Mean Free Path

Setting up the geometric argument. Consider a system of identical molecules, each of diameter $d$, with $n$ molecules per unit volume. To make the counting tractable, imagine that only one molecule is actually moving (with average speed $v$) while every other molecule is momentarily frozen in place (Figure 9.8). If this one molecule moves for a time $t$, it sweeps out a path of length $vt$, and it will collide with any stationary molecule whose centre lies within an imaginary cylinder of length $vt$ and cross-sectional area $\pi d^2$ traced out around its path (a collision occurs whenever the two molecular centres come within one diameter $d$ of each other). The number of such collisions in time $t$ therefore equals the number of molecules inside that cylinder's volume, $\pi d^2 vt$, which is $\pi d^2 vtn$.

Mean free path (first approximation). The mean free path is the total distance travelled divided by the number of collisions in that time:

Correcting for the fact that every molecule is actually moving. The assumption that only one molecule moves while the rest sit still is not realistic -- in an actual gas, every molecule moves randomly at once, so what matters is the relative speed between the colliding molecules, not the speed of one molecule relative to a fixed background. Accounting for this properly (a calculation left for higher classes) introduces an extra factor of $\sqrt2$, giving the corrected and final expression

This says the mean free path is inversely proportional to the number density: as $n$ increases, molecular collisions become more frequent, so each molecule travels a shorter distance, on average, before its next collision.

Rewriting in terms of mass density. Using $\rho=mn$ (mass density), equation (9.26) can be rewritten as

Rewriting in terms of pressure and temperature. From the ideal gas equation $PV=NkT$, the number density is $n=N/V=P/kT$. Substituting into (9.26) gives

$\lambda = \frac{kT}{\sqrt2\,\pi d^2 P}. \qquad (9.28)$ …