Black Body Radiation Adiabatic Expansion

The question describes a spherical cavity filled with black body radiation undergoing an adiabatic expansion. We are given the initial volume and temperature and need to find the final temperature after the volume doubles.

For black body radiation, the energy density $\rho$ is related to temperature $T$ by the Stefan-Boltzmann law, $\rho = a T^4$, where $a$ is a constant. The total internal energy $U$ in a volume $V$ is $U = \rho V = a T^4 V$.

Black body radiation exerts a pressure $P = \frac{1}{3} \rho = \frac{1}{3} a T^4$.

An adiabatic process is one where there is no heat exchange with the surroundings. For such a process, the first law of thermodynamics states $dU + P dV = 0$.

Substituting the expressions for $U$ and $P$: $d(aT^4V) + \left(\frac{1}{3} a T^4\right) dV = 0$

Expanding the first term using the product rule $d(uv) = u dv + v du$: $a(4T^3 dT \cdot V + T^4 dV) + \frac{1}{3} a T^4 dV = 0$

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

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

Rearrange the terms to separate variables $T$ and $V$: $\frac{dT}{T} = -\frac{1}{3} \frac{dV}{V}$

Integrate both sides: $\int \frac{dT}{T} = \int -\frac{1}{3} \frac{dV}{V}$ $\ln(T) = -\frac{1}{3} \ln(V) + \text{constant}$ $\ln(T) + \frac{1}{3} \ln(V) = \text{constant}$ $\ln(T) + \ln(V^{1/3}) = \text{constant}$ $\ln(TV^{1/3}) = \text{constant}$

This gives the adiabatic relation for black body radiation: $TV^{1/3} = \text{constant}$.

Now, we apply this relation to the given initial and final conditions:

• Initial state: Volume $V_1 = V$, Temperature $T_1 = 300$ K
• Final state: Volume $V_2 = 2V$, Temperature $T_2$ (unknown)

Using the adiabatic relation:

$\qquad T_1 V_1^{1/3} = T_2 V_2^{1/3}$ $\qquad 300 \times V^{1/3} = T_2 \times (2V)^{1/3}$ $\qquad 300 \times V^{1/3} = T_2 \times 2^{1/3} \times V^{1/3}$

Divide both sides by $V^{1/3}$ (assuming $V \neq 0$): $\qquad 300 = T_2 \times 2^{1/3}$

Solve for $T_2$: $\qquad T_2 = \frac{300}{2^{1/3}}$

Now, we calculate the value: $\qquad 2^{1/3} \approx 1.2599$ $\qquad T_2 \approx \frac{300}{1.2599} \approx 238.11$ K

We need to find the option closest to 238.11 K. Let's check the given options:

• 150 K
• 300 K
• 250 K
• 240 K

Comparing 238.11 K with the options, 240 K is the closest value.

Thus, after the adiabatic expansion to a volume 2V, the temperature of the black body radiation will be closest to 240 K.