JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 9)
Calculate the value of mean free path ($$\lambda$$) for oxygen molecules at temperature 27$$^\circ$$C and pressure 1.01 $$\times$$ 105 Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. (k = 1.38 $$\times$$ 10$$-$$23 JK$$-$$1)
32 nm
58 nm
86 nm
102 nm
Explanation
$${I_{mean}} = {{RT} \over {\sqrt 2 \pi {d^2}{N_A}P}}$$
$$ = {{1.38 \times 300 \times {{10}^{ - 23}}} \over {\sqrt 2 \times 3.14 \times {{(0.3 \times {{10}^{ - 9}})}^2} \times 1.01 \times {{10}^5}}}$$
$$ = 102 \times {10^{ - 9}}$$ m
$$ = 102$$ nm
$$ = {{1.38 \times 300 \times {{10}^{ - 23}}} \over {\sqrt 2 \times 3.14 \times {{(0.3 \times {{10}^{ - 9}})}^2} \times 1.01 \times {{10}^5}}}$$
$$ = 102 \times {10^{ - 9}}$$ m
$$ = 102$$ nm
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