In this problem you will model helium-3 as a non-interacting Fermi gas. Although ${}^3\mathrm{He}$liquefies at low temperatures, the liquid has an unusually low density and behaves in many ways like a gas because the forces between the atoms are so weak. Helium-3 atoms are spin-1/2 fermions, because of the unpaired neutron in the nucleus.

(a) Pretending that liquid 3He is a non-interacting Fermi gas, calculate the Fermi energy and the Fermi temperature. The molar volume (at low pressures) is $37{\mathrm{cm}}^3$•

(b)Calculate the heat capacity for $\mathrm{T}<<{\mathrm{T}}_{\mathrm{f}}$, and compare to the experimental result ${\mathrm{C}}_{\mathrm{V}}=(2.8{\mathrm{K}}^{-1})\mathrm{NkT}$(in the low-temperature limit). (Don't expect perfect agreement.)

(c)The entropy of solid ${}^{3}He$below 1 K is almost entirely due to its multiplicity of nuclear spin alignments. Sketch a graph S vs. T for liquid and solid ${}^{3}He$at low temperature, and estimate the temperature at which the liquid and solid have the same entropy. Discuss the shape of the solid-liquid phase boundary shown in Figure 5.13.

Step by step solution

01

Part (a) Step 1: Given information

We have given, the He-3 a non interacting Fermi gas model.

We have to find the Fermi energy and temperature.

02

Simplify

The Fermi energy is given by consider the mass as three proton mass as Helium-3 is,

$$\begin{gathered} {\mathrm{\epsilon }}_{\mathrm{f}}=\frac{{\mathrm{h}}^2}{8\mathrm{m}}{\left(\frac{3\mathrm{N}}{\mathrm{\pi V}}\right)}^{\frac{2}{3}} \\ {\mathrm{\epsilon }}_{\mathrm{f}}=\frac{{(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s})}^2}{8(3\times 1.67\times {10}^{-27}\mathrm{kg})}{\left(\frac{3(6.02\times {10}^{23})}{\mathrm{\pi }(37\times {10}^{-6}{\mathrm{m}}^3}\right)}^{\frac{2}{3}} \\ {\mathrm{\epsilon }}_{\mathrm{f}}=6.8\times {10}^{-23}\mathrm{J} \end{gathered}$$

Then the Fermi temperature is

$$\begin{gathered} \mathrm{T}=\frac{{\mathrm{\epsilon }}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{B}}} \\ \mathrm{T}=\frac{6.8\times {10}^{-23}\mathrm{J}}{1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}} \\ \mathrm{T}=5\mathrm{K} \end{gathered}$$

03

Part (b) Step 1: Given information

We have given,

$\mathrm{T}<<{\mathrm{T}}_{\mathrm{f}}$

we have to calculate the heat capacity.

04

Simplify

The heat capacity is given by as in the terms of the given Fermi temperature T is,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{v}}=\frac{{\mathrm{\pi }}^2{\mathrm{Nk}}_{\mathrm{B}}^2\mathrm{T}}{2{\mathrm{\epsilon }}_{\mathrm{f}}} \\ \frac{{\mathrm{C}}_{\mathrm{v}}}{{\mathrm{Nk}}_{\mathrm{B}}\mathrm{T}}=\frac{{(1.34)}^2(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K})}{2\times 6.8\times {10}^{-23}\mathrm{J}} \\ \frac{{\mathrm{C}}_{\mathrm{v}}}{{\mathrm{Nk}}_{\mathrm{B}}\mathrm{T}}=1.0{\mathrm{K}}^{-1} \end{gathered}$$

This given capacity is greater than found capacity.

05

Part (c) Step 1: Given information

We have given,

$\mathrm{T}=1\mathrm{K}$

We have to plot the graph between the entropy S and T.

06

Simplify

The entropy of the liquid ,

$$\begin{gathered} \mathrm{S}={\int }_0^{\mathrm{T}}\frac{{\mathrm{C}}_{\mathrm{V}}}{\mathrm{T}}\mathrm{dT} \\ \mathrm{S}={\int }_0^{\mathrm{T}}\frac{2.8\mathrm{NKT}}{\mathrm{T}}\mathrm{dT} \\ \mathrm{S}=2.8\mathrm{NKT}{\mathrm{K}}^{-1} \end{gathered}$$

Since,

$\mathrm{S}=\mathrm{NK}\mathrm{ln\Omega }$

By comparing we can say,

$$\begin{gathered} \mathrm{T}=\frac{\ln \left(2\right)}{2.8} \\ \mathrm{T}=0.2475\mathrm{K} \end{gathered}$$

here the entropy of liquid will be same for the solids also.

They are connected at one point where the entropy of the system for solid and liquid phase are same as showing in the figure.

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