Mean Free Path

Representing the average distance a molecule travels between successive collisions, the mean free path is $\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}P}$, where $d$ is the molecular diameter and $P$ the pressure. This relation follows from treating molecules as hard spheres sweeping out a collision cylinder of cross-section $\pi d^2$, with the factor $\sqrt{2}$ accounting for the relative motion of all the other molecules rather than a single moving target. Substituting $k_B=1.38\times 10^{-23}$ J/K, $T=300$ K, $d=3\times 10^{-10}$ m, and $P=1.0\times 10^5$ Pa: the numerator is $4.14\times 10^{-21}$ and the denominator is $\sqrt{2}\pi (3\times 10^{-10}{)}^2(10^5)\approx 4.0\times 10^{-14}$, giving $\lambda \approx 1.04\times 10^{-7}$ m. The value $1.0\times 10^{-9}$ m wrongly uses the molecular diameter as the free path. The value $1.0\times 10^{-5}$ m omits the $\pi d^2$ cross-section. The value $1.0\times 10^{-8}$ m results from dropping the $\sqrt{2}$ factor and misplacing a power of ten. This applies the NCERT collision model of kinetic theory, where lower pressure or smaller molecules lengthen the free path, and the mean free path also sets the collision frequency that controls transport properties such as viscosity, diffusion, and thermal conduction in gases. As a check, the mean free path at atmospheric pressure is typically around $10^{-7}$ m, far larger than the molecular size yet far smaller than the container, confirming our answer.