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Mean Free Path


The mean free path is the average distance traveled by a particle between collisions with other particles. It is dependent on the temperature, pressure, and composition of the gas, as well as the size and shape of the particles.

The mean free path can be defined as the inverse of the product of the number density of particles and their collision cross-section \begin{align} \lambda=\frac{1}{\sqrt{2}\pi d^2 n} \end{align} where $d$ is diameter of the particle and $n$ is the number density of the particle.

Problems from IIT JEE

Problem (JEE Mains 2020): An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true?

  1. the mean free path decreases.
  2. the mean time between collisions decreases.
  3. the mean free path remains unchanged.
  4. the mean time between collisions remains unchanged.

Problem (JEE Mains 2020): In a dilute gas at pressure $p$ and temperature $T$, the mean time between successive collisions of a molecule varies with $T$ as

  1. $T$
  2. $1/T$
  3. $\sqrt{T}$
  4. $1/\sqrt{T}$

Problem (JEE Mains 2021): Calculate the value of mean free path $\lambda$ for oxygen molecules at temperature 27 deg C and pressure $1.01\times{10}^{5}$ Pa. Assume the molecular diameter 0.3 nm and the gas is ideal.

  1. 32 nm
  2. 58 nm
  3. 86 nm
  4. 102 nm

Problem (IIT JEE 1995): From the following statements concerning ideal gas at any given temperature $T$, select the correct one(s)

  1. The coefficient of volume expansion at constant pressure is the same for all ideal gases.
  2. The average translational kinetic energy per molecule of oxygen gas is $3\ kT$, $k$ being Boltzmann constant.
  3. The mean-free path of molecules increases with decrease in the pressure.
  4. In a gaseous mixture, the average translational kinetic energy of the molecules of each component is different.

Solution: ifferentiate ideal gas equation, $pV=nRT$, to get \begin{align} p\mathrm{d}V+V\mathrm{d}p=nR\mathrm{d}T. \end{align} This reduces to $p\mathrm{d}V=nR\mathrm{d}T$ at constant pressure ($\mathrm{d}p=0$). The coefficient of volume expansion at constant pressure, \begin{align} \gamma & =\frac{1}{V}\frac{\mathrm{d}V}{\mathrm{d}T} \\ &=\frac{1}{T}, \end{align} is the same for all ideal gases.

The average translational kinetic energy per molecule of ideal gas is $\frac{3}{2}kT$.

At a given temperature, the decrease in pressure leads to increase in volume. Thus, the mean free path increases with decrease in pressure.


  1. Kinetic Theory of Gases
  2. RMS speed of gas molecules
  3. Law of equipartition of energy
  4. Specific Heats Cv and Cp for Monatomic and Diatomic Gases New
JEE Physics Solved Problems in Mechanics