Question

For a molecule of an ideal gas. the number density is $2\sqrt{2}\times 10^{5}c{m}^{-3}$ and the mean free path is $\frac{10^{-2}}{\pi }cm$ . The diameter of the gas molecule is

• A. $5\times 10^{-4}cm$
• B. $0.5\times 10^{-4}cm$
• C. $2.5\times 10^{-4}cm$
• D. $4\times 10^{-4}cm$

Answer 1

Answer: (A)

Mean free path,
$\lambda =\frac{1}{\sqrt{2}\pi n{d}^2}$
$\Rightarrow d^2=\frac{1}{\sqrt{2}\pi n\lambda }=\frac{1\times \pi }{\sqrt{2}\times \pi \times 2\sqrt{2}\times 10^8\times 10^{-2}}$
$\Rightarrow d^2=\frac{1}{4\times 10^6}$
$\Rightarrow d=\frac{1}{2}\times 10^{-3}cm$
$\Rightarrow d=5\times 10^{-4}cm$