Kinetic Theory

JEE Main level, single correct, medium

23 of 168

Question 23 of 168

The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are 3×10−7 m3 \times 10^{-7} \mathrm{~m} and 600 m/s600 \mathrm{~m} / \mathrm{s} respectively. Find the frequency of its collisions.

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Answer: (C) 2×109/s2 \times 10^9 / \mathrm{s}

 Frequency =1T=Vavg λ=6003×157=2×109sec−1\begin{aligned} & \text { Frequency }=\frac{1}{T}=\frac{V_{\text {avg }}}{\lambda} \\ & =\frac{600}{3 \times 157}=2 \times 10^9 \mathrm{sec}^{-1}\end{aligned}

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