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1. KINETIC THEORY OF
GASES
PRESENTED BY:
MR. B.SESHA SAI,
PGT, (PHY)
K.V. EOI Kathmandu

2. CONTENTS
ASSUMPTIONS
RMS VELOCITY
RELATION BETWEEN K.E. AND TEMP.
BOYLE’S LAW
CHARLE’S LAW
PERFECT GAS EQUATION
GAY LUSSAC’S LAW
AVOGADRO’S LAW

3. CONTENTS
ABSOLUTE ZERO TEMP.
DERIVATION OF GAS PRESSURE
DEGREES OF FREEDOM
LAW OF EQUIPARTITION OF ENERGY
Cp/Cv OF MONOATOMIC GAS
Cp/Cv OF TRIATOMIC(LINEAR) GAS
Cp/Cv OF DIATOMIC GAS
Cp/Cv OF TRIATOMIC(NON-LINEAR) GAS

4. ASSUMPTIONS
 All the molecules of a gas are identical as regards their
shape and mass. The molecules of different gases are
different.
 The gas molecules behave as rigid, elastic and smooth
spheres.
 The size of gas molecule is extremely small as
compared to intermolecular distance.
 The molecules of gas are in continuous random motion.
They move with all possible velocities in all possible
directions. They obey Newton’s Laws of Motion.
 The molecules collide with one another and also with the
walls of the container in perfectly elastic manner.
 The number of molecules per unit volume of gas is very
large.

5.  Between two successive collisions molecule moves with
uniform velocity. The distance between two successive
collisions is called free path. The average distance
traveled by molecule between successive collisions is
called mean free path.
 Their energy is wholly kinetic.
 The collisions are instantaneous. The time elapsed in
each collision is very small as compared to the time
elapsed between two successive collisions.

6. DERIVATION EXPRESSION OF
GAS PRESSURE
 Consider a gas enclosed in a cubical vessel of side l
having perfectly elastic walls.
Volume of gas=l3
No. of gas molecules=n
Mass of each molecule=m
Mass of gas M=mn
Let O be the origin.
Consider a molecule moving with velocity C1 as in fig.
u1, v1, w1 , are reaction component of C1 are along OX,OY
OZ respectively.

7.  Let the molecule strike the face A1 of the vessel.
Only the component u1(perpendicular to A1) is effective.
 Initial momentum of molecule along X-axis=mu1
 Collision is perfectly elastic. So K.E. is conserved. So
molecule rebound with velocity =-u1
 Final momentum of molecule along X-axis=-mu1
 Change in momentum =-mu1-mu1=-2 mu1
 After rebound the molecule strikes A2, it rebounds with
velocity u1 and again collide against A1. The molecule
travels distance 2l before colliding with face A1 again.
 Time interval between successive collisions on face A1
 Number of collision on A1 by molecule in 1 second
Contd.

8.  Change in momentum of molecule in 1 sec=
 From Newton’s IInd Law
 Force exerted on molecule by face A1 =
 From Newton’s IIIrd Law,
Force exerted by molecule on face A1 =f1=
 The forces due to other molecules
 Total force exerted by the n molecules on face A1
Contd.
Contd.

9.  Pressure on face A1
 Let Px and Py be pressures exerted by gas on faces
perpendicular to Y and Z axis.
Contd.

10. A gas exerts the same pressure in all directions. Thus;

11. Where;
C=Mean Square Velocity

12. RELATION BETWEEN K.E AND
TEMPERATURE
 Let us consider one mole of a gas. Let M and V be its
mass and volume respectively. The pressure exerted by
gas is given by:
But, PV=RT

13. BOYLE’S LAW
Which is Boyle’s Law.

14. BOYLE’S LAW GRAPH

15. CHARLE’S LAW
 The volume of given mass of gas is directly proportional
to its absolute temperature provided the pressure
remains constant.
According to Kinetic theory of gases,
At constant pressure for given mass of gas.
Which is Charle’s Law.

17. GAY LUSSAC’S LAW
The pressure P of a given mass of gas is directly
proportional to its absolute temperature T, provided the
volume V remains constant.
According to Kinetic theory of gases,
At constant volume for given mass of gas:
Which is Gay Lussac’s Law.

18. PERFECT GAS EQUATION
 According to Kinetic theory of gases,
Where R=universal gas constant.

20. AVOGADRO’S LAW
Equal volumes of all gases at the same temperature and
pressure contain equal number of molecules.
Proof: Consider two gases having same temperature,
pressure and volume. Let one gas contain n1 molecules
each of mass m1. Let second gas contain n2 contain
molecules each of mass m2. Let c1 and c2 be rms
velocities of two gases.
and
Contd.

21. Since both gases are at equal temperature;
Thus, average translational K.E. per molecule is same
for each gas.
Thus, from I and II; n1= n2
This proves the Avogadro’s Law.
Contd.

22. ABSOLUTE ZERO OF
TEMPERATURE
 Absolute zero of temperature is that temperature at
which the velocities of the gas molecule becomes zero.
We know that; average translational K.E. per molecule is
When T=0, c=0.
The absolute zero for an ideal gas is -273
o
C.

23. RMS VELOCITY OF GAS
MOLECULES
 RMS speed of molecules is the square root of the mean
of the squares of the velocities of individual molecules of
the gas.

24. DEGREE OF FREEDOM
 The total number of co-ordinates or independent
quantities required to completely specify the position and
configuration of a dynamical system is called degree of
freedom.
If A be number of particles in system R is number of
independent relations between them, then, degree of
freedom (N)= 3A-R.
For monoatomic gas, N=3(1)-0=3
For diatomic gas, N=3(2)-1=5
For linear triatomic gas, N=3(3)-2=7
For non-linear triatomic gas, N=3(3)-3=6

25. LAW OF EQUIPARTITION OF
ENERGY
 For a dynamical system in thermal equilibrium the
energy of system is equally distributed among the
various degrees of freedom and energy associated with
each degree of freedom=1/2KT.
 This law holds good for all degrees of freedom whether
translational, rotational or vibrational.
Proof : The average translational K.E. of a gas molecule
But for gas;

26. Cp/Cv FOR MONOATOMIC GASES
 Total energy associated with one monoatomic gas
molecule (U)=(3/2)RT.

27. Cp/Cv FOR DIATOMIC GASES
 Total energy associated with one diatomic gas molecule
(U)=(5/2)RT.

28.  Total energy associated with one triatomic gas molecule
(U)=(7/2)RT.
Cp/Cv FOR TRIATOMIC(LINEAR) GASES

29. Cp/Cv FOR TRIATOMIC(NON-LINEAR)
GASES
 Total energy associated with
one triatomic gas molecule
(U)=(6/2)RT=3RT.