Physics - Heat & Thermo Question with Solution | TestHub

$\because$in an adiabatic process.

$Q=0$

So by first law of thermodynamics

$Q=dU+W$

$\Rightarrow 0=dU+W$

$\Rightarrow W=-dU$

$\therefore W=PdV$

$\Rightarrow PdV=-dU$....(i)

Given that$\frac{U}{V}\propto T^4\Rightarrow U=kV{T}^4$

$\Rightarrow dU=kd\left(V{T}^4\right)=K\left(T^{4}dV+4T^{3}V dT\right)$

Also,$P=\frac{1}{3}\frac{U}{V}=\frac{1}{3}\frac{kV{T}^4}{V}=\frac{K{T}^4}{3}$

Putting these values in equation (i)

$\Rightarrow \frac{K{T}^4}{3}dV=-k\left(T^{4}dV+4T^{3}V dT\right)$

$\Rightarrow \frac{TdV}{3}=-TdV-4VdT$

$\Rightarrow \frac{4T}{3}dV=-4VdT$

$\Rightarrow \frac{1}{3}\frac{dV}{V}=-\frac{dT}{T}$

$\Rightarrow \frac{1}{3}\ln V=-\ln T \Rightarrow \ln V=\ln T^{-3}$

$\Rightarrow V{T}^3=\text{constant}$

$\frac{4}{3}\pi R^3 T^3=\text{constant}$

$RT=\text{constant}$

$\Rightarrow T\propto \frac{1}{R}$