Question

Calculate the mean free path of one $\text{mole}$ of air molecules at standard temperature and pressure. Assume radius of molecule is $1.66\times {10}^{-10}\text{m}$. (Given: $\frac{1}{6.022}=0.166$)

• A. $\frac{6.6}{\pi \sqrt{2}}\times {10}^{-5}\text{mm}$
• B. $\frac{0.066}{\pi \sqrt{2}}\times {10}^{-5}\text{mm}$
• C. $\frac{0.033}{\pi \sqrt{2}}\times {10}^{-5}\text{mm}$
• D. $\frac{33.7}{\pi \sqrt{2}}\times {10}^{-5}\text{mm}$

Answer 1

The correct option is D $\frac{33.7}{\pi \sqrt{2}}\times {10}^{-5}\text{mm}$

We know that,
At STP, volume of one mole of any ideal gas is $22.4\text{L}$ or $22.4\times {10}^{-4}{\text{m}}^3$.
Also,
Mean free path,
${\lambda }_{mean}=\frac{V/N}{\sqrt{2}\pi d^2}$
$\therefore {\lambda }_{mean}=\frac{22.4\times {10}^{-3}}{\sqrt{2}\pi \times 6.023\times {10}^{23}\times (2\times 1.66\times {10}^{-10}{)}^2}$
$\therefore {\lambda }_{mean}=\frac{33.7}{\pi \sqrt{2}}\times {10}^{-8}\text{m}$
$\therefore {\lambda }_{mean}=\frac{33.7}{\pi \sqrt{2}}\times {10}^{-5}\text{mm}$