Question

Consider a gas of noninteracting spin-0 bosons at high temperatures, when $\mathit{T}\mathbf{\gg }{\mathit{T}}_{\mathbf{c}}$. (Note that “high” in this sense can still mean below 1 K.)

1. Show that, in this limit, the Bose-Einstein function can be written approximately as
   ${\overline{\mathbf{n}}}_{\mathbf{B}\mathbf{E}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\left(\in -\mu \right)\mathbf{/}\mathbf{k}\mathbf{T}}\left[1+e^{-\left(\in -\mu \right)/kT}+\cdots \right]$.
   
2. Keeping only the terms shown above, plug this result into equation 7.122 to derive the first quantum correction to the chemical potential for gas of bosons.
   
3. Use the properties of the grand free energy (Problems 5.23 and 7.7) to show that the pressure of any system is given by In $\mathit{P}\mathbf{=}\left(kT/V\right)$, where $\mathit{Z}$is the grand partition function. Argue that, for gas of noninteracting particles, In $\mathit{Z}$can be computed as the sum over all modes (or single-particle states) of In ${\mathit{Z}}_{\mathbf{i}}$, where ${\mathit{Z}}_{\mathbf{i}}$; is the grand partition function for the $\mathit{i}\mathbf{th}$mode.
   
4. Continuing with the result of part (c), write the sum over modes as an integral over energy, using the density of states. Evaluate this integral explicitly for gas of noninteracting bosons in the high-temperature limit, using the result of part (b) for the chemical potential and expanding the logarithm as appropriate. When the smoke clears, you should find
   $\mathit{p}\mathbf{=}\frac{\mathbf{N}\mathbf{k}\mathbf{T}}{\mathbf{V}}\left(1-\frac{N{v}_Q}{4\sqrt{2}V}\right)$,
   again neglecting higher-order terms. Thus, quantum statistics results in a lowering of the pressure of a boson gas, as one might expect.
5. Write the result of part (d) in the form of the virial expansion introduced in Problem 1.17, and read off the second virial coefficient, $\mathit{B}\left(T\right)$. Plot the predicted $\mathit{B}\left(T\right)$for a hypothetical gas of noninteracting helium-4 atoms.
6. Repeat this entire problem for gas of spin-1/2 fermions. (Very few modifications are necessary.) Discuss the results, and plot the predicted virial coefficient for a hypothetical gas of noninteracting helium-3 atoms.

Short answer

1. The Bose-Einstein distribution is given as ${\overline{n}}_{BE}=e^{-\left(\epsilon -\mu \right)/kT}\left[1+e^{-\left(\epsilon -\mu \right)/kT}\right]$.
2. The chemical potential is $\mu =-kT\ln \left(\frac{V}{K{v}_Q}\right)-\frac{kTN{v}_Q}{\sqrt{8}V}$.
3. The pressure of a gas is given as $P=\left(\frac{kT}{V}\right)\ln Z$and the logarithm of the grand partition function is computed as a sum over all modes.
4. The pressure of a gas is $P=\frac{NkT}{V}\left(1-\frac{N{v}_Q}{4\sqrt{2}V}\right)$.
5. The value of the virial coefficient for helium is given as $\beta \left(T\right)=-\left(7.089\times {10}^{-5}{\text{\hspace{0.17em}m}}^{\text{3}}\cdot K^{3/2}/\text{mol}\right)T^{-3/2}$.
6. For spin, half particle number density is given as $N=\frac{2V}{v_Q}{e}^{\mu /kT}\left[1-\frac{e^{\mu /kT}}{\sqrt{8}}\right]$. The viral coefficient is given as $\beta \left(T\right)=\frac{N_A}{8\sqrt{2}}{\left(\frac{h^2}{2\pi mkT}\right)}^{3/2}$and the value of the virial coefficient is given as $\beta \left(T\right)=\left(5.457\times {10}^{-5}{\text{\hspace{0.17em}m}}^{\text{3}}\cdot K^{3/2}/\text{mol}\right)T^{-3/2}$.

Step-by-step solution

Part a. Step 1. Given.

There are non-interacting spin-0 bosons at high temperatures.

Part a. Step 2. Formula used.

Write the expression for Bose-Einstein Distribution.

${\overline{n}}_{BE}=\frac{1}{e^{\left(\epsilon -\mu \right)/kT}-1}$ …… (1)

Here, ${\overline{n}}_{BE}$ is the Bose-Einstein Distribution, $\epsilon$ is the energy of bosons, $\mu$ is the chemical potential, $k$ is the Boltzmann constant and $T$is the temperature.

Part a. Step 3. Calculation.

Multiply and divide the RHS by $e^{-\left(\epsilon -\mu \right)/kT}$of equation (1).

${\overline{n}}_{BE}=\frac{e^{-\left(\epsilon -\mu \right)/kT}}{1-e^{-\left(\epsilon -\mu \right)/kT}}$

At high temperatures, the factor $e^{-\left(\epsilon -\mu \right)/kT}$is less than 1, so using binomial expansion.

${\overline{n}}_{BE}=e^{-\left(\epsilon -\mu \right)/kT}\left[1+e^{-\left(\epsilon -\mu \right)/kT}\right]$

Part a. Step 4. Conclusion.

Thus, the Bose-Einstein distribution is given as ${\overline{n}}_{BE}=e^{-\left(\epsilon -\mu \right)/kT}\left[1+e^{-\left(\epsilon -\mu \right)/kT}\right]$.

Part b. Step 1. Given.

There are non-interacting spin-0 bosons at high temperatures.

Part b. Step 2. Formula used.

Write the expression for Bose-Einstein Distribution.

$N=\underset{0}{\overset{\infty }{\int }}g\left(\epsilon \right){\overline{n}}_{BE}d\epsilon$ …… (2)

Here, $N$is the number density, $g\left(\epsilon \right)$is the energy density and ${\overline{n}}_{BE}$is Bose-Einstein Distribution.

Write the expression for Bose-Einstein Distribution.

${\overline{n}}_{BE}=e^{-\left(\epsilon -\mu \right)/kT}\left[1+e^{-\left(\epsilon -\mu \right)/kT}\right]$

Here, ${\overline{n}}_{BE}$is the Bose-Einstein Distribution, $\epsilon$is the energy of bosons, $\mu$is the chemical potential, $k$is the Boltzmann constant and $T$is the temperature.

Write the expression for density of states.

$\begin{array}{l}g\left(\epsilon \right)=g_0\sqrt{\epsilon }\\ =\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V\sqrt{\epsilon }\end{array}$

Here, role="math" localid="1651740877830" $g\left(\epsilon \right)$is the density of states, role="math" localid="1651740907072" $g_0$is the constant of the density of states, $m$is the mass of particles, role="math" localid="1651740949903" $h$is the Planck’s constant, $V$is the volume of gas, and $\epsilon$is the energy of particles.

Write the expression for quantum volume.

$v_Q={\left(\frac{h^2}{2\pi mkT}\right)}^{3/2}$

Here, $v_Q$ is the quantum volume.

Part b. Step 3. Calculation.

Substitute $e^{-\left(\epsilon -\mu \right)/kT}\left[1+e^{-\left(\epsilon -\mu \right)/kT}\right]$for ${\overline{n}}_{BE}$in equation (2).

$N=\underset{0}{\overset{\infty }{\int }}g\left(\epsilon \right)e^{-\left(\epsilon -\mu \right)/kT}\left[1+e^{-\left(\epsilon -\mu \right)/kT}\right]d\epsilon$

Substitute $g_0\sqrt{\epsilon }$for $g\left(\epsilon \right)$in the above equation.

$N=g_0\underset{0}{\overset{\infty }{\int }}{e}^{-\left(\epsilon -\mu \right)/kT}\left[1+e^{-\left(\epsilon -\mu \right)/kT}\right]\sqrt{\epsilon }d\epsilon$

$N=g_0\underset{0}{\overset{\infty }{\int }}\left[e^{-\left(\epsilon -\mu \right)/kT}+e^{-2\left(\epsilon -\mu \right)/kT}\right]\sqrt{\epsilon }d\epsilon$ …… (3)

Substitute $x$for $\sqrt{\frac{\epsilon }{kT}}$and $dx$for $\frac{1}{2}kTxd\epsilon$in the first term of equation (3).

$\begin{array}{l}{g}_0\underset{0}{\overset{\infty }{\int }}{e}^{-\left(\epsilon -\mu \right)/kT}\sqrt{\epsilon }d\epsilon =g_0{e}^{\mu /kT}\underset{0}{\overset{\infty }{\int }}{e}^{-\epsilon /kT}\sqrt{\epsilon }d\epsilon \\ =2g_0{\left(kT\right)}^{3/2}{e}^{\mu /kT}\underset{0}{\overset{\infty }{\int }}{x}^2{e}^{-x^2}dx\end{array}$

Substitute $\frac{\sqrt{\pi }}{4}$for $\underset{0}{\overset{\infty }{\int }}{x}^2{e}^{-x^2}dx$in the above equation.

$\begin{array}{l}{g}_0\underset{0}{\overset{\infty }{\int }}{e}^{-\left(\epsilon -\mu \right)/kT}\sqrt{\epsilon }d\epsilon =2g_0{\left(kT\right)}^{3/2}{e}^{\mu /kT}\frac{\sqrt{\pi }}{4}\\ =\frac{\sqrt{\pi }}{2}{g}_0{\left(kT\right)}^{3/2}{e}^{\mu /kT}\end{array}$

Substitute $x$for $\sqrt{\frac{\epsilon }{kT}}$and $dx$for $\frac{1}{2}kTxd\epsilon$in the second term of equation (3).

$\begin{array}{l}{g}_0\underset{0}{\overset{\infty }{\int }}{e}^{-2\left(\epsilon -\mu \right)/kT}\sqrt{\epsilon }d\epsilon =g_0{e}^{2\mu /kT}\underset{0}{\overset{\infty }{\int }}{e}^{-2\epsilon /kT}\sqrt{\epsilon }d\epsilon \\ =2g_0{\left(kT\right)}^{3/2}{e}^{2\mu /kT}\underset{0}{\overset{\infty }{\int }}{x}^2{e}^{-2x^2}dx\end{array}$

Substitute $\frac{\sqrt{\pi }}{4\sqrt{8}}$for $\underset{0}{\overset{\infty }{\int }}{x}^2{e}^{-2x^2}dx$in the above equation.

$\begin{array}{l}{g}_0\underset{0}{\overset{\infty }{\int }}{e}^{-2\left(\epsilon -\mu \right)/kT}\sqrt{\epsilon }d\epsilon =2g_0{\left(kT\right)}^{3/2}{e}^{2\mu /kT}\frac{\sqrt{\pi }}{4\sqrt{8}}\\ =\frac{\sqrt{\pi }}{2\sqrt{8}}{g}_0{\left(kT\right)}^{3/2}{e}^{2\mu /kT}\end{array}$

Substitute $\frac{\sqrt{\pi }}{2}{g}_0{\left(kT\right)}^{3/2}{e}^{\mu /kT}$for $g_0\underset{0}{\overset{\infty }{\int }}{e}^{-\left(\epsilon -\mu \right)/kT}\sqrt{\epsilon }d\epsilon$and $\frac{\sqrt{\pi }}{2\sqrt{8}}{g}_0{\left(kT\right)}^{3/2}{e}^{2\mu /kT}$for $g_0\underset{0}{\overset{\infty }{\int }}{e}^{-2\left(\epsilon -\mu \right)/kT}\sqrt{\epsilon }d\epsilon$in equation (3).

$\begin{array}{l}N=\frac{\sqrt{\pi }}{2}{g}_0{\left(kT\right)}^{3/2}{e}^{\mu /kT}+\frac{\sqrt{\pi }}{2\sqrt{8}}{g}_0{\left(kT\right)}^{3/2}{e}^{2\mu /kT}\\ =\frac{\sqrt{\pi }}{2}{g}_0{\left(kT\right)}^{3/2}{e}^{\mu /kT}\left[1+\frac{e^{\mu /kT}}{\sqrt{8}}\right]\end{array}$

Substitute $\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V$for $g_0$in the above equation.

$\begin{array}{l}N=\frac{\sqrt{\pi }}{2}\frac{2}{\sqrt{\pi }}{\left(\frac{2\pi m}{h^2}\right)}^{3/2}V{\left(kT\right)}^{3/2}{e}^{\mu /kT}\left[1+\frac{e^{\mu /kT}}{\sqrt{8}}\right]\\ ={\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}V{e}^{\mu /kT}\left[1+\frac{e^{\mu /kT}}{\sqrt{8}}\right]\end{array}$

Substitute $v_Q$for ${\left(\frac{h^2}{2\pi mkT}\right)}^{3/2}$in the above equation.

$\begin{array}{l}N=\frac{V}{v_Q}{e}^{\mu /kT}\left[1+\frac{e^{\mu /kT}}{\sqrt{8}}\right]\\ e^{-\mu /kT}=\frac{V}{N{v}_Q}\left[1+\frac{e^{\mu /kT}}{\sqrt{8}}\right]\end{array}$

Substitute $N{v}_Q$for $e^{-\mu /kT}$in bracket term of the above equation.

$e^{-\mu /kT}=\frac{V}{N{v}_Q}\left[1+\frac{N{v}_Q}{\sqrt{8}}\right]$

Take logarithm on both sides and solve for$\mu$.

$\begin{array}{l}\frac{-\mu }{kT}=\ln \left(\frac{V}{N{v}_Q}\left[1+\frac{N{v}_Q}{\sqrt{8}}\right]\right)\\ \mu =-kT\ln \left(\frac{V}{N{v}_Q}\left[1+\frac{N{v}_Q}{\sqrt{8}}\right]\right)\\ =-kT\ln \left(\frac{V}{K{v}_Q}\right)-kT\ln \left[1+\frac{N{v}_Q}{\sqrt{8}V}\right]\end{array}$

Expand the logarithm term as $\ln \left(1+x\right)\approx x$in the above equation.

$\mu =-kT\ln \left(\frac{V}{K{v}_Q}\right)-\frac{kTN{v}_Q}{\sqrt{8}V}$

The first term is classical chemical potential and the second term is the correction term.

Part b. Step 4. Conclusion.

Thus, the chemical potential is $\mu =-kT\ln \left(\frac{V}{K{v}_Q}\right)-\frac{kTN{v}_Q}{\sqrt{8}V}$.

Part c. Step 1. Given.

There are non-interacting spin-0 bosons at high temperatures.

Part c. Step 2. Formula used.

Write the expression for grand potential.

$\Phi =-PV$ …… (4)

Here, $\Phi$is the grand potential, $P$is the pressure of gas and $V$is the volume of gas.

Write the expression for grand potential in form of a grand partition function.

$\Phi =-kT\ln \left(Z\right)$ ……. (5)

Here, $k$is the Boltzmann constant, $T$is the temperature and $Z$is the partition function.

Consider a system of non-interacting particles in a box. Write the expression for the grand partition function.

$\begin{array}{l}{Z}_{tot}=Z_1{Z}_2{Z}_3\mathrm{.....}\\ =\prod _n{Z}_n\end{array}$ …… (6)

Part c. Step 3. Calculation.

Equate equation (4) and equation (5).

$\begin{array}{l}PV=kT\ln \left(Z\right)\\ P=\frac{kT}{V}\ln \left(Z\right)\end{array}$

Take the logarithm on both sides of equation (6).

$\begin{array}{l}\ln \left(Z_{tot}\right)=\ln \left(Z_1{Z}_2{Z}_3\mathrm{...}\right)\\ =\ln \left(Z_1\right)+\ln \left(Z_2\right)+\ln \left(Z_3\right)+\mathrm{...}\\ =\sum _n\ln \left(Z_n\right)\end{array}$

Part c. Step 4. Conclusion.

Thus, the pressure of a gas is given as $P=\left(\frac{kT}{V}\right)\ln Z$and the logarithm of the grand partition function is computed as the sum over all modes.

Part d. Step 1. Given.

There are non-interacting spin-0 bosons at high temperatures.

Part d. Step 2. Formula used.

Write the expression for the logarithm of the grand partition function.

$\ln \left(Z_{tot}\right)=\sum _n\ln \left(Z_n\right)$ …… (7)

Here, $Z_{tot}$is the total sum of the grand partition function and $Z_n$is the grand partition function for $n$particles.

Write the expression for the grand partition function.

$Z_n=\frac{1}{1-e^{-\left(\epsilon -\mu \right)/kT}}$ …… (8)

Here, $\mu$ is the chemical potential, $k$ is the Boltzmann constant and $T$is the temperature.

Write the expression for quantum volume.

$v_Q={\left(\frac{h^2}{2\pi mkT}\right)}^{3/2}$

Here, $v_Q$ is the quantum volume.

Write the expression for the pressure of the gas.

$P=\left(\frac{kT}{V}\right)\ln Z$ …… (9)

Here, $Z$ is the grand partition function, $P$ is the pressure of gas and $V$is the volume of gas.

Part d. Step 3. Calculation.

Change the summation into integral and multiply it with the spherical co-ordinate factor $n^2\sin \theta$in equation (7).

$\ln \left(Z_{tot}\right)=\underset{0}{\overset{\pi /2}{\int }}d\Phi \underset{0}{\overset{\pi /2}{\int }}\sin \theta d\theta \underset{0}{\overset{\infty }{\int }}{n}^2\ln \left(Z_n\right)dn$

$\ln \left(Z_{tot}\right)=\frac{\pi }{2}\underset{0}{\overset{\infty }{\int }}{n}^2\ln \left(Z_n\right)dn$ …… (9)

Take logarithm of equation (8).

$\ln \left(Z_n\right)=-\ln \left(1-e^{-\left(\epsilon -\mu \right)/kT}\right)$

Expand the logarithm of the above equation.

$\ln \left(Z_n\right)=e^{-\left(\epsilon -\mu \right)/kT}+\frac{1}{2}{e}^{-2\left(\epsilon -\mu \right)/kT}$

Substitute $e^{-\left(\epsilon -\mu \right)/kT}+\frac{1}{2}{e}^{-2\left(\epsilon -\mu \right)/kT}$for $\ln \left(Z_n\right)$in equation (9).

$\begin{array}{c}\ln \left(Z_n\right)=\frac{\pi }{2}\underset{0}{\overset{\infty }{\int }}{n}^2\left[e^{-\left(\epsilon -\mu \right)/kT}+\frac{1}{2}{e}^{-2\left(\epsilon -\mu \right)/kT}\right]dn\\ =\frac{\pi }{2}\left[\underset{0}{\overset{\infty }{\int }}{n}^2{e}^{-\left(\epsilon -\mu \right)/kT}dn+\frac{1}{2}\underset{0}{\overset{\infty }{\int }}{n}^2{e}^{-2\left(\epsilon -\mu \right)/kT}dn\right]\\ =\frac{\pi }{2}{e}^{\mu /kT}\left[\underset{0}{\overset{\infty }{\int }}{n}^2{e}^{-\epsilon /kT}dn+\frac{1}{2}\underset{0}{\overset{\infty }{\int }}{n}^2{e}^{-2\epsilon /kT}dn\right]\end{array}$

Substitute $x$for $\frac{\epsilon }{kT}$, $\sqrt{\frac{h^2}{8m{L}^{2}kT}}$for $x$and $dx$for $\frac{d\epsilon }{kT}$in the above equation.

$\ln \left(Z_{tot}\right)=\frac{\pi }{2}{\left(\frac{8m{L}^{2}kT}{h^2}\right)}^{3/2}{e}^{\mu /kT}\left[\underset{0}{\overset{\infty }{\int }}{x}^2{e}^{-x^2}dx+\frac{1}{2}{e}^{\mu /kT}\underset{0}{\overset{\infty }{\int }}{x}^2{e}^{-2x^2}dx\right]$

Substitute $\frac{\sqrt{\pi }}{4}$for $\underset{0}{\overset{\infty }{\int }}{x}^2{e}^{-x^2}dx$and $\frac{\sqrt{\pi }}{4\sqrt{8}}$for $\underset{0}{\overset{\infty }{\int }}{x}^2{e}^{-2x^2}dx$in the above equation.

$\ln \left(Z_{tot}\right)=\frac{\pi }{2}{\left(\frac{8m{L}^{2}kT}{h^2}\right)}^{3/2}{e}^{\mu /kT}\left[\frac{\sqrt{\pi }}{4}+e^{\mu /kT}\frac{\sqrt{\pi }}{8\sqrt{8}}\right]$

Substitute $v_Q$for ${\left(\frac{h^2}{2\pi mkT}\right)}^{3/2}$and $V$ for $L^3$ in the above equation.

$\ln \left(Z_{tot}\right)=\frac{V}{v_Q}{e}^{\mu /kT}\left[1+e^{\mu /kT}\frac{1}{2\sqrt{8}}\right]$

Substitute $\frac{V}{N{v}_Q}\left[1-\frac{N{v}_Q}{\sqrt{8}}\right]$for $e^{\mu /kT}$in the above equation.

role="math" localid="1651745720233" $\begin{array}{l}\ln \left(Z_{tot}\right)=\frac{V}{v_Q}\left[1+\frac{N{v}_Q}{V}\frac{1}{2\sqrt{8}}\right]\frac{N{v}_Q}{V}\left[1-\frac{N{v}_Q}{\sqrt{8}V}\right]\\ =N\left[1-\frac{N{v}_Q}{\sqrt{8}V}\right]\left[1+\frac{N{v}_Q}{V}\frac{1}{2\sqrt{8}}\right]\end{array}$

Simplify the above term and neglect the higher terms.

$\ln \left(Z_{tot}\right)=N\left[1-\frac{N{v}_Q}{4\sqrt{2}V}\right]$

Substitute $N\left[1-\frac{N{v}_Q}{4\sqrt{2}V}\right]$for $\ln \left(Z\right)$in equation (9).

$P=\frac{NkT}{V}\left[1-\frac{N{v}_Q}{4\sqrt{2}V}\right]$

Part d. Step 4. Conclusion.

Thus, the pressure of a gas is $P=\frac{NkT}{V}\left(1-\frac{N{v}_Q}{4\sqrt{2}V}\right)$.

Part e. Step 1. Given.

There are non-interacting spin-0 bosons at high temperature

Part e. Step 2. Formula used.

Write the expression of pressure in terms of virial expansion.

$P=\frac{NkT}{V}\left[1+\frac{\beta \left(T\right)}{V/n}\right]$ …… (10)

Here, $P$is the pressure of the gas, $V$is the volume of gas, $N$is the number of particles, $k$is the Boltzmann constant, $T$is the temperature of the gas, $n$is the number of a mole in gas and $\beta \left(T\right)$is a viral coefficient.

Write the expression for pressure of the gas.

$P=\frac{NkT}{V}\left(1-\frac{N{v}_Q}{4\sqrt{2}V}\right)$ …… (11)

Here, $v_Q$ is the quantum volume.

Part e. Step 3. Calculation.

Equate the equation (10) and equation (11) to get the value of the virial coefficient.

$\beta \left(T\right)=-\frac{N_A{v}_Q}{4\sqrt{2}}$

Substitute ${\left(\frac{h^2}{2\pi mkT}\right)}^{3/2}$for $v_Q$ in the above equation.

$\beta \left(T\right)=-\frac{N_A}{4\sqrt{2}}{\left(\frac{h^2}{2\pi mkT}\right)}^{3/2}$

Substitute $6.022\times {10}^{-23}{\text{\hspace{0.17em}mol}}^{-1}$for $N_A$, $6.626\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}$for $h$, $6.64\times {10}^{-27}\text{\hspace{0.17em}kg}$for$m$ and $1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}$for $k$in the above equation.

$\begin{array}{l}\beta \left(T\right)=-\frac{6.022\times {10}^{-23}{\text{\hspace{0.17em}mol}}^{-1}}{4\sqrt{2}}{\left(\frac{{\left(6.626\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}\right)}^2}{2\left(3.14\right)\left(6.64\times {10}^{-27}\text{\hspace{0.17em}kg}\right)\left(1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}\right)T}\right)}^{3/2}\\ =-\left(7.089\times {10}^{-5}{\text{\hspace{0.17em}m}}^{\text{3}}\cdot K^{3/2}/\text{mol}\right)T^{-3/2}\end{array}$

Plot $\beta \left(T\right)$versus $T\left(K\right)$.

Part e. Step 4. Conclusion.

Thus, the value of the virial coefficient for helium is given as $\beta \left(T\right)=-\left(7.089\times {10}^{-5}{\text{\hspace{0.17em}m}}^{\text{3}}\cdot K^{3/2}/\text{mol}\right)T^{-3/2}$.

Part f. Step 1. Given.

There are non-interacting spin-0 bosons at high temperatures.

Part f. Step 2. Formula used.

Write the expression of number density.

$N=\frac{V}{v_Q}{e}^{\mu /kT}\left[1-\frac{e^{\mu /kT}}{\sqrt{8}}\right]$ …… (12)

Here, $N$ is the number density of particles, $v_Q$ is the quantum volume, $V$ is the volume of gas, $\mu$ is the chemical potential, $k$ is the Boltzmann constant and $T$is the temperature.

Write the expression for the grand partition function.

$Z=1+e^{-\left(\epsilon -\mu \right)/kT}$

Write the expression for the viral coefficient.

$\beta \left(T\right)=\frac{N_A{v}_Q}{4\sqrt{2}}$ …… (13)

Here, $\beta \left(T\right)$ is the virial coefficient, $N_A$ is the Avogadro number, and $v_Q$is the quantum volume

Part f. Step 3. Calculation.

Multiply by the factor of 2 with equation (12) for spin half particles.

$N=\frac{2V}{v_Q}{e}^{\mu /kT}\left[1-\frac{e^{\mu /kT}}{\sqrt{8}}\right]$

Multiply the factor of $\frac{1}{2}$with equation (13) for spin half particles.

$\beta \left(T\right)=-\frac{N_A{v}_Q}{8\sqrt{2}}$

Substitute ${\left(\frac{h^2}{2\pi mkT}\right)}^{3/2}$for $v_Q$ in the above equation.

$\beta \left(T\right)=\frac{N_A}{8\sqrt{2}}{\left(\frac{h^2}{2\pi mkT}\right)}^{3/2}$

Substitute $6.022\times {10}^{-23}{\text{\hspace{0.17em}mol}}^{-1}$for $N_A$, $6.626\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}$for $h$, $3\times 1.66\times {10}^{-27}\text{\hspace{0.17em}kg}$for$m$ and $1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}$for $k$in the above equation.

$\begin{array}{l}\beta \left(T\right)=\frac{6.022\times {10}^{-23}{\text{\hspace{0.17em}mol}}^{-1}}{8\sqrt{2}}{\left(\frac{{\left(6.626\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}\right)}^2}{2\left(3.14\right)\left(3\times 1.66\times {10}^{-27}\text{\hspace{0.17em}kg}\right)\left(1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}\right)T}\right)}^{3/2}\\ =\left(5.457\times {10}^{-5}{\text{\hspace{0.17em}m}}^{\text{3}}\cdot K^{3/2}/\text{mol}\right)T^{-3/2}\end{array}$

Plot $\beta \left(T\right)$versus $T\left(K\right)$.

Part f. Step 4. Conclusion.

Thus, for a spin, half particle number density is given as $N=\frac{2V}{v_Q}{e}^{\mu /kT}\left[1-\frac{e^{\mu /kT}}{\sqrt{8}}\right]$. The viral coefficient is given as $\beta \left(T\right)=\frac{N_A}{8\sqrt{2}}{\left(\frac{h^2}{2\pi mkT}\right)}^{3/2}$and the value of virial coefficient is given as $\beta \left(T\right)=\left(5.457\times {10}^{-5}{\text{\hspace{0.17em}m}}^{\text{3}}\cdot K^{3/2}/\text{mol}\right)T^{-3/2}$.