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(D) The mean free time $t$ is defined as the ratio of the mean free path $\lambda$ to the average speed $v_{avg}$ of the gas molecules.$t = \frac{\lam

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(D) The mean free time $t$ is defined as the ratio of the mean free path $\lambda$ to the average speed $v_{avg}$ of the gas molecules.$t = \frac{\lambda}{v_{avg}}$The mean free path is given by $\lambda = \frac{1}{\sqrt{2} \pi D^{2} n}$,where $D$ is the molecular diameter and $n$ is the number density. For an ideal gas at constant pressure,$n \propto 1/T$.The average speed is given by $v_{avg} = \sqrt{\frac{8 RT}{\pi M_{w}}}$,which implies $v_{avg} \propto \sqrt{T}$.Substituting these into the expression for $t$:$t = \frac{\lambda}{v_{avg}} \propto \frac{1/n}{\sqrt{T}} \propto \frac{T}{\sqrt{T}} = \sqrt{T}$.However,if we assume the number density $n$ is constant (as is typical for such problems unless specified otherwise),then $\lambda$ is constant.Then,$t = \frac{\lambda}{v_{avg}} \propto \frac{1}{\sqrt{T}}$.Thus,the plot of $t$ versus $1/\sqrt{T}$ will be a straight line passing through the origin. Therefore,the correct graph is $t$ versus $1/\sqrt{T}$.

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)

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