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

Answer: (C)

Fro an adiabatic process $T{v}^{r-1}=$constant, we know that average time of collision between molecules
$\tau =\frac{1}{n\pi \sqrt{2}{V}_{rms}{d}^2}$
where $n=$ no, of molecules per unit volume
$V_{rms}=rms$ velocity of molecules
As $n\propto \frac{1}{V}$ and $V_{rms}\propto \sqrt{t}$
$\tau \propto \frac{V}{\sqrt{T}}$
Thus we can write:
$n=k_1{v}^{-1}$ and $V_{rms}=K_2{T}^{1/2}$
where $k_1$ and $k-2$ are constant
for adiabatic process $T{V}^{r-1}=$ constant
Thus we can write
$\tau \propto V{T}^{-1/2}\propto V(v^{1-r}{)}^{-1/2}$
or $\tau \propto V^{\frac{r+1}{2}}$
Average time between collision $=\frac{meansfreepath}{V_{rms}}$
$t=\frac{1}{\frac{\pi d^{2}N/V}{\sqrt{\frac{3RT}{M}}}};t=\frac{CV}{\sqrt{T}}$ where $C=\frac{\sqrt{M}}{\pi d^{2}B\sqrt{3R}}$
$\Rightarrow T\propto \frac{V^2}{t^2}$
For adiabatic
$T{V}^{\gamma -1}=k$
$\frac{V^2}{t^2}{V}^{\gamma -1}=k$
$\frac{V^{\gamma -1}}{t^2}=k,t\propto \frac{\gamma +1}{2}$
so $q=\frac{\gamma +1}{2}$