(i) We know from postulates of kinetic theory that the molecules of a gas are in random motion and they collide with each other.
(ii) Between two successive collisions, a molecule moves along a straight path with uniform velocity.
(iii) This path is called mean free path. Consider a system of molecules each with diameter d. Let n be the number of molecules per unit volume.
(iv) Assume that only one molecule is in motion,and all others are at rest.

(v) If a molecule moves with average speed v in a time t, the distance travelled is vt.
(vi) In this time t, consider the molecule to move in an imaginary cylinder of volume nd2vr.
(vii) It collides with any molecule. whose center is within this cylinder. Therefore, the number of collisions is equal to the number of molecules in the volume of the imaginary cylinder.
(viii) It is equal to \(\pi\)d²vtn. The total path length divided by the number of collisions in time t is the mean free path.
Mean free pat, \(\lambda =\frac{distance \ travelled}{Number \ of \ collisions}\)
\(\lambda =\frac { vt }{ n{ \pi d }^{ 2 }vt } =\frac { 1 }{ n{ \pi d }^{ 2 } } \)   ...(1)
(ix) Though we have assumed that only one molecule is moving at a time and other molecules are at rest, in actual practice all the molecules are in random motion.
(x) So the average relative speed of one molecule with respect to other molecules has to be taken into account. After some detailed calculations (you will learn in higher classes) the correct expression for mean free path .
\(\therefore \ \lambda =\frac { 1 }{ \sqrt { 2 } n{ \pi d }^{ 2 } } \)   ...(2)
(xi) The equation (1) implies that the mean free path is inversely proportional to number density.
(xii) When the number density increases the molecular collisions increases and it decreases the distance travelled by the molecule before collisions:
Case1: Rearranging the equation (2) using 'm' (mass of the molecule)
\(\therefore \ \lambda =\frac { 1 }{ \sqrt { 2 } n{ \pi d }^{ 2} mm } \)
But mn = mass per unit volume = p (density of the gas)
\(\therefore \ \lambda =\frac { 1 }{ \sqrt { 2 } n{ \pi d }^{ 2 } p} \)
Also we know that PV = NkT
P =\(\frac{N}{V}\)KT= nKT
\(\therefore n =\frac{P}{KT}\)
Substituting n = \(\frac{P}{KT}\) in equation, we get
\(\lambda =\frac { 1 }{ \sqrt { 2 } n{ \pi d }^{ 2 }P } \)

Monoatomic molecule
Average kinetic energy of a molecule
\(=\left[\frac{3}{2} k T\right]\)
Total energy of a mole of gas \(=\frac{3}{2} k T \times N_{A}=\frac{3}{2} R T\)
For one mole, the molar specific heat at constant volume
\(\mathrm{C}_{\mathrm{V}} =\frac{d U}{d t}=\frac{d}{d t}\left[\frac{3}{2} R T\right]  \)
\(\mathrm{C}_{\mathrm{V}} =\left[\frac{3}{2} R\right]  \)
\(\mathrm{C}_{\mathrm{P}} =\mathrm{C}_{\mathrm{V}}+\mathrm{R}  \)
\(=\frac{3}{2} R+R=\frac{5}{2} R\)
The ratio of specific heats,
\(=\frac{C_{p}}{C_{v}}=\frac{\frac{5}{2} R}{\frac{3}{2} R}=\frac{5}{3}=1.67\)
Diatomic molecule
Average kinetic energy of a diatomic molecule at low temperature \(=\frac{5}{2} k T\). Total energy of one mole of gas
\(=\frac{5}{2} k T \times N_{A}=\frac{5}{2} R T\)
(Here, the total energy is purely kinetic)
For one mole specific heat at constant volume
\(\mathrm{C}_{\mathrm{V}}=\frac{d U}{d T}=\left[\frac{5}{2} R T\right]=\frac{5}{2} R\)
But \(C_{P}=C_{v}+R\)
\(=\frac{5}{2} R+R=\frac{7}{2} R\)
\(\therefore \gamma=\frac{C_{p}}{C_{v}}=\frac{\frac{7}{2} R}{\frac{5}{2} R}=\frac{7}{5}=1.40\)
Energy of a diatomic molecule at high temperature is equal to \(\frac{7}{2} \mathrm{RT}\)
\(C_{v} =\frac{d U}{d t}=\left[\frac{7}{2} R T\right]=\frac{7}{2} R  \)
\(\therefore C_{p} =C_{v}+R=\frac{7}{2} R+R  \)
\(C_{P} =\frac{9}{2} R\)
Note that the C_V and C_P are higher for diatomic molecules than the mono atomic molecules. It implies that to increase the temperature of diatomic gas molecules by \(1^{\circ} \mathrm{C}\) it require more heat energy than mono atomic molecules.
\(\therefore \gamma=\frac{C_{P}}{C_{V}}=\frac{\frac{9}{2} R}{\frac{7}{2} R}=\frac{9}{7}=1.28\)
Triatomic molecule
a) Linear molecule
\(\text {Energy of one mole } =\frac{7}{2} k T \times N_{A}=\frac{7}{2} R T  \)
\(C_{v} =\frac{d U}{d T}  \)
\(=\frac{d}{d t}\left[\frac{7}{2} R T\right]  \)
\(C_{v} =\frac{7}{2} R  \)
\(C_{P} =C_{v}+R=\frac{7}{2} R+R=\frac{9 R}{2}\)
\(\therefore \gamma=\frac{C_{P}}{C_{V}}=\frac{\frac{9}{2} R}{\frac{7}{2} R}=\frac{9}{7}\)
= 1.28
b) Non-linear molecule
\(\text {Energy of a mole } =\frac{6}{2} k T \times N_{A}=\frac{6}{2} R T=3 R T  \)
\(C_{V} =\frac{d U}{d T}=3 R  \)
\(C_{V} =C_{V}+\mathrm{R}  \)
\(=3 R+R=4 R \)
\(\therefore \gamma =\frac{C_{p}}{C_{v}}=\frac{4 R}{3 R}=\frac{4}{3}=1.33\)
Note that according to kinetic theory model of gases the specific heat capacity at constant volume and constant pressure are independent of temperature. But in reality it is not sure. The specific heat capacity varies with the temperature.

To understand the microscopic origin of temperature in the same way.
Rewrite the equations
\(\mathrm{P}  =\frac{1}{3} n m \overline{v^{2}} \text { or } P \frac{1}{3} \frac{N}{V} m \overline{v^{2}} \quad \text { as }\left[n=\frac{N}{V}\right] \\ \)
\(\mathrm{P}  =\frac{1}{3} n m \overline{v^{2}} \text { or } P \frac{1}{3} \frac{N}{V} m \overline{v^{2}} \\ \)
\(\mathrm{PV}  =\frac{1}{3} N m \overline{v^{2}}\)   ....(1)
Comparing the equation (1) with ideal gas equation PV = Nkt
\(\mathrm{NkT} =\frac{1}{3} N m \overline {v^{2} } \)
\(\mathrm{kT} =\frac{1}{3} m v\overline v^{2}\)   ....(2)
Multiply the above equation by 3 / 2 on both sides,
\(\frac{3}{2} k T=\frac{1}{2} m \overline v^{2}\)
R.H.S of the equation is called average kinetic energy of a single molecule \((\overline{KE})\)
The average kinetic energy per molecule
\(\overline{K E}=\frac{3}{2} k T  \)
\(\frac{3}{2} k T =\frac{1}{2} \overline{m v^{2}}\)
Implies that the temperature of a gas is a measure of the average translational kinetic energy per molecule of the gas.

A molecule of mass m moving with a velocity \(\vec{v}\) having components \(\left(v_{x}, v_{y}, v_{z}\right)\) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with same speed and its x-component is reversed. The components of velocity of the molecule after collision are \(\left(-v_{x}, v_{y}, v_{z}\right)\)
The x-component of momentum of the molecule before collision = mv_x
The x-component of momentum of the molecule after collision = mv_x
The change in momentum of the molecule in x direction
= Final momentum - initial momentum 
= \(-\mathrm{mv}_{\mathrm{x}}-\mathrm{mv}_{\mathrm{x}} \)
= \(-2 \mathrm{mv}_{\mathrm{x}}\)
According to law of conservation of linear momentum, the change in momentum of the wall \(=2 \mathrm{mv}_{\mathrm{x}}\)
The number of molecules hitting the right side wall in a small interval of time ∆t is calculated as follows.
The molecules within the distance of v_x∆t  from the right side wall and moving towards the right will hit the wall in the time interval ∆t. The number of
molecules that will hit the right side wall in a time interval ∆t is equal to the product of volume \(\left(\mathrm{Av}_{x} \Delta t\right)\)and number density of the molecules (n). Here A is area of the wall and n is number of molecules per unit volume \(\left(\frac{N}{V}\right)\). We have assumed that the number density is the same throughout the cube.
Not all the n molecules will move to the right, therefore on an average only half of the n molecules move to the right and the other half moves towards left side. The number of molecules that hit the right side wall in a time interval
\(\Delta t=\frac{n}{2} A v_{x} \Delta t\)   .....(1)
In the same interval of time ∆t, the total momentum transferred by the molecules.
\(\Delta \dot{p} =\frac{n}{2} A v_{x} \Delta t \times 2 m v_{x}  \)
\(=A v_{x}^{2} m n \Delta t\)   ....(2)
From Newton's second law, the change in momentum in a small interval of time gives rise to force.
The force exerted by the molecules on the wall (in magnitude)

\(\mathrm{F} =\frac{\Delta p}{\Delta t} \)
\(=n m A v_{x}^{2} \)   ...(3)
Pressure, P = force divided by the area of the wall.

\(\mathrm{P} =\frac{F}{A}  \)
\(=n m v_{x}{ }^{2}\)
Since all the molecules are moving completely in random manner, they do not have same speed. So we can replace the term v_x² by the average \(\overline{v_{x}^{2}}\)
\(\mathrm{P}=\frac{F}{A}=n m v_{x}^{2}  \)
\(\mathrm{P}=n m \overline{v_{x}^{2}}\)
Since the gas is assumed to move in random direction, it has no preferred direction of motion. (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So.\( \overline{v_{x}^{2}}=\overline{v_{y}^{2}}=\overline{v_{x}^{2}}\).
The mean square speed is written as
\(\overline{v^{2}}=\overline{v_{\dot{x}}^{2}}+\overline{v_{y}^{2}}+\overline{v_{z}^{2}}=\overline{3 v_{x}^{2}}  \)
\(\overline{v_{x}^{2}}=\frac{1}{3} \overline{v^{2}}  \)
\(\mathrm{P}=n m \overline{v_{x}^{2}}  \)
\(\mathrm{P}=\frac{1}{3} n m \overline{v^{2}} \text { or } P \frac{1}{3} \frac{N}{V} m \overline{v^{2}} \quad \text { as }\left[n=\frac{N}{V}\right]\)

(i) All the molecules of a gas are identical, elastic spheres.
(ii) The molecules of different gases are different.
(iii) The number of molecules in a gas is very large and the average separation between them is larger than size of the gas molecules.
(iv) The molecules of a gas are in a state of continuous random motion.
(v) The molecules collide with one another and also with the walls of the container.
(vi) These collisions are perfectly elastic so that there is no loss of kinetic energy during collisions.
(vii) Between two successive collisions, a molecule moves with uniform velocity.
(viii) The molecules do not exert any force of attraction or repulsion on each other except during collision. The molecules do not possess any potential energy and the energy is wholly kinetic.
(ix) The collisions are instantaneous. The time spent by a molecule in each collision is very small compared to the time elapsed between two consecutive collisions.
(x) These molecules obey Newton's laws of motion even though they move randomly.