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 = U \over V\, \propto \,T^4 and pressure p = 1 \over 3( U \over V ) . If the shell now undergoes an adiabatic expansion the relation between T and R is:

Heat and Thermodynamics PYQ — JEE Main 2015 · Shift 0 · Q62 with Solution

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Correct answer: A

For black body radiation inside the spherical shell, $u = \frac{U}{V} \propto T^4$ and $p = \frac{1}{3}u = \frac{1}{3}\frac{U}{V}.$

Let $u = aT^4$ for some constant $a$ . Then $U = uV = aT^4V.$

Also, $p = \frac{1}{3}aT^4.$

For an adiabatic process, $dQ = 0.$ From the first law of thermodynamics, $dU + p\,dV = 0.$

Now, $U = aT^4V.$ So, $dU = aV\,d(T^4) + aT^4\,dV.$

Since $d(T^4)=4T^3\,dT,$ we get $dU = 4aVT^3\,dT + aT^4\,dV.$

Substitute into $dU + p\,dV = 0:$ $4aVT^3\,dT + aT^4\,dV + \frac{1}{3}aT^4\,dV = 0.$

Combine the last two terms: $4aVT^3\,dT + \frac{4}{3}aT^4\,dV = 0.$

Divide by $4aT^3V$ : $\frac{dT}{T} + \frac{1}{3}\frac{dV}{V} = 0.$

Thus, $\frac{dT}{T} = -\frac{1}{3}\frac{dV}{V}.$

Integrating, $\ln T = -\frac{1}{3}\ln V + \text{constant}$ $T \propto V^{-1/3}.$

For a sphere, $V = \frac{4}{3}\pi R^3 \propto R^3.$

Therefore, $T \propto (R^3)^{-1/3} = R^{-1}.$

So, $T \propto \frac{1}{R}.$

Hence the correct option is A.