Q.Define mean free path. Derive the expression $\lambda = \dfrac{1}{\sqrt{2}\,\pi d^2 n}$ for the mean free path of a gas molecule, explaining the role of the factor $\sqrt{2}$.

A molecule of diameter $d$ collides with another molecule whenever their centres come within distance $d$ of each other, so as it moves it effectively sweeps out a cylindrical tube of cross-sectional area $\pi d^2$. Moving at average speed $\bar{v}$ for a time $t$, it sweeps a volume $\pi d^2\bar{v}t$, and in a gas of number density $n$, the number of collisions suffered in this time is the number of other molecules whose centres lie in this swept volume: $n\pi d^2\bar{v}t$.

A first, simplified estimate divides the distance travelled by the number of collisions, giving $\lambda \approx \bar{v}t/(n\pi d^2\bar{v}t) = 1/(n\pi d^2)$. This ignores that the OTHER molecules being collided with are themselves moving, not stationary. Accounting properly for the relative velocity between two randomly moving molecules (rather than treating the target molecules as fixed) shows that the correct average RELATIVE speed exceeds the average individual molecular speed by a factor of $\sqrt{2}$ -- so the true collision rate is $\sqrt{2}$ times th …

Find the collision rate from the swept-cylinder volume $\pi d^2\bar{v}t$, then correct the simplified stationary-target estimate by the $\sqrt …

• Stopping at the simplified $\lambda=1/(n\pi d^2)$ (omitting the $\sqrt{2}$ factor), which treats the other molecules as motionless targets -- physically incorrect for a gas where every molecule is moving. …