The mean free path $\lambda$ of a gas molecule of diameter $d$ at number density $n$ is approximately:
A$\lambda = \dfrac{1}{\sqrt{2}\pi d^2 n}$
B$\lambda = \sqrt{2}\pi d^2 n$
C$\lambda = \dfrac{1}{n d}$
D$\lambda = \pi d n$
Answer & Solution
Correct answer: A. $\lambda = \dfrac{1}{\sqrt{2}\pi d^2 n}$
The mean free path is the average distance a molecule travels between collisions. Standard kinetic-theory derivation gives:
$\lambda = \dfrac{1}{\sqrt{2}\pi d^2 n}$
where $d$ is the molecular diameter and $n$ is the number density.
Note the $d^2$ (cross-sectional area), $\sqrt{2}$ factor from accounting for the relative motion of two molecules colliding (both move, not just one), and the inverse-density dependence (more molecules around means more collisions, shorter path).
At STP, $\lambda$ for air is about 70 nm — small but huge compared to molecular size (a few Å).
Related questions
Molecules in a gas move at speeds of the order of the speed of:Compared with solids, atoms in gases are much:The mean free path in gases is of the order of:Argon is an example of which kind of gas?A gas like argon has which kind of degrees of freedom only?Motion of a body as a whole from one point to another is called:A molecule confined to a plane has how many degrees of freedom?A molecule free to move in space needs how many coordinates?