Q.Derive the expression $P = \tfrac{1}{3}\rho\overline{v^2}$ for the pressure exerted by an ideal gas, where $\rho$ is the density of the gas and $\overline{v^2}$ is the mean square speed of its molecules.

Consider $N$ molecules of mass $m$ in a cubical box of side $l$ (volume $V=l^3$), and one wall of area $A=l^2$ perpendicular to $x$. A molecule with $x$-velocity $v_x$ striking this wall rebounds elastically with velocity $-v_x$, so it transfers momentum $\Delta p_x = 2mv_x$ to the wall. It returns to strike the same wall again after travelling a round trip of $2l$, taking time $2l/v_x$, giving an average force from this molecule of $f = 2mv_x/(2l/v_x) = mv_x^2/l$.

Summing over all $N$ molecules and replacing the sum by $N$ times the average $\overline{v_x^2}$,

Since molecular motion is random with no preferred direction, $\overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}$, and since $\overline{v^2}=\overline{v_x^2}+\overline{v_y^2}+\overline{v_z^2}$, this gives $\overline{v_x^2}=\tfrac{1}{3}\overline{v^2}$. Substituting,

where $\rho=Nm/V$ is the gas density.

Find the impulse and collision frequency for one molecule striking a wall, sum over $N$ molecules, then use the random-direction assumption to relate $\overline{v_x^2}$ to the full mean square speed $\overline{v^2}$.

• Forgetting the factor of $\tfrac{1}{3}$, which comes specifically from splitting $\overline{v^2}$ equally among the three perpendicular directions -- omitting this step gives $P=\rho\overline{v_x^2}$ instead, an easy but incorrect shortcut.
• Using the round-trip time $l/v_x$ instead of $2l/v_x$ (forgetting the molecule must travel to the far wall AND back before striking the same wall again).