1. Consider a system that may be unoccupied with energy zero, or occupied by one particle in either of two states, one of energy zero and one of energy \(\varepsilon\). Find the Gibbs sum for this system is in terms of the activity \(\lambda\equiv e^{\beta\mu}\). Note that the system can hold a maximum of one particle.

2. Solve for the thermal average occupancy of the system in terms of \(\lambda\).

3. Show that the thermal average occupancy of the state at energy \(\varepsilon\) is \begin{align} \langle N(\varepsilon)\rangle = \frac{\lambda e^{-\frac{\varepsilon}{kT}}}{\mathcal{Z}} \end{align}

4. Find an expression for the thermal average energy of the system.

5. Allow the possibility that the orbitals at \(0\) and at \(\varepsilon\) may each be occupied each by one particle at the same time; Show that \begin{align} \mathcal{Z} &= 1 + \lambda + \lambda e^{-\frac{\varepsilon}{kT}} + \lambda^2 e^{-\frac{\varepsilon}{kT}} \\ &= (1+\lambda)\left(1+e^{-\frac{\varepsilon}{kT}}\right) \end{align} Because \(\mathcal{Z}\) can be factored as shown, we have in effect two independent systems.

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.

Let us imagine a new mechanics in which the allowed occupancies of an orbital are 0, 1, and 2. The values of the energy associated with these occupancies are assumed to be \(0\), \(\varepsilon\), and \(2\varepsilon\), respectively.

1. Derive an expression for the ensemble average occupancy \(\langle N\rangle\), when the system composed of this orbital is in thermal and diffusive contact with a resevoir at temperature \(T\) and chemical potential \(\mu\).

2. Return now to the usual quantum mechanics, and derive an expression for the ensemble average occupancy of an energy level which is doubly degenerate; that is, two orbitals have the identical energy \(\varepsilon\). If both orbitals are occupied the toal energy is \(2\varepsilon\). How does this differ from part (a)?

In carbon monoxide poisoning the CO replaces the \(\textsf{O}_{2}\) adsorbed on hemoglobin (\(\text{Hb}\)) molecules in the blood. To show the effect, consider a model for which each adsorption site on a heme may be vacant or may be occupied either with energy \(\varepsilon_A\) by one molecule \(\textsf{O}_{2}\) or with energy \(\varepsilon_B\) by one molecule CO. Let \(N\) fixed heme sites be in equilibrium with \(\textsf{O}_{2}\) and CO in the gas phases at concentrations such that the activities are \(\lambda(\text{O}_2) = 1\times 10^{-5}\) and \(\lambda(\text{CO}) = 1\times 10^{-7}\), all at body temperature \(37^\circ\text{C}\). Neglect any spin multiplicity factors.

1. First consider the system in the absence of CO. Evaluate \(\varepsilon_A\) such that 90 percent of the \(\text{Hb}\) sites are occupied by \(\textsf{O}_{2}\). Express the answer in eV per \(\textsf{O}_{2}\).

2. Now admit the CO under the specified conditions. Fine \(\varepsilon_B\) such that only 10% of the Hb sites are occupied by \(\textsf{O}_{2}\).