Since oxygen is a diatomic molecule with 5 degrees of freedom. Degrees of freedom of molecules in 2 moles of oxygen = f₁
\(=2 \mathrm{~N} \times 5=10 \mathrm{~N}\)
Since argon is a mono atomic molecule with 3 degrees of freedom. Degrees of freedom of molecules in 4 moles of argon = f₂
\(=4 \mathrm{~N} \times 3=12 \mathrm{~N}\)
∴ Total degrees of freedom of the mixture = f
\(=\mathrm{f}_{1}+\mathrm{f}_{2}=22 \mathrm{~N}\)
According to the principle of law of equipartition energy, energy associated with each degree of freedom of a molecule
\(=\frac{1}{2} k T\)
∴ Total energy of the system \(=\frac{1}{2} k T \times 22\)
But k = R
∴ Total energy of the system \(=\frac{1}{2} R T \times 22=11 \mathrm{RT}\)

Molar mass of methane gas \(=16.04 \times 10^{-3} \mathrm{~kg} / \mathrm{mole}\)
RMS speed of methane gas \(v_{m s}=471.8 \mathrm{~ms}^{-1}\)
Universal gas constant \(\mathrm{R}=8.31 \mathrm{~J} / \mathrm{mol} / \mathrm{k}\)
\(v_{m s}=\sqrt{\frac{3 R T}{m}}\)
\(\therefore\) Surface temperture of Jupiter
\(\mathrm{T} =\frac{v_{r m s}^{2} m}{3 R}  \)
\(=\frac{(471.8)^{2} \times 16.04 \times 10^{-3}}{3 \times 8.31}  \)
\(=\frac{3.549456 \times 10^{-3}}{24.93}  \)
\(\therefore \mathrm{T} =143 \mathrm{~K}=143-273=-130^{\circ} \mathrm{C}\)

The average distance travelled by the molecule between two successive collisions is called mean free path
Mean free path \(\lambda=\frac{k T}{\sqrt{2} \pi d^{2} P}\)
Where K - Boltzmann's constant
T - Temperature
d - diameter of the molecule
P - Pressure

Law of equipartition energy states that the average kinetic energy of system of molecules in thermal equilibrium at temperature T is uniformly distributed to all degrees of freedom (x or y or z) directions of motion so that each degree of freedom will get \(\frac{1}{2}\)kT of energy.

The minimum number of independent coordinates needed to specify the position and configuration of a thermodynamic system in space is called degree of freedom of the system.

The escape speed of gases on the surface of Moon is much less than the root mean square speeds of gases due to low gravity. Due to this all the gases escape from the surface of the Moon.

The average kinetic energy of the molecule is directly proportional to the absolute temperature of the gas.

Pressure arises due to momentum transfer to the wall of the container.

The average speed
\(\overset { - }{ v } =\frac { 2+3+4+5+5+5+6+6+7+9 }{ 10 } =5.2{ ms }^{ -1 }\)
To find the rms speed, first calculate the mean square speed \(\overset { - }{ { v }^{ 2 } } \)
\(\overset { - }{ { v }^{ 2 } } =\frac { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 }+{ 5 }^{ 2 }+{ 5 }^{ 2 }+6^{ 2 }+{ 6 }^{ 2 }+{ 7 }^{ 2 }+{ 9 }^{ 2 } }{ 10 } \)
= 30.6ms²s^-2
The rms speed
\({ v }_{ rms }\sqrt { \overset { - }{ { v }^{ 2 } } } =\sqrt { 30.6 } =5.53{ ms }^{ -1 }\)
The most probable speed is 5 m s^-1 because three of the particles have that speed.

The internal energy of ideal gas = \(\frac{3}{2}\)NKT
The number of air molecules is given in terms of number of moles so, rewrite the expression as follows 
U = \(\frac{3}{2}\)\(\mu\)RT
Since Nk = μR. Here μ is number of moles.
Gas constant R = 8.31\(\frac{J}{molK}\)
Temperature T = 273 + 27 =300K
U =\(\frac{3}{2}\)\(\times\)0.5\(\times\)8.31\(\times\)300 = 1869.75J
This is approximately equivalent to the kinetic energy of a man of 57 kg running with a speed of 8 m s^-1.

From the graph we get T₁ > T₂

The area under the graph will give total member of gas molecules in the system
\(\mathrm{n} =\frac{P V}{R T} \quad \mathrm{R}=k \)
\(\therefore \quad \mathrm{n} =\frac{P V}{k T} \)

Carbon dioxide
\(v_{r m s}=1.73 \sqrt{\frac{k T}{m}}\)

\(C_{p}-C_{v}=R\)
For diatomic gas (N₂) No of degrees of freedom = 5
\(\therefore S_{p}-S_{v}=R / 28\)

For mono atomic molecule No of degrees of freedom = 3
For diatomic molecule of degrees of freedom = 5
For triatomic molecule No of degree of freedom = 7
\(\text { Total }=\left[3 x_{1}+7\left(\mu_{2}+\mu_{3}\right)\right] N_{A}\)

Pressure is directly proportional to density.

Mean free path is independent of temperature and pressure

\(\mathrm{V}=\frac{f}{2} R T \quad \mathrm{C}_{\mathrm{V}}=\frac{f}{2} R \)
\(\mathrm{C}_{\mathrm{P}}=\left(1+\frac{f}{2}\right) R \)
\(\gamma=\frac{C_{p}}{C_{v}}=1+\frac{2}{f} \)
\(=\frac{f+2}{f} \)

\(\gamma=\frac{27}{17}\)
Number of moles of helium
\(\mathrm{n}=\frac{8}{4}=2\)
Number of moles of oxygen
\(n^{\prime}=\frac{16}{32}=\frac{1}{2}\)
For mono atomic Helium gas
\(\mathrm{f} =3 \)
\(\mathrm{C}_{\mathrm{V}} =\frac{f}{2} R \)
\(=\frac{3}{2} R \)
For diatomic oxygen gas
f = 5
\(\mathrm{C}_{\mathrm{V}} =\frac{f}{2} R \)
\(=\frac{5}{2} R \)
\(\mathrm{C}_{\mathrm{V}} \text { mixture } =\frac{n c_{v}+n^{\prime} C_{v}^{\prime}}{n+n^{\prime}} \)
\(=\frac{2 \times \frac{3}{2} R+\frac{1}{2} \times \frac{5}{2} R}{2+\frac{1}{2}} \)
\(C_{V} =\frac{3 R+\frac{5}{4} R}{\frac{5}{2}} \)
\(=\frac{17 R}{10} \)
\(\gamma =\frac{C_{p}}{C_{v}} \)
\(=1+\frac{R}{C_{V}} \)
\(=1+\frac{R}{\frac{17 R} {10}}\)
\(=1+\frac{10}{17} \)
\(=\frac{27}{17} \)

Pressure is independent of internal energy.

Average K.E of each degree of freedom
\(=\frac{1}{2} k T=\frac{1}{2} N T[k=N]\)
N - no. of moles
T - temperature

As Temperature of room A is less than that of room B evidently Room A has more air in it.

\(v_{m s}=1.73 \sqrt{\frac{k T}{m}}\)
\(\text { AT Increased by } 10 \text { times }\)
\(v_{\mathrm{ms}} \propto \Delta T\)
\(\text { RMS speed increases by } 10 \text { times. }\)

As it moves with respect to x axis
\(\Delta p_{x}=-m u, \Delta p_{y}=0\)

In 1827, Robert Brown, a botanist reported that grains of pollen suspended in a liquid moves randomly from one place to other. The random (Zig - Zag path) motion of pollen suspended in a liquid is called Brownian motion. In fact we can observe the dust particle in water moving in random directions. This discovery puzzled scientists for long time. There were a lot of explanations for pollen or dust to move in random directions were found adequate. After a systematic study, Wiener and Gouy proposed that Brownian motion is to the bombardment of suspended particles by bombardment of suspended particles by molecules of the surrounding fluid. But during 19⁺⁺⁺ century people did not accept that every matter is made up of small atoms or molecules. In the year 1905, Einstein gave systematic theory of Brownian motion based on kinetic theory and he deduced the average size of molecules.
According to kinetic theory any particle suspended in a liquid or gas is continuously bombarded from all the directions so that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner as shown in Figure. But when we put our hand in water it causes no random motion because the mass of our hand is so large that the momentum transferred. by the molecular collision is not enough to move our hand.

Factors affecting Brownian Motion:
(i) Brownian motion increases with increasing temperature.
(ii) Brownian motion decreases with bigger particle size, high viscosity and density of the liquid (or) gas.

According to kinetic theory any particle suspended in a liquid or gas is continuously bombarded from all the directions so that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner which is called Brownian motion.

(i) Temperature of the gas
(ii) Pressure of the gas
(iii) Diameter of the gas molecules

This law states that at constant temperature and pressure, equal volumes of all gases contain same number of molecules. For two different gases at the same temperature and pressure, according to kinetic theory of gases. We get
\(\mathrm{P}=\frac{1}{3} n m \overline{v^{2}} \ or \ \mathrm{P}=\frac{1}{3} \frac{N}{V} m \overline{v^{2}}\)
\(\mathrm{P}=\frac{1}{3} \frac{N_{1}}{V} m_{1} v_{1}^{2} \)
\(=\frac{1}{3} \frac{N_{2}}{V} m_{2} v_{2}^{2}\)   ......(1)
where \(\overline{v_{1}^{2}} \ and \ \overline{v_{2}^{2}}\) are the mean square speed for two gases and \(\mathrm{N}_{1} \ and \ \mathrm{N}_{2}\) are the number of gas molecules in two different gases.
At the same temperature, average kinetic energy per molecule is the same for two gases.
\(\frac{1}{2} m_{1} \overline{v_{1}^{2}}=\frac{1}{2} m_{2} \overline{v_{2}^{2}}\)   ........(2)
Dividing the equation (1) by (2) we get
N₁ = N₂

We get \(P V=\frac{2}{3} U\) But the internal energy of an ideal gas is equal to N times the average kinetic energy \((\epsilon)\) of each molecule.
\(\mathrm{U}=\mathrm{N} \in\)
For a fixed temperature, the average translational kinetic energy \(\in\) will remain constant It implies that
\(\mathrm{PV} =\frac{2}{3} \mathrm{~N} \in  \)
\(\text {Thus } \mathrm{PV} =\text { constant }\)
Therefore, pressure of a given gas is inversely proportional to its volume provided the temperature remains constant. This is Boyle's law.

We get \(P V=\frac{2}{3} U\) For a fixed pressure, the volume of the gas is proportional to internal energy of the gas or average kinetic energy of the gas and the average kinetic energy is directly proportional to absolute temperature. It implies that.
\(\mathrm{V} \propto \mathrm{T} \text { or }  \)
\(\frac{V}{T}=\text { constant }\)
This is Charles' law

Relation between the average kinetic energy and pressure is
\(\mathrm{P}=P=\frac{2}{3} K . E\)
P - Pressure
K.E - kinetic energy

The rms speed of gas molecules
\(v_{r m}=\sqrt{\frac{3 k T}{m}}=1.73 \sqrt{\frac{k T}{m}}\)
The average speed of gas molecule
\(\bar{v}=\sqrt{\frac{8 k T}{\pi m}}=1.6 \sqrt{\frac{k T}{m}}\)
The most probable speed of gas molecular
\(v_{m p}=\sqrt{\frac{2 k T}{m}}=1.4 .1 \sqrt{\frac{k T}{m}}\)

(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
\(\lambda =\frac { 1 }{ \sqrt { 2 } n{ \pi d }^{ 2 }P } \)

In general our interest our interest is to find how many gas molecules have the range of speed from v to v + dv. This is given by Maxwell's speed distribution function.
\({ N }_{ v }=4\pi N{ \left( \frac { m }{ 2\pi KT } \right) }^{ \frac { 3 }{ 2 } }{ v }^{ 2 }{ e }^{ \frac { { mv }^{ 2 } }{ 2KT } }\)   ....(1)
The above expression is graphically shown as follows

From the figure it is clear that, for a given temperature the number of molecules having lower speed increases parabolically but decreases exponentially after reaching most probable speed. The rms speed, average speed and most probable speed are indicated in the figure. It can be seen that the rms speed is greatest among the three. To Know the number of molecules in the range of speed between \(50 \mathrm{~m} \mathrm{~s}^{-1} \ and \ 60 \mathrm{~m}\mathrm{s}^{-1}\), we need to integrate \(\int_{50}^{60} 4 \pi N\left(\frac{m}{2 \pi k T}\right)^{\frac{3}{2}} v^{2} e^{\frac{m v^{2}}{2 k T}} d v=\mathrm{N}\left(50\right.\ to \ \left.60 \mathrm{~ms}^{-1}\right)\). In general the number of molecules within the range of speed v and v + dv is given by
\(\int_{v}^{v+d v} 4 \pi N\left(\frac{m}{2 \pi k T}\right)^{\frac{3}{2}} v^{2} e^{\frac{m v^{2}}{2 k T}} d v=N(v \text { to } v+d v) \)
The exact integration is beyond the scope of the book. But we can infer the behaviour of gas molecules from the graph.
(i) The area under the graph will give the total number of gas molecules in the system.
(ii) Figure shows the speed distribution graph for two different temperatures. As temperature increases, the peak of the curve is shifted to the right. It implies that the average speed of each molecule will increase. But the area under each graph is same since it represents the total number of gas molecules.

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.

Monoatomic molecule
A monoatomic molecule by virtue of its nature has only three translational degrees of freedom. Therefore f = 3
Example: Helium, Neon, Argon

Diatomic temperature
At Normal temperature.
A molecule of a diatomic gas consists of two atoms bound to each other by a force of attraction. Physically the molecule can be regarded as a system of two point masses fixed at the ends of a massless elastic spring.
The center of mass lies in the center of the diatomic molecule. so, the motion of the center of mass requires three translational degrees of freedom. In addition, the diatomic can rotate about three mutually perpendicular axes. But the moment of inertia about its own axis of rotation is negligible. Therefore, it has only two rotational degrees of freedom (one rotation is about Z axis and another rotation is about Y axis). Therefore totally there are five degrees of freedom. f = 5
At High Temperature
At a very high temperature such as 5000 K, the diatomic molecules possess additional two degrees of freedom due to vibrational motion [one due to kinetic energy of vibration and the other is due to potential energy]. So, totally are seven degrees of freedom. f = 7.
Examples: Hydrogen, Nitrogen, Oxygen
Triatomic molecules
There are two cases.
Linear triatomic molecule
In this type, two atoms lie on either side of the central atom.

Linear triatomic molecule has three translational degrees of freedom. It has two rotational degrees of freedom because it is similar to diatomic molecule except there is an addtional atom at the center. At normal temperature, linear triatomic molecule will have five degrees of freedom. At high temperature it has two additional vibrational degrees of freedom. So a linear triatomic molecule has seven degrees of freedom.
Example: Carbon dioxide
Non-Linear triatomic molecule
In this case, the three atoms lie at the vertices of a triangle.
In has three translational degrees of freedom and three rotational degrees of freedom about three mutually orthogonal axes. The total degrees of freedom f = 6

Example: Water, Sulphur dioxide

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.