Derive an expression for pressure exerted by a gas on the basis of kinetic theory of gases.

Answer: \(P=\dfrac{1}{3}\dfrac{N}{V}m\overline{v^2}=\dfrac{1}{3}\rho v_{rms}^2\)

Consider an ideal gas of \(N\) molecules, each of mass \(m\), in a cubical container of side \(L\) (volume \(V=L^3\)). [Figure: cube with edges along X, Y, Z axes; a molecule moving with velocity components \(v_x\), \(v_y\), \(v_z\) hits the face perpendicular to the X axis.]

Momentum change: A molecule with velocity components \((v_x,v_y,v_z)\) strikes the wall perpendicular to the X axis and rebounds elastically with \(-v_x\). Change in its momentum \(=-mv_x-mv_x=-2mv_x\); the momentum given to the wall is \(2mv_x\).

Rate of collisions: The molecule travels \(2L\) between successive collisions with the same wall, so time between collisions \(\Delta t=\dfrac{2L}{v_x}\). Force exerted by one molecule \(f=\dfrac{2mv_x}{2L/v_x}=\dfrac{mv_x^2}{L}\).

Total force: \(F=\dfrac{m}{L}(v_{1x}^2+v_{2x}^2+\dots+v_{Nx}^2)=\dfrac{mN}{L}\overline{v_x^2}\), where \(\overline{v_x^2}\) is the mean of \(v_x^2\).

Pressure: \(P=\dfrac{F}{L^2}=\dfrac{mN}{L^3}\overline{v_x^2}=\dfrac{mN}{V}\overline{v_x^2}\).

Since the motion is random, \(\overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}\) and \(\overline{v^2}=\overline{v_x^2}+\overline{v_y^2}+\overline{v_z^2}=3\overline{v_x^2}\), so \(\overline{v_x^2}=\dfrac{1}{3}\overline{v^2}=\dfrac{1}{3}v_{rms}^2\).

\(\therefore P=\dfrac{1}{3}\dfrac{N}{V}m\,v_{rms}^2\).

As \(Nm=M\) (total mass) and \(\dfrac{M}{V}=\rho\), \(P=\dfrac{1}{3}\rho\,v_{rms}^2\).