Statistical Mechanics -- Practice Problems

Statistical Mechanics — Practice Problems

10 MCQ questions covering the Boltzmann distribution, partition functions, statistical entropy, thermodynamic properties from partition functions, and ensembles.

Worked Examples

Example 1: Boltzmann Distribution — Two-Level System

Problem: A system has two energy levels: $\varepsilon_0 = 0$ and $\varepsilon_1 = 100$ cm⁻¹. Calculate the fraction of molecules in the excited state at 300 K and 1000 K.

Solution:

The fraction in the excited state:

$\frac{n_1}{n_0 + n_1} = \frac{e^{-\varepsilon_1/k_BT}}{1 + e^{-\varepsilon_1/k_BT}}$

At 300 K:

$\frac{\varepsilon_1}{k_BT} = \frac{hc\tilde{\nu}}{k_BT} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{10})(100)}{(1.381 \times 10^{-23})(300)} = 0.481$

$\frac{n_1}{n_0 + n_1} = \frac{e^{-0.481}}{1 + e^{-0.481}} = \frac{0.618}{1.618} = 0.382$

At 1000 K:

$\frac{\varepsilon_1}{k_BT} = 0.481 \times \frac{300}{1000} = 0.144$

$\frac{n_1}{n_0 + n_1} = \frac{e^{-0.144}}{1 + e^{-0.144}} = \frac{0.866}{1.866} = 0.464$

Key insight: At higher temperature, the population becomes more uniform (approaching 50:50 at $T \to \infty$ ). At $T \to 0$ , all molecules collapse into the ground state.

Example 2: Partition Function — Calculating Thermodynamic Properties

Problem: A molecule has three energy levels: $\varepsilon_0 = 0$ , $\varepsilon_1 = 200$ cm⁻¹, $\varepsilon_2 = 500$ cm⁻¹. Calculate the internal energy per mole at 500 K.

Solution:

Step 1: Calculate the partition function

$q = \sum_i e^{-\varepsilon_i/k_BT}$

Using $k_B/hc = 0.6950$ cm⁻¹/K:

$\frac{\varepsilon_1}{k_BT} = \frac{200}{0.6950 \times 500} = 0.576$

$\frac{\varepsilon_2}{k_BT} = \frac{500}{0.6950 \times 500} = 1.439$

$q = e^0 + e^{-0.576} + e^{-1.439} = 1 + 0.562 + 0.237 = 1.799$

Step 2: Calculate internal energy

$U = N_A k_B T^2 \frac{\partial \ln q}{\partial T} = \frac{N_A \sum_i \varepsilon_i e^{-\varepsilon_i/k_BT}}{q}$

$U = \frac{N_A(0 + 200 \times 0.562 + 500 \times 0.237)}{1.799}$

$U = \frac{N_A(112.4 + 118.5)}{1.799} = \frac{N_A \times 230.9}{1.799} = 128.4 N_A \text{ cm}^{-1}$

Converting to kJ/mol:

$U = 128.4 \times (6.626 \times 10^{-34})(3.00 \times 10^{10}) \times 6.022 \times 10^{23} = 1.54 \text{ kJ/mol}$

Example 3: Maxwell-Boltzmann Distribution

Problem: Calculate the most probable speed, mean speed, and rms speed of N₂ at 300 K.

Solution:

Molar mass of N₂: $M = 28.0$ g/mol = 0.028 kg/mol

Most probable speed:

$v_p = \sqrt{\frac{2RT}{M}} = \sqrt{\frac{2 \times 8.314 \times 300}{0.028}} = \sqrt{178,157} = 422 \text{ m/s}$

Mean speed:

$\langle v \rangle = \sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8 \times 8.314 \times 300}{\pi \times 0.028}} = \sqrt{226,655} = 476 \text{ m/s}$

RMS speed:

$v_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.028}} = \sqrt{267,236} = 517 \text{ m/s}$

Relationship: $v_p < \langle v \rangle < v_{\text{rms}}$

Key insight: The distribution is asymmetric — it has a long tail toward high speeds. This means there are always some molecules moving much faster than average, which is important for understanding reaction rates (only fast molecules can overcome activation barriers).

Boltzmann Distribution and Partition Functions

Q1. In the Boltzmann distribution, the probability of finding a particle in state $i$ with energy $\\varepsilon_i$ at temperature $T$ is $p_i = e^\{-\\varepsilon_i/k_BT\}/q$ . What is the molecular partition function $q$ ?

A. $q = \\sum_i e^\{-\\varepsilon_i/k_BT\}$ (sum over all states) B. $q = \\prod_i e^\{-\\varepsilon_i/k_BT\}$ C. $q = e^\{-\\varepsilon_0/k_BT\}$ D. $q = \\varepsilon_i e^\{-\\varepsilon_i/k_BT\}$ ’,

Answer: A

Answer: A — The partition function $q = \\sum_i e^\{-\\varepsilon_i/k_BT\}$ is the normalization constant for the Boltzmann distribution, ensuring $\\sum_i p_i = 1$ . It encodes all thermodynamic information: $U = Nk_B T^2 (\\partial \\ln q / \\partial T)_V$ , $S = Nk_B(\\ln q + T(\\partial \\ln q / \\partial T)_V)$ , $A = -Nk_BT\\ln q$ . Higher temperatures give larger $q$ (more states accessible).

Q2. For a system with energy levels $\\varepsilon_0 = 0$ , $\\varepsilon_1 = \\epsilon$ , $\\varepsilon_2 = 2\\epsilon$ (each non-degenerate), what is the partition function at temperature $T$ ?

A. $q = 1 + e^\{-\\epsilon/k_BT\} + e^\{-2\\epsilon/k_BT\}$ B. $q = e^\{-\\epsilon/k_BT\} + e^\{-2\\epsilon/k_BT\}$ C. $q = 3e^\{-\\epsilon/k_BT\}$ D. $q = 1 + 2e^\{-\\epsilon/k_BT\} + 3e^\{-2\\epsilon/k_BT\}$ ’,

Answer: A

Answer: A — $q = \\sum_i g_i e^\{-\\varepsilon_i/k_BT\}$ where $g_i$ is the degeneracy. Since all states are non-degenerate ( $g_i = 1$ ): $q = e^\{-0/k_BT\} + e^\{-\\epsilon/k_BT\} + e^\{-2\\epsilon/k_BT\} = 1 + e^\{-\\epsilon/k_BT\} + e^\{-2\\epsilon/k_BT\}$ . At high $T$ ( $k_BT \\gg \\epsilon$ ), all three terms approach 1 and $q \\to 3$ . At low $T$ , $q \\to 1$ .

Q3. The translational partition function for a single ideal gas molecule in a container of volume $V$ at temperature $T$ is:

A. $q_\{\\text\{trans\}\} = V\\left(\\frac\{2\\pi m k_B T\}\{h^2\}\\right)^\{3/2\}$ B. $q_\{\\text\{trans\}\} = V k_B T / h$ C. $q_\{\\text\{trans\}\} = (2\\pi m k_B T)^\{3/2\} / h^3$ D. $q_\{\\text\{trans\}\} = N k_B T / P$ ’,

Answer: A

Answer: A — From quantum mechanics, the translational energy levels in a 3D box give $q_\{\\text\{trans\}\} = \\sum_\{n_x,n_y,n_z\} \\exp\\left(-\\frac\{h^2\}\{8mk_B T V^\{2/3\}\}(n_x^2+n_y^2+n_z^2)\\right)$ . In the classical limit (high $T$ , large $V$ ), this sum becomes the integral $q_\{\\text\{trans\}\} = V(2\\pi m k_B T/h^2)^\{3/2\}$ . The $V$ dependence leads directly to the ideal gas equation of state.

Q4. The rotational partition function for a heteronuclear diatomic molecule (treated as a rigid rotor) at sufficiently high temperature is:

A. $q_\{\\text\{rot\}\} = \\frac\{T\}\{\\sigma \\Theta_r\} = \\frac\{k_B T\}\{\\sigma h B\}$ where $\\sigma$ is the symmetry number B. $q_\{\\text\{rot\}\} = e^\{-\\Theta_r/T\}$ C. $q_\{\\text\{rot\}\} = \\frac\{\\Theta_r\}\{T\}$ D. $q_\{\\text\{rot\}\} = (T/\\Theta_r)^2$ ’,

Answer: A

Answer: A — At high $T$ ( $T \\gg \\Theta_r = \\hbar^2/2Ik_B$ ), $q_\{\\text\{rot\}\} = T/(\\sigma \\Theta_r)$ . The symmetry number $\\sigma = 1$ for heteronuclear diatomics (e.g., HCl) and $\\sigma = 2$ for homonuclear (e.g., N $_2$ , accounting for indistinguishable orientations). This factor prevents overcounting rotational states by the indistinguishable rotations.

Thermodynamic Properties and Ensembles

Q5. The vibrational partition function for a harmonic oscillator with frequency $\ u$ is $q_\{\\text\{vib\}\} = \\frac\{e^\{-h\ u/2k_BT\}\}\{1 - e^\{-h\ u/k_BT\}\}$ . At room temperature ( $T \\ll \\Theta_\{\\text\{vib\}\} = h\ u/k_B$ ), the vibrational contribution to the internal energy is approximately:

A. $U_\{\\text\{vib\}\} \\approx \\frac\{1\}\{2\}N h\ u$ (essentially the zero-point energy) B. $U_\{\\text\{vib\}\} \\approx 0$ C. $U_\{\\text\{vib\}\} \\approx Nk_BT$ (full equipartition) D. $U_\{\\text\{vib\}\} \\approx 3Nk_BT$ ’,

Answer: A

Answer: A — At $T \\ll \\Theta_\{\\text\{vib\}\}$ , most molecules are in the ground vibrational state ( $v = 0$ ), so $U_\{\\text\{vib\}\} \\approx N\\varepsilon_0 = N(h\ u/2)$ . The classical equipartition prediction ( $k_BT$ per vibrational mode) is not reached because the energy spacing $h\ u$ is much larger than $k_BT$ . Only at $T \\gg \\Theta_\{\\text\{vib\}\}$ does $U_\{\\text\{vib\}\} \\to Nk_BT$ .

Q6. The Sackur-Tetrode equation gives the molar entropy of a monatomic ideal gas: $S_m = R\\ln\\left[\\frac\{e^\{5/2\} V\}\{N_A \\Lambda^3\}\\right]$ . What is the thermal de Broglie wavelength $\\Lambda$ ?

A. $\\Lambda = \\frac\{h\}\{\\sqrt\{2\\pi m k_B T\}\}$ B. $\\Lambda = \\frac\{h\}\{mc\}$ C. $\\Lambda = \\frac\{h\}\{mv\}$ D. $\\Lambda = \\sqrt\{\\frac\{h\}\{2\\pi m k_B T\}\}$ ’,

Answer: A

Answer: A — The thermal de Broglie wavelength $\\Lambda = h / \\sqrt\{2\\pi m k_B T\}$ represents the quantum wavelength associated with a particle at temperature $T$ . When $V/N\\Lambda^3 \\gg 1$ , the gas behaves classically (Boltzmann statistics). When $V/N\\Lambda^3 \\lesssim 1$ , quantum effects (Bose-Einstein or Fermi-Dirac statistics) become important.

Q7. The canonical ensemble describes a system at fixed $N$ , $V$ , and $T$ . The canonical partition function is $Q = q^N/N!$ (for distinguishable translational states). The Helmholtz free energy is:

A. ” $A = -k_BT \\ln Q = -Nk_BT(\\ln q - \\ln N + 1)$ using Stirling’s approximation”, ’ $A = k_BT \\ln Q$ B. $A = -Nk_BT \\ln q$ C. $A = E - TS + PV$ ’,

Answer: A

Answer: A — $A = -k_BT \\ln Q$ . For an ideal gas: $Q = q^N/N!$ , so $\\ln Q = N\\ln q - \\ln(N!) \\approx N\\ln q - N\\ln N + N$ (Stirling). Thus $A = -Nk_BT(\\ln(q/N) + 1)$ . From $A$ , all other thermodynamic properties follow: $S = -(\\partial A/\\partial T)_V$ , $P = -(\\partial A/\\partial V)_T$ , $U = A + TS$ .

Q8. The Maxwell-Boltzmann speed distribution gives the probability density for molecular speeds as $f(v) \\propto v^2 e^\{-mv^2/2k_BT\}$ . The most probable speed is:

A. $v_p = \\sqrt\{2k_BT/m\}$ B. $v_p = \\sqrt\{8k_BT/\\pi m\}$ C. $v_p = \\sqrt\{3k_BT/m\}$ D. $v_p = k_BT/m$ ’,

Answer: A

Answer: A — The Maxwell-Boltzmann distribution gives three characteristic speeds: most probable $v_p = \\sqrt\{2k_BT/m\}$ , mean $\\langle v \\rangle = \\sqrt\{8k_BT/\\pi m\}$ , and root-mean-square $v_\{\\text\{rms\}\} = \\sqrt\{3k_BT/m\}$ . The relationship is $v_p < \\langle v \\rangle < v_\{\\text\{rms\}\}$ . All scale as $\\sqrt\{T/m\}$ .

Q9. The Boltzmann entropy formula $S = k_B \\ln W$ relates entropy to the number of microstates $W$ . For a system of $N$ distinguishable particles with two energy levels ( $\\varepsilon_0$ and $\\varepsilon_1$ ), if $n$ particles are in the excited state, $W = \\binom\{N\}\{n\}$ . What is the maximum entropy configuration?

A. Maximum $S$ when $n = N/2$ (equal populations of both levels) B. Maximum $S$ when $n = N$ (all particles excited) C. Maximum $S$ when $n = 0$ (all particles in ground state) D. Maximum $S$ when $n = 1$ (single excitation)’,

Answer: A

Answer: A — $W = \\binom\{N\}\{n\}$ is maximized when $n = N/2$ , giving the most microstates. $S = k_B \\ln \\binom\{N\}\{n\}$ is therefore maximized at equal populations. This is the Second Law: spontaneous processes tend toward configurations with more microstates (higher $W$ , higher $S$ ). At finite $T$ , the actual equilibrium population depends on the energy gap via the Boltzmann distribution.

Q10. The equipartition theorem states that each quadratic degree of freedom contributes $\\frac\{1\}\{2\}k_BT$ to the internal energy. For a diatomic gas at moderate temperatures (translational and rotational modes fully excited, vibrational modes frozen), what is $C_V$ ?

A. $C_V = \\frac\{5\}\{2\}R$ per mole (3 translational + 2 rotational degrees of freedom) B. $C_V = \\frac\{3\}\{2\}R$ (translational only) C. $C_V = 3R$ (including vibration) D. $C_V = \\frac\{7\}\{2\}R$ ’,

Answer: A

Answer: A — A diatomic molecule has 3 translational degrees of freedom (contributes $3/2 R$ to $C_V$ ), 2 rotational (linear molecule rotates about 2 axes, contributes $2 \\times 1/2 R = R$ ), and 1 vibrational mode. At moderate temperatures ( $T_\{\\text\{rot\}\} \\ll T \\ll T_\{\\text\{vib\}\}$ ), vibration is not excited. Total $C_V = 3/2 R + R = 5/2 R$ . At very high $T$ , vibration adds $R$ more (kinetic + potential), giving $7/2 R$ .

Intuition

Statistical mechanics bridges the microscopic and macroscopic worlds: A single molecule has no “temperature” or “entropy” — these are emergent properties of vast numbers of particles. Statistical mechanics uses probability to connect the behaviour of individual atoms to the bulk properties we measure in the lab.

Why it matters: It explains why thermodynamic laws work, predicts material properties from first principles, and is essential for understanding phase transitions, protein folding, and climate models.

The key insight: Entropy is really about the number of microstates — the more ways a system can arrange itself while looking the same macroscopically, the higher its entropy.

Common Mistakes

Confusing $C_V$ and $C_P$ : For an ideal gas, $C_P = C_V + R$ . At constant volume, heat goes entirely into temperature change ( $C_V$ ). At constant pressure, some energy does expansion work, so $C_P > C_V$ . Using $C_V$ when the process is at constant pressure (or vice versa) gives wrong heat calculations.

Assuming equipartition applies at all temperatures: Each degree of freedom only contributes $\frac{1}{2}k_BT$ when the temperature is high enough to excite that mode ( $k_BT \gg \Delta E$ ). Vibrational modes are “frozen out” at low temperatures because $\hbar\omega \gg k_BT$ . The classical equipartition theorem fails at low temperatures.

Confusing microstates with macrostates: A microstate specifies the exact quantum state of every particle. A macrostate is specified by bulk properties ( $T$ , $P$ , $V$ ). Many microstates correspond to the same macrostate — entropy $S = k_B \ln \Omega$ counts the number of microstates $\Omega$ for a given macrostate. Confusing these concepts leads to incorrect entropy calculations.

Cross-References

Statistical Mechanics: Detailed notes on partition functions, Boltzmann distribution, and entropy.
Thermodynamics: Covers Gibbs free energy and entropy from the macroscopic perspective.
Practice Physical Chemistry: Interactive practice problems covering quantum chemistry and statistical mechanics.

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