91Ó°ÊÓ

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

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

Where;

$k_B→$Boltzmann's constant

$h→$Planck's constant

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

$\mathit{T}\mathbf{→}$Temperature

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

(a)

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

Since the expression for the heat capacity per unit is

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

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

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

That is then just inserted in equation (2)

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

(b)

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

$\begin{array}{c}\frac{1}{V}\frac{∂U}{∂T}=\left(\frac{32{\mathrm{Ï€}}^5{\mathrm{κ}}_B^4}{15h^3{c}^3}\right)T^3\\ \frac{1}{V}\frac{∂U}{∂T}=\frac{32{\mathrm{Ï€}}^5{(1.38×{10}^{−23}//K)}^4{(300\text{K})}^3}{15(6.63×{10}^{−34}\text{J}.\text{s}){(3×{10}^8\text{m}/\text{s})}^3}\\ \frac{1}{V}\frac{∂U}{∂T}=8.12×{10}^{−8}\text{J}/\text{K}â‹\dots {\text{m}}^3\end{array}$