- 1(current)
Q. No.The mean free path for a gas, with molecular diameter \(d\) and number density \(n,\) can be expressed as:
| 1. | \( \dfrac{1}{\sqrt{2} n \pi {d}^2} \) | 2. | \( \dfrac{1}{\sqrt{2} n^2 \pi {d}^2} \) |
| 3. | \(\dfrac{1}{\sqrt{2} n^2 \pi^2 d^2} \) | 4. | \( \dfrac{1}{\sqrt{2} n \pi {d}}\) |
Subtopic: Â Mean Free Path |
 84%
Level 1: 80%+
NEET - 2020
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If the mean free path of atoms is doubled, then the pressure of the gas will become:
1. \(\frac{P}{4}\)
2. \(\frac{P}{2}\)
3. \(\frac{P}{8}\)
4. \(P\)
Subtopic: Â Mean Free Path |
 72%
Level 2: 60%+
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If the pressure in a closed vessel is reduced by drawing out some gas, the mean free path of the molecules:
| 1. | decreases |
| 2. | increases |
| 3. | remains unchanged |
| 4. | increases or decreases according to the nature of the gas |
Subtopic: Â Mean Free Path |
 77%
Level 2: 60%+
Hints
- 1(current)
Q. No.
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