Question:
The plot that depicts the behavior of the mean free time \(t\) (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature (\(T\)), qualitatively, is:
(Graphs are schematic and not drawn to scale)
The plot that depicts the behavior of the mean free time \(t\) (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature (\(T\)), qualitatively, is:
(Graphs are schematic and not drawn to scale)
(Graphs are schematic and not drawn to scale)
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Mean free time \(=\) mean free path \(/\) average speed, and the average speed varies as \(\sqrt{T}\).
Updated On: Nov 5, 2026
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The Correct Option is B
Solution and Explanation
Mean free time \(t=\dfrac{\lambda}{v_{avg}}\) where \(\lambda=\dfrac{1}{\sqrt{2}\,\pi d^2 n}\) (for a fixed gas sample, fixed number density) and \(v_{avg}=\sqrt{\dfrac{8k_BT}{\pi m}}\propto\sqrt{T}\).
Hence \(t\propto\dfrac{1}{\sqrt{T}}\).
So a plot of \(t\) against \(\dfrac{1}{\sqrt{T}}\) is a straight line through the origin with positive slope. This is the graph in option (2) (option (1) is t vs 1/T, option (3) is a falling line against \(\sqrt{T}\), option (4) is t vs T).
Answer: (2) t versus \(1/\sqrt{T}\), a straight line with positive slope
Hence \(t\propto\dfrac{1}{\sqrt{T}}\).
So a plot of \(t\) against \(\dfrac{1}{\sqrt{T}}\) is a straight line through the origin with positive slope. This is the graph in option (2) (option (1) is t vs 1/T, option (3) is a falling line against \(\sqrt{T}\), option (4) is t vs T).
Answer: (2) t versus \(1/\sqrt{T}\), a straight line with positive slope
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