Q.The mean free path of a gas molecule is $2.0 \times 10^{-7}\ \text{m}$ at a certain pressure and temperature. If the pressure is halved while the temperature is kept constant, find the new mean free path.

The mean free path lambda = kT / (root-2 x pi x d^2 x P), so at constant pressure it is directly proportional to temperature T.

The mean free path is given by lambda = 1 / (root-2 x pi x d^2 x n), where d is the molecular diameter and n is the number density. Using the ideal gas relation n = P/(kT), this becomes lambda = kT / (root-2 x pi x d^2 x P). At constant pressure, lambda is therefore directly proportional to the absolute temperature T. By contrast, lambda is inversely related to pressure, inversely rel …

Mean free path follows from the molecular collision cross-section.

A molecule of diameter $d$ moving among others of number density $n$ sweeps out a cylindrical volume of cross-section $\pi d^{2}$ per unit path length. Since every molecule is in motion, the relative speed between two colliding molecules is on average $\sqrt{2}$ times a single molecule's mean speed, so the number of collisions per unit path length is $\sqrt{2}\,\pi d^{2} n$. …