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 \(U\).

Consider two noninteracting systems \(A\) and \(B\). We can either treat these systems as separate, or as a single combined system \(AB\). We can enumerate all states of the combined by enumerating all states of each separate system. The probability of the combined state \((i_A,j_B)\) is given by \(P_{ij}^{AB} = P_i^AP_j^B\). In other words, the probabilities combine in the same way as two dice rolls would, or the probabilities of any other uncorrelated events.

1. Show that the entropy of the combined system \(S_{AB}\) is the sum of entropies of the two separate systems considered individually, i.e. \(S_{AB} = S_A+S_B\). This means that entropy is extensive. Use the Gibbs entropy for this computation. You need make no approximation in solving this problem.
2. Show that if you have \(N\) identical non-interacting systems, their total entropy is \(NS_1\) where \(S_1\) is the entropy of a single system.

Note

In real materials, we treat properties as being extensive even when there are interactions in the system. In this case, extensivity is a property of large systems, in which surface effects may be neglected.

Suppose that a system of \(N\) atoms of type \(A\) is placed in diffusive contact with a system of \(N\) atoms of type \(B\) at the same temperature and volume.

1. Show that after diffusive equilibrium is reached the total entropy is increased by \(2Nk\ln 2\). The entropy increase \(2Nk\ln 2\) is known as the entropy of mixing.

2. If the atoms are identical (\(A=B\)), show that there is no increase in entropy when diffusive contact is established. The difference has been called the Gibbs paradox.

3. Since the Helmholtz free energy is lower for the mixed \(AB\) than for the separated \(A\) and \(B\), it should be possible to extract work from the mixing process. Construct a process that could extract work as the two gasses are mixed at fixed temperature. You will probably need to use walls that are permeable to one gas but not the other.

Note

This course has not yet covered work, but it was covered in Energy and Entropy, so you may need to stretch your memory to finish part (c).

Consider a three-state system with energies \((-\epsilon,0,\epsilon)\).

1. At infinite temperature, what are the probabilities of the three states being occupied? What is the internal energy \(U\)? What is the entropy \(S\)?
2. At very low temperature, what are the three probabilities?
3. What are the three probabilities at zero temperature? What is the internal energy \(U\)? What is the entropy \(S\)?
4. What happens to the probabilities if you allow the temperature to be negative?