Assuming the expression for the pressure exerted by the gas on the walls of the container, it can be shown that pressure is

The correct option is A ${\left[\frac{2}{3}\right]}^{rd}$ kinetic energy per unit volume of a gas
Kinetic energy of the gas $K=\frac{1}{2}m{v}_{rms}^2$

Rms velocity of gas molecules $v_{rms}=\sqrt{\frac{3RT}{m}}$

$\therefore$ $K=\frac{1}{2}m.\frac{3RT}{m}=\frac{3}{2}RT$

From ideal gas equation, $RT=PV$

$⟹$ $K=\frac{3}{2}PV$

Thus we get pressure of the gas $P=\frac{2}{3}\frac{K}{V}$