The Gibbs free energy, \(G\), is given by \begin{align*} G = U + pV - TS. \end{align*}

1. Find the total differential of \(G\). As always, show your work.
2. Interpret the coefficients of the total differential \(dG\) in order to find a derivative expression for the entropy \(S\).
3. From the total differential \(dG\), obtain a different thermodynamic derivative that is equal to \[ \left(\frac{\partial {S}}{\partial {p}}\right)_{T} \]

You are given the following Gibbs free energy: \begin{equation*} G=-k T N \ln \left(\frac{a T^{5 / 2}}{p}\right) \end{equation*} where \(a\) is a constant (whose dimensions make the argument of the logarithm dimensionless).

1. Compute the entropy.

2. Work out the heat capacity at constant pressure \(C_p\).

3. Find the connection among \(V\), \(p\), \(N\), and \(T\), which is called the equation of state (Hint: find the volume as a partial derivative of the Gibbs free energy).

4. Compute the internal energy \(U\).

The internal energy is of any ideal gas can be written as \begin{align} U &= U(T,N) \end{align} meaning that the internal energy depends only on the number of particles and the temperature, but not the volume.^*

The ideal gas law \begin{align} pV &= Nk_BT \end{align} defines the relationship between \(p\), \(V\) and \(T\). You may take the number of molecules \(N\) to be constant. Consider the free adiabatic expansion of an ideal gas to twice its volume. “Free expansion” means that no work is done, but also that the process is also neither quasistatic nor reversible.

1. What is the change in entropy of the gas? How do you know this?

2. What is the change in temperature of the gas?

1. Find an expression for the free energy as a function of \(T\) of a system with two states, one at energy 0 and one at energy \(\varepsilon\).

2. From the free energy, find expressions for the internal energy \(U\) and entropy \(S\) of the system.

3. Plot the entropy versus \(T\). Explain its asymptotic behavior as the temperature becomes high.

4. Plot the \(S(T)\) versus \(U(T)\). Explain the maximum value of the energy \(U\).

A one-dimensional harmonic oscillator has an infinite series of equally spaced energy states, with \(\varepsilon_n = n\hbar\omega\), where \(n\) is an integer \(\ge 0\), and \(\omega\) is the classical frequency of the oscillator. We have chosen the zero of energy at the state \(n=0\) which we can get away with here, but is not actually the zero of energy! To find the true energy we would have to add a \(\frac12\hbar\omega\) for each oscillator.

1. Show that for a harmonic oscillator the free energy is \begin{equation} F = k_BT\log\left(1 - e^{-\frac{\hbar\omega}{k_BT}}\right) \end{equation} Note that at high temperatures such that \(k_BT\gg\hbar\omega\) we may expand the argument of the logarithm to obtain \(F\approx k_BT\log\left(\frac{\hbar\omega}{kT}\right)\).

2. From the free energy above, show that the entropy is \begin{equation} \frac{S}{k_B} = \frac{\frac{\hbar\omega}{kT}}{e^{\frac{\hbar\omega}{kT}}-1} - \log\left(1-e^{-\frac{\hbar\omega}{kT}}\right) \end{equation}

   

   

   This entropy is shown in the nearby figure, as well as the heat capacity.

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 \(E=0\).

1. Find the Helmholtz free energy \(F = U-TS\) for a crystal containing \(N\) nuclei which do not interact with each other.

2. Find an expression for the entropy as a function of temperature for this system. (Hint: use results of part a.)

3. Indicate what your results predict for the entropy at the extremes of very high temperature and very low temperature.

Consider the bottle in a bottle problem in a previous problem set, summarized here.

A small bottle of helium is placed inside a large bottle, which otherwise contains vacuum. The inner bottle contains a slow leak, so that the helium leaks into the outer bottle. The inner bottle contains one tenth the volume of the outer bottle, which is insulated.

The volume of the small bottle is 0.001 m²³ and the volume of the big bottle is 0.01 m³. The initial state of the gas in the small bottle was \(p=106\) Pa and its temperature \(T=300\) K. Approximate the helium gas as an ideal gas of equations of state \(pV=Nk_BT\) and \(U=\frac32 Nk_BT\).

1. How many molecules of gas does the large bottle contain? What is the final temperature of the gas?

2. Compute the integral \(\int \frac{{\mathit{\unicode{273}}} Q}{T}\) and the change of entropy \(\Delta S\) between the initial state (gas in the small bottle) and the final state (gas leaked in the big bottle).

3. Discuss your results.

Consider a paramagnet, which is a material with \(n\) spins per unit volume each of which may each be either “up” or “down”. The spins have energy \(\pm mB\) where \(m\) is the magnetic dipole moment of a single spin, and there is no interaction between spins. The magnetization \(M\) is defined as the total magnetic moment divided by the total volume. Hint: each individual spin may be treated as a two-state system, which you have already worked with above.

1. Find the Helmholtz free energy of a paramagnetic system (assume \(N\) total spins) and show that \(\frac{F}{NkT}\) is a function of only the ratio \(x\equiv \frac{mB}{kT}\).

2. Use the canonical ensemble (i.e. partition function and probabilities) to find an exact expression for the total magentization \(M\) (which is the total dipole moment per unit volume) and the susceptibility \begin{align} \chi\equiv\left(\frac{\partial M}{\partial B}\right)_T \end{align} as a function of temperature and magnetic field for the model system of magnetic moments in a magnetic field. The result for the magnetization is \begin{align} M=nm\tanh\left(\frac{mB}{kT}\right) \end{align} where \(n\) is the number of spins per unit volume. The figure shows what this magnetization looks like.

3. Show that the susceptibility is \(\chi=\frac{nm^2}{kT}\) in the limit \(mB\ll kT\).

Consider a hanging rectangular rubber sheet. We will consider there to be two ways to get energy into or out of this sheet: you can either stretch it vertically or horizontally. The distance of vertical stretch we will call \(y\), and the distance of horizontal stretch we will call \(x\).

If I pull the bottom down by a small distance \(\Delta y\), with no horizontal force, what is the resulting change in width \(\Delta x\)? Express your answer in terms of partial derivatives of the potential energy \(U(x,y)\).

As discussed in class, we can consider a black body as a large box with a small hole in it. If we treat the large box a metal cube with side length \(L\) and metal walls, the frequency of each normal mode will be given by: \begin{align} \omega_{n_xn_yn_z} &= \frac{\pi c}{L}\sqrt{n_x^2 + n_y^2 + n_z^2} \end{align} where each of \(n_x\), \(n_y\), and \(n_z\) will have positive integer values. This simply comes from the fact that a half wavelength must fit in the box. There is an additional quantum number for polarization, which has two possible values, but does not affect the frequency. Note that in this problem I'm using different boundary conditions from what I use in class. It is worth learning to work with either set of quantum numbers. Each normal mode is a harmonic oscillator, with energy eigenstates \(E_n = n\hbar\omega\) where we will not include the zero-point energy \(\frac12\hbar\omega\), since that energy cannot be extracted from the box. (See the Casimir effect for an example where the zero point energy of photon modes does have an effect.)

Note

This is a slight approximation, as the boundary conditions for light are a bit more complicated. However, for large \(n\) values this gives the correct result.

1. Show that the free energy is given by \begin{align} F &= 8\pi \frac{V(kT)^4}{h^3c^3} \int_0^\infty \ln\left(1-e^{-\xi}\right)\xi^2d\xi \\ &= -\frac{8\pi^5}{45} \frac{V(kT)^4}{h^3c^3} \\ &= -\frac{\pi^2}{45} \frac{V(kT)^4}{\hbar^3c^3} \end{align} provided the box is big enough that \(\frac{\hbar c}{LkT}\ll 1\). Note that you may end up with a slightly different dimensionless integral that numerically evaluates to the same result, which would be fine. I also do not expect you to solve this definite integral analytically, a numerical confirmation is fine. However, you must manipulate your integral until it is dimensionless and has all the dimensionful quantities removed from it!

2. Show that the entropy of this box full of photons at temperature \(T\) is \begin{align} S &= \frac{32\pi^5}{45} k V \left(\frac{kT}{hc}\right)^3 \\ &= \frac{4\pi^2}{45} k V \left(\frac{kT}{\hbar c}\right)^3 \end{align}

3. Show that the internal energy of this box full of photons at temperature \(T\) is \begin{align} \frac{U}{V} &= \frac{8\pi^5}{15}\frac{(kT)^4}{h^3c^3} \\ &= \frac{\pi^2}{15}\frac{(kT)^4}{\hbar^3c^3} \end{align}

(modified from K&K 4.6) We discussed in class that \begin{align} p &= -\left(\frac{\partial F}{\partial V}\right)_T \end{align} Use this relationship to show that

1. \begin{align} p &= -\sum_j \langle n_j\rangle\hbar \left(\frac{d\omega_j}{dV}\right), \end{align} where \(\langle n_j\rangle\) is the number of photons in the mode \(j\);

2. Solve for the relationship between pressure and internal energy.

In this entire problem, keep results to first order in the van der Waals correction terms \(a\) and $b.

1. Show that the entropy of the van der Waals gas is \begin{align} S &= Nk\left\{\ln\left(\frac{n_Q(V-Nb)}{N}\right)+\frac52\right\} \end{align}

2. Show that the energy is \begin{align} U &= \frac32 NkT - \frac{N^2a}{V} \end{align}

3. Show that the enthalpy \(H\equiv U+pV\) is \begin{align} H(T,V) &= \frac52NkT + \frac{N^2bkT}{V} - 2\frac{N^2a}{V} \\ H(T,p) &= \frac52NkT + Nbp - \frac{2Nap}{kT} \end{align}

Einstein condensation temperature Starting from the density of free particle orbitals per unit energy range \begin{align} \mathcal{D}(\varepsilon) = \frac{V}{4\pi^2}\left(\frac{2M}{\hbar^2}\right)^{\frac32}\varepsilon^{\frac12} \end{align} show that the lowest temperature at which the total number of atoms in excited states is equal to the total number of atoms is \begin{align} T_E &= \frac1{k_B} \frac{\hbar^2}{2M} \left( \frac{N}{V} \frac{4\pi^2}{\int_0^\infty\frac{\sqrt{\xi}}{e^\xi-1}d\xi} \right)^{\frac23} T_E &= \end{align} The infinite sum may be numerically evaluated to be 2.612. Note that the number derived by integrating over the density of states, since the density of states includes all the states except the ground state.

Note: This problem is solved in the text itself. I intend to discuss Bose-Einstein condensation in class, but will not derive this result.