Question
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: $$\left( {\gamma = \frac{{{C_p}}}{{{C_v}}}} \right)$$
A.
$$\frac{{\gamma + 1}}{2}$$
B.
$$\frac{{\gamma - 1}}{2}$$
C.
$$\frac{{3\gamma + 5}}{6}$$
D.
$$\frac{{3\gamma - 5}}{6}$$
Answer :
$$\frac{{\gamma + 1}}{2}$$
Solution :
$$\eqalign{
& \tau = \frac{1}{{\sqrt 2 \pi {d^2}\left( {\frac{N}{V}} \right)\sqrt {\frac{{3RT}}{M}} }} \cr
& t\mu \frac{V}{{\sqrt T }} \cr
& {\text{As, }}T{V^{\gamma - 1}} = K \cr
& {\text{So, }}\tau \, \propto \,{V^{\gamma + \frac{1}{2}}} \cr
& {\text{Therefore, }}q = \frac{{g + 1}}{2} \cr} $$