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Physics · Kinetic Theory of Gases

JEE Main 2025 — 4 April, Morning Shift — Question 49

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.

  1. Option A:

    9×105/s9 \times 10^{5} / \mathrm{s}

  2. Option B:

    5×108/s5 \times 10^{8} / \mathrm{s}

  3. Option C:

    2×1010/s2 \times 10^{10} / \mathrm{s}

  4. Option D:

    2×109/s2 \times 10^{9} / \mathrm{s}

    Correct

Answer: D

Step-by-step solution

Frequency of collision = average speed  mean free path =\frac{\text { average speed }}{\text { mean free path }}

=6003×10−7=2×109/s\begin{aligned} & =\frac{600}{3 \times 10^{-7}} \\ & =2 \times 10^{9} / \mathrm{s} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Morning Shift
Subject
Physics
Chapter
Kinetic Theory of Gases
Topic
Energy of Gas and Gas Laws and Miscellaneous Problems
The mean free path and the average speed of oxygen molecules at 300 K… | JEE Main 2025 PYQ with Solution · DhiX AI