KTGMedium
Question
The mean free path of a molecule of diameter $5 \times 10^{- 10}\text{ }m$ at the temperature $41^{\circ}C$ and pressure $1.38 \times 10^{5}\text{ }Pa$, is given as $\_\_\_\_$ m.
(Given $k_{B} = 1.38 \times 10^{- 23}\text{ }J/K$ ).
Options
A.$2\sqrt{2} \times 10^{- 10}$
B.$10\sqrt{2} \times 10^{- 8}$
C.$2\sqrt{2} \times 10^{- 8}$
D.$2 \times 10^{- 8}$
Solution
$\lambda = \frac{k_{B}T}{\sqrt{2}\pi\sigma^{2}P}$
$$= \frac{1.38 \times 10^{- 23} \times (273 + 41) \times 100}{\sqrt{2} \times 3.14 \times \left( 5 \times 10^{- 10} \right)^{2} \times 1.38 \times 10^{5}} = 2\sqrt{2} \times 10^{- 8}$$
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