Statistical Thermodynamics Questions and Answers

Answer: A

The heat capacity is 3R per mole.

Only the temperature dependence of q matters. Collecting the powers of T: \(q \propto T^{3/2} \cdot T^{3/2} = T^{3}\). The molar internal energy and heat capacity follow from \(\ln q = 3\ln T + \text{const}\):

\[ \begin{aligned} U &= RT^{2}\frac{d\ln q}{dT} = RT^{2} \cdot \frac{3}{T} = 3RT \\ C_{V} &= \frac{dU}{dT} = 3R \end{aligned} \]

The first factor looks like a rotational term and the second like a translational one, each contributing 3R/2. 3R/2 counts only one of them, 9R/2 adds an extra 3R/2 that is not there, and 6R doubles the result.

Answer: A

A canonical ensemble keeps N, V, T constant.

In the canonical ensemble each system has a fixed number of particles and a fixed volume, and exchanges energy with a large heat bath, so its temperature is fixed while its energy fluctuates. That is why the canonical partition function is \(Q = \sum e^{-E_{i}/kT}\) over states of given N and V.

Fixed N, V, E describes the microcanonical ensemble, an isolated system. Fixed N, P, T is the isothermal-isobaric ensemble, where the volume fluctuates. Fixed μ, V, T is the grand canonical ensemble, where particles as well as energy are exchanged.

Answer: B

For ortho-hydrogen the nuclear wavefunction is symmetric and the rotational quantum number is odd.

Protons are fermions, so the total wavefunction of H₂ must change sign when the two nuclei are exchanged. In ortho-hydrogen the nuclear spins are parallel (total nuclear spin I = 1), and the three spin functions of a triplet are symmetric under exchange. The rotational part must then be antisymmetric, and rotational functions are antisymmetric under exchange only for odd J (1, 3, 5, ...).

Para-hydrogen is the opposite case: antisymmetric nuclear spin (I = 0) paired with even J. 'Antisymmetric and even' therefore describes para, not ortho, and the two mixed combinations (symmetric with even, antisymmetric with odd) would give a total wavefunction that is symmetric under exchange, which the Pauli principle forbids.

Answer: D

The partition function proportional to V is the translational partition function.

For a particle of mass m free to move in a volume V,

\[ \begin{aligned} q_{trans} &= \frac{V}{\Lambda^{3}} \\ \Lambda &= \frac{h}{\sqrt{2\pi m kT}} \end{aligned} \]

so it grows in direct proportion to the volume available. The vibrational, rotational and electronic partition functions depend on the molecule's internal energy levels (vibrational frequencies, moments of inertia, electronic states) and on temperature, but not on the size of the container.

Answer: D

\(\ln Q_{t,m} = N\ln q_{t} - N\ln N + N\).

The molecules of a gas are indistinguishable, so the molar partition function is

\[ \begin{aligned} Q_{t,m} &= \frac{q_{t}^{N}}{N!} \\ \ln Q_{t,m} &= N\ln q_{t} - \ln N! \\ &\approx N\ln q_{t} - N\ln N + N \end{aligned} \]

using Stirling's approximation \(\ln N! \approx N\ln N - N\). Dropping the +N term gives the first option, writing \(\ln N\) for \(\ln N!\) gives the second, and the third has the wrong sign on \(N\ln N\).

Answer: B

The answer is \(q_{trans}(T, V)\), \(q_{vib}(T)\), \(q_{rot}(T)\).

Translational energy levels are those of a particle in a box, so they depend on the size of the container:

\[ \begin{aligned} q_{trans} &= \frac{V}{\Lambda^3} \\ \Lambda &= \frac{h}{\sqrt{2\pi m k_B T}} \end{aligned} \]

Rotational levels depend on the moment of inertia and vibrational levels on the bond's frequency. Both are properties of the molecule alone, so \(q_{rot}\) and \(q_{vib}\) depend only on T.

The other options make vibration or rotation depend on volume, which an ideal gas molecule's internal motion does not, or drop the volume from \(q_{trans}\).

Answer: A

The equation of state is \(P(V - Nb) = Nk_BT\).

Pressure follows from the canonical partition function by

\[ P = k_B T \left( \frac{\partial \ln Q}{\partial V} \right) \]

at constant N and T.

Only the factor \((V - Nb)^N\) depends on V, so

\[ \begin{aligned} \ln Q &= N \ln(V - Nb) + \ldots \\ P &= \frac{N k_B T}{V - Nb} \end{aligned} \]

This is the van der Waals equation without the attraction term: b is the excluded volume per molecule. \(PV = Nk_BT\) ignores that excluded volume, \(P(V - Nb) = Nk_B\) has lost the temperature and is dimensionally wrong, and \(PV^{(N-b)} = k_BT\) is not what differentiation gives. Of the four, only one equation follows, so this multiple-select question has a single correct choice.

Answer: C

The function is the Gibbs free energy.

Each thermodynamic potential is natural to a particular set of fixed variables. With particle number N, pressure P and temperature T held fixed (the isothermal-isobaric ensemble), the potential that is minimised at equilibrium is \(G = H - TS\), with \(dG = -SdT + VdP + \mu dN\).

The Helmholtz free energy goes with fixed N, V, T (the canonical ensemble). Enthalpy is natural to fixed S and P, and entropy to fixed N, V, E (the microcanonical ensemble).

Answer: C

The correct statement is a = +1 for Fermi-Dirac statistics.

The general distribution for indistinguishable particles is

\[ N_{j} = \frac{g_{j}}{e^{(\epsilon_{j}-\mu)/kT} + a} \]

For fermions, which obey the Pauli principle, a = +1, so N_j/g_j can never exceed 1. For bosons a = −1, which lets many particles crowd into one state. With a = 0 the expression reduces to the classical Maxwell-Boltzmann distribution.

So options A and B have the signs swapped between the two statistics, and a = 1/2 (option D) corresponds to no standard statistics.

Answer: D

In a grand canonical ensemble, X can exchange both energy and particles with Y.

The three standard ensembles differ in what the system shares with its surroundings. A microcanonical ensemble (fixed N, V, E) is isolated. A canonical ensemble (fixed N, V, T) exchanges energy with a heat reservoir but not particles. A grand canonical ensemble (fixed μ, V, T) exchanges both energy and particles with a reservoir, which fixes the temperature and the chemical potential while N and E fluctuate.

Exchange of energy only is the canonical case, exchange of neither is the microcanonical case, and particle exchange without energy exchange is not one of the standard ensembles.