INTRODUCTION

Matter can be classified into three categories depending upon its physical state namely solid,
liquid and gaseous states. Solids have a definite volume and shape; liquids also have a definite
volume but no definite shape; gases have neither a definite volume nor a definite shape.
DISTINCTION BETWEEN THREE STATES OF MATTER
Sl. No.

Solids

Liquids

Gases

Particles are very Closely

packed

Particles are loosely packed

Particles are very loosely

packed

Voids are extremely small

Voids are relatively larger

Voids are very large

Inter particle forces are large

Inter

Particle motion is restricted

to vibratory motion.

Particle motion is very slow

particle forces
intermediate

are

Intermediate
forces
negligible

Particle motion is very rapid

and also random.

MEASURABLE PROPERTIES OF GASES

Mass, volume, temperature are the important measurable properties of gases.
Mass: The mass of the gas is related to the number of moles as
w
n= M

are

Where n = number of moles

w = mass of gas in grams
M = molecular mass of the gas
Volume: Since gases occupy the entire space available to them, therefore the gas
volume means the volume of the container in which the gas is enclosed.
Units of Volume: Volume is generally expressed in litre or cm 3 or dm3 1m3 = 103 litre
= 103 dm3 = 106 cm3.
Pressure: The force exerted by the gas per unit area on the walls of the container is
equal to its pressure.
Units of Pressure: The pressure of a gas is expressed in atm, Pa, Nm 2, bar or,
lb/In2 (psi).
760 mm = 1 atm = 10132.5 KPa = 101325 Pa = 101325 Nm2
760 mm of Hg = 1.01325 bar = 1013.25 milli bar = 14.7 lb/2n 2 (psi)
Temperature: Temperature is defined as the degree of hotness. The SI unit of
temperature is Kelvin. On the Celsius scale water freezes at 0C and boils at 100C
where as in the Kelvin scale water freezes at 273 K and boils at 373 K.

GAS LAWS
The state of a sample of gas is defined by 4 variables i.e. P, V, n & T. Gas laws are the simple
relationships between any two of these variables when the other two are kept constant.

Boyles Law: The changes in the volume of a gas by varying pressure at a constant temperature
of a fixed amount of gas was quantified by Robert Boyle in 1662. The law was named after his
name as Boyles law. It states that:
The volume of a given mass of a gas is inversely proportional to its pressure at a constant
temperature.
Mathematically
1
P V (n, T constant)
1
P
V
(n, T constant)
K
i.e. P = V (where K is the constant proportionality)

or PV = K (constant)
Let V1 be the volume of a given mass of the gas having pressure P 1 at temperature T. Now, if the
pressure is changed to P2 at the same temperature, let the volume changes to V 2. The
quantitative relationship between the four variables P 1, V1, P2 and V2 is:
P1V1 = P2V2 (temperature and mass constant)
Graphical Representation of Boyles Law

V (litre)

PV (atm litre)

T2 >T1

P (atm)
(a) Plot of V against P

T2
T1

P (atm)
(b) Plot of PV against P

Fig. (a) show the plot of V vs P at a particular temperature. It shows that P increases
V decreases.
Plot (b) shows the plot of PV vs P at particular temperature. It indicates that PV value
remains constant inspite of regular increase in P.

Illustration 1. A sample of gas occupies 100 litres at 1 atm pressure and at 0C. If the volume
of the gas is to be reduced to 5 litres at the same temperature, what additional
pressure must be applied?
Solution:

Here P1 = 1 atm
P2 = ?
V1 = 100 litre
V2 = 5 litre
T1 = 273 K
T2 = 273 K
As temperature is constant
Then from Boyles law
P1V1 = P2V2
or 1 100 = P2 5
P2 = 20 atm
Additional pressure that should be applied = P2 P1 = 20 1 = 19 atm

Exercise 1.
A sample of hydrogen gas occupies 10 litre at 190 mm presence. What would be the
volume of this sample at 1 atm pressure, if the temperature is kept constant?
Exercise 2.
5g of ethane is confined in a bulb of one litre capacity. The bulb is so weak that it will
burst if the pressure exceeds 10 atm. At what temperature will the pressure of the gas
reach the bursting value?
Charles Law: The French Scientist, Jacques Charles in 1787 found that for a fixed amount of a
gas at constant pressure, the gas expands as temperature increases. The law can be stated as
the volume of a given mass of a gas increases or, decrease by 1/273 of its volume at 0C for
each degree rise or, fall of temperature, provided pressure is kept constant.
Charles also found that for a given mass of a gas if pressure is kept constant, the volume
increases linearly with temperature.
V = V0 (1 + t) or V V0 = V0 t
If the temperature is measured in the Celsius scale and V 0 is the volume at 0C, it is found that
= 1/273. The volume at temperature T is then;
t
V0

273 t
T
1 273
273
= V0
VT = 273
Vt = V0

Volume (cc)

Where T = 273 + t is the temperature on the Kelvin scale, which has 273C as its zero point.
Let V1 be the volume of a certain mass of a gas at temperature T 1 and at pressure P. If
temperature is changed to T2 keeping pressure constant, the volume changes to V 2. The
relationship between for variables V1, T1, V2 and T2 is:
V1 V2

T1 T2
(Pressure and Mass Constant)
Graphical Representation of Charles Law

V0

-273

-200

-100

100

200

73

173

273

373

473 (K)

300C

COMBINED GAS EQUATION

The Boyles and Charles law can be combined to give a relationship between the three variables
P, V and T. Let a certain amount of a gas in a vessel have a volume V 1, pressure P1 and
temperature T1. On changing the temperature and pressure to T 2 and P2 respectively, the gas
occupies a volume V2.
Then we can write

P1V1 P2 V2

T1
T2
The above relation is called the combined gas law.
Avogadros Law: The Avogadros law states that at a given temperature and pressure, the
volume of a gas is directly proportional to the amount of gas i.e.
V n (P and T constant)
or V = constant n
Where n is the amount of the substance
It was said that 1 mol of any gas at 0C and under 1 atm pressure occupies 22.4 103 m3 or
22.4 litre.
Avogadro further generalised the statement that a mole of any substance contains 6.022 1023
particles (molecules, atoms or any other entities).
IDEAL GAS EQUATION
A gas that would obey Boyles and Charles law under the conditions of temperature and pressure
is called an ideal gas.
Here, we combine four measurable variables P, V, T and n to give a single equation.
V n [P, T constant]
Avogadros law
V T [n, P constant]
Charles law
1
V P [n, T constant] Boyles law
The combined gas law can be written as

nT
P or PV nT
PV = nRT

this is called ideal gas equation

where R is the constant of proportionality or universal gas constant
The value of R was found out to be
R = 8.314 J mol1 K1
R = 0.0821 litre atm K1 mol1
R = 2 cal K1 mol1
Relation between molar mass and density (From ideal gas equation)
W
M
nRT
P
V
W RT
P
V M
RT
PM
Pd
d
M
RT
n

Illustration 2. What is the increase in volume when the temperature of 600 ml of air increases
from 27C to 47C under constant pressure?
Solution:

Charles law is applicable as the pressure and amount remains constant.

V1 V2

T1 T2

T1
V2
T2

or V1 =
320
600
V1 = 300
= 640 ml
Increase in volume of air = 640 600 = 40
Illustration 3. At which of the four conditions, the density of nitrogen will be the largest?
(A) STP
(B) 273 K and 2 atm
(C) 546 K and 1 atm
(D) 546 K and 2 atm

Solution:

PM
P
Density of a gas is given = RT . Obviously the choice that has greater T would
have greater density.
Hence, (B) is correct.

Exercise 3.
An open bulb containing air at 19C was cooled to a certain temperature at which the
no. of moles of gaseous molecules increased by 25%. What is the final temperature?
Exercise 4.
o
A certain mixture of helium and argon weighing 5 g occupies a volume of 10 L at 25 C
and 1 atm pressure. What is the composition of the mixture in mass percentage?

Gay Lussacs Law (temperature pressure law)

It state that pressure of the given of mass a gas is directly proportional to the Kelvin temperature
at constant volume
P T
(n.v. are constant)
P1 P2
P

K
T
T2
T
1
PAY-LOAD
When a balloon is buoyant, it can take along some weight into the upper atmosphere. The
maximum weight a balloon can carry along is called its pay-load.
Pay load = weight of the air displaced (weight of the balloon + weight of the gas it contains)
Illustration 4. A flask of 2 dm3 capacity contains O2 at 101.325 kPa and 300 K. The gas
pressure is reduced to 0.1 Pa. Assuming ideal behaviour, answer the following:
(i) What will be the volume of the gas which is left behind?
(ii) What amount of O2 and the corresponding number of molecules are left
behind in the flask?
(iii) If now 2g of N2 is introduced, what will be the pressure of the flask?
Solution:

Given that
V1 = 2dm3
Pt = 101.325 kPa
P2 = 0.1 Pa T = 300 K
We have the following results.
(i) The volume of O2 left behind will be the same, i.e. 2dm3

(ii) The amount of O2 left behind is given by

V1
104 kpa 2 dm

R T 8.314kpadm3k 1mol1 300K

n = P2
8.019 10 8 mol
(iii) 2g of N2 = 1/14 mol
Total amount of gases in flask, 1/ 14 mol + 8.125 108 mol 1/14 mol.
Thus, the pressure of the flask is given by

1
3 1
1

8.315kpadm k mol 300k

14
mol

nRT
2dm3
P= V =
= 89.08 kPa.

Illustration 5. When 2 g of gaseous substance A is introduced into an initially evacuated flask

o
kept at 25 C, the pressure is found to be 1 atm. The flask is evacuated and 3 g
o
of B is introduced. The pressure is found to be 0.5 atm at 25 C. Calculate the
ratio

Solution:

MA
MB

(M Molecular weight).

m
RT
M
PA m A MB

PB MA mB

PV

(because V, R and T are constants)

1
2 MB

0.5 MA
3
MA 1

MB 3
Illustration 6. A balloon of diameter 20 m weights 100 kg. Calculate its pay load, if it is filled
o
3
with He at 1 atm and 27 C. Density of air is 1.2 kg m .
Solution:

Weight of balloon = 100 kg

Volume of balloon

10 10 4 g

4 3 4 22 20

100
3
3 7 2

4190 103 L
Weight of gas (He) in balloon

3
PVm 1 4190 10 4

68.13 104 g
0.082 300
RT

4
4
4
Total weight of gas and balloon 68.13 10 10 10 78.13 10 g

1.2 4190 106

502.8 10 4 g
3
10
Weight of air displaced
Payload = weight of air displaced (weight of balloon + weight of gas)
502.8 10 4 78.13 10 4 424.67 10 4 g
Pay load

DALTONS LAW OF PARTIAL PRESSURE

The relation between the pressures of the mixture of nonreacting gases enclosed in a vessel to
their individual pressure is described in the law. The law was given by John Dalton in 1807. It
states that.
At constant temperature, the pressure exerted by a mixture of two or more nonreacting gases
enclosed in a definite volume, is equal to the sum of the individual pressures which each gas
would exert if present alone in the same volume.
The individual pressures of gases are known as partial pressures.
If P is the total pressure of the mixture of nonreacting gases at temperature T and volume V, and
P1, P2, P3 . represent the partial pressures of the gases, then
P = P1 + P2 + P3+ (T, V are constant)
Partial Pressure in terms of Mole Fraction
Mole fraction defines the amount of a substance in a mixture as a fraction of total amount of all
substances. If n=1 moles of way substance is present in n moles of the mixture, then mole
n1
fraction of the substance, 1 = n
If

PN2

is the partial pressure of nitrogen the mixture of SO 2 and N2. Then

RT
PN2 nN2
V ( nN2 = no. of moles of N2)
RT
nN2 nSO2
V
and Pmixture =

dividing we get,

PN2
nN2

N2
Pmixture nN2 nSO2

or,

pN2 N2 Pmixture

Illustration 7. A 2.5 litre flask contains 0.25 mole each of sulphur dioxide and nitrogen gas at
27C. Calculate the partial pressure exerted by each gas and also the total
pressure.
Solution:

Partial pressure of SO2

nRT 0.25 8.314 300
PSO2

V
2.5 10 3
= 2.49 105 Nm2 = 2.49 105 Pa
P
Similarly N2 = 2.49 105 Pa
Following Daltons Law
P
P
PTotal = N2 + SO2
= 2.49 105 Pa + 2.49 105 Pa = 4.98 105 Pa

Illustration 8.

At an under water depth of 250 ft; the pressure is 8.38 atm. What should be the
mole percent of oxygen in the diving gas for the partial pressure of oxygen in the
mixture to be 0.21 atm, the same as it is in air at 1 atm?

Solution:

Since

PO2 XO2 Ptotal then XO2

PO2
Ptotal

0.21
0.025
8.38

X O2

So, the diving gas should contain 2.5% of O2 .

Exercise 5.
A gaseous mixture containing 8g of O 2 and 224 ml ofN2 at STP is enclosed in a flask
of 5 litre capacity of 0C. Find the partial pressure of each gas and also calculate the
total pressure in the vessel.
GRAHAMS LAW OF DIFFUSION/EFFUSION
The ability of a gas to spread and occupy the whole available volume irrespective of other gases
present in the container is called diffusion.
Effusion is the process by which a gas escapes from one chamber of a vessel through a small
volume diffused
V
r=

time taken
T
opening or an orifice.
Thomas Graham in 1831 proposed the law of gaseous diffusion. The law states under similar
conditions of temperature and pressure, the rates of diffusion of gases are inversely proportional
to the square roots of their densities.
1
r
d where is the rate of diffusion and d is the density of the gas.
Now, if there are two gases A and B having r 1 and r2 as their rates of diffusion and d 1 and d2 their
densities respectively. Then
1
1
r1
d2
r1
r2

r
d1
d1
d2
and
or, 2
(at same T and P)
M2
r1
M2
V T
2
1 1
M1
r2
M1 V2 T2
2
Here M1 and M2 are the molecular masses of the gases having densities d 1 and d2 respectively.
Grahams law of diffusion also holds good for effusion.
Effect of pressure on state of diffusion
The rate of diffusion (r) of a gas at constant temperature is directly preoperational to its pressure

r P
r

at cons tan t temperature

r1 P1
M2

r2 P2
M1
Illustration 9. Which of the two gases ammonia and hydrogen chloride will diffuse faster and by
what factor?
rNH3
Solution:

rHCl

MHCl

MNH

1/ 2

1/ 2

36.5

r
= 17
= 1.46 or NH3 = 1.46 rHCl
Thus ammonia will diffuse 1.46 times faster than hydrogen chloride gas
Illustration 10. A certain gas diffuses from two different vessels A and B. The vessel A has a
circular orifice while vessel B has a square orifice of length equal to the radius of
the orifice of vessel A. The ratio of the rates of diffusion of the gas from vessel A
to vessel B, assuming same temperature and pressure is;
(A)
(B) 1/
(C) 1:1
(D) 2:1
Solution:
The rate of diffusion is directly proportional to the area of orifice.
dA r2
dB r2
dA
dB

=
Hence, (A) is correct.
Exercise 6.
A gaseous mixture of O2 and X containing 20% of X diffused through a small hole in
134 secs while pure O2 takes 124 secs to diffuse through the same hole. Find
molecular weight of X?
Exercise 7.
A straight glass tube of 200 cm length has two inlets X and Y at the two ends. HCl gas
through inlet X and NH3 gas through inlet Y are allowed to enter in the tube at the
same time and same pressure. White fumes first appear at a point P inside the tube.
Calculate the distance of P from X.
o
Exercise 8. For 10 minute each, at 27 C, from two identical holes nitrogen and an
unknown gas are leaked into a common vessel of 3 L capacity. The resulting pressure
is 4.18 bar and the mixture contains 0.4 mole of nitrogen. What is the molar mass of
unknown gas?

KINETIC MOLECULAR THEORY OF GASES

The various gas laws which we have studied so far were based an experimental facts.
The theory that provides an explanation for the various experimental observations about a gas is
based on the Kinetic Molecular model. It is assumed that all gases are made up of identical
molecules, which are in a state of constant random motion.
Postulates of the Model
The following are the postulates of the model:

A gas consists of a large number of identical molecules of mass m. The dimensions of

these molecules are very small compared to the space between them. Hence the
molecules are treated as point masses.

There are practically no attractive forces between the molecules. The molecules
therefore more independently.

The molecules are in a state of ceaseless and random motion, colliding with each other
and with the walls of the container. The direction of their motions changes only on
collision. These collisions are known as elastic collisions in which the energy and
momenta of the molecules are concerned. In nonelastic collisions these quantities are
not conserved.

The pressure of a gas is the result of collision of molecules with the wall of the container.

The average kinetic energy of the colliding molecules is directly proportional to its
temperature.

Velocities of gas molecules

Average Velocity
As per kinetic theory of gases, each molecule is moving with altogether different velocity. Let n
molecules be present in a given mass of gas, each one moving with velocity v 1, v2, v3 vn. The
average velocity or UaV = average of all such velocity terms.
n1u1 n2u2 n3u3
u1 u2 u2 L un
n1 n2 n3 L
n
Average velocity =
=
8RT
8PV
8P

M
b
Uav =
Root Mean Square Velocity
Maxwell proposed the term Urms as the square root of means of square of all such velocities.
n1u12 n2u22 n3u32
u12 u22 u32 L
2
Urms
n1 n2 n3
n
=
=
3RT
3PV
3P

M
M
d
Also Urms
Most probable velocity
It is the velocity which is possessed by maximum no. of molecules.
2RT
2PV
2P

M
M
d
Vmp =
2RT
8RT
M
Furthermore UMP : UAV : Urms : :
: M :
8
2: : 3

3RT
M

1 : 1.128 : 1.224
Also Uav = Urms 0.9213
Exercise 9.
The average speed at T1k and the most probable speed at T2k of CO2 gas is 9 104 cm
sec1. Calculate the value of T1 and T2.
Exercise 10.

Calculate the rms speed of ozone kept in a closed vessel at 20C and 82 cm Hg
pressure.
Exercise 11.
Calculate total energy of one mole of an ideal monatomic gas at 27 C?
Exercise 12.
Calculate Vrms of O2 at 273K and 1 105 Pa pressure. The density of O 2 under these
conditions is 1.42 kg/m3?
Exercise 13.
Calculate the average and total kinetic energies of 0.6 mol of an ideal gas at 0C.
Kinetic Energy of Gas
1
2
mnurms
3
As per kinetic equation PV =
For 1 mole m n = Molecular Mass (M)
1 2
2 1
2
K.E. 3
Murms Mu2 K.E./ mole
RT
3 2
3
PV = 3
or mole 2
.
K.E.
3 RT
3

molecule
2
N
Also
= 2 kT
R

k N

Where k is the Boltzmann constant

Illustration 11. Calculate rms speed of O2 at 273 K and 1 105Pa pressure. The density of O 2
under these conditions is 1.42 kg m3.
Solution:

Data are given in SI units

3P
3 105

d
1.42 = 459.63 m sec1

Illustration 12. At what temperature will the r.m.s. velocity of oxygen be one and half times of its
value at N.T.P.?
Solution:

1
3
mc 2 KT
2
2
3
Suppose the temperature required is T when the velocity will be 2 C

3 / 2C

T
9
T or, T = 4 273 = 614.25K

Kinetic Gas Equation

The kinetic gas equation has been derived on the basis of postulates of the kinetic theory of
gases. The equation is:
1
nNu2
PV = 3
Where,

P = Pressure
V = Volume

m = Mass of gas molecules

N = No. of molecules in Volume V
u = Root mean square velocity (rms)
Kinetic Energy of Gas Molecules
The kinetic energy of gas molecules can be easily calculated on the basis of kinetic gas equation.
1
Average K.E. of one molecule = 2 mu2
3 1
1
mNu2
2
2
3
2
K.E. of N molecules = mNu =
1
3
3
2
PV RT Q PV mNu
3
2
= 2
[
and PV = RT fare 1 mole of gas]
R is constant
KE of a gas T
Illustration 13. Calculate K.E. of 4g of N2 at 13C.
Solution:

3 4
3

8.314 260
nRT
2
EK =
= 2 28
= 463.21 J mol1

Exercise 14.
At what temperature rms of SO2 is equal to that of CH4 at 27C
DEVIATION FROM IDEAL GAS BEHAVIOUR
A gas which obeys the gas laws and the gas equation PV = nRT strictly at all temperatures and
pressures is said to be an ideal gas. The molecules of ideal gases are assumed to be volume
less points with no attractive forces between one another. But no real gas strictly obeys the gas
equation at all temperatures and pressures. Deviations from ideal behaviour are observed
particularly at high pressures or low temperatures. The deviation from ideal behaviour is
expressed by introducing a factor Z known as compressibility factor in the ideal gas equation. Z
PV
Z
nRT
may be expressed as

In case of ideal gas, PV = nRT Z = 1

In case of real gas, PV nRt Z 1
Thus in case of real gases Z can be 1 or 1
(i) When Z 1, it is a negative deviation. It shows that the gas is more compressible
than expected from ideal behaviour.
(ii) When Z 1, it is a positive deviation. It shows that the gas is less compressible than
expected from ideal behaviour.

Causes of deviation from ideal behaviour

The causes of deviations from ideal behaviour may be due to the following two assumptions of
kinetic theory of gases. There are

The volume occupied by gas molecules is negligibly small as compared to the volume
occupied by the gas.
The forces of attraction between gas molecules are negligible.

The first assumption is valid only at low pressures and high temperature, when the volume
occupied by the gas molecules is negligible as compared to the total volume of the gas. But at
low temperature or at high pressure, the molecules being in compressible the volumes of
molecules are no more negligible as compared to the total volume of the gas.
The second assumption is not valid when the pressure is high and temperature is low. But at high
pressure or low temperature when the total volume of gas is small, the forces of attraction
become appreciable and cannot be ignored.
Van Der Waals Equation
The general gas equation PV = nRT is valid for ideal gases only Van der Waal is 1873 modified
the gas equation by introducing two correction terms, are for volume and the other for pressure to
make the equation applicable to real gases as well.
Volume correction
Let the correction term be v
Ideal volume vi = (V v)
Now v n or v = nb
[n = no. of moles of real gas; b = constant of proportionality called Van der Waals constant]
Vi = V nb
b = 4 volume of a single molecule.
Pressure Correction
Let the correction term be P
Ideal pressure Pi = (P + p)
2

n
an2
P
V = V 2
Now,
Where a is constant of proportionality called another Van der Waals constant.
Hence ideal pressure

an2
P

V2
Pi =
Here,
n = Number of moles of real gas
V = Volume of the gas
a = A constant whose value depends upon the nature of the gas
Substituting the values of ideal volume and ideal pressure, the modified equation is obtained as

an2
P 2 (v nb) nRT
V

Illustration 14. 1 mole of SO2 occupies a volume of 350 ml at 300K and 50 atm pressure.
Calculate the compressibility factor of the gas.
Solution:

P = 50 atm
n = 1 mole

V = 350 ml = 0.350 litre

PV
nRT
T = 300L Z =

50 0.350
0.711
Z = 1 0.082 300
Thus SO2 is more compressible than expected from ideal behaviour.
Exercise 15.
Out of NH3 and N2 which will have
(a) Larger value of a
(b) Larger value of b
Exercise 16.
2 moles of NH3 occupied a volume of 5 litres at 27C. Calculate the pressure if the gas
obeyed Vander Waals equation. Given a = 4.17 litre2 atm mole2, b = 0.037 litre/mole.
Exercise 17.
Calculate the percentage of free volume available in 1 mole gaseous water at 1 atm
o
H O at 100o C
and 100 C. Density of liquid 2
is 0.958g / mL.
Vander Waals equation, different forms

At low pressures: V is large and b is

negligible in comparison with V. The Vander
Waals equation reduces to:
a

a
P 2 V RT
V

V
;
PV +
= RT

N2

Ideal gas

H2
He

a
0
P
PV = RT - V or PV RT
Deviation of gases from ideal behaviour
This accounts for the dip in PV vs P isotherm
with pressure.
at low pressures.
0C
At fairly high pressures
50 C
100C
a
100C
2
V may be neglected in comparison with P.

The Vander Waals equation becomes

P (V b) = RT
PV Pb = RT
PV = RT + Pb or PV RT
This accounts for the rising parts of the PV vs P
isotherm at high pressures.

CO2

Z
1.0

0C

50C

P (atm)

The plot of Z vs P for N2 gas at

different temperature is shown here.

a
2
At very low pressures: V becomes so large that both b and V become negligible and
the Vander Waals equation reduces to PV = RT. This shows why gases approach ideal
behaviour at very low pressures.
Hydrogen and Helium: These are two lightest gases known. Their molecules have very
small masses. The attractive forces between such molecules will be extensively small. So
a
V 2 is negligible even at ordinary temperatures. Thus PV RT. Thus Vander Waals

equation explains quantitatively the observed behaviour of real gases and so is an

improvement over the ideal gas equation.

Vander Waals equation accounts for the behaviour of real gases. At low pressures, the
gas equation can be written as,

a
Vm
a
1
P 2 Vm RT
V
RT
VmRT

m
or Z =
Where Z is known as compressibility factor. Its value at low pressure is less than 1 and it
decreases with increase of P. For a given value of V m, Z has more value at higher
temperature.
At high pressures, the gas equation can be written as
P (Vm b) = RT
PVm
Pb
Z = RT = 1 + RT
Here, the compressibility factor increases with increase of pressure at constant
temperature and it decreases with increase of temperature at constant pressure. For the
gases H2 and He, the above behaviour is observed even at low pressures, since for
these gases, the value of a is extremely small.
Illustration 15. One litre of a gas at 300 atm and 473 K is compressed to a pressure of 600 atm
and 273 K. The compressibility factors found are 1.072 & 1.375 respectively at
initial and final states. Calculate the final volume.
Solution:

P1V1 = Z1nRT1 and P2V2 = Z2nRT2

P2 V2
T
Z
1 2
T2
P1V1 Z1
or V2 =

Z2 T2 P1V1

Z1 T1 P2

1.375 273 300 1

600 = 370.1 ml
= 1.072 473

Illustration 16. The behaviour of a real gas is usually depicted by plotting compressibility factor
Z versus P at a constant temperature. At high temperature and high pressure, Z
is usually more than one. This fact can be explained by van der Waals equation
when
(A) the constant a is negligible and not b
(B) the constant b is negligible and not a
(C) both constants a and b are negligible
(D) both the constants a and b are not negligible.

P
Solution:

n2 a

V 2

(V nb) = nRT
At low pressures, b can be ignored as the volume of the gas is very high. At high
temperatures a can be ignored as the pressure of the gas is high.
P (Vb) = RT
PV - Pb = RT PV = RT + Pb

PV
Pb
Z 1
RT
RT

Hence, (A) is correct.

Exercise 18.

The compressibility factor for CO 2 at 273K and 100atm pressure is 0.2005.

Calculate the volume occupied by 0.2 mole of CO 2 gas at 100 atm assuming
ideal behaviour

Some other important definitions

Relative Humidity (RH)
At a given temperature it is given by equation
partial pressure of water in air
vapour pressure of water
RH =
Boyles Temperature (Tb)
Temperature at which real gas obeys the gas laws over a wide range of pressure is called Boyles
Temperature. Gases which are easily liquefied have a high Boyles temperature [T b(O2)] = 46 K]
whereas the gases which are difficult to liquefy have a low Boyles temperature [T b(He) = 26K].
a
1
Ti
Boyles temperature Tb = Rb 2
where Ti is called Inversion Temperature and a, b are called van der Waals constant.

Critical Constants
Critical Temperature (Tc): It (Tc) is the maximum temperature at which a gas can be
liquefied i.e. the temperature above which a gas cant exist as liquid.
8a
Tc
27Rb
Critical Pressure (Pc): It is the minimum pressure required to cause liquefaction at Tc
a
Pc
27b2
Critical Volume: It is the volume occupied by one mol of a gas at Tc and Pc
Vc = 3b
Molar heat capacity of ideal gases: Specific heat c, of a substance is defined as the amount of
heat required to raise the temperature of is defined as the amount of heat required to raise the
temperature of 1 g of substance through 1 0C, the unit of specific heat is calorie g -1 K-1. (1 cal is
defined as the amount of heat required to raise the temperature of 1 g of water through 1 0C)
Molar heat capacity C, is defined as the amount of heat required to raise the temperature of 1
mole of a gas trough 10C. Thus,
Molar heat capacity = Sp. Heat molecular wt. Of the gas
For gases there are two values of molar heats, i.e., molar heat at constant pressure and molar
heat at constant molar heat at constant volume respectively denoted by C p and Cv. Cp is greater
than Cv and Cp-R = 2 cal mol-1 K-1.
From the ratio of Cp and Cv, we get the idea of atomicity of gas.
For monatomic gas Cp = 5 cal and Cv =3 cal

4
1.67
3

is poisson' s ratio

for diatomic gas Cp = 7 cal and Cv = 5 cal

7
1.40
5

Cp

Cv

For polyatomic gas Cp = 8 cal and Cv= cal

8
1.33
6
C Cv m
also Cp = Cp m, v
where, Cp and Cv are specific heat and m, is molecular weight.
Illustration 17. Calculate Vander Waals constants for ethylene
TC = 282.8 k; PC = 50 atm

Solution:

b=

1 RTC 1 0.082 282.8

8 PC
8
50

= 0.057 litres/mole
27 2 TC2 27
1
R

(0.082)2 (282.8)2
64
PC 64
50
a=
= 4.47 lit2 atm mole2

Gas Eudiometry
The relationship amongst gases, when they react with one another, is governed by two laws,
namely Gay-Lussac law and Avogadros law.
Gaseous reactions for investigation purposes are studied in a closed graduated tube open at one
end and the other closed end of which is provided with platinum terminals for the passage of
electricity through the mixture of gases. Such a tube is known as Eudiometer tube and hence
the name Eudiometry also used for Gas analysis.
During Gas analysis, the Eudiometer tube filled with mercury is inverted over a trough containing
mercury. A known volume of the gas or gaseous mixture to be studied is next introduced, which
displaces an equivalent amount of mercury. Next a known excess of oxygen is introduced and the
electric spark is passed, whereby the combustible material gets oxidised. The volumes of carbon
dioxide, water vapour or other gaseous products of combustion are next determined by absorbing
them in suitable reagents. For example, the volume of CO 2 is determined by absorption in KOH
solution and that of excess of oxygen in an alkaline solution of pyrogallol. Water vapour produced
during the reaction can be determined by noting contraction in volume caused due to cooling, as
by cooling the steam formed during combustion forms liquid (water) which occupies a negligible
volume as compared to the volumes of the gases considered. The excess of oxygen left after the
combustion is also determined by difference if other gases formed during combustion have
already been determined. From the data thus collected a number of useful conclusions regarding
reactions amongst gases can be drawn.

Volume-volume relationship amongst Gases or simple Gaseous reactions.

Composition of Gaseous mixtures.
Molecular formulae of Gases.
Molecular formulae of Gaseous Hydrocarbons.
The various reagents used for absorbing different gases are
O3 turpentine oil

O2 alkaline pyrogallol
NO FeSO4 solution
CO2,SO2 alkali solution (NaOH, KOH, Ca(OH)2, HOCH2CH2NH2, etc.)
NH3 acid solution or CuSO4 solution
Cl2
water
Equation for combustion of hydrocarbons.
y

y
x 4

2
CxHy +
O2

xCO2 + H2O
General Assumptions: In all problems, it is assumed that the sparking occurs at room
temperature. This implies that water formed would be in liquid state and that nitrogen gas is inert
towards oxidation.
Illustration 18. A gaseous hydrocarbon requires 6 times its own volume of O 2 for complete
oxidation and produces 4 times its volume of CO2. What is its formula?
Solution:

The balanced equation for combustion

y

y
x 4

2
CxHy +
O2

xCO2 + H2O

y
y
x
x 4

4
1 volume
volume
= 6 (by equation)
or 4x + y = 24
(1)
Again x = 4 since evolved CO2 is 4 times that of hydrocarbon
16 + y = 24 or y = 8 formula of hydrocarbon C4H8

Exercise 19.
50 ml. of a mixture of NH 3 and H2 was completely decomposed by sparking into
nitrogen and hydrogen. 40 ml of oxygen was then added and the mixture was
sparked again. After cooling to the room temperature the mixture was shaken with
alkaline pyrogallol and a contraction of 6 ml. was observed. Calculate the % of NH 3 in
the original mixture. (Assuming that nitrogen does not react with oxygen.)
Exercise 20.
When a certain quantity of oxygen was ozonised in suitable apparatus, the volume
decreases by 4 ml. On addition of turpentine oil, the volume decreases further by 8
ml. Find the formula of ozone, assuming all measurements are made at same
temperature and pressure.
Liquid State

The liquid molecules are relatively close together.

The intermolecular forces of attraction in case of liquids are much larger than in gases.
Unlike gases, liquids have a definite volume although no definite shape (similarity with
gases).
The molecules are in constant random motion.
The average kinetic energy of molecules in a given sample is proportional to the absolute
temperature.

(T )
Guldbergs rule: Normal boiling point b of the liquid is nearly two third of its critical
(T )
temperature c when both are expressed on the absolute scale.
2
Tb Tc
3
Troutons rule : The molar heat of vapourisation of a liquid expressed in Joules divided
by normal b.p. of the liquid on the absolute scale is approximately equal to 88.
H
88 JK 1 mol1
Tb.p.

Evaporation
Evaporation is the spontaneous change in which a liquid changes into vapours at the surface of
liquid. Evaporation occurs at all temperatures. Evaporation increases with increase in surface
area, increase in temperature and decrease in intermolecular attractive forces. In contrast to
evaporation, boiling takes place at a definite temperature and it involves bubble formation below
the surface. Evaporation produces cooling.
Vapour Pressure
Vapour pressure of a liquid at any temperature may be defined as the pressure exerted by the
vapour present above the liquid in equilibrium with the liquid at that temperature. The magnitude
of vapour pressure depends upon the nature of liquid and temperature.
Non polar or less polar liquids have fairly high vapour pressure due to weak forces of
CCl4 , CHCl3 ,
attraction (e.g.
ether etc). Polar liquids (e.g. water, alcohols, etc.) have
lower vapour pressure because of strong dipole dipole interaction between their
molecules.
Vapour pressure of a liquid is constant at a given temperature. Further the vapour
pressure of a liquid increases with increase in temperature. When the vapour pressure of
the liquid is equal to the external pressure (normal pressure or 1 atm pressure) acting
upon the surface of the liquid, the bubbles increase in size and escape freely; the
temperature at which this happens is called the boiling point of the liquid. In case the
external pressure is more than the atmospheric pressure, more heat will be required to
make the vapour pressure equal to the external pressure and hence higher will be the
boiling point. In case, the external pressure is low (as on the top of a mountain), the
boiling point of the liquid decreases. This explains why a liquid boils at a lower
temperature on the top of a mountain. (where pressure in low) than on the sea shore.
Substances having high vapour pressure (e.g. petrol) evaporate more quickly than
substances of low vapour pressure. (e.g. motor oil)
SURFACE TENSION
Surface Tension of a liquid is defined as the force acting at right angles to the surface along one
centimeter length of the surface. It is represented by the greek letter gamma, .
Due to surface tension molecules tend to leave the surface, i.e. the surface of the liquid tends to
contract to the smallest possible area for a given volume of the liquid. Further for a given volume
of the liquid, sphere has the minimum surface area. This explains why the drops of a liquid are
spherical.
Thus it is apparent that in order to increase its surface area, force must be exerted to overcome
the surface tension. In other words, work will have to be done to increase the surface area. Thus

the surface tension of a liquid is defined as the work (energy) required to expand the surface of a
liquid by unit area. Mathematically,
Work done
surface tension
Change in area
Thus surface tension of a liquid may also be defined as the force in dynes necessary to rupture
its surface along one centimeter length. In SI units it is defined as the force in newtons
(1 newton 105 dynes)
required to rupture 1 meter length of the surface of a liquid.
1
Thus the units of surface tension are dynes per cm (or Newtons per metre, Nm in SI system).
Surface tension of a liquid is measured with the help of apparatus called stalgmometer.
The surface tension of a liquid decreases with increase of temperature and becomes zero at its
critical temperature (where the surface of separation between liquid and its vapour disappears).
The decrease in surface tension with increase of temperature is due to the fact that with increase
of temperature, the kinetic energy of the molecule (and hence the speed of molecules) increases
and hence the intermolecular forces of attraction decreases.
Surface tension in everyday life

Cleansing action of soap and detergents is due to their property of lowering the interfacial
tension between water and greasy substances. Thus soap solution due to its lower
surface tension can penetrate into the fibre to surround the greasy substances and wash
them away.
Efficacy of tooth pastes, mouth washes and nasal jellies is partly due to the presence of
substances having lower surface tension. Lowering of interfacial tension helps these
preparations to spread evenly over the surface with which they come in contact thereby
increasing the efficiency of their antiseptic action.

VISCOSITY
It is well known that all liquids do not flow with the same speed. Some liquids like water, alcohol,
ether, etc. flow very rapidly while some one like glycerine, honey, castor oil, etc. flow slowly. This
indicates that every liquid has some internal resistance to flow. This internal resistance to flow
possessed by a liquid is called its viscosity. The liquids which flow rapidly have low internal
resistance and hence are said to be low viscous, i.e. their viscosity is less. On the other hand, the
liquids which flow slowly have high internal resistance and hence are said to be more viscous, i.e.
their viscosity is high. This force of friction which one part of the liquid offers to another part of the
liquid is called Viscosity.
Coefficient of viscosity may be defined as the force per unit area required to maintain unit
difference of velocity between two parallel layers in the liquid, one unit apart. In C.G.S. units, it is
(dynes cm2 )
expressed in dynes per square centimeter
. This unit is called a poise after the name
of Poiseulle who pioneered the study of viscosity. Low values of viscosity are expressed in
(102 poise)
(103 poise)
Nm2 s
centipoise
and millipoise
. In S.I. units, viscosity is expressed in
or Pas (pascal second).
1 Pas 1 Nm2 s (105 dyn)(10 2 cm)2 s 10 dyn cm 2 s 10 poise
The viscosity of a liquid generally decreases with rise in temperature. With increase of
temperature, the kinetic energy of the molecules of the liquid increases and hence the liquid
starts flowing faster, i.e. the viscosity decreases. The decrease in viscosity is found to be about
3% per degree rise of temperature.

Viscosity in everyday life

Lubricating oils are graded according to their viscosity. A good quality or all weather

lubricating is one whose viscosity and hence lubricating property does not change much
with increase of temperature. Such oils are obtained by adding long chain coiling
polymers to the oil. As temperature rises, the polymer particles tend to uncoil and thus
increase the viscosity of oil thereby compensating for the decrease of viscosity of the oil
with rise of temperature.
The condition of high blood pressure and thus strain on heart may also be explained on

the property of viscosity. In arteriosclerosis (hardening of arteries), arterial walls contract

resulting in decrease of diameter of capillaries.
Narrow capillaries offer resistance to the flow of blood due to viscosity with the result
greater force is needed to make blood flow through capillaries. This results in a condition
of high blood pressure and strain on the heart.
The increased blood circulation required during fever is supplied by its temperature

dependence properly (recall that rise of every degree centigrade temperature decreases
viscosity of blood by about 3%). Thus lowering of viscosity results in a more rapid flow of
blood without any extra strain on the heart.
In case of asphyxia, concentration of CO in blood increases resulting in swelling of
corpuscles which then increases viscosity of blood.

LIQUID CRYSTALS
In a temperature range just above the melting point, crystals of certain substances can exist in a
definite pattern as in solid but can flow like a liquid. Such crystals are called liquid crystals. When
white light falls on a liquid crystal, it reflects only one colour, and as the temperature is changed it
reflects different light. Thus, liquid crystals can be used to detect even small temperature
changes. There are two important types of liquid crystals namely nematic liquid crystals (needle
like) and smectic (soap like) liquid crystals.
THE SOLID STATE
Solids are rigid and have definite shapes. They also possess definite volume which is
independent of the volume of the vessel.
Solids possess higher densities and are almost incompressible. All these characteristic properties
are due to strong inter particle forces, smaller inter particle spares and restricted motion of the
particles in the solid state.
The temperature at which solid changes into liquid state at normal pressure is called melting point
of the substance.
Classification of Solids
Solids are broadly classified into two types: Crystalline solids and amorphous solids.
Distinction between Crystalline and Amorphous Solids
Sl. No.

Crystalline Solids

Amorphous Solids

1.

The internal arrangement of particles is

The internal arrangement of particles is not

regular

regular

2.

They have sharp melting points.

They do no have sharp melting points.

3.

Crystalline solids are regarded as true solid

Amorphous solids are regarded as super

cooled liquids or pseudo solids

4.

Crystalline solids give a regular cut when cut

with sharp edged knife.

Amorphous solids give irregular cut.

5.

Crystalline solids are anisotropic.

Amorphous solids are isotropic.

Classification of Solids based on different binding forces

Crystalline solids can be classified into different categories depending upon the type of
constituent particles and the nature of attractive forces operating between them. Various
categories are:

Atomic solids
Molecular solids
Ionic solids
Covalent solids
Metallic solids

Atomic Solids
In these solids the constituent particles are atoms. These closely packed atoms are held up by
London dispersion forces. Some examples are crystals of noble gases. Such solids are very soft,
possess very low melting points and poor conductors of heat and electricity.
Molecular Solids
In these solids, the constituent particles which pack up together are molecules of the substance.
CO2 , I2 , CCl4
These molecules may be non polar (dipole moment = 0) such as
etc. or they may
H2O, HCl, HF
be polar (dipole moment > 0) like
etc.
In case of non polar molecules, the attractive forces operating between the molecules

are Vander Waal forces (also called dispersion forces). The example of such solids are :
CO2 )
dry ice (Solid
, iodine (crystals).
In case of polar molecules, the attractive forces operating between the molecules in solid
SO 2
state are dipole dipole forces. The examples of such solids are : solid
, solid HCl.
However, in some solids with polar molecules, the interparticle forces are hydrogen
(H O)
bonds. The examples of such solids are ice 2 ; solid hydrogen fluoride (HF); solid
(NH3 )
ammonia
, etc.

Characteristics of Molecular Solids

Some of the general characteristics of molecular solids are :
They are generally soft.

Their melting points are low to moderately high. The melting points of solids with non
polar molecules are relatively low whereas solids with polar molecules have moderately

high melting points.

They are generally bad conductors of heat and electricity.

They have generally low density.

Ionic Solids
In ionic solids, the constituent particles are ions of opposite charges. Each ion is surrounded by a
definite number of ions of opposite charge. The number of ions that surround a particular ion of
opposite charge its called co ordination number of the ion. For example, in sodium chloride
(Na )
crystal each sodium ion
is surrounded by six chloride ions. Hence coordination number of
Na is 6. At the same time each chloride ion is surrounded by six Na ions. Therefore the co

2
ordination number of Cl ion is also 6. However, in calcium fluoride crystal each Ca ion is

(F)
surrounded by eight fluoride
ions and each F ion is surrounded by four Ca ions. Thus, in
2

CaF2
crystal co ordination numbers of Ca
and F ions are respectively 4 and 8. The

interparticle forces in ionic solids are ionic bonds operating between the ions of opposite charges
some examples of ionic solids are : sodium chloride (NaCl) ; ceasium chloride (CsCl), zinc
(CaF2 )
sulphide (ZnS), calcium fluoride
, etc.
Characteristics of Ionic Solids
Some common characteristics of ionic solids are as follows:
They are hard, brittle and have low volatility.

They have high melting points.

They are poor conductors of electricity in solid state, however they become good

conductors of electricity in molten state or in dissolved state.

They are generally soluble in polar solvents like water.

Covalent Solids
In these types of solids the constituent particles are atoms of same or different elements
connected to each other by covalent bond network. For example, in diamond only carbon atoms
constitute the covalent network while carborundum covalent bond network is constituted by
silicon and carbon atoms. Obviously, the interparticle forces operating in these solids are covalent
bonds. These solids are also called network solids because the covalent bonds extend in three
dimensions forming a giant interlocking structure. Some examples of covalent solids are :
Diamond, silicon carbide, aluminium nitrite etc.
Characteristics of Covalent Solids
Some common characteristics of covalent solids are :
They are very hard. Diamond is the hardest naturally occurring substance.
They have very high melting points.
They are poor conductors of heat and electricity.
They have high heats of fusion.
Metallic Solids

In these type of solids, the constituent particles are metal atoms. The interparticle forces in these
solids are metallic bonds. In the metallic crystals the metal atoms occupy the fixed positions but
their valence electrons are mobile. The close packed assembly of metal kernels (part of metal
atom without valence electrons) remain immersed in the sea of mobile valence electrons. The
attractive force between the kernels and mobile valence electrons is termed as metallic bond.
Characteristics of Metallic Solids
The common characteristics of metallic solids are as follows:
They generally range from soft to very hard.
They are malleable and ductile.
They are good conductors of heat and electricity.
They possess bright lustre.
They have high melting and boiling points.
They have moderate heats of fusion.
The summary of classification of solids on the basis of interparticle forces is given in
Classification of Solids on the Basis of Binding Forces
Crystal
Classification
Atomic

Unit Particles

Binding Forces

Properties

Examples

Atoms

London
dispersion forces

Noble gases

Molecular

Polar or
non polar
molecules

Ionic

Positive and
negative ions

Vander Waals
forces (London
dispersion, dipole
dipole forces
hydrogen bonds)
Ionic bonds

Soft, very low melting,

poor thermal and electrical
conductors
Fairly soft, low to
moderately high melting
points, poor thermal and
electrical conductors

NaCl, ZnS

Covalent

Atoms that are

connected in
covalent bond
network
Cations in
electron cloud

Hard and brittle, high

melting points, high heats
of fusion, poor thermal and
electrical conductors
Very hard, very high
melting points, poor
thermal and electrical
conductors
Soft to very hard, low to
very high melting points,
excellent thermal and
electrical conductors,
malleable and ductile

Metallic

Covalent bonds

Metallic bonds

Dry ice (solid

CO2 )
,
methane
(CH4 )

Diamond,
quartz,
silicon
All metallic
elements, for
example,
Cu, Fe, Zn

Intermolecular forces or Vander Waals forces

Intermolecular forces or vander Waals forces originate from the following three types of
interactions.
Dipole Dipole interactions: In case of polar molecules, the vander waals forces are
mainly due to electrical interaction between oppositively charged ends of molecules (Fig.
NH3 ,SO2 , HCl, HF
1. a) called dipole dipole interactions. For example, gases such as

etc.have permanent dipole moments as a result of which there is appreciable dipole

dipole interactions between the molecules of these gases. The magnitude of this type of
interaction depends upon the dipole moment of the molecule concerned. Evidently,
greater the dipole moment, stronger is the dipole dipole interactions. Because of these
attractive forces, these gases can be easily liquefied.

(a )
D ip o le - D ip o le In te r a c tio n s

(b )
D ip o le - in d u c e d d ip o le

(c )
M o m e n t a r y d ip o le - in d u c e d d ip o le

F ig . 1 . In te r a c tio n s b e tw e e n m o le c u le s

Dipole Induced dipole Interactions: A polar molecule may sometimes polarize a non

polar molecule which lies in its vicinity and thus induces polarity in that molecule just as
a magnet induces magnetic polarity in a neutral piece of iron lying close by. The induced
dipole then interacts with the dipole moment of the first molecule and thereby the two
molecules are attracted together (Fig. 1. b). The magnitude of this interaction, evidently
depends upon the magnitude of the dipole moment of the polar molecule and the
HO
polarizability of the non polar molecule. The solubility of inert gases in 2 increases
from He to Rn due to a corresponding increase in magnitude of the dipole induced
dipole interactions as the polarizability of the noble gas increases with increase in size
from He to Rn.
Momentary dipole induced dipole interactions: The electrons of neutral molecules
keep on oscillating w.r.t. the nuclei of atoms. As a result, at a given instant, one side of
the molecule may have a slight excess of electrons relative to the opposite side. Thus a
non polar molecule may become momentarily self polarized. This polarized molecule
may induce a dipole moment in the neighbouring molecule. These two induced dipoles
then attract each other (Fig. 1. c). These momentary dipole induced dipole attractions
are also called London forces or dispersive forces. The magnitude of these forces
depends upon the following:
(i) Size or molecular mass: The melting points and boiling points of non polar
molecules increase as the size or molecular mass of the molecule increases. For
example, the m.p. and b.p. of alkanes, halogens, noble gases etc. increase as the
molecular mass of the molecule increases.
(ii) Geometry / Shape : For example, isomer n pentane has higher boiling point than
neo pentane because the former is zig zag chain with larger sites of contact and
hence large intermolecular forces whereas the latter is nearly spherical and hence
has less contact and weaker forces.

Illustration19. Why liquid have a definite volume but no definite shape?

Solution:

This is because the intermolecular forces are strong enough to hold the molecule
together but not so strong as to fix them into definite positions (as in solids)
instead, they possess fluidity and hence no definite shape.

Illustration 20. Why diethyl ether has higher vapour pressure than ethyl alcohol at a particular
temperature.
Solution:

This is because the intermolecular forces of attraction in ethyl alcohol are

stronger than those present in diethyl ether.

Illustration 21. Why falling liquid drops are spherical?

Solution:

This is due to property of surface tension possessed by the liquids. This makes
the surface area minimum. For a given volume, sphere has minimum surface
area.
Illustration 22. Arrange the following in the increasing order of melting point of different types of
crystalline solid (i) Covalent solid (ii) Metallic solid (iii) Molecular (iv) Ionic
Solution:

Covalent > Ionic > Metallic > Molecular

( : 4000 K) ( : 1000 K) (880 1000 K) (273 600 K)

Illustration 23. Explain

(a) Aerated water bottles are kept under water during summer.
(b) A bottle of liquid ammonia is cooled before opening the seal.
Solution:

CO2
(a) Aerated water contains
dissolved in water. The solubility of gas
decreases with rise in temperature. Thus, amount of free gas in bottle can
increase in summer, which may result into increase in pressure and
ultimately explosion. To avoid this they are kept under cold water.
(b) If the seal is opened without cooling, the liquid ammonia suddenly vaporises
and gushes out of the bottle with force. This may lead to serious accidents.

ANSWER TO EXERCISE
Exercise 1:
2.5 L
Exercise 2:
730.8 K
Exercise 3:
39.4C
Exercise 4:
He = 25.24%, Ar = 74.76%
Exercise 5:
PO2

= 1.12 atm ,

PN2

= 0.448 atm, Ptotal = 1.568 atm

Exercise 6:
58.84
Exercise 7:
81.1 cm
Exercise 8:
428
Exercise 9:
T1 = 168K, T2 = 2143 K
Exercise 10:
3.9 104 cm/sec
Exercise 11:
900 cal
Exercise 12:
459.63 m / s
Exercise 13.
3.405 KJ mol1 , 2.043 KJ
Exercise 14.
927C
Exercise 15:
(a)
(b)

NH3
N2

Exercise 16:
9.33 atm
Exercise 17:
99.9386%
Exercise 18.
Exercise 19:
72%
Exercise 20:
O3

0.448 litre

MISCELLANEOUS EXERCISES
Exercise 1:

Can we apply Daltons law of partial pressure to a mixture of carbon monoxide

and oxygen?

Exercise 2:

Out of dry air and wet air, which is heavier?

Exercise 3:

Carbon dioxide is heavier than oxygen and nitrogen but is does not form the
lower layer of the atmosphere. Explain.

Exercise 4:

The molecular speeds of gaseous molecules are analogous to those of rifle

bullets; why is then odour of the gaseous molecule not detected so fast?

Exercise 5:

Liquid ammonia bottle is cooled before opening the seal. Explain.

Exercise 6:

100 mL of O2 is collected over water at 23oC and 800 torr. Compute the standard
volume of the dry oxygen. Vapour pressure of water at 23oC is 21.1 torr.

Exercise 7:

A mixture of Na2CO3 and NaHCO3 has a mass of 22 g. Treatment with excess

HCl solution liberates 6 L of CO 2 at 25oC and 0.947 atm pressure. Determine the
percent Na2CO3 in the mixture.

Exercise 8:

A jar contains a gas and a few drops of water. The pressure in the jar is 830 mm of
Hg. The temperature of the jar is reduced by 1%. The vapour pressure of water at
two temperatures are 30 and 25 mm of Hg. Calculate the new pressure in the jar.

Exercise 9:

The radius of an Xe atom is 1.3 10-8 cm. A 100 cm3 container is filled with Xe at a
pressure of 1 atm and a temperature of 273 K. Calculate the fraction of the volume
that is occupied by Xe atoms.

Exercise 10:

A container with a volume of 3.0 L holds N 2 (g) and H2O (I) at 29 C. The pressure
is found to be 1 atm. The water is then split into hydrogen and oxygen by
electrolysis, according to the reaction, H2O (l)
H2 (g) +1/2 O2 (g). After the
reaction is complete, the pressure is 1.86 atm. What mass of water was present in
the container? The aqueous tension of water at 29 C is 0.04 atm.

ANSWERS TO MISCELLANEOUS EXERCISES

Exercise 1:

No, the law cannot be applied because these gases readily combine to form
carbon dioxide. The law can be applied only to the non reacting gases.

Exercise 2:

Dry air mainly contains gases like nitrogen and oxygen along with small
proportions of the other gases. The wet air on the other hand, also contains a
small amount of water vapours. Now, vapour density of water vapours is less
than that of nitrogen and oxygen because molar mass of H 2O (18) is less than
that of N2 (28) and O2 (32). This means that heavier molecules or nitrogen and
oxygen have been replaced by lighter molecules of water vapours when the dry
air changes to we air. Therefore, dry air is heavier than wet air.

Exercise 3:

The diffusion of the gases is quite independent of the gravitational pull by the
earth. The molecules of carbon dioxide remain distributed throughout the
atmosphere. Therefore, carbon dioxide does not form the lower layer of the
atmosphere.

Exercise 4:

No, doubt molecules in the gaseous state move almost with the speed of rifle
bullets. But these molecules donot follow a straight path. Since they undergo
collisions with one another at a very fast rate, the path followed is zig-zag. Thus,
the odour cannot be detected at the same speed at which the molecules move.

Exercise 5:

Liquid ammonia bottle contains the gas under very high pressure. If the bottle is
opened as such, then the sudden decrease in pressure will lead to a large
increase in the volume of the gas. As a result, the gas will come out of the bottle
all of a sudden with force. This will lead to the breakage of the bottle and also
some serious accident. However, if the bottle is cooled under tap water for
sometime, this will lead to a decrease in the volume of the gas to a large extent.
If the seal is opened now, the gas will come out of the bottle at a slow rate, thus,
reducing the chances of accident.

Exercise 6:

The gas collected is actually a mixture of oxygen and water vapour. The partial
pressure of water in the mixture at 23oC is 21.1 torr. Hence
Pressure of dry oxygen = (total pressure) (vapour pressure of water) = 799 torr
Thus, for the dry oxygen V1 = 100 mL, T1 = 23 + 273 = 296 K
P1 = 799 torr converting to STP
T P
273 K 779 torr

V2 V1 2 1 100 mL

94.5 mL
T
P
296
K
760
torr

1 2

Exercise 7:

Na2CO3 2HCl CO 2 H2O 2NaCl

NaHCO3 HCl CO2 H2 O NaCl
The number of mol of CO2 produced is equal to the number of mol of Na 2CO3
plus number of mol of NaHCO3 initially present
PV
0.947 6
n

0.232 mol
RT 0.0821 298
Let x = number of g Na2CO3, then 22 x = number of g NaHCO 3. The number of
mol of each reactant is equal to the mass divided by the formula weight.
x
22 x

0.232
106
84
x = 12.1 g
22 x = 9.9 g

Exercise 8:

12.1
100 55% Na2CO3
22
At T K
Pgas = Pdry gas + Pmoisture
Pdry gas = 830 30 = 800 mm

Now at new temperature

T1 T

T
0.99 T
100

P1 P2
P

constant
T
T2
Since, V1 = V2; T
or 1
800 0.99 T
Pdry gas
792 mm
T
Pgas = Pdry gas + Pmoisture
= 792 + 25
= 817 mm of Hg
Exercise 9:

We have
n = PV/RT =

(1atm) (100 x 10 -3dm3 )

= 4.467 x 10 -3 mol
(0.082atm dm 3 K -1 mol-1 (273K)

Now, the number of Xe atoms is N = nNA

= (4.467 10-3 mol)( 6.023 1023 mol-1) = 2.69 1021
Volume occupied by atoms = {4/3 r3} N
= 4/3 3.14 (1.3 10-8 cm)3 (2.69 1021) = 2.47 10-2 cm3
2.47x10 -2
=
x100 = 2.47x10 -2
100
Percentage of volume occupied by Xe atoms

Exercise 10:

nN2

0.96 3
0.116
0.0821 302

0.04 3
0.004834
0.0821 302
Total number of moles after water has split up
1.86 3
nT
0.225
0.0821 302
H and O2
Total number of moles of 2
that have come into vapour state
= 0.225 0.120834 = 0.104218
0.104218 2

0.06947
3
So, number of moles water initially
nH2O vapour

Mass of

H2O 0.06947 18 1.25 g

SOLVED PROBLEM
Subjective:
Board Type Questions
Prob 1.

A gas occupies one litre under atmospheric pressure. What will be the volume of the
same amount of gas under 750 mm of Hg at the same temperature?

Sol.

Given V1 = 1 litre
V2 = ?
P1V1 = P2V2
750
V2
1 1 = 760

P1 = 1 atm
750
P2 = 760 atm

V2 = 1.0133 litre = 1013.3 ml

Prob 2.

Sol.

How large a balloon could you fill with 4g of He gas at 22C and 720 mm of Hg?
720
Given, P = 760 atm, T = 295K, w = 4g
and m = 4 for He
w
RT
M
PV =
720
4
V
0.0821 295
760
= 4
V = 25.565 litre

Prob 3.

Calculate the temperature at which 28g N 2 occupies a volume of 10 litre at

2.46 atm

Sol.

w = 28g, P = 2.46 atm, V = 10 litre, m = 28

w
RT
Now, PV = M
(R = 0.0821 litre atm K1 mole1)
T = 299.6 K

Prob 4.

A gas occupies 300 ml at 27C and 730 mm pressure what would be its volume at
STP

Sol.

300
730
1000
V2 =
litre, P2 = 760 atm, T2 = 300K
At STP, V1= ? P1= 1 atm, T1= 273K
P2 V2 P1V1

T2
T1
or V1 = 0.2622 litre
Volume at STP = 262.2 ml

Prob 5.

In Victor Meyers experiment, 0.23g of a volatile solute displaced air which measures
112 ml at NTP. Calculate the vapour density and molecular weight of the substance.

Sol.

Volume occupied by solute at NTP = Volume of air displaced at NTP

= 112 ml
w
RT
For volatile solute
PV = M
at NTP, P = 1 atm, T = 273 K
112
0.23
1
0.0821 273
1000 = m
m = 46.02 and V.D. = 23.01

IIT Level Questions

Prob 6.

A cylindrical balloon of 21 cm diameter is to be filled up with H 2 at NTP from a

cylinder containing the gas at 20 atm at 27C. The cylinder can hold 2.82 litre of
water at NTP. Calculate the number of balloons that can be filled up.
3

Sol.

4 21

3
2
Volume of 1 balloon which has to be filled =
= 4851 ml
= 4.851 litre
Let n balloons be filled, then volume of H2 occupied by balloons = 4.851 n
Also, cylinder will not be empty and it will occupy volume of H 2 = 2.82 litre.
Total volume occupied by H2 at NTP = 4.851 n + 2.82 litre
At STP
P2 = 1 atm Available H2
V1= 4.851 n + 2.82
P2 = 20 atm
T1 = 273 K T2 = 300K
P1V1 P2 V2
T2
T
= 2
V2= 2.82 litre
1 (4.851n 2.82) 20 2.82

273
300
or
n = 10

Prob 7.

A 20g chunk of dry ice is placed in an empty 0.75 litre wire bottle tightly closed what
would be the final pressure in the bottle after all CO 2 has been evaporated and
temperature reaches to 25C?

Sol.

w = 20g dry CO2 which will evaporate to develop pressure

m = 44, V = 0.75 litre, P = ? T = 298K
w
RT
m
PV =
20
P 0.75 = 44 0.0821 298
P = 14.228 atm
Pressure inside the bottle = P + atm pressure
= 14.828 + 1 = 15.828 atm

Prob 8.

The pressure of the atmosphere is 2 106 mm at about 100 mile from the earth and
temperature is 180C. How many moles are there in 1 ml gas at this attitude?

Sol.

2 10 6
Given, P = 760 atm
T = 180 + 273 = 93 K
1
1000
V = 1 ml =
litre
PV = nRT
2 106
4 103
760
= n 0.0821 93

n = 3.45 1013 mole

Prob 9.

50 litre of dry N2 is passed through 36g of H 2O at 27C. After passage of gas, there is
a loss of 1.20g in water. Calculate vapour pressure of water.

Sol.

The water vapours occupies the volume of N2 gas i.e. 50 litre

For H2O vapour V = 50 litre, w = 1.20g, T = 300K, m = 18
1.2
w
0.0821 300
PV RT
m
or P 50 = 18
P = 0.03284 atm
= 24.95 mm

Prob 10.

A mixture of CO and CO2 is found to have a density of 1.50g/litre at 30C and 730
mm. What is the composition of the mixture?

Sol.

For a mixture of CO and CO2, d = 1.50 g/litre

730
atm,
P = 760
T = 303K
w
w
RT PV
RT
Vm
PV = m
;

730 1.5
0.0821 303
760 = m
38.85
i.e. molecular weight of mixture of CO and CO2 = 38.85
Let % of mole of CO be a in mixture then
a 28 (100 a)44
100
Average molecular weight =
28a 4400 44a
100
38.85 =
a = 32.19
Mole % of CO = 32.19
Mole % of CO2 = 67.81
Prob 11.

Sol.

The average speed of an ideal gas molecule at 27C is 0.3 m sec 1. Calculate
average speed at 927C.
8RT
M

uav=
Given uav = 0.3 m sec1 at 300K

8R 300
M

u1= 0.3 =
at T = 273 + 927 = 1200K
8R 1200
M
u2 =

Prob 12.

u2
1200

0.3
300 u2 = 0.6 m sec1
Pure O2 diffuses through an aperture in 224 seconds, whereas mixture of O 2 and
another gas containing 80% O 2 diffuses from the same in 234 sec. What is the
molecular weight of gas?

Sol.

For gaseous mixture 80% O2, 20% gas

32 80 20 m
100
Average molecular weight Mm =
Now, for diffusion of gaseous mixture and pure O2
rO2
VO2 Em
Mm
Mm

rm
MO2
EO2 Vm
MO2
or
Mm
1
234
32 80 20 m

224
1
32
100

Mm= 34.92
= 34.92
m = 46.6

Prob 13.

Calculate the pressure exerted by 10 23 gas molecules, each of mass 10 22 g in a

container of volume one litre. The rms speed is 105 cm sec1.

Sol.

Given, n = 1023 m = 1022g V = 1 litre = 103cc

urms = 105cm sec1
1
2
mnurms
PV = 3
1
10 22 1023 (105 )2
3
P 10 =
P = 3.3 107dyne cm2
3

Prob 14.

Sol.

Calculate the temperature at which CO 2 has the same rms speed to that of O 2 at
STP.

rrms of O2 =

3RT
M

at STP, urms of O2 =
3RT
44
For CO2 urms CO2 =
3R 273 3RT

32
44
Given both are same;

3R 273
32

T =375.38 K = 102.38C
Prob 15.

Calculate the compressibility factor for CO 2, if one mole of it occupies 0.4 litre at
300K and 40 atm. Comment on the result.

Sol.

PV
Compressibility factor (Z) = nRT
40 0.4
Z = 1 0.0821 300 = 0.65

Z value is lesser than 1 and thus, nRT PV. In order to have Z = 1, volume of CO 2
must have been more at same P and T or CO2 is more compressible than ideal gas.

Objective:
Prob 1.

An LPG cylinder contains 15 kg of butane gas at 27C and 10 atmospheric pressure.

It was leaking and its pressure fell down to 8 atmospheric pressure after one day.
The gas leaked is:
(A)1kg
(B) 2 kg
(C) 3 kg
(D) 4 kg

Sol.

For the cylinder V = constant

Hence P1V = n1RT and P2V = n2RT
w1
P1 n1
w

M 1
w2
P2 n2
w2
M

10 15

8
w2

Hence
w2 = 12 kg
Gas leaked out = 15 12 = 3 kg
(C)
Prob 2.

How much should the pressure be increased in order to decrease the volume of the
gas by 5% at constant temperature?
(A) 5%
(B) 5.26%
(C) 10%
(D) 4.26%

Sol.

PV= constant at constant temperature

New volume = 0.95V
P1V = P2 0.95V
P1
P2 = 0.95 = 1.0525P1
Increase in pressure = 1.0526 P1 P1 = 0.0526 P1 = 5.26% of P1
(B)

Prob 3.

Sol.

In what ratio by mass carbon monoxide and nitrogen should be mixed so that partial
pressure exerted by each is same?
(A)1:1
(B) 1:2
(C) 2:1
(D) 3:4
w1
w2
CO = w1g = 28 mol, N2 = w2g = 28 mol
w2
w1
28
PCO

w1 w 2 w1 w 2

28 28
w2
PN2
P
w1 w 2

as P = N2
Hence w1 = w2
w1 : w2= 1:1

Prob 4.

(A)
A box of 1 litre capacity is divided into two equal compartments by a thin partition
which are filled with 6g H2 and 16g CH4 respectively. The pressure in each
compartment is recorded as P atm. The total pressure when partition is removed will
be
(A) P
(B) 2P
(C) P/2
(D) P/4

Sol.

As pressure in each compartment is P atm, after the partition is removed the total
pressure will be P atm.
(A)

Prob 5.

At 27C, a gas is compressed to half of its volume. To what temperature it must now
be heated so that gas occupies just its original volume?
(A) 54C
(B) 600C
(C) 327C
(D) 327K

Sol.

V
2
PV =
P2 = 2P now; initial volume = V2
Final volume = V
V1 V2

T
T2
Keeping pressure constant 1
V
2 V
300 T2

, T2 = 600K = 327C
(C)
P2

Prob 6.

The vapour density of a gas is 11.2. The volume occupied by 11.2g of this gas at
N.T.P. is:
(A) 1 litre
(B) 11.2 litre
(C) 22.4 litre
(D) 20 litre

Sol.

Vapour Density = 11.2

Molecular weight = 22.4
Hence 22.4g = 22.4 litre
11.2g = 11.2 litre
(B)

Prob 7.

A gas is found to have a density of 1.80g litre 1 at 760 mm of pressure and 27C. The
gas will be
(A) O2
(B) CO2
(C) NH3
(D) SO2

Sol.

Prob 8.

dRT 1.8 0.082 300

760
P
760
M=
= 44
Hence the gas is CO2
(B)
A mixture of He and Ar contains 8 mole of He for every 2 mole of Ar. The partial
pressure of Ar is:

Sol.
Prob 9.

(A)2/3 the total pressure

(C) 2/5 the total pressure
2
1
P P
5
PArgon= 2 8
(D)

The vapour densities of CH 4, O2 are in the ratio 1:2. The ratio of diffusions of O 2 and
CH4 at same P and T is:
(A) 1:2
(B) 2:1
(C) 1:1.424
(D) 1.414 : 1
rO2

Sol.

(B) 1/3 total pressure

(D) 1/5 the total pressure

rCH4

16

32

1
1

2 1.424

(C)

Prob 10.

A per weighed vessel was filled with oxygen at NTP and weighed. It was then
evacuated, filled with SO2 at the same temperature and pressure and again weighed.
The weight of oxygen is:
(A)The same as that of SO2
(B) 1/2 that of SO2
(C) Twice that of SO2
(D) 1/4 that of SO2

Sol.

PV = nRT n1 = n2
w1 w 2
w2 w2

m2 m 2

or 32 64
1
w2
2
w1 =
(B)

Prob 11.

Sol.

The gases are at absolute temperature 300K and 350 K respectively. The ratio of
average kinetic energy of their molecules is:
(A) 7:6
(B) 6:7
(C) 36:49
(D) 49:36
KE1 T1 300 6

KE2 T2 350 7
(B)

Prob 12.

Sol.

Prob 13.

5g of each of the following gases at 87C and 750 mm pressure are taken. Which of
them will have the least volume?
(A)HF
(B) HCl
(C)
HBr(D) Hi
w
1
RT
V
m
PV = m
. If other factors are same,
Volume will be least in case of Hi because HI low the maximum molecular weight.
(D)

The rms speed of hydrogen is

temperature of the gas then
T TN2
(A) H2

7 times the rms speed of nitrogen. If T is the

(B)

TH2 TN2

(C)
Sol.

TH2 TN2

urms (H2) =
2rTH2
or,

2 =
TN2 2TH2

(B)

(D)

TH2 7TN2

7urms (N2 )

7
or

3RTN2
28

TN2 TH2

Prob 14.

A gaseous mixture contains 1g of H 2, 4g of He, 7g of N2 and 8g O2. The gas having

the highest partial pressure is:
(A)H2
(B) O2
(C) He
(D) N2

Sol.

P = mole fraction Pm
The gas having highest mole fraction will have highest partial pressure
Helium
(C)

Prob 15.

Which sample contains least number of molecules?

(A)1g of CO2
(B) 1g of N2
(C) 1g of H2
(D) 1g of CH4

Sol.

1
1g of CO2 = 44 = least number of moles
= least number of molecules
(A)

True and False

Prob 16.

In the van der Waals equation

n 2 a
P
+

V - nb = nRT
V2

The constant a represents the actual volume of the gas molecules.

Sol.

False

Prob 17.

A mixture of ideal gases is cooled upto liquid helium temperature (4.22 K) to form an
ideal solution.

Sol.

False

Fill in the Blanks

Prob 18.

A graph between volume (in L) vs. temperature (in oC) at constant pressure on
extrapolation cuts the temperature axis at..

Sol.

273oC

Prob 19.

Aqueous tension is the vapour pressure of . and depends only upon

Sol.

water vapours, temperature

Prob 20.

Boiling point is the temperature at which the vapour pressure becomes equal to

Sol.

external pressure (atmospheric pressure)

ASSIGNMENT PROBLEMS
Subjective:
Level - O
1. Account for the following properties of gases on the basis of kinetic molecular theory
(i) High compressibility
(ii) Gases occupy whole of the volume available to them.
2. What would have been the effect of gas pressure if the collisions between gas molecules
were not elastic?
3. A sample of a mixture of gases contains 2 moles of N 2, 3 moles of H2 and 5 moles of O2.
What is the mole fraction of each gas in the mixture?
4. A gas diffuses 1.27 times as rapidly as Cl 2 under the same conditions of temperature and
pressure. Answer the following:
(i) Is the gas heavier or lighter than Cl2.
(ii) What is the ratio of the molecular mass of the gas to that of Cl 2?
5. 1 mole of SO2 occupies a volume of 350 ml at 300K and 50 atm pressure. Calculate the
compressibility factor of the gas.
6. Calculate the pressure exerted by 8.5g of ammonia (NH 3) contained in 0.5 litre vessel of
300K. For ammonia a = 4 atm L2 mol2, b = 0.036 L mol1.
7. Explain following observation
Aerated water bottles are kept under water during summer.
8. Draw the plot of log P vs log V for Boyles law.
9. Is it possible to cool the gas to 0 Kelvin.
10. How is partial pressure of a gas related to the total pressure in a mixture of gases?
11. A gas at 20C occupies the volume 140 ml. Calculate the temperature at which the volume
of gas becomes 65 ml. Assume P as constant.
12. Calculate the volume occupied by 7g N2 at 27C and 750 mm of Hg.
13. O2 is present in 1 litre flask at a pressure of 7.6 1010 mm of Hg. Calculate no. of molecules
at 0C.
14. Find the volume of vessel containing 16 g O2 gas at NTP.
15. A gas at 0C and 1 atm pressure occupies 2.5 litres. What change in temperature would be
necessary if the pressure is to be adjusted to 1.5 atm and the gas has been transferred to a
2.0 litre container?

Level I
1. O2 is collected over water at 20C. The pressure inside shown by the gas is 750 mm of Hg.
What is the pressure due to O2 if vapour pressure of H2O is 18 mm at 20C?
2. The volume of the average adult lung when expanded is about 6 litre at 98.4F if the pressure
of oxygen in inhaled air is 168 mm of Hg, calculate the mass of O 2 required to occupy the
lung at 98.4F.
3. Calculate the total pressure in a mixture of 4g of O 2 and 2g of H2 confined in a bulb of 1 litre
at 0C
4. A gaseous mixture of O2 and N2 are in the ratio of 1:4 by weight. Calculate the ratio of
molecules.
5. The best vacuum so far attained in laboratory in laboratory is 10 10 mm of Hg. How many
molecules of gas remain per cc at 20C in this vacuum?
6. Calculate root mean square speed of O3 at 20C and 82 cm Hg
7. The mass of molecule A is twice the mass of molecule B. The rms speed of A is twice the rms
speed of B. If the samples of A and B contain the same number of molecules, what will be the
ratio of pressure of two samples in separate containers of equal volume?
8. Compressibility factor (Z) for N2 at 50C and 800 atm pressure is 1.95. Calculate mole of N 2
gas required to fill a gas cylinder of 100 litre capacity under the given conditions.
9. Calculate the temperature of gas if it obeys Van der Waals equation from the following data.
A flask of 25 litre contains 10 mole of agas under 50 atm.
Given: a = 5.46 atm lit2 mol2 and b = 0.031 lit mol1
3 RT
10. The kinetic molecular theory attributes an average translational kinetic energy of 2 N to
each particle. What rms speed would a mist particle of mass 10 12g have at room temperature
(27C) according to kinetic theory of gases.

11. (a) Two flasks of equal volume connected by a narrow tube (of negligible volume) are at
27C and contain 0.7 moles of H 2 at 0.5 atm. One of the flask is then immersed into a
bath kept at 127C, while the other remains at 27C. Calculate the final pressure and the
number of moles of H2 is each flask.
(b) 1 gm of an alloy of Al and Mg reacts with excess HCl to form AlCl 3, MgCl2 and H2. The
evolved H2 collected over mercury at 0C occupied 1200 ml at 699 mm Hg. What is the
composition of alloy?
12. Ladenburg found that a sample of ozonised oxygen containing 86.16% of O 3 by weight
required 430 secs to diffuse under conditions where pure O 2 required 369.5 secs. Determine
the vapour density of O3.

13. Assuming O2 molecule spherical in shape and occupying the radius 150 pm. Calculate
(i) the volume of single molecule of gas
(ii) the percentage of empty space in one mole of O2 at NTP
14. 9 volumes of gaseous mixture consisting of gaseous organic compound A and just sufficient
amount of O2 required for complete combustion yielded on burning 4 volumes of CO 2, 6
volumes of water vapour and 2 volumes of N 2 all volumes measured at the same temperature
and pressure. If the compound A contained only C, H and N
(i) how many volumes of O2 required for complete combustion?
(ii) What is the molecular formula of the compound A?
15. Two gases in adjoining vessels are brought into contact by opening a stop cock between
them. The one vessel measured 0.25 lit and contained NO at 800 torr and 220 K, the other
measured 0.1 lit and contained O2 at 600 torr and 220 K. The reaction to form N 2O4 (solid)
exhausts the limiting reactant completely.
(a) Neglecting the vapour pressure of N 2O4 what is the pressure of the gas remaining at 220
K after completion of the reaction?
(b) What weight of N2O4 is formed?

Level II
1. (a) How large a balloon could you fill with 4.0 g of He gas at 22 oC and 720 mm
of Hg?
(b) Calculate the density of CO2 at 100 0C and 800 mm Hg pressure.
2. (a) A container has 3.2 g of a certain gas at NTP. What would be the mass of the same
gas contained in the same vessel at 200C and 16 atm. Pressure?
(b) A certain quantity of a gas measured 500 mL at a temperature of 15C and 750 mm Hg.
What pressure is required to compress this quantity of gas into a 400 mL vessel at a
temperature of 50C?
3. (a) Three footballs are respectively filled with N 2, H2 and He to the same pressure. In what
order are these footballs to be reinflated?
(b) The ratio of rates of diffusion of two gases A and B under same pressure is 1:4. If the
ratio of their masses present in the mixture is 2:3. What is the ratio of their mole fraction
in mixture?
4. (a) Calculate the rms speed in cm/sec at 25C of a free electron and a molecule of UF 6.
(b) The flask (A) and (B) have equal volumes. Flask (A) contains H 2 gas at 300K, while (B)
contains equal mass of CH4 at 900K. Calculate the ratio of average speed of molecules
in flask (A) and (B)/
5. (a) Calculate total energy of one mole of an ideal monatomic gas at 27C?
(b) How much thermal energy should be added to 3.45 g Ne in a 10 litre flask to raise the
temperature from 0C to 100C. (Atomic weight Ne = 20.18)?
6. The critical temperature and pressure of CO 2 gas are 304.2K and 72.9 atm respectively.
What is the radius of O2 molecule assuming it to behave as Vander Waals gas?
7. Argon has TC= 122C PC = 48 atm. What is the radius of the argon atm?
8. 0.22g of a sample of a volatile compound containing C, H and Cl only yielded on combustion
in O2 0.195g of CO2 and 0.0804g of H2O. 0.12g of the compound occupied a volume of 37.24
ml at 105C and 768 mm Hg pressure. Calculate the molecular formula of the compound.
9. 10 ml of mixture of CH4, C2H4 and CO2 are exploded with excess O2. A contraction of
17 ml was observed after explosion. After the treatment with KOH solution there was a 2 nd
contraction of 14 ml. Find the % of gases in mixture?
10. 25cc of a gaseous mixture containing N 2 and N2O is passed through a hot Cu tube. The
resulting gas has a volume of 20 ml. find the percentage of N 2 and N2O in the original gas
mixture (The volumes have been measured under same conditions of temperature &
pressure)?
11. Calculate the pressure exerted by 1 mole of methane (CH 4) in a 250 ml container at 300 K
using van der Waals equation. What pressure will be predicted by ideal gas equation?
a = 2.253L2 atm. mol-2,
b = 0.04281 lit mol-1
R = 0.0821 L atm mol-1K.
12. To what minimum degree centigrade the temperature of earth will have to be raised to get
atmosphere free earth like moon?
2gR e
Given Vesp =
When Re = 6.37 106 m
Assume: atmospheric air chiefly contains N2 and O2 only.

13. (a) A 10 litre box contained 41.4 gm of a mixture of C xH8 and CxH12. At 44C the total
pressure is 1.56 atm. Analysis of the gas mixture shows 86% of C & 14% of hydrogen by
weight of hydrocarbon sample.
What gases are in the box?
How many moles of each gas are in the box?
(b) 24 ml. of water gas containing only hydrogen and carbon monoxide in equal proportions
by volume, are exploded with 80 ml. of air, 20% by volume of which is oxygen. If all
gaseous volumes are measured at room temperature and pressure, calculate the
composition by volume of the resulting gaseous mixture.
14. An open vessel at 27C is heated until three-fifths of the air in it has been expelled. Assuming
the volume of the vessel remains constant, find the temperature to which the vessel has to be
heated.
15. At 60C the density of N2O4 gas was found to be 30.2. Calculate the percentage of NO 2
molecules by weight and by volume.

Objective:
Level I
1. According to the kinetic theory of gases
(A) There are intermolecular attractions
(B) Molecules have considerable volume
(C) No intermolecular attractions
(D) The velocity of molecules decreases after each collision
2. The Van der Waals equation explains the behaviour of
(A) Ideal gases
(B) Real gases
(C) Vapour
(D) Nonreal gases
3. Which of the following is not a crystalline solid?
(A) KCl
(C) Glass

(B) CsCl
(D) Rhombic sulphur

4. The density of neon will be highest at

(A) STP
(C) 273C, 1 atm

(B) 0C, 2 atm

(D) 273C, 2 atm

5. Absolute zero is defined as the temperature

(A) at which all molecular motion ceases
(C) at which ether boils

(B) at which liquid helium boils

(D) all of the above

6. At constant temperature, in a given mass of an ideal gas

(A) The ratio of pressure and volume always remains constant
(B) Volume always remains constant
(C) Pressure always remains constant
(D) The product of pressure and volume always remains constant
7. At the same temperature and pressure, which of the following gases will have the highest
kinetic energy per mole?
(A) Hydrogen
(B) Oxygen
(C) Methane
(D) All are same
8. Which of the following mixture of gases does not obey Daltons law of Partial pressure?
(A) O2 and CO2
(B) N2 and O2
(C) Cl2 and O2
(D) NH3 and HCl
9. The temperature at which real gases obey the ideal gas laws over a wide range of pressure
is called
(A) Critical temperature
(B) Boyle temperature
(C) Inversion temperature
(D) Reduced temperature
10. The compressibility factor for an ideal gas is:
(A) 1.5
(C) 2.0

(B) 1.0
(D)

11. According to Grahams law at a given temperature, the ratio of the rates of diffusion r A/rB of
gases A and B is given by
PA M A

P
M B
(A) B

1/ 2

MA P
A

M
P
(B) B B

1/ 2

PA M B

P
M A
(C) B

1/ 2

MA P
B

M
P
(D) B A

1/ 2

12. 4.4 gm of a gas at STP occupies a volume of 2.24 litre, the gas can be
(A) O2
(B) CO
(C) NO2
(D) CO2
13. A certain sample of gas has a volume of 0.2 litre measured to 1 atm. Pressure and 0C. At
the same pressure but at 273C, its volume will be
(A) 0.4 litre
(B) 0.8 litre
(C) 27.8 litre
(D) 55.6 litre
14. If the four tubes of a car are fitted to the same pressure with N 2, O2, H2 and Ne separately,
then which one will be filled first?
(A) N2
(B) O2
(C) H2
(D) Ne
15. If pressure becomes double at the same absolute temperature of 2 litre CO 2, then the volume
of CO2 becomes
(A) 2 litre
(B) 4 litre
(C) 5 litre
(d) 1 litre
16. A gaseous mixture contains 56g of N 2, 44g of CO2 and 16g of CH4. The total pressure of the
mixture is 720 mm Hg. The partial pressure of CH 4 is:
(A) 180 mm
(B) 360 mm
(C) 540 mm
(D) 720 mm
17. Same mass of CH4 and H2 is taken in container. The partial pressure caused by H2 is
8
1
(A) 9
(B) 9
1
2
(C)

(D) 1

18. The major binding force of diamond, silicon and quartz is

(A) Electrostatic force
(B) Electrical attraction
(C) Covalent bond force
(D) Van der Waals force
19. The density of air is 0.00130 g/ml. The vapour density of air will be
(A) 0.0065
(B) 0.65
(C) 14.4816
(D) 14.56
20. Molecular weight of a gas that diffuses twice as rapidly as the gas will molecular weight 64 is:
(A) 16
(B) 8
(C) 64
(D) 6.4

Level II
1. If P, V, T represent pressure, volume and temperature of a gas, the correct representation of
Boyles law is
1
V
T (at constant P)
(A)
(B) PV = RT
1
V
P (at constant T)
(C)
(D) PV = nRT
2. Five grams each of the following gases at 87C and 750mm pressure are taken. Which of
them will have the least volume?
(A) HF
(B) HCl
(C) HBr
(D) HI
3. A real gas most closely approaches the behaviour of an ideal gas at
(A) 15 atm and 200K
(B) 1 atm and 273K
(C) 0.5 atm and 500K
(D) 15 atm and 500K
4. Which one of the following statements is true?
(A) The gas equation is not valid at high pressure and low temperature
(B) The product of pressure and volume of a fixed amount of gas is independent of
temperature
(C) Molecules of different gases have the same kinetic energy at a give temperature
(D) The gas constant per molecule is called as Boltzmanns constant
5. Two separate bulbs contain ideal gas A and B. The density of gas A is twice that of B. The
molecular mass of A is half that of B. The two gases are at the same temperature. The ratio
of pressure A to that of gas B is:
1
(A) 2
(B) 2
1
(C) 4
(D) 4
6. Use of hot air balloons in sports and metrological observations is an application of
(A) Boyles law
(B) Kelvins law
(C) Charles law
(D) Browns law
7. For an ideal gas, the number of moles per litre in terms of its pressure P, gas constant R and
temperature T is:
PT
(A) R
(B) PRT
P
PT
RT
(C)
(D) P
8. The Van der Waals equation reduces itself to the ideal gas equation at
(A) High pressure and low temperature
(B) low pressure and low temperature
(C) low pressure and high temperature
(D) high pressure alone
9. Air at sea level is dense. This is a practical application of
(A) Boyles law
(B) Charles law

(C) Avogadros law

10. When the product of pressure and volume is plotted
gas, the line obtained is:
(A) Parallel to Xaxis
(C) Linear with positive slope

(D) Daltons law

against pressure for a given amount of a
(B) Parallel to Yaxis
(D) Linear with negative slope

11. The rms velocity of CO2 at a temperature T (T in Kelvin) is X cm sec 1. At what temperature
(in
Kelvin)
the
rms
velocity
of
N 2O
would
be
4x
cm
sec 1?
(A) 16T
(B) 2T
(C) 4T
(D) 32T
12. The following graph illustrates

(A) Daltons law

(C) Boyles law
13. The ratio of rates diffusion of SO2, O2 and CH4 is
(A) 1 : 2 : 2
(C) 2 : 2 : 1

(B) Charless Law

(D) Gay Lussacs Law

(B) 1 : 2 : 4
(D) 1 : 2 :

14. A bottle of cold drink contains 200 ml liquid in which CO 2 is 0.1 molar. Suppose CO2behaves
like an ideal gas, the volume of the dissolved CO2 at STP is
(A) 0.224 litre
(B) 0.448 litre
(C) 22.4 litre
(D) 2.24 litre
15. In the equation PV = nRT, which one cannot be the numerical value of R?
(A) 8.31 107 erg K1 mol1(b) 8.31 107 dyne cm K1 mol1
1
1
(C) 8.31 J K mol
(D) 8.31 atm K1 mol1
16. A bottle of ammonia and a bottle of dry hydrogen chloride connected through a long tube are
opened simultaneously at both ends, the white ammonium chloride ring first formed will be
(A) at the centre of the tube
(B) near the hydrogen chloride bottle
(C) near the ammonia bottle
(D) throughout the length of the tube
17. Root mean square velocity of a gas molecule is proportional to
(A) m1/2
(B) m0
1/2
(C) m
(D) m

18. A sample of a given mass of a gas at a constant temperature occupies 95 cc under a

pressure of 9.962 104 Nm2. At the same temperature, its volume at a pressure of
10.13 104 N m2 is:
(A) 190cc
(B) 93cc
(C) 46.5 cc
(D) 47.5 cc
19. Equal volumes of the gases which do not react together are enclosed in separate vessels.
This pressures of 100 mm, and 400 mm respectively. If the two vessels are joined together,
then what will be the pressure of the resulting mixture (temperature remaining constant)?
(A) 125 mm
(C) 1000 mm

(B) 500mm
(D) 250 mm

20. There are 6.02 1022 molecules each of N2, O2, and H2 which are mixed together at 760 mm
and 273 K. The mass of the mixture in grams is
(A) 6.02
(B) 4.12
(C) 3.09
(D) 7

ANSWERS TO ASSIGNMENT PROBLEMS

Subjective:
Level - O
3.

0.2, 0.3, 0.5

4. Lighter, 1:1.61

5.

0.711

6. 21.51 atm

11.

155.54C

12. 6.239 litre

13.

2.68 10

14. 11.2 L

15.

54.6C

10

Level I
1.

732mm

2. 1.67g

3.

25.215 atm

4. 7:32

5.

9.081 1042 molecules

6. 3.9 104 cm sec1

7.

PA = 8PB

8. 2240.8

9.

1256.93C

10. 0.35 cm sec1

11.

(a) 0.571 atm, 0.3 and 0.4 moles

(b) 44 % Mg, 56 % Al

12. 23.3

13.

1.41 1023 cc/molecule, 99.96%

14. 7volume, C2H6N2

15.

0.30 atm, 0.402g

Level - II
1.

(a) 25.65 litre

(b) 1.5124 g/litre

2. (a) 29.55 g
(b) 1051 mm

3.

(a) H2 He N2
(b) 1:24

4. (a) 1.16 107 cm/sec

(b) 1.632

5.

(a) 900 cals (b) 51.75 cal

6.

1.62

7. 1.47 108 cm

8.

C2H4Cl2

9. 45% CH4, 40% C2H4, 15% CO2

10.

60% N2, 40% NO

11.

82.836 atm, 98.52 atm

12.

81.12 cm

13.

(a) C5H8, C5H12 0.432, 0.167 moles

14.

750 K

(b) O2 = 4ml, N2 = 64ml, CO2 = 12 ml

15. 52.3%, 68.7%

Objective:
Level I

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

Level II