Heat capacity (at constant volume) is defined as$\mathbf{\partial }\mathit{U}\mathbf{/}\mathbf{\partial }\mathit{T}$. (a) Using a result derived in Example 9.6. obtain an expression for the heat capacity per unit volume, in$\mathit{J}\mathbf{/}\mathbf{\text{K}}\mathbf{\cdot }{\mathbf{\text{mi}}}^{\mathbf{3}}$, of a photon gas. (b) What is its value at$\mathbf{300}\mathbf{}\mathbf{\text{K}}$?

Expression for energy of photons in a container$\mathit{U}$is given by-

$\mathit{U}\mathbf{=}\frac{\mathbf{8}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{V}{\mathbf{k}}_{\mathbf{B}}^{\mathbf{4}}{\mathbf{T}}^{\mathbf{4}}}{\mathbf{15}{\mathbf{h}}^{\mathbf{3}}{\mathbf{c}}^{\mathbf{3}}}\mathbf{\dots }\mathbf{.}\mathbf{..}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Where;

$k_B\to$Boltzmann's constant

$h\to$Planck's constant

$\mathit{c}\mathbf{\to }$Speed of light in vacuum

$\mathit{T}\mathbf{\to }$Temperature

$\mathit{V}\mathbf{\to }$Volume of container

(a)

It is given that Heat capacity$=\frac{\partial U}{\partial T}$

Since the expression for the heat capacity per unit is

$\frac{1}{V}=\frac{\partial U}{\partial T}\dots \dots \mathrm{..}\left(2\right)$

The derivative of equation (1) needs to be taken with respect toT

$\begin{array}{c}U=\frac{8{\pi }^{2}V{k}_B^4{T}^4}{15h^3{c}^3}n\partial \\ U=4\left(\frac{8{\pi }^{5}V{k}_B^4}{15h^3{c}^3}\right)T^3\partial T\\ \partial U=\frac{32{\pi }^{5}V{k}_B^4{T}^3}{15h^3{c}^3}\partial T\\ \partial U=\frac{32{\pi }^{5}V{k}_B^4{T}^3}{15h^3{c}^3}\end{array}$

That is then just inserted in equation (2)

$\begin{array}{c}\frac{1}{V}\frac{\partial U}{\partial T}=\frac{1}{V}\left(\frac{32{\pi }^{5}V{k}_B^4{T}^3}{15h^3{c}^3}\right)\\ \frac{1}{V}\frac{\partial U}{\partial T}=\frac{32{\pi }^{5}V{k}_B^4{T}^3}{15h^3{c}^3}\end{array}$

(b)

To find the heat capacity per unit volume at$300\text{K}$, substitute$1.38\times {10}^{-32}\text{J}/\text{K}$ for$k_B,6.63\times {10}^{-34}\text{J}.\text{s}$for h, $3\times {10}^8\text{m}/\text{s}$for $\text{c}$, and$300\text{K}$for $T$in equation

$\begin{array}{c}\frac{1}{V}\frac{\partial U}{\partial T}=\left(\frac{32{\pi }^5{\kappa }_B^4}{15h^3{c}^3}\right)T^3\\ \frac{1}{V}\frac{\partial U}{\partial T}=\frac{32{\pi }^5{(1.38\times {10}^{-23}//K)}^4{(300\text{K})}^3}{15(6.63\times {10}^{-34}\text{J}.\text{s}){(3\times {10}^8\text{m}/\text{s})}^3}\\ \frac{1}{V}\frac{\partial U}{\partial T}=8.12\times {10}^{-8}\text{J}/\text{K}\cdot {\text{m}}^3\end{array}$