JEE Main 2015 — Thermodynamics Question with Solution

For an adiabatic process $T{V}^{\gamma -1}=$ constant.
We know that average time of collision between molecules

$\tau =\frac{1}{n\pi \sqrt{2}{v}_{\mathrm{rms}}{d}^2}$

where, $n=$ number of molecules per unit volume $V_{\mathrm{rms} }=$ rms velocity of molecules

As $n\propto \frac{1}{V}$ and $v_{\mathrm{rms}}\propto \sqrt{T}$

$\tau \propto \frac{V}{\sqrt{T}}$

Thus, we can write

$n=K_1{V}^{-1}$ and $V_{\mathrm{rms}}=K_2{T}^{\frac{1}{2}}$.

where, $K_1$ and $K_2$ are constants.

For adiabatic process, $T{V}^{\gamma -1}=$ constant. Thus, we can write

$\tau \propto V{T}^{-\frac{1}{2}}\propto V{\left(V^{1-\gamma }\right)}^{\frac{-1}{2}}$

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