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Question: Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be c...

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}(\dfrac{U}{V}). If the shell now undergoes an adiabatic expansion the relation between T and R is:
A.T∝e−RA.T\propto {{e}^{-R}}
B.T∝e−3RB.T\propto {{e}^{-3R}}
C.T∝1RC.T\propto \dfrac{1}{R}
D.T∝1R3D.T\propto \dfrac{1}{{{R}^{3}}}

Explanation

Solution

Hint: The basic thermodynamic laws and equations are required to solve this problem. Basic differentiation and integration will be required too.

Step by step solution:
We have a spherical shell of radius R at temperature T. Internal energy per unit volume u=UV∝T4.u=\dfrac{U}{V}\propto {{T}^{4}}. This can be simplified to UV=CT4\dfrac{U}{V}=C{{T}^{4}} where C is a constant.

Hence, U=CVT4U=CV{{T}^{4}}

Next, P=13(UV).P=\dfrac{1}{3}(\dfrac{U}{V}). Substituting the value of U in this equation, P=13(CVT4V)P=\dfrac{1}{3}(\dfrac{CV{{T}^{4}}}{V})

P=CT43P=\dfrac{C{{T}^{4}}}{3}

Now, we will use the adiabatic expansion condition, dQ=dU+dW=0dQ=dU+dW=0

dU=−dWdU=-dW

We know that, dW=PdVdW=PdV

Substituting the values of U and P into the above equation, we get

d(CVT4)=−(CT43)dVd(CV{{T}^{4}})=-(\dfrac{C{{T}^{4}}}{3})dV

Using differentiation by parts d(uv)=vdu+udvd(uv)=vdu+udv

4CVT3dT+CT4dV=−CT43dV4CV{{T}^{3}}dT+C{{T}^{4}}dV=\dfrac{-C{{T}^{4}}}{3}dV
4CVT3dT=−CT43dV−CT4dV4CV{{T}^{3}}dT=\dfrac{-C{{T}^{4}}}{3}dV-C{{T}^{4}}dV
4VdT=−(T3+T)dV4VdT=-(\dfrac{T}{3}+T)dV
4VdT=−4T3dV4VdT=\dfrac{-4T}{3}dV
1TdT=−13VdV\dfrac{1}{T}dT=\dfrac{-1}{3V}dV
ln⁡T=−13ln⁡V+ln⁡C\ln T=\dfrac{-1}{3}\ln V+\ln C
ln⁡T+ln⁡(V13)=ln⁡C\ln T+\ln ({{V}^{\dfrac{1}{3}}})=\ln C
ln⁡(TV13)=ln⁡C\ln (T{{V}^{\dfrac{1}{3}}})=\ln C
TV13=CT{{V}^{\dfrac{1}{3}}}=C

We know that the volume of a spherical shell is, V=43πR3.V=\dfrac{4}{3}\pi {{R}^{3}}.

Therefore, T(43πR3)13=CT{{(\dfrac{4}{3}\pi {{R}^{3}})}^{\dfrac{1}{3}}}=C
TR(4π3)13=CTR{{(\dfrac{4\pi }{3})}^{\dfrac{1}{3}}}=C

Hence, TR=D,TR=D, where D=C(4π3)13.D=\dfrac{C}{{{(\dfrac{4\pi }{3})}^{\dfrac{1}{3}}}}. That means D is a constant value.

Therefore, T∝1RT\propto \dfrac{1}{R}

Note: Remembering the basic thermodynamic equations is necessary and how they change during different conditions such as Isothermal expansion, Isobaric expansion etc.
The differentiation by parts is easy only when you remember the formula well d(uv)=vdu+udv.d(uv)=vdu+udv.