Thermodynamics

JEE Main level, single correct, medium

160 of 190

Question 160 of 190

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=UV∝T4u=\dfrac{U}{V}\propto {T}^{4} and pressure p=13(UV)p=\dfrac{1}{3}\left(\dfrac{U}{V}\right) . If the shell now undergoes an adiabatic expansion the relation between T and R is:

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Answer: (D) T∝1RT\propto \dfrac{1}{R}

∵∵ in an adiabatic process.

Q=0Q=0

So by first law of thermodynamics

Q=dU+WQ=dU+W

⇒0=dU+W\Rightarrow 0=dU+W

⇒W=−dU\Rightarrow W=-dU

∴W=PdV∴W=PdV

⇒PdV=−dU\Rightarrow PdV=-dU ....(i)

Given that UV∝T4⇒U=kVT4\dfrac{U}{V}\propto {T}^{4}\Rightarrow U=kV{T}^{4}

⇒dU=kd(VT4)=K(T4dV+4T3VdT)\Rightarrow dU=kd\left(V{T}^{4}\right)=K\left({T}^{4}dV+4{T}^{3}VdT\right)

Also, P=13UV=13kVT4V=KT43P=\dfrac{1}{3}\dfrac{U}{V}=\dfrac{1}{3}\dfrac{kV{T}^{4}}{V}=\dfrac{K{T}^{4}}{3}

Putting these values in equation (i)

⇒KT43dV=−k(T4dV+4T3VdT)\Rightarrow \dfrac{K{T}^{4}}{3}dV=-k\left({T}^{4}dV+4{T}^{3}VdT\right)

⇒TdV3=−TdV−4VdT\Rightarrow \dfrac{TdV}{3}=-TdV-4VdT

⇒4T3dV=−4VdT\Rightarrow \dfrac{4T}{3}dV=-4VdT

⇒13dVV=−dTT\Rightarrow \dfrac{1}{3}\dfrac{dV}{V}=-\dfrac{dT}{T}

⇒13ln⁡⁡V=−ln⁡⁡T⇒ln⁡⁡V=ln⁡⁡T−3\Rightarrow \dfrac{1}{3}\ln ⁡V=-\ln ⁡T\Rightarrow \ln ⁡V=\ln ⁡{T}^{-3}

⇒VT3=constant\Rightarrow V{T}^{3}=\text{constant}

43πR3T3=constant\dfrac{4}{3}\pi {R}^{3}{T}^{3}=\text{constant}

RT=constantRT=\text{constant}

⇒T∝1R\Rightarrow T\propto \dfrac{1}{R}

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