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(B) For an adiabatic process,the first law of thermodynamics is $dQ = dU + PdV = 0$,so $dU = -PdV$.Given $U = V \cdot E = V \cdot (aT^4)$,where $a$ is

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

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(B) For an adiabatic process,the first law of thermodynamics is $dQ = dU + PdV = 0$,so $dU = -PdV$.Given $U = V \cdot E = V \cdot (aT^4)$,where $a$ is a constant.Then $dU = d(aVT^4) = a(T^4 dV + 4VT^3 dT)$.Substituting into $dU = -PdV$ with $P = \frac{1}{3} aT^4$:$aT^4 dV + 4aVT^3 dT = -\frac{1}{3} aT^4 dV$.Rearranging terms: $4aVT^3 dT = -\frac{4}{3} aT^4 dV$.Dividing by $4aVT^3$: $\frac{dT}{T} = -\frac{1}{3} \frac{dV}{V}$.Integrating both sides: $\ln T = -\frac{1}{3} \ln V + C$,which implies $T \propto V^{-1/3}$.Since $V = \frac{4}{3} \pi R^3$,we have $V \propto R^3$.Therefore,$T \propto (R^3)^{-1/3} = \frac{1}{R}$.

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