Mean Free Path Dependence

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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?

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The mean free path of a gas obeys $\lambda=\dfrac{k_BT}{\sqrt{2}\,\pi d^{2}P}$, so for a fixed molecular diameter it is directly proportional to absolute temperature and inversely proportional to pressure, $\lambda\propto T/P$. The cleanest way to see the combined effect is to recognise that the path ultimately depends on number density $n=P/(k_BT)$ through $\lambda\propto1/n$. When the temperature is doubled, $\lambda$ would double; when the pressure is halved, $\lambda$ would double again. Combining the two independent effects multiplies the path by $2	imes2=4$. 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 $\lambda\propto T	imes P$. The value 0.25 inverts the dependence, treating $\lambda$ 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 $n=P/(k_BT)$ falls to one-quarter of its original value under these changes, and since $\lambda\propto1/n$, the mean free path must rise four-fold, in exact agreement with the result.

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