Question

The mean free path for a gas, with molecular diameter $d$ and number density $n$ can be expressed as $(T$ is temperature and $P$ is pressure of the gas$)$

• A. $\frac{1}{\sqrt{2}\pi n^2{d}^2}$
• B. $\frac{1}{\sqrt{2}\pi n{d}^2}$
• C. $\frac{K_{B}T}{\sqrt{2}\pi P{d}^2}$
• D. Both (B) and (C)

Answer 1

The correct option is D Both (B) and (C)

$\lambda =\frac{1}{\sqrt{2}\pi n{d}^2}$
Where $n=\frac{N}{V}$
From $PV=N{K}_{B}T$
$\frac{N}{V}=\frac{P}{K_{B}T}$
$\lambda =\frac{1}{\sqrt{2}\pi \left(\frac{P}{K_{B}T}\right)d^2}$
$\lambda =\frac{K_{B}T}{\sqrt{2}\pi P{d}^2}$