Hard-sphere fluid A hard-sphere fluid (or gas) is just like and ideal gas except that the gas (fluid) particles are not point particles, but instead have a finite volume. The following equations approximately describe the hard-sphere fluid.

\[U=\frac{3}{2} N k_B T\]

\[\eta=\frac{N}{V} \frac{4 \pi}{3} R^3\]

\[\frac{F}{N k_B T}=\ln \left( \frac{N}{V} \left( \frac{3 h^2}{4 \pi m} \frac{N}{U} \right)^\frac32 \right) + \frac{4 \eta - 3 \eta^2}{\left( 1-\eta \right)^2}-\frac12\]

where $k_B$ is a constant, and you may consider the number of spheres $N$ as a constant for this problem. Similarly the mass $m$, the radius $R$, and Plank's constant $h$ are constants. In these equations $\eta$ is called the packing fraction, and represents the fraction of the volume that is occupied by hard spheres (which has no impact whatsoever on how you solve this problem).

The isothermal compressibility is defined as \begin{equation} K_{T}=-\frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_{T} \end{equation} $K_T$ is be found by measuring the fractional change in volume when the the pressure is slightly changed with the temperature held constant. In contrast, the adiabatic compressibility is defined as \begin{equation} K_{S}=-\frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_{S} \end{equation} and is measured by making a slight change in pressure without allowing for any heat transfer. This is the compressibility, for instance, that would directly affect the speed of sound. Show that \begin{equation} \frac{K_{T}}{K_{S}} = \frac{C_{p}}{C_{V}} \end{equation} Where the heat capacities at constant pressure and volume are given by \begin{align} C_{p} &= T \left(\frac{\partial S}{\partial T}\right)_{p} \\ C_{V} &= T \left(\frac{\partial S}{\partial T}\right)_{V} \end{align}

The equation of state of a gas that departs from ideality can be approximated by \[ p=\frac{NkT}{V}\left(1+\frac{NB_{2}(T)}{V}\right) \] where $B_{2}(T)$ is called the second virial coefficient which increases monotonically with temperature. Find $\left( \frac{\partial U}{\partial V}\right)_{T}$ and determine its sign.

For the three thermodynamic potentials defined as

\begin{quote} \(F=U-T S\) \hfill Helmholtz free energy

\(H=U+P V\) \hfill Enthalpy

\(U-T S+P V\) \hfill Gibbs free energy \end{quote}

A two level system A system consists of $N$ identical non-interacting (independent) objects. Each object has two energy eigenstates labeled $A$ and $B$, with energies $E_A=E_0$ and $E_B=-E_0$ ($E_0=\left|E_0 \right|$).

At low temperatures, a diatomic molecule can be well described as a rigid rotor. The Hamiltonian of such a system is simply proportional to the square of the angular momentum \begin{align} H &= \frac{1}{2I}L^2 \end{align} and the energy eigenvalues are \begin{align} E_{lm} &= \hbar^2 \frac{l(l+1)}{2I} \end{align}

Nuclei of a particular isotope species contained in a crystal have spin $I=1$, and thus, $m = \{+1,0,-1\}$. The interaction between the nuclear quadrupole moment and the gradient of the crystalline electric field produces a situation where the nucleus has the same energy, $E=\varepsilon$, in the state $m=+1$ and the state $m=-1$, compared with an energy $E=0$ in the state $m=0$, i.e. each nucleus can be in one of 3 states, two of which have energy $E=\varepsilon$ and one has energy $E=0$.