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

$U=\underset{n}{鈭�}\frac{{蔚}_n}{\left.e^{\frac{\left({蔚}_n-渭\right)}{k_{B}T}-1}\right)}$

Here,

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

$k_B$= Boltzmann's constant,

$T$= temperature.

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

Substitute the value of $蔚g\left(蔚\right)\text{=}{蔚}_n\text{in the equation}U=\underset{n}{鈭�}\frac{{蔚}_n}{(e^{\frac{\left({蔚}_n-渭\right)}{k_{B}T}}-1)}$

$U={鈭�}_0^{鈭�}\frac{蔚g\left(蔚\right)}{\left.e^{\frac{\left({蔚}_n-渭\right)}{k_{B}T}}-1\right)}d蔚$

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

$U=\frac{2}{\sqrt{蟺}}{\left(\frac{2蟺m}{h^2}\right)}^{3/2}V{鈭�}_0^{鈭�}\frac{{蔚}^{3/2}}{\left(e^{(蔚-渭)/k_{B}T}-1\right)}d蔚$

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

For $T<T_c$set $渭=0$and $x=\frac{蔚}{k_{B}T}$in the expression $U=\frac{2}{\sqrt{蟺}}{\left(\frac{2蟺m}{h^2}\right)}^{3/2}V{鈭�}_0^{鈭�}\frac{{蔚}^{3/2}}{\left(e^{(蔚-渭)/k_{B}T}-1\right)}d蔚$

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

We know,

${鈭�}_0^{鈭�}\left(\frac{x^{3/2}}{e^x-1}\right)dx=\frac{3}{4}\sqrt{蟺}味\left(\frac{5}{2}\right)$

$=1.783$

so,

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

Specific heat at constant volume of system=

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

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

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

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

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

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

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

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 =

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

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

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

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

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