B.Sc. Sem II Kinetic Theory of Gases

1 | U I , S e m e s t e r - I I ( P A N a g p u r e )
KINETIC THEORY OF GASES
Assumptions of Kinetic Theory of Gases
The kinetic theory of gases based on following assumptions:
1) A gas consists of very large number of extremely small molecules.
2) The molecules are rigid and perfectly elastic spheres of very small radius.
3) All the molecules of same gas are identical.
4) The molecules are always in a state of random motion.
5) Due to their random motion, the molecules collide with each other and with the walls of the
container.
6) The collision of the gas molecules is perfectly elastic.
7) The intermolecular forces of attraction between gas molecules are negligible.
8) The actual volume of occupied by the molecules is very small as compared to the total volume of
the gas.
9) The molecules travel in a straight line with constant velocity between two successive collisions.
10) The average number of molecules per unit volume of the gas remains constant.
11) At constant temperature, the average kinetic energy of the gas molecules remains constant. The
average kinetic energy of the gas molecules is directly proportional to the absolute temperature.
A gas which satisfies all the above assumptions is called perfect gas or ideal gas.
Free Path:
The distance covered by the molecule between two successive collisions is called its free path.
Mean Free Path (λ):
The average distance covered by a molecule between two successive collisions is called its mean free
path.
Mean Velocity:
The average of the velocities of all the molecules is called mean velocity.
For N molecules having velocities C1, C2, C3, …….. CN , the mean velocity is given by
1 2 3 ..... NC C C C
C
N
+ + + +
=
Mean Square Velocity:
The average of the squares of the velocities of all the molecules is called mean squares velocity.
2 2 2 2
2 1 2 3 ..... NC C C C
C
N
+ + + +
= ( )21
i
i
C
N
= 

Root Mean Square (RMS) Velocity:
The square root of the average of the squares of the velocities of all the molecules is called root mean
squares velocity. It is denoted by Crms.
2 2 2 2
2 1 2 3 ..... N
rms
C C C C
C C
N
+ + + +
= =
Pressure Exerted by a Gas:
An expression for pressure exerted by an ideal gas is given by
2
1
3
mNC
P
V
=
Where: m is the mass of gas molecule, V is the volume of the gas, N is the total number of gas
molecules
and
2
C is the mean square velocity of gas molecules.
The density of gas is given by
mN M
V V
 = =
Where M = mN is the total mass of the gas;
21
3
P C =
Kinetic Interpretation of Temperature:
Consider one mole of gas, the pressure exerted by the gas is given by
(for one mole of gas, N =NA, Avagadro’s number)
2
2
2
2
2
1
3
1
3
1 1
2
3 2
2 1
3 2
1 3
2 2
A
A
A
A
A
mN C
P
V
PV mN C
PV N mC
PV N mC
PV
mC
N
=
 =
 
 =  
 
 
 =  
 
 
 = 
 
Ideal gas equation for one mole of gas is PV RT=

2
2
1 3
2 2
1 3
2 2
A
RT
mC
N
mC kT
 
 = 
 
 
 = 
 
Where
A
R
k
N
= is called Boltzmann constant.
21
2
mC T 
 
 
Thus, average kinetic energy of a gas molecule is directly proportional to the absolute temperature.
In other words, higher the temperature of the gas, more will be the average kinetic energy possessed
by the gas molecules.
From above expression we may write
2
or rmsC T C T 
Thus, the root mean square velocity of the gas molecules is directly proportional to the square root f
the absolute temperature of the gas.
Derivation of Gas Laws:
Boyle’s law: If the temperature of the given mass of a gas is kept constant, pressure exerted by a gas
is inversely proportional to its volume.
i.e.
1
P
V
 Or constantPV = , at constant temperature.
Consider a gas having N molecules each of mass m. Let P and V be the pressure and volume of the
gas at constant temperature T. From kinetic theory of gases, the pressure exerted by gas is given by
2
2
2
2
1
3
1
3
1 1
2
3 2
2 1
3 2
mNC
P
V
PV mNC
PV N mC
PV N mC
=
 =
 
 =  
 
 
 =  
 

But 21 3
2 2
mC kT
 
= 
 
is the average kinetic energy of gas molecule and it is constant at constant
temperature.
For given mass of gas N is also constant.
Therefore,
constant, at constant temperaturePV T=
This is the Boyle’s law.
Charle’s Law:
If the pressure of the given mass of a gas is kept constant, volume of a gas is directly proportional to
its absolute temperature.
Consider a gas having N molecules each of mass m. Let P and V be the pressure and volume of the
gas at constant temperature T. From kinetic theory of gases, the pressure exerted by gas is given by
2
2
1
3
1
3
mNC
P
V
mNC
V
P
=
 =
22 1
3 2
N
V mC
P
 
 =  
 
But 21 3
2 2
mC kT
 
= 
 
is the average kinetic energy of gas molecule.
2 3
3 2
N
V kT
P
 
 =  
 
If pressure is constant, then for given mass of gas, we have T, at constant pressureV P
This is the Charle’s law.
Degrees of Freedom
The number of degrees of freedom of a dynamical system is defined as the total number of co-
ordinates or independent variables required to describe the position and configuration of the system.
The number of degrees of freedom of a dynamical system may also be defined as the number of
squared terms occurs in the expression of total energy of the system.

(a) A particle moving in a straight line along any one of the axes has one translational degree of
freedom.
(b) A particle moving in a plane (X and Y axes) has two translational degrees of freedom.
(c) A particle moving in space (X, Y and Z axes) has three translational degrees of freedom.
The rotational motion also can have three co-ordinates in space, like translational motion. Therefore
a rigid body will have six degrees of freedom ; three due to translational motion and three due to
rotational motion.
Monoatomic molecule
Since a monoatomic molecule consists of only a single atom of point mass it has
three degrees of freedom of translational motion along the three co-ordinate axes as
shown in figure.
Examples: molecules of rare gases like helium, argon, etc.
Diatomic molecule
The diatomic molecule can rotate about any axis at right angles to its own axis.
Hence it has two degrees of freedom of rotational motion in addition to three
degrees of freedom of translational motion along the three axes. So, a diatomic
molecule has five degrees of freedom as shown in figure.
Examples: molecules of O2, N2, CO, Cl2, etc.
Triatomic molecule (Linear type)
In the case of triatomic molecule of linear type, the centre of mass lies at the central
atom. It, therefore, behaves like a diamotic moelcule with three degrees of freedom
of translation and two degrees of freedom of rotation, totally it has five degrees of
freedom as shown in figure.
Examples: molecules of CO2, CS2, etc.
Triatomic molecule (Non-linear type)
A triatomic non-linear molecule may rotate, about the three mutually
perpendicular axes, as shown in figure. Therefore, it possesses three degrees of
freedom of rotation in addition to three degrees of freedom of translation along
the three co-ordinate axes Hence it has six degrees of freedom
Examples: molecules of H2O, SO2, etc.
In all the above cases, only the translational and rotational motion of the molecules have been
considered. The vibrational motion of the molecules has not been taken into consideration.

Law of Equipartition of Energy
Law of equipartition of energy states that for a dynamical system in thermal equilibrium the total
energy of the system is shared equally by all the degrees of freedom. The energy associated with each
degree of freedom per moelcule is
1
2
𝑘𝑇, where k is the Boltzmann’s constant.
Let us consider one mole of a monoatomic gas in thermal equilibrium at temperature T. Each
molecule has 3 degrees of freedom due to translational motion. According to kinetic theory of gases,
the mean kinetic energy of a gas molecule is
3
2
𝑘𝑇.
∴
3
2
𝑘𝑇 =
1
2
𝑚𝐶̅2
As,
1
2
𝑚𝐶̅2
=
1
2
𝑚𝐶 𝑥
2
+
1
2
𝑚𝐶 𝑦
2
+
1
2
𝑚𝐶𝑧
2
∴
3
2
𝑘𝑇 =
1
2
𝑚𝐶 𝑥
2
+
1
2
𝑚𝐶 𝑦
2
+
1
2
𝑚𝐶𝑧
2
Since molecules move at random, the average kinetic energy corresponding to each degree of freedom
is the same.
∴
1
2
𝑘𝑇 =
1
2
𝑚𝐶 𝑥
2
=
1
2
𝑚𝐶 𝑦
2
=
1
2
𝑚𝐶𝑧
2
Thus, mean kinetic energy per molecule per degree of freedom is
1
2
𝑘𝑇. Therefore mean kinetic
energy per mole per degree of freedom is
1
2
𝑁𝑘𝑇 =
1
2
𝑅𝑇.
Molar Specific Heat of Gas at Constant Volume: (CV)
The quantity of heat required to raise the temperature of one mole of gas through 1K (1o
C) when the
volume is kept constant is called molar specific heat at constant volume. Its SI unit is JK-1
mole-1
.
Molar Specific Heat of Gas at Constant Pressure: (CP)
The quantity of heat required to raise the temperature of one mole of gas through 1K (1o
C) when the
pressure is kept constant is called molar specific heat at constant pressure. Its SI unit is JK-1
mole-1
.
Relation between CP and CV:
CP - CV = R
Where R is called gas constant = 8.314 J K−1
mol−1
.
This relation is known as the Mayor’s relation between the two molar specific heats of a gas.
Specific Heat Ratio (γ):
The specific heat ratio for the gas is defined as the ratio of its specific heat at constant pressure (CP) to
specific heat at constant volume (CV).

𝛾 =
𝐶 𝑃
𝐶 𝑉
According to the law of equipartition of energy, mean kinetic energy per mole per degree of freedom
of a gas is
𝑈 =
1
2
𝑅𝑇
Hence for gas having f degrees of freedom, the total energy for one mole of gas given by,
𝐸 = 𝑓𝑈 =
1
2
𝑓𝑅𝑇
Therefore molar specific heat at constant volume CV is given by,
𝐶 𝑉 =
𝑑𝐸
𝑑𝑇
=
𝑑
𝑑𝑇
(
1
2
𝑓𝑅𝑇)
∴ 𝐶 𝑉 =
1
2
𝑓𝑅 ………………(1)
According to Mayor’s relation,
𝐶 𝑃 − 𝐶 𝑉 =𝑅
𝐶 𝑃 =𝑅+ 𝐶 𝑉
From equation (1)
𝐶 𝑃 =𝑅+
1
2
𝑓𝑅 = 𝑅(1+
1
2
𝑓) ……………………. (2)
Now from equation (1) & (2), we have
𝐶 𝑃
𝐶 𝑉
=
(1 +
1
2
𝑓)
1
2 𝑓
=
𝑓 + 2
𝑓
∴ 𝛾 =
𝑓 + 2
𝑓
Using this expression we can calculate the specific heat ratios for the gas.
1. For monoatomic gases: f = 3 → 𝛾 =
3+2
3
=
5
3
= 1.67
2. For diatomic gases: f = 5 → 𝛾 =
5+2
5
=
7
5
= 1.4
3. For triatomic or polyatomic gases: f = 6 → 𝛾 =
6+2
6
=
8
6
= 1.34
Expression for mean free path ():
To derive the expression for mean free path of a gas molecule, we use the following
assumptions:
1) A gas molecule is perfectly spherical in shape.
2) All the gas molecules except a molecule under consideration are at rest.

Consider the gas, which contains n number of
molecules per unit volume. Let σ be the diameter of each
molecule. Consider a molecule A moving with velocity c.
The molecule A collides with the other molecules, whose
centers lie within the distance σ from its path. Thus all
such molecules with which molecule A will collide in time
t, lie inside a cylinder of length (ct) and area of cross section ( σ2
), as shown in figure. Thus the
number of collisions of molecule O in time t,
N = volume of cylinder x number molecules per unit volume.
N = (ct *  σ2
) n =  σ 2
n ct
Now mean free path of a gas molecule,
2
Distance covered by the molecule in time
Number of collisions in time
t
t
ct
nvt



=
=
2
1
n


 =
This expression is not correct, since it is based on the assumption that only one molecule
under consideration is in motion and all other are at rest. But in actual all the molecules are in motion.
Hence the modified expression using Maxwell’s law of distribution of velocities of the
molecules is,
2
1
2 n


=
Effect temperature and pressure on mean free path λ:
The expression for mean free path λ is given by
2
1
2 n


= ----------------(1)
For one mole of ideal gas, we have
PV RT=
Dividing by N (Avogadro’s number) on both sides, we have
PV RT
N N
=
P
kT
n
=
Where,
N
n
V
= (number of molecules per unit volume) and
R
k
N
= (Boltzmann’s constant)


P
n
kT
=
Using in equation (1), we have

( )2
1
2 /P kT


=
 2
2
kT
P


= -------------- (2)
From equation (2) we can say that
1) Mean free path λ increases with increase in temperature.
2) Mean free path λ decreases with increase in pressure.
Transport phenomenon:
Consider a gas in a container, where there is concentration gradient, energy gradient, etc. Due
to this, there is non-uniformity in the density, non-uniformity in the temperature, etc. In such situation
some phenomenon’s singly or jointly come into play, trying to destroy the cause of non-uniformity.
Thus when there is concentration gradient, mass transport is takes place to destroy concentration
gradient, when there is energy gradient, energy transport is takes place to destroy temperature
gradient.
These phenomenon in which some physical quantity is transported to established equilibrium
(uniformity), are called as transport phenomenon.
There are three transport phenomenon’s involving transports of one type of physical quantity,
1) Viscosity, it involves momentum transport.
2) Conduction, it involves energy transport.
3) Diffusion, it involves mass transport.
Viscosity and coefficient of viscosity of a gas:
When there is velocity gradient in the gaseous system, different layers move with different
velocities. Then at the surface of contact between two layers, the force comes into play which
increases the speed of the slower moving layer and decreases the speed of the faster moving layer.
This force gives rise to phenomenon of the viscosity in the gas.
Consider two layers having separation dz, moving with
velocities u and u + du. The velocity gradient is
du
dz
. Then the
force at the surface of contact between two layers is directly
proportional to the area of contact and also to the velocity gradient.

i.e,
du
F dA
dz


du
F dA
dz
=
The proportionality constant  is called the coefficient of viscosity of the gas; hence it can be defined
as the retarding force per unit area per unit velocity gradient.
Expression for coefficient of viscosity of a gas:
Consider the coordinate axes in the gaseous system. Let the molecules are moving in a
direction parallel to x-axis. Consider the molecules in the layer A, at height z (called as reference
layer) are moving with velocity u. The molecules in the layer B, at height ( )z + are moving with
velocity
du
u
dz

 
+ 
 
. The molecules in the layer C, at height ( )z − are moving with velocity
du
u
dz

 
− 
 
. Where,
du
dz
is the velocity gradient & λ is the mean free path.
If n be the number of molecules per unit volume, and then we can say that, (n/3) molecules
per unit volume move along x-axis, (n/3) molecules per unit volume move along y-axis and (n/3)
molecules per unit volume move along z-axis. Hence the number of molecules moving in the positive
direction of z-axis per unit volume is (n/6).
If c is the average velocity of gas molecules, then the number of molecules per unit volume
per second passing through area dA of layer A from above is
6
n
c dA
 
 
 
.
If m is the mass of each gas molecule, then the momentum of molecules in the layer B is
du
m u
dz

 
+ 
 
.
Hence transfer of momentum per second from layer B in downward direction through area dA
of layer A is,
6
n du
c dA m u
dz

   
  +   
   
Similarly transfer of momentum per second from layer B in upward direction through area dA
of layer A is,
6
n du
c dA m u
dz

   
  −   
   
Net transfer of momentum per second through area dA
=
6 6
n du n du
c dA m u c dA m u
dz dz
 
       
  + −   −       
       
2
6
ncm du
dA
dz

=

3
nmc du
dA
dz

=
This gives rise to the viscous force, which is given by
du
F dA
dz
=

3
du nmc du
dA dA
dz dz

 =

3
nmc
 =
 where,
3
c
nm
 
 = =
This is the expression for the coefficient of viscosity of a gas.
Effect temperature and pressure on coefficient of viscosity of a gas:
The expression for the coefficient of viscosity of a gas is given by
3
nmc
 =
We have, 2
1
2 n


= and
8kT
c
m
=

2
2
8 1
3 2
2 1
3
mn kT
m n
mkT

 

 
=
=
This equation shows that, the coefficient of viscosity of a gas is
1) directly proportional to the square root of the absolute temperature, and
2) independent of pressure.

Expression for thermal conductivity of a gas:
If there is temperature gradient in the gaseous system,
there is also an energy gradient. Let the energy gradient is
dE
dz
along z-axis.
Consider the coordinate axes in the gaseous system.
Let the molecules are moving in a direction parallel to x-axis.
Consider the molecules in the layer A, at height z (called as
reference layer) having energy E. The molecules in the layer
B, at height ( )z + having energy
dE
E
dz

 
+ 
 
. The
molecules in the layer C, at height ( )z − having energy
dE
E
dz

 
− 
 
.
If n be the number of molecules per unit volume, and then we can say that, (n/3) molecules
per unit volume move along x-axis, (n/3) molecules per unit volume move along y-axis and (n/3)
molecules per unit volume move along z-axis. Hence the number of molecules moving in the positive
direction of z-axis per unit volume is (n/6).
If c is the average velocity of gas molecules, then the number of molecules per unit volume
per second passing through area dA of layer A from above is
6
n
c dA
 
 
 
.
Hence transfer of energy per second from layer B in downward direction through area dA of
layer A is,
6
n dE
c dA E
dz

   
  +   
   
.
Similarly transfer of energy per second from layer C in upward direction through area dA of
layer A is,
6
n dE
c dA E
dz

   
  −   
   
Net transfer of energy per second through area dA
=
6 6
n dE n dE
c dA E c dA E
dz dz
 
       
  + −   −       
       
2
6
nc dE
dA
dz

=
3
nc dE
dA
dz

= ……….. (1)
If the temperature of the reference layer A is T, then
E = mCvT

v
dE dT
mC
dz dz
 = ……. (2)
According to the conduction law, transfer of energy per second through area dA,
dT
K dA
dz
= ……… (3)

3
v
dT nc dT
K dA mC dA
dz dz

=

3
v
nmc
K C

=
This is the expression for the thermal conductivity K.
Since,
3
nmc
 = ,  vK C=
Effect temperature and pressure on thermal conductivity K:
The expression for the thermal conductivity K of a gas is given by
3
v
nmc
K C

=
Since,
3
nmc
 = ,  vK C=
We have, 2
1
2 n


= and
8kT
c
m
=

2
2
8 1
3 2
2 1
3
mn kT
m n
mkT

 

 
=
=
 2
2 1
3
v
mkT
K C
 
=
This equation shows that, the thermal conductivity is
1) Directly proportional to the square root of the absolute temperature, and
2) Independent of pressure.

Andrews' Experiment on Carbon Dioxide: (Andrews' Experimental Isotherm)
In the attempt of liquefying some of the imperfect gases, Andrews in 1869, was led to study the
isotherms of carbon dioxide. The curves obtained by Andrews are shown in figure. The main
conclusions are discussed below.
1. Let us consider the isotherm corresponding to low temperature 13.1O
C. As we increase the
pressure, the volume decreases considerably (portion AB) and finally liquefaction of the gas begins
(at point B). As long as liquefaction continues the pressure remains constant and the volume
continuously decreases (horizontal portion BC), more and more gas condensed into liquid. When all
the gas has condensed into liquid (at point C), after that the curve rises almost vertically (portion CD)
corresponds to the fact that liquids are only slightly compressible.
2. The isotherm corresponding to temperature 21.5 O
C is of the same form but the horizontal portion
B'C' is shorter than BC. It means in this isotherm, the volume of the vapour when condensation
begins, is smaller while volume of the liquid, when condensation has finished is greater in comparison
to the isotherm at 13.1 O
C. As temperature is raised further, isotherms behaves in the same manner, till
at 31.4 O
C.
3. At temperature 31.4 O
C the horizontal portion has just disappeared and the two volumes have
become the same and a kink is observed in isotherm denoted by K and known as critical point. This
isotherm is called the critical isotherm for carbon dioxide.
4. Above this temperature the horizontal portion is absent for all the isotherms such as for 35.5 O
C,
and there is no formation of liquid, but volume decreases rapidly till it becomes equal to that of the
liquid at that temperature.
5. This peculiar behaviour of the isotherms also disappears at higher temperature as for 48.1 O
C.

Discussion of the Results:
We conclude that, above the critical isotherm no liquid state is at all possible even under the greatest
pressure, while below it there are three separate regions. In the region enclosed by the dotted curve
whose highest point P, called the critical point, lies on the critical isotherm, both liquid and gaseous
states coexist. To the left of the line PC and below the critical isotherm there is the liquid region while
to the right of PB and below the critical isotherm there is the gaseous region. Thus, there exists a
continuity of the liquid and gaseous states.
The Critical Constants:
Consider the critical isotherm and critical point as shown in figure. If gaseous CO2 is compressed
above 31.4O
C no liquid state is possible even under the greatest pressure. This temperature is known
as critical temperature for CO2.
We may define critical temperature TC as the maximum temperature at which a gas can be liquefied
by pressure alone.
The pressure just sufficient to liquefy the gas at the critical temperature is called the critical pressure
PC.
The volume of the gas at critical temperature and pressure is called the critical volume VC. These
three quantities are called the critical constants of the gas.
Vander-Waals' Equation of State:
The equation of state for perfect gas was deduced theoretically from the kinetic theory of gases. Two
important assumptions were made in the deduction of this equation:
1. The molecules were considered to be negligibly small in size and considered as point object.
2. The gas was supposed to have only kinetic energy and have no potential energy due to
intermolecular attractions.
It is now known that both these assumptions are only approximate and very nearly hold in case of
extremely rarefied gases only, i.e., when the pressure is very small.
1. Clausius pointed out that the molecules, though extremely small in size, cannot be regarded as
geometrical points, and hence we must consider their size. This will leads to volume correction in real
gas equation.

2. Hirn pointed out that the molecules must exert force of attraction on one another in the same way as
solid or liquid does. Hence potential energy due to forces of cohesion must be taken into account. This
will leads to pressure correction in real gas equation.
Vander-Waals work out a systematic theory, taking into account both these factors and proposed
another equation of state for real gases.
Deduction of Vander-Waals' Equation of State
Correction for finite size: (Volume Correction)
Suppose one mole of a real gas is enclosed in a vessel of volume V. It is evident that the space
available for the free motion of the molecule becomes less when molecular size is taken into account,
because one molecule cannot lie within the space occupied by other molecules. Thus
the volume available for the free motion of the molecule is less than the volume V of
the vessel. Therefore V in the ideal gas equation has to be replaced by V – b.
Now to find the magnitude of b, assume that the molecules are hard sphere of radius r.
At the instance of collision, the centre-to-centre distance of the two colliding
molecules will be 2r. This implies that around any molecule, other molecules cannot
lie in a spherical volume Vs. This volume is called the sphere of exclusion and is eight times the
volume of a molecule, Vm, i,e. 3 34 4
(2 ) 8 8
3 3
S mV r r V = =  =
Let us imagine that the vessel is filled one by one molecule up to N molecules. Then,
The volume available to the first molecule = V.
The volume available to the second molecule = V – Vs
The volume available to the third molecule = V – 2Vs
In the same way, the volume available to the Nth molecule = V – (N – 1)Vs.
Hence, the average volume available to each molecule, obtained by taking arithmetic mean is
( ) ( 2 ) ...... ( ( 1) )S S SV V V V V V N V
V
N
+ − + − + + − −
=
1
[ (1 2 ....... ( 1)]SV NV V N
N
= − + + + −

1 ( 1) ( 1)
[ ]
2 2
4
2
S S
S m
N N N
V NV V V V
N
N
V V V V NV
− −
= − = −
= − = −
V V b= −
Where 4 mb NV=
Correction for intermolecular attraction: (Pressure Correction)
A molecule in the interior of the gas is surrounded by molecules from all sides therefore net force of
attraction would be zero on it. Hence, it will behave as if there were no intermolecular attractions.
However, a molecule close to the surface experiences a net inward force because the molecules are
distributed only on inner side. Thus whenever a molecule of this kind collides with the walls of the
vessel, it will exert pressure on the wall lesser than expected for perfect gas shown in figure. This
drop in pressure is known as cohesive pressure.
This pressure will be proportional to the number of molecules per unit volume (N/V) in the surface
layer and the number of molecules per unit volume in the layer just below the surface layer. Thus
where a' is a constant of proportionality. If we put a’N2
= a, we can write
Hence, in the perfect gas equation we must replace p by the sum of the observed pressure for any real
gas p and the drop caused by intermolecular attractions Δ p. Hence, the equation of state for an ideal
gas modifies to
2
( )
a
p V b RT
V
 
+ − = 
 
This is Vander Waal’s equation of state.
The quantities a and b are known as Vander Waals constants because they are considered to be
constant for any gas irrespective of its temperature but differ for different gases. At sufficiently large

volume, the term a/V2
becomes negligible in comparison with p, and b becomes negligible in
comparison with V. The Vander-Waals equation then reduces to the equation of state of an ideal gas.
Determination of Critical Constants:
Consider critical isotherm MKN as shown in figure for critical temperature TC. At critical point K,
the tangent to the curve is horizontal. 0
p
V
 
 = 
 
The curve is concave upward on the left of K and concave downward
on the right of K. Therefore K is called point of inflexion and at this
point
2
2
0
p
V
 
= 
 
.
We have Vander Waal’s equation of state
2
( )
a
p V b RT
V
 
+ − = 
 
2
( )
RT a
p
V b V
 = −
−
…… (1)
2 3
2
( )
p RT a
V V b V
 
 = − + 
 − 
…….. (2)
2
2 3 4
2 6
( )
p RT a
V V b V
 
 = − 
 − 
……. (3)
But at critical point p → pC, V → Vc and T → Tc , 0
p
V
 
= 
 
,
2
2
0
p
V
 
= 
 
,
2
( )
C
C
C C
RT a
p
V b V
 = −
−
……. (4)
2 3 2 3
2 2
0
( ) ( )
C C
C C C C
RT RTa a
V b V V b V
= − +  =
− −
……. (5)
3 4 3 4
2 26 6
0
( ) ( )
C C
C C C C
RT RTa a
V b V V b V
= −  =
− −
……. (6)
Dividing equation (5) by (6), we have

2
3
C CV b V− =
3CV b= ……. (7)
Using this value in equation (5)
2 3
2
(2 ) (3 )
CRT a
b b
=
8
27
C
a
T
bR
= ……. (8)
Using equation (7) and (8) in equation (4), we have
( )
2
8
27
(2 ) 3
C
a
R
abR
p
b b
 
 
  = −
2 2 2
4
27 9 27
C
a a a
p
b b b
 = − = ……. (9)
Constants of Vander Waal’s equation:
We have the relations
3CV b= ……. (1)
8
27
C
a
T
bR
= ……. (2)
2
27
C
a
p
b
= ……. (3)
Consider
2
8 27
27
8
8
C
C
C
C
C
C
T a b
p bR a
T b
p R
RT
b
p
=
=
 =
Consider
22 2
2
2
2 2
8 27
27
64
27
27
64
C
C
C
C
C
C
T a b
p bR a
T a
p R
R T
a
p
 
=  
 
 
=  
 
 =

Numericals:
1. Ideal gases in a closed container initially have volume V and pressure P. If the final pressure is 4P
and the volume is kept constant, what is the ratio of the initial kinetic energy with the final kinetic
energy Ans: 1:4
2. What is the average translational kinetic energy of molecules in an ideal gas at 57o
C. Given
Boltzmann‘s constant (k) = 1.38 x 10-23
Joule/Kelvin. Ans: 6.831 x 10-21
J
3. Three moles of gas are in a 36 liters volume space. Each gas molecule has a kinetic energy of
5 x 10- 21
Joule. Universal gas constant = 8.315 J/mole.K and Boltzmann’s constant = 1.38 x 10-23
J/K.
What is the gas pressure in the container. Ans: 1.67 x 105
Pascal or 1.67 atmospheres
4. Calculate the total rotational kinetic energy of all the molecules in one mole of air at 25.0ºC.
Ans: 2480 J
5. The mass of a helium atom is 6.66×10-27
kg. Compute the specific heat at constant volume for
helium gas (in J/kg.K) from the molar heat capacity at constant volume. Given Molecular mass of
helium is 4.00 g. Ans: 3120 J/kg. K
6. If one mole of a monatomic gas (γ=5/3) is mixed with one mole of a diatomic gas (γ=7/5). Find the
value of γ for the mixture. Ans: 3/2
7. Carbon dioxide gas (1.00 mole) at 373 K occupies 536 mL at 50.0 atmosphere pressure. What is the
calculated value of the pressure using (i) Ideal gas equation (ii) Van der Waals equation?
[Data - Van der Waals constants for carbon dioxide: a = 3.61 L2
atm mol-2
; b = 0.0428 L mol-1
]
Ans : ( i)57.1 atm, (ii) 49.6 atm
8. Predict which of the substances, NH3, N2, CH2Cl2, Cl2, CCl4 has
(i) the smallest van der Waals "a" constant
(ii) the largest "b" constant.
9. (i) Using Van der Waals equation, calculate the temperature of 20.0 mole of helium in a 10.0 litre
cylinder at 120 atmosphere pressure. [Data - Van der Waals constants for helium: a = 0.0341 L2
at
mol-2
; b = 0.0237 L mol-1
] (ii) Compare this value with the temperature calculated from the ideal gas
equation.

B.Sc. Sem II Kinetic Theory of Gases

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