If the pressure in a closed vessel is reduced by drawing out some gas, the mean free path of the gas molecules will

Mean free path of gas molecules is the average distance travelled by a molecule between two successive collisions. It is represented by λ

$\lambda=\sqrt{2} \pi d^{2} n$

where d = diameter of molecule and n = number of molecules per unit volume of the gas.

Also

$\lambda=\frac{\mathrm{K}_{\mathrm{B}} \mathrm{T}}{\sqrt{2} \pi \mathrm{d}^{2} P}$

where $\mathrm{K}_{\mathrm{B}}$ is Boltzmann constant; P is the pressure and T is temperature of the gas. So Mean

free path of gas molecules is inversely proportional to the pressure of the gas.

As per the question, the pressure in the closed

vessel is reduced by drawing out some gas. So

as per the equation, the mean free path of the

gas molecules will increase.