expansion the gas is cooled.
Problem 57. A cyclic process ABCA is shown in the V-T diagram. Process on the P-V diagram is
V
C
B
A
(a)
(b)
(c)
A
V
P
A
(d)
C
V
B
V
genius
Thermodynamic Processes 120