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 $u=\dfrac{U}{V}\propto {{T}^{4}}$ and pressure $P=\dfrac{1}{3}(\dfrac{U}{V}).$ If the shell now undergoes an adiabatic expansion the relation between T and R is:
\[A.T\propto {{e}^{-R}}\]
\[B.T\propto {{e}^{-3R}}\]
\[C.T\propto \dfrac{1}{R}\]
$D.T\propto \dfrac{1}{{{R}^{3}}}$
Answer
Verified
486.6k+ views
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 $u=\dfrac{U}{V}\propto {{T}^{4}}.$ This can be simplified to $\dfrac{U}{V}=C{{T}^{4}}$ where C is a constant.
Hence, $U=CV{{T}^{4}}$
Next, $P=\dfrac{1}{3}(\dfrac{U}{V}). $Substituting the value of U in this equation, $P=\dfrac{1}{3}(\dfrac{CV{{T}^{4}}}{V})$
$P=\dfrac{C{{T}^{4}}}{3}$
Now, we will use the adiabatic expansion condition, $dQ=dU+dW=0$
$dU=-dW$
We know that, $dW=PdV$
Substituting the values of U and P into the above equation, we get
$d(CV{{T}^{4}})=-(\dfrac{C{{T}^{4}}}{3})dV$
Using differentiation by parts $d(uv)=vdu+udv$
$4CV{{T}^{3}}dT+C{{T}^{4}}dV=\dfrac{-C{{T}^{4}}}{3}dV$
$4CV{{T}^{3}}dT=\dfrac{-C{{T}^{4}}}{3}dV-C{{T}^{4}}dV$
$4VdT=-(\dfrac{T}{3}+T)dV$
\[4VdT=\dfrac{-4T}{3}dV\]
\[\dfrac{1}{T}dT=\dfrac{-1}{3V}dV\]
\[\ln T=\dfrac{-1}{3}\ln V+\ln C\]
$\ln T+\ln ({{V}^{\dfrac{1}{3}}})=\ln C$
$\ln (T{{V}^{\dfrac{1}{3}}})=\ln C$
$T{{V}^{\dfrac{1}{3}}}=C$
We know that the volume of a spherical shell is, $V=\dfrac{4}{3}\pi {{R}^{3}}.$
Therefore, $T{{(\dfrac{4}{3}\pi {{R}^{3}})}^{\dfrac{1}{3}}}=C$
$TR{{(\dfrac{4\pi }{3})}^{\dfrac{1}{3}}}=C$
Hence, $TR=D,$ where $D=\dfrac{C}{{{(\dfrac{4\pi }{3})}^{\dfrac{1}{3}}}}.$ That means D is a constant value.
Therefore, $T\propto \dfrac{1}{R}$
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 $d(uv)=vdu+udv.$
Step by step solution:
We have a spherical shell of radius R at temperature T. Internal energy per unit volume $u=\dfrac{U}{V}\propto {{T}^{4}}.$ This can be simplified to $\dfrac{U}{V}=C{{T}^{4}}$ where C is a constant.
Hence, $U=CV{{T}^{4}}$
Next, $P=\dfrac{1}{3}(\dfrac{U}{V}). $Substituting the value of U in this equation, $P=\dfrac{1}{3}(\dfrac{CV{{T}^{4}}}{V})$
$P=\dfrac{C{{T}^{4}}}{3}$
Now, we will use the adiabatic expansion condition, $dQ=dU+dW=0$
$dU=-dW$
We know that, $dW=PdV$
Substituting the values of U and P into the above equation, we get
$d(CV{{T}^{4}})=-(\dfrac{C{{T}^{4}}}{3})dV$
Using differentiation by parts $d(uv)=vdu+udv$
$4CV{{T}^{3}}dT+C{{T}^{4}}dV=\dfrac{-C{{T}^{4}}}{3}dV$
$4CV{{T}^{3}}dT=\dfrac{-C{{T}^{4}}}{3}dV-C{{T}^{4}}dV$
$4VdT=-(\dfrac{T}{3}+T)dV$
\[4VdT=\dfrac{-4T}{3}dV\]
\[\dfrac{1}{T}dT=\dfrac{-1}{3V}dV\]
\[\ln T=\dfrac{-1}{3}\ln V+\ln C\]
$\ln T+\ln ({{V}^{\dfrac{1}{3}}})=\ln C$
$\ln (T{{V}^{\dfrac{1}{3}}})=\ln C$
$T{{V}^{\dfrac{1}{3}}}=C$
We know that the volume of a spherical shell is, $V=\dfrac{4}{3}\pi {{R}^{3}}.$
Therefore, $T{{(\dfrac{4}{3}\pi {{R}^{3}})}^{\dfrac{1}{3}}}=C$
$TR{{(\dfrac{4\pi }{3})}^{\dfrac{1}{3}}}=C$
Hence, $TR=D,$ where $D=\dfrac{C}{{{(\dfrac{4\pi }{3})}^{\dfrac{1}{3}}}}.$ That means D is a constant value.
Therefore, $T\propto \dfrac{1}{R}$
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 $d(uv)=vdu+udv.$
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success
Master Class 11 Computer Science: Engaging Questions & Answers for Success
Master Class 11 Maths: Engaging Questions & Answers for Success
Master Class 11 Social Science: Engaging Questions & Answers for Success
Master Class 11 Economics: Engaging Questions & Answers for Success
Master Class 11 Business Studies: Engaging Questions & Answers for Success
Trending doubts
10 examples of friction in our daily life
What problem did Carter face when he reached the mummy class 11 english CBSE
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
State and prove Bernoullis theorem class 11 physics CBSE
The sequence of spore production in Puccinia wheat class 11 biology CBSE