Statistical Thermodynamics Questions and Answers

Answer: C

The probability is exp(−E/k_BT) / (1 + exp(−E/k_BT)).

By the Boltzmann distribution, each state's population is proportional to exp(−E/k_BT), divided by the partition function, the sum of these factors over all states. With levels at 0 and E,

\[ \begin{aligned} q &= 1 + e^{-E/k_{B}T} \\ p_{\mathrm{exc}} &= \frac{e^{-E/k_{B}T}}{1 + e^{-E/k_{B}T}} \end{aligned} \]

Option D is the probability of the ground state, option B is the partition function itself (which exceeds 1, so it cannot be a probability), and 0 holds only at absolute zero.

Answer: C

Fermi-Dirac statistics assumes the particles are indistinguishable and that only one particle may occupy a given state.

Fermi-Dirac statistics applies to fermions, particles of half-integer spin such as electrons, protons and neutrons. Identical quantum particles cannot be labelled, so they are indistinguishable, and fermions obey the Pauli exclusion principle, which allows at most one particle in each quantum state. The resulting occupancy of a state of energy E never exceeds 1:

\[ f(E) = \frac{1}{e^{(E-\mu)/kT} + 1} \]

Distinguishable particles with any number per level is Maxwell-Boltzmann statistics. Indistinguishable particles with any number per state is Bose-Einstein statistics, which applies to bosons such as photons. Distinguishable particles with single occupancy belongs to none of the three standard statistics.

Answer: A

In a canonical ensemble X can exchange only energy with Y.

The canonical ensemble describes a system with fixed number of particles N, fixed volume V and fixed temperature T. The temperature is held constant by contact with a large heat reservoir, and the system and reservoir exchange energy as heat, so the system's energy fluctuates. The walls are closed to matter, so the number of particles cannot change.

Exchange of neither energy nor particles defines the microcanonical ensemble (fixed N, V, E), an isolated system. Exchange of both energy and particles defines the grand canonical ensemble (fixed μ, V, T). An ensemble exchanging particles but not energy is not one of the standard three.

Answer: D

The partition function is 1 + 2e^−E/kT.

The molecular partition function sums the Boltzmann factor over all levels, each weighted by its degeneracy g:

\[ \begin{aligned} q &= \sum_{i} g_{i}\,e^{-E_{i}/kT} \\ &= 1\cdot e^{0} + 2\,e^{-E/kT} = 1 + 2e^{-E/kT} \end{aligned} \]

The ground level at zero energy is non-degenerate and contributes exactly 1. The doubly degenerate upper level contributes 2e^−E/kT.

2e^−E/kT leaves out the ground state, 2 + e^−E/kT puts the degeneracy on the wrong level, and 3e^−E/kT treats all three states as if they had energy E.

Answer: A

At low temperature C_V is proportional to T³.

In the Debye model of a solid, the heat capacity comes from lattice vibrations (phonons). At temperatures well below the Debye temperature θ_D, only low-frequency acoustic modes are excited, and their number grows as T³, giving the Debye T³ law:

\[ \begin{aligned} C_{V} &= \frac{12\pi^{4}}{5}R \\ &\times \left(\frac{T}{\theta_{D}}\right)^{3} \end{aligned} \]

The heat capacity falls to zero as T approaches 0, as the third law requires. (1/T)³ and (R/T)² would make C_V grow without limit as T falls, and (RT)² does not match the Debye result. At high temperature C_V instead approaches the constant 3R (Dulong and Petit).

Answer: C

Each system behaves as an isolated system.

In the canonical ensemble each system has fixed N, V and T: it exchanges energy with a heat bath but not matter, so it is closed. If constant temperature is replaced by constant energy, the systems can exchange neither energy nor matter with their surroundings. That is the microcanonical (N, V, E) ensemble, and each member is an isolated system.

'Closed' describes the canonical case being replaced. 'Open' systems exchange matter as well, as in the grand canonical ensemble (μ, V, T), and 'condensed' is a phase description, not a type of ensemble system.

Answer: B

The temperature is 500 K.

For the vibrational levels (each non-degenerate) the Boltzmann ratio is set by the energy gap hcν̄ between v = 1 and v = 0:

\[ \frac{N_{1}}{N_{0}} = e^{-hc\bar{\nu}/kT} = \frac{1}{2} \]

\[ \begin{aligned} \frac{346.5}{T} &= \ln 2 = 0.693 \\ T &= \frac{346.5}{0.693} = 500\ \text{K} \end{aligned} \]

773 K and 227 K are 500 K with 273 added or subtracted, a slip between kelvin and degrees Celsius, and 300 K is simply room temperature.

Answer: A

The number of microstates is 56.

Distributing q identical quanta among N distinguishable particles is a stars-and-bars count: the number of ways is

\[ \begin{aligned} W &= \frac{(q + N - 1)!}{q!\,(N - 1)!} \\ &= \frac{8!}{5!\,3!} = \frac{8 \times 7 \times 6}{6} = 56 \end{aligned} \]

126 is what you get with N = 5 and q = 5 (9!/5!4!), mixing up the roles of particles and quanta, and 35 comes from 7!/4!3!, one too few in the numerator. 14 has no basis in the count.