Pressure of a Gas: Kinetic Interpretation

Consider an ideal gas of $N$ identical molecules, each of mass $m$, confined inside a cubical container of side $l$ (so volume $V = l^3$), and focus on one wall of the cube, of area $A = l^2$, taken perpendicular to the $x$-axis. A single molecule moving with velocity component $v_x$ toward this wall strikes it, and since the collision is perfectly elastic and the wall is rigid and far more massive than the molecule, the molecule rebounds with its $x$-component of velocity exactly reversed, from $v_x$ to $-v_x$, while its $y$- and $z$-components are unchanged.

The change in the molecule's momentum along $x$ in this one collision is therefore

so by Newton's third law, the wall receives an impulse of $+2mv_x$ from this molecule. Between successive collisions with THIS SAME wall, the molecule must travel to the opposite wall and back, a round-trip distance of $2l$, taking a time $\Delta t = 2l/v_x$. So this one molecule delivers an impulse $2mv_x$ to the wall once every $2l/v_x$ seconds, giving an average force from this single molecule of

Summing this over all $N$ molecules in the box, and replacing the sum of individual $v_x^2$ values by $N$ times their average, $\overline{v_x^2}$, the total average force on the wall is $F = Nm\overline{v_x^2}/l$, and the pressure is this force divided by the wall's area $A = l^2$:

Since the molecules' motion is assumed to be completely random with no preferred direction (assumption 1, Section 9.2), the average of $v_x^2$, $v_y^2$ and $v_z^2$ must all be equal, and since $v^2 = v_x^2+v_y^2+v_z^2$ for every molecule, this means $\overline{v_x^2} = \tfrac{1}{3}\overline{v^2}$, where $\overline{v^2}$ is the mean of the SQUARE of the full molecular speed (the mean square speed). Substituting,

$\boxed{P = \frac{1}{3}\frac{Nm}{V}\overline{v^2} = \frac{1}{3}\rho\overline{v^2}}$ …