Question:
An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true?
(A) the mean free path of the molecules decreases.
(B) the mean collision time between the molecules decreases.
(C) the mean free path remains unchanged.
(D) the mean collision time remains unchanged.
An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true?
(A) the mean free path of the molecules decreases.
(B) the mean collision time between the molecules decreases.
(C) the mean free path remains unchanged.
(D) the mean collision time remains unchanged.
(A) the mean free path of the molecules decreases.
(B) the mean collision time between the molecules decreases.
(C) the mean free path remains unchanged.
(D) the mean collision time remains unchanged.
Show Hint
In a closed (rigid) container the number density is fixed, so \(\lambda = \frac{1}{\sqrt{2}\pi d^2 n}\) does not change, while the molecular speed grows with temperature.
Updated On: Nov 5, 2026
- (C) and (D)
- (A) and (D)
- (B) and (C)
- (A) and (B)
Show Solution
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The Correct Option is C
Solution and Explanation
Mean free path: \(\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}\) where \(n\) is the number of molecules per unit volume.
The container is closed, so \(N\) and \(V\) are fixed and \(n\) does not change on heating. Hence \(\lambda\) remains unchanged: (C) is true and (A) is false.
Mean collision time \(\tau = \frac{\lambda}{v_{avg}}\) with \(v_{avg}\propto\sqrt{T}\). As \(T\) rises, \(v_{avg}\) increases while \(\lambda\) is constant, so \(\tau\) decreases: (B) is true and (D) is false.
True statements: (B) and (C).
Answer: (3) (B) and (C)
The container is closed, so \(N\) and \(V\) are fixed and \(n\) does not change on heating. Hence \(\lambda\) remains unchanged: (C) is true and (A) is false.
Mean collision time \(\tau = \frac{\lambda}{v_{avg}}\) with \(v_{avg}\propto\sqrt{T}\). As \(T\) rises, \(v_{avg}\) increases while \(\lambda\) is constant, so \(\tau\) decreases: (B) is true and (D) is false.
True statements: (B) and (C).
Answer: (3) (B) and (C)
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