JEE MAIN - Physics (2015 (Offline) - No. 21)
Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as $${V^q},$$ where $$V$$ is the volume of the gas. The value of $$q$$ is: $$\left( {\gamma = {{{C_p}} \over {{C_v}}}} \right)$$
$${{\gamma + 1} \over 2}$$
$${{\gamma - 1} \over 2}$$
$${{3\gamma + 5} \over 6}$$
$${{3\gamma - 5} \over 6}$$
Explanation
$$\tau = {1 \over {\sqrt 2 \pi {d^2}\left( {{N \over V}} \right)\sqrt {{{3RT} \over M}} }}$$
$$\tau \propto {V \over {\sqrt T }}$$
As, $$\,\,\,\,T{V^{\gamma - 1}} = K$$
So, $$\,\,\,\,\tau \propto {V^{\gamma + 1/2}}$$
Therefore, $$q = {{\gamma + 1} \over 2}$$
$$\tau \propto {V \over {\sqrt T }}$$
As, $$\,\,\,\,T{V^{\gamma - 1}} = K$$
So, $$\,\,\,\,\tau \propto {V^{\gamma + 1/2}}$$
Therefore, $$q = {{\gamma + 1} \over 2}$$
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