In a vessel containing gas, the gaseous molecules are moving in all possible directions with very high velocity and are colliding with the walls of the vessel. Due to the continuous bombardment of the gaseous molecules to the walls of the vessel, the gas exerts pressure.

## Expression for Pressure Exerted by a Gas

Consider a cubical vessel of each side of length $l$ containing gas. Let there be $n$ number of gaseous molecules inside the vessel and let $m$ be the mass of each molecule. Let the $n$ molecules be moving with velocities $c_1, c_2, c_3, …, c_n$ respectively. Then, the velocity $c_1$ of $\text{molecule}_1$ can be resolved into three components $u_1$, $v_1$ and $w_1$ along $X$, $Y$ and $Z$-axis respectively as shown in figure.

Then, \[{c_1}^2={u_1}^2+{v_1}^2+{w_1}^2\] Similarly, for other gas molecules, \[{c_2}^2={u_2}^2+{v_2}^2+{w_2}^2\] \[{c_3}^2={u_3}^2+{v_3}^2+{w_3}^2\] \[—————–\] \[{c_n}^2={u_n}^2+{v_n}^2+{w_n}^2\]

Let the $\text{molecule}_1$ strike the surface of the cube along $X-$axis. Then, the momentum with which the $\text{molecule}_1$ strikes is $mu_1$. Since the collision is perfectly elastic in nature, the $\text{molecule}_1$ rebounds back with the same velocity $u_1$ after striking the surface and the momentum becomes $-mu_1$.

Change of momentum of the $\text{molecule}_1$ along $X-$axis \[=mu_1-(-mu_1)\] \[=2mu_1\] Here, the $\text{molecule}_1$ strikes the surface and covers a distance $l$. When it rebounds back and strikes the other opposite surface, it again covers the same distance $l$. So, the $\text{molecule}_1$ covers the total distance of $2l$.

Time between two successive collision \[=\frac{\text{Distance covered}}{\text{Velocity}}\] \[=\frac{2l}{u_1}\]

Rate of change in momentum of $\text{molecule}_1$ along $X-$axis \[=\frac{\text{Change in Momentum}}{\text{Time Taken}}\] \[=2mu_1×\frac{u_1}{2l}\] \[=\frac{m{u_1}^2}{l}\]

Thus, force exerted by $\text{molecule}_1$ along $X-$axis \[=\frac{m{u_1}^2}{l}\] Force exerted by all the gas molecules along $X-$axis \[F_x=\frac{m{u_1}^2}{l}+\frac{m{u_2}^2}{l}+…+\frac{m{u_n}^2}{l}\] \[F_x=\frac{m}{l}({u_1}^2+{u_2}^2+…+{u_n}^2)\]

Pressure exerted by the gas molecules in the direction of $X-$axis \[P_x=\frac{F_x}{l^2}\] \[P_x=\frac{m}{l^3}({u_1}^2+{u_2}^2+…+{u_n}^2)\]

Similarly, pressure exerted in the direction of $Y-$axis and $Z-$axis, \[P_y=\frac{m}{l^3}({v_1}^2+{v_2}^2+…+{v_n}^2)\] \[P_z=\frac{m}{l^3}({w_1}^2+{w_2}^2+…+{w_n}^2)\]

Thus, average pressure exerted by all gas molecules, \[P=\frac{1}{3}(P_x+P_y+P_z)\] \[P=\frac{1}{3}\frac{m}{l^3}[({u_1}^2+{u_2}^2+…+{u_n}^2)+({v_1}^2+{v_2}^2+…+{v_n}^2)+({w_1}^2+{w_2}^2+…+{w_n}^2)]\] \[P=\frac{1}{3}\frac{m}{l^3}[({u_1}^2+{v_1}^2+{w_1}^2)+({u_2}^2+{v_2}^2+{w_2}^2)+…+({u_n}^2+{v_n}^2+{w_n}^2)]\] \[P=\frac{1}{3}\frac{m}{l^3}[{c_1}^2+{c_2}^2+…+{c_n}^2]\] \[P=\frac{1}{3}\frac{mn}{l^3}\left[\frac{{c_1}^2+{c_2}^2+…+{c_n}^2}{n}\right]\text{___(1)}\]

Now, \[\left[\frac{{c_1}^2+{c_2}^2+…+{c_n}^2}{n}\right]=\bar{c}^2\] \[\bar{c}^2=\text{Mean Square Velocity of gas}\] \[mn=M\;\;\;[M=\text{Total mass of the gas}]\] \[l^3=V \;\;\;[V=\text{Volume of the gas}]\] Then, equation $(1)$ becomes, \[P=\frac{1}{3}\frac{M}{V}\bar{c}^2\text{___(2)}\]

Also, \[\frac{M}{V}=ρ\;\;\;\text{[Density of the gas]}\] ∴Equation $(2)$ becomes, \[P=\frac{1}{3}ρ\bar{c}^2\] This is the pressure exerted by the gas molecules on the walls of the vessel.