The mean free path of a molecule of diameter $5 imes 10^{-10} \mathrm{~m}$ at the temperature $41^

Question

The mean free path of a molecule of diameter $5 	imes 10^{-10} \mathrm{~m}$ at the temperature $41^{\circ} \mathrm{C}$ and pressure $1.38 	imes 10^5 \mathrm{~Pa}$, is given as $\_\_\_\_$ m. (Given $k_B=1.38 	imes 10^{-23} \mathrm{~J} / \mathrm{K}$ ).

Professor PC Thomas Classes · Answer

The mean free path is given by $\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$. From the ideal gas law $P = n k_B T$, we substitute the number density $n = \frac{P}{k_B T}$ to get $\lambda = \frac{k_B T}{\sqrt{2}\pi d^2 P}$. Substituting the given values: $T = 41^\circ	ext{C} + 273 = 314 	ext{ K}$, $d = 5 	imes 10^{-10} 	ext{ m}$, $P = 1.38 	imes 10^5 	ext{ Pa}$. This gives $\lambda = \frac{1.38 	imes 10^{-23} 	imes 314}{\sqrt{2} 	imes 3.14 	imes (25 	imes 10^{-20}) 	imes 1.38 	imes 10^5} = \frac{100 	imes 10^{-23}}{\sqrt{2} 	imes 25 	imes 10^{-15}} = \frac{4}{\sqrt{2}} 	imes 10^{-8} = 2\sqrt{2} 	imes 10^{-8} 	ext{ m}$.