Question

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes 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
$(\gamma =\frac{C_p}{C_v})$

• A. $\frac{3\gamma +5}{6}$
• B. $\frac{3\gamma -5}{6}$
• C. $\frac{\gamma +1}{2}$
• D. $\frac{\gamma -1}{2}$

Answer 1

The correct option is C $\frac{\gamma +1}{2}$

Average time of collision $\tau =\frac{1}{n\pi \sqrt{2}{v}_{rms}{d}^2}$----(1)
Since $v_{rms}\propto \sqrt{T}$ and$n\propto \frac{1}{V}$

For an adiabatic process
$T{V}^{\gamma -1}$ = constant
$\sqrt{T}\propto V^{\frac{1-\gamma }{2}}$
so, $\tau \propto \frac{V}{\sqrt{T}}$(from 1)

$\therefore \tau \propto V^{\frac{1+\gamma }{2}}$