Question

For a molecule of an ideal gas. the number density is $2 \sqrt{2} \times 10^5 \, cm^{-3}$ 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,
$\lambda=\frac{1}{\sqrt{2} \pi n d^{2}}$
$\Rightarrow d^{2} =\frac{1}{\sqrt{2} \pi n \lambda}=\frac{1 \times \pi}{\sqrt{2} \times \pi \times 2 \sqrt{2} \times 10^{8} \times 10^{-2}}$
$\Rightarrow d^{2}=\frac{1}{4 \times 10^{6}}$
$\Rightarrow d=\frac{1}{2} \times 10^{-3}\, cm$
$\Rightarrow d=5 \times 10^{-4}\, cm$