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JEE MAIN 2021
16-03-2021 S2
Question
Calculate the value of mean free path ( $\lambda$ ) for oxygen molecules at temperature $27^{\circ} \mathrm{C}$ and pressure $1.01 \times 10^5 \mathrm{~Pa}$. Assume the molecular diameter 0.3 nm and the gas is ideal. $$ \left(\mathrm{k}=1.38 \times 10^{-23} \mathrm{JK}^{-1}\right) $$
Select the correct option:
A
58 nm
B
32 nm
C
86 nm
D
102 nm
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \lambda=\frac{\mathrm{RT}}{\sqrt{2} \pi \mathrm{~d}^2 \mathrm{~N}_{\mathrm{A}} \mathrm{P}} \\ & \lambda=102 \mathrm{~nm}\end{aligned}$
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