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# 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 $u = \dfrac{U}{V}\alpha {T^4}$ and pressure $P = \dfrac{1}{3}\left( {\dfrac{U}{V}} \right)$ . If the shell now undergoes an adiabatic expansion the relation between T and R is:a. $T\alpha {e^{ - R}}$b. $T\alpha {e^{ - 3R}}$c. $T\alpha \dfrac{1}{R}$d. $T\alpha \dfrac{1}{{{R^3}}}$

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Hint: Diffusing thermal energy within one material or between materials in contact is called thermal conductivity. In thermal conduction, the heat directly is proportional to the area of cross section. Using this we have to find the relation between the temperature and R.

Complete step by step answer:
Diffusing thermal energy within one material or between materials in contact is called thermal conductivity.
As the temperature increases in an object then the kinetic energy also increases. This results in the collisions between molecules distributions.
The main mode of heat transfer in between the solids is conduction. In thermal conduction, the heat directly is proportional to the area of cross section.
It is proportional to the change in temperature. Greater the molecular agitation greater is the heat conduction.

A substance is said to be a good conductor of heat if the value of coefficient of thermal conductivity is large.
Silver is the best conductor of heat among all the materials.
Gases are less efficient conductors than the liquid. Conductor increases when the resistance of the body decreases.
Conductors are the main mode of heat transfer in case of solids.
In a spherical shell, the internal energy per unit volume is given by $\dfrac{U}{V}\alpha {T^4}$
$\Rightarrow U = CV{T^4}$
Here C is a constant

Next the value of P,
$\Rightarrow P = \dfrac{1}{3}\left( {\dfrac{U}{V}} \right) = \dfrac{1}{3}\left( {\dfrac{{CV{T^4}}}{V}} \right)$
From adiabatic expansion, $dQ = 0$ and $dU = - dW$
$\Rightarrow d{\left( {CVT} \right)^4} = - PdV$
We get,
$\Rightarrow 4VdT = - \dfrac{4}{3}TdV$
$\Rightarrow \dfrac{{dT}}{T} = \dfrac{{dV}}{{3V}}$
On integrating,
$\Rightarrow T{V^{\dfrac{1}{3}}} = {C^1}$
$\Rightarrow T{\left( {\dfrac{4}{3}\pi {R^3}} \right)^{\dfrac{1}{3}}} = {C^1}$
$TR =$ Constant
$\Rightarrow T\alpha \dfrac{1}{R}$

Hence, the correct answer is option (C).

Note: For heat exchangers, the thermal conductors are very essential and they allow heat to be exchanged between liquids without mixing them.
The heat produced is transferred by thermal conduction. A substance is said to be a good conductor of heat if the value of coefficient of thermal conductivity is large.