The formula for the mean free path $\lambda$ in an ideal gas (number density $n$, molecular diameter $d$) is:

1. A $\lambda = \frac{1}{\sqrt{2}\,n\pi d^2}$ Correct answer
2. B $\lambda = n\pi d^2$
3. C $\lambda = RT/(n\pi d^2)$
4. D $\lambda = d/n$

Solution

Standard derivation gives $\lambda = 1/(\sqrt{2}\,n\pi d^2)$.

About this question

Subject: Physics · Chapter: Kinetic Theory of Gases · Topic: Mean Free Path

This question is part of PrepWiser's free JEE Main question bank. 15 more solved questions on Kinetic Theory of Gases are available — start with the harder ones if your accuracy is >70%.

More Kinetic Theory of Gases questions

• Easy TRUE_FALSE

  The mean free path increases as the number density of gas molecules decreases.

• Easy MCQ

  The mean free path of a gas molecule depends on:

• Medium MCQ

  The molar heat capacity at constant volume $C_V$ for a diatomic ideal gas (ordinary temperatures) is:

• Easy MCQ

  The total number of degrees of freedom of a diatomic ideal gas molecule at ordinary temperatures is:

• Easy MCQ

  The number of translational degrees of freedom of a monatomic gas molecule is: