Figure 7.37 shows the heat capacity of a Bose gas as a function of temperature. In this problem you will calculate the shape of this unusual graph.

(a) Write down an expression for the total energy of a gas of $N$bosons confined to a volume $V$, in terms of an integral (analogous to equation 7.122).

(b) For $T<T_c$you can set $\mu =0$. Evaluate the integral numerically in this case, then differentiate the result with respect to $T$to obtain the heat capacity. Compare to Figure 7.37.

(c) Explain why the heat capacity must approach $\frac{3}{2}Nk$in the high- $T$limit.

(d) For $T>T_c$you can evaluate the integral using the values of $\mu$calculated in Problem 7.69. Do this to obtain the energy as a function of temperature, then numerically differentiate the result to obtain the heat capacity. Plot the heat capacity, and check that your graph agrees with Figure 7.37.

Figure 7.37. Heat capacity of an ideal Bose gas in a three-dimensional box.

The energy expression for the gas that satisfy the Bose- Einstein's statistics is

$U=\sum _n\frac{{\epsilon }_n}{\left.e^{\frac{\left({\epsilon }_n-\mu \right)}{k_{B}T}-1}\right)}$

Here,

${\epsilon }_n$= energy of the $n^{th}$particle,

$k_B$= Boltzmann's constant,

$T$= temperature.

The energy of the $n^{th}$particle is ${\epsilon }_n=\epsilon g\left(\epsilon \right)$where $g\left(\epsilon \right)$is the density of states

Substitute the value of $\epsilon g\left(\epsilon \right)\text{=}{\epsilon }_n\text{in the equation}U=\sum _n\frac{{\epsilon }_n}{(e^{\frac{\left({\epsilon }_n-\mu \right)}{k_{B}T}}-1)}$

$U={\int }_0^{\infty }\frac{\epsilon g\left(\epsilon \right)}{\left.e^{\frac{\left({\epsilon }_n-\mu \right)}{k_{B}T}}-1\right)}d\epsilon$

Substituting the value of $\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V\sqrt{\epsilon }\text{=}g\left(\epsilon \right)$

$U=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\int }_0^{\infty }\frac{{\epsilon }^{3/2}}{\left(e^{(\epsilon -\mu )/k_{B}T}-1\right)}d\epsilon$

Thus, the total energy of a gas of N bosons confined to a volume V= $U=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\int }_0^{\infty }\frac{{\epsilon }^{3/2}}{\left(e^{(\epsilon -\mu )/k_{B}T}-1\right)}d\epsilon$

For $T<T_c$set $\mu =0$and $x=\frac{\epsilon }{k_{B}T}$in the expression $U=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\int }_0^{\infty }\frac{{\epsilon }^{3/2}}{\left(e^{(\epsilon -\mu )/k_{B}T}-1\right)}d\epsilon$

$U=\frac{2}{\sqrt{\pi }}\left(\frac{2\pi m}{h^2}\right)V{\left(k_{B}T\right)}^{5/2}{\int }_0^{\infty }\left(\frac{x^{3/2}}{e^x-1}\right)dx$

We know,

${\int }_0^{\infty }\left(\frac{x^{3/2}}{e^x-1}\right)dx=\frac{3}{4}\sqrt{\pi }\zeta \left(\frac{5}{2}\right)$

$=1.783$

so,

$U=\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\left(k_{B}T\right)}^{5/2}(1.783)$

Specific heat at constant volume of system=

$C_V={\left(\frac{\partial U}{\partial T}\right)}_V$

$={\left(\frac{\partial }{\partial T}\left(\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\left(k_{B}T\right)}^{5/2}(1.783)\right)\right)}_V$

$=\left(\frac{5}{2}\right)(1.783)\left(\frac{2}{\sqrt{\pi }}\right){\left(\frac{2\pi m}{h^2}\right)}^{\frac{3}{2}}V{\left(k_{B}T\right)}^{3/2}{k}_B$

$=5.031{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\left(k_{B}T\right)}^{3/2}{k}_B$

Using the expression $k_B{T}_C=0.527\left(\frac{h^2}{2\pi m}\right){\left(\frac{N}{V}\right)}^{2/3}$

$\frac{C_V}{N{k}_B}=\left(\frac{5.031}{2.612}\right){\left(\frac{T}{T_C}\right)}^{3/2}$

$=1.926{\left(\frac{T}{T_C}\right)}^{3/2}$

This expression shows the nature of slope in the figure 7.37.

For $T=T_c$, the constant $\frac{C_V}{N{k}_B}$is,

$\frac{C_V}{N{k}_B}=1.926{\left(\frac{T_C}{T_C}\right)}^{3/2}$

$=1.926$

The system ought to behave like a standard substance gas with 3 degrees of freedom at the upper temperature. By the equipartition theorem, the heat capacity should be$\frac{3}{2}N{K}_B$

The energy of the system =

$U=\frac{2}{\sqrt{\pi }}\left(\frac{2\pi m}{h^2}\right)V{\left(k_{B}T\right)}^{5/2}{\int }_0^{\infty }\left(\frac{x^{3/2}}{e^{\frac{x-c}{t}}-1}\right)dx$

$=(0.432)N{k}_B{T}_C{\int }_0^{\infty }\left(\frac{x^{3/2}}{e^{\frac{x-c}{t}}-1}\right)dx$

$\frac{U}{N{k}_B{T}_C}=(0.432){\int }_0^{\infty }\left(\frac{x^{3/2}}{e^{\frac{x-c}{t}}-1}\right)dx$

$\text{At}t=\frac{T}{T_C}\text{, the above equation is numerically equal to}\frac{C_V}{N{k}_B}\text{.}$

Given below is the graph between $\frac{C_V}{N{k}_B}\text{and}\frac{T}{T_C}$