Question

Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume $u=\frac{U}{V}\propto T^4$ and pressure $P=\frac{1}{3}\left(\frac{U}{V}\right)$. If the shell now undergoes an adiabatic expansion the relation between T and R is:

• A. $T\propto \frac{1}{R}$
• B. $T\propto \frac{1}{R^3}$
• C. $T\propto e^{-R}$
• D. $T\propto e^{-3R}$

Answer 1

The correct option is A $T\propto \frac{1}{R}$

Since the ideal gas is taken in the spherical ball, so we can use $PV=nRT$
According to the given data
$P=\frac{1}{3}\left(\frac{U}{V}\right)$...(i)
and
$u=\frac{U}{V}\propto T^4$
$\frac{U}{V}=k{T}^4$, where k is proportionality constant.
$\Rightarrow U=kV{T}^4$...(ii)
Using equation (i) and (ii)
$P=\frac{1}{3}\left(\frac{kV{T}^4}{V}\right)$
$P=\frac{1}{3}k{T}^4$...(iii)
Since the given process is adiabatic
$dU=w$
$=dU=-PdV$
Using equation (ii) and (iii)
$=d\left(kV{T}^4\right)=-\frac{1}{3}k{T}^{4}dV$
$=4kV{T}^{3}dT+k{T}^{4}dV=-\frac{1}{3}k{T}^{4}dV$
$=4kV{T}^{3}dT=-\frac{4}{3}k{T}^{4}dV$
$=VdT=-\frac{1}{3}TdV$
$=\frac{dT}{T}=-\frac{1}{3V}dV$
Integrating both side :
$=lnT=-\frac{1}{3}lnV$
$\Rightarrow T{V}^{\frac{1}{3}}=$ constant....(iv)
Volume of sphere $\left(V\right)$ is $\frac{4}{3}\pi R^3$.....(v)
Using equation (iv) and (v) we get
$TR=$ constant
$\Rightarrow T\propto \frac{1}{R}$