The mean free path of a molecule of diameter 5 × 10 ? 10 ? m 5×10 ?10 m at the temperature 41 ? C 41 ? C and pressure 1.38 × 10 5 ? Pa 1.38×10 5 Pa, is given as

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Solution

Q.
The mean free path of a molecule of diameter $5 \times 10^{- 10} m$ at the temperature $41 ° C$ and pressure $1.38 \times 10^{5} Pa$ is given as ______ m.

(Given $k_{B} = 1.38 \times 10^{- 23} J/K$ ) [2026]

1 $10 \sqrt{2} \times 10^{- 8}$

2 $2 \sqrt{2} \times 10^{- 8}$

3 $2 \times 10^{- 8}$

4 $2 \sqrt{2} \times 10^{- 10}$

Ans.
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