Mean Free Path Dependence
A dilute gas initially in equilibrium has its absolute temperature doubled while its pressure is simultaneously halved; by what factor does the molecular mean free path change?
Select the correct option:
Solution
4
The mean free path of a gas obeys λ=2πd2PkBT, so for a fixed molecular diameter it is directly proportional to absolute temperature and inversely proportional to pressure, λ∝T/P. The cleanest way to see the combined effect is to recognise that the path ultimately depends on number density n=P/(kBT) through λ∝1/n. When the temperature is doubled, λ would double; when the pressure is halved, λ would double again. Combining the two independent effects multiplies the path by 2×2=4. Hence the mean free path increases by a factor of four. The value 2 accounts for only one of the two changes. The value 1 incorrectly assumes the effects cancel, as if λ∝T×P. The value 0.25 inverts the dependence, treating λ as proportional to P/T. This applies the NCERT kinetic-theory expression for mean free path, where heating spreads molecules apart at fixed pressure and lowering pressure reduces number density. As a plausibility check, the molecular number density n=P/(kBT) falls to one-quarter of its original value under these changes, and since λ∝1/n, the mean free path must rise four-fold, in exact agreement with the result.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
About This Question
- Subject
- physics
- Chapter
- kinetic theory of gases
- Topic
- mean free path dependence
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
4
The mean free path of a gas obeys λ=2πd2PkBT, so for a fixed molecular diameter it is directly proportional to absolute temperature and inversely proportional to pressure, λ∝T/P. The cleanest way to see the combined effect is to recognise that the path ultimately depends on number density n=P/(kBT) through λ∝1/n. When the temperature is doubled, λ would double; when the pressure is halved, λ would double again. Combining the two independent effects multiplies the path by 2×2=4. Hence the mean free path increases by a factor of four. The value 2 accounts for only one of the two changes. The value 1 incorrectly assumes the effects cancel, as if λ∝T×P. The value 0.25 inverts the dependence, treating λ as proportional to P/T. This applies the NCERT kinetic-theory expression for mean free path, where heating spreads molecules apart at fixed pressure and lowering pressure reduces number density. As a plausibility check, the molecular number density n=P/(kBT) falls to one-quarter of its original value under these changes, and since λ∝1/n, the mean free path must rise four-fold, in exact agreement with the result.
This hard difficulty physics question is from the chapter kinetic theory of gases, covering the topic of mean free path dependence. It appeared in the 2025 exam.
Looking for more practice? Explore all physics questions or browse kinetic theory of gases questions on RankGuru.