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Kinetic Theory
Mean Free Path
3 questions from 3 papers, each with its answer.
- questions
- 3
- papers
- 3
- years asked, 2017–2026
- 2 of 10
- of Kinetic Theory
- 8.82%
Appears in
Concepts tested
- expression for mean free path
- Mean free path
- Mean free path and molecular size
All Questions
2026 Re-NEET 1 question
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The mean free path of molecules in an ideal gas A is half that of another ideal gas B. The diameter of the spherical molecules of gas A is twice the diameter of the molecules of B. If number densities of the gases A and B are $n_A$ and $n_B$, respectively, the correct option is:$\lambda=\frac{1}{\sqrt{2}\pi d^2n} \Rightarrow n\propto\frac{1}{\lambda d^2}$. $\frac{n_A}{n_B}=\frac{\lambda_B}{\lambda_A}\times\left(\frac{d_B}{d_A}\right)^2=2\times\left(\frac{1}{2}\right)^2=\frac{1}{2}$, so $n_A=\frac{n_B}{2}$.
2020 1 question
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The mean free path for a gas, with molecular diameter $d$ and number density $n$ can be expressed as:Mean free path $\lambda = \frac{1}{\sqrt{2}\,n\pi d^2}$
2014 AIPMT 1 question
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The mean free path of molecules of a gas, (radius 'r') is inversely proportional to:Mean free path $\lambda_m = \frac{1}{\sqrt2\,\pi d^2n}$, where d = diameter of the molecule
So $\lambda_m \propto \frac{1}{r^2}$