Practice

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}

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). Difficulty: hard

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}

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 $T$, $q \to 1$. Difficulty: hard

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}

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. Difficulty: hard

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

A. $q_{\text{rot}

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. Difficulty: hard

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}

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 $U_\{\text\{vib\}\} \to Nk_BT$. Difficulty: hard

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}

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. Difficulty: hard

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

• $A = -k_BT \ln Q$. 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: $S = -(\partial A/\partial T)_V$, $P = -(\partial A/\partial V)_T$, $U = A + TS$. Difficulty: medium

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}

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 $\sqrt\{T/m\}$. Difficulty: hard

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

• $W = \binom\{N\}\{n\}$ 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. Difficulty: hard

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}

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 $7/2 R$. Difficulty: hard

Q3. For a two-level system with energy gap epsilon, the population ratio n2/n1 equals e to the power minus epsilon over kT. As temperature approaches infinity, this ratio approaches:

A. 1 B. 0 C. infinity D. e

Answer: A

• As T grows the exponent approaches zero, so the ratio approaches e^0 = 1: both levels become equally populated. Absolute zero gives ratio 0 (all particles in the ground state). Difficulty: medium

Q4. The molar entropy of a perfect crystal at absolute zero is zero. This statement is:

A. the Third Law of Thermodynamics B. the First Law of Thermodynamics C. Hess’s Law D. Raoult’s Law

Answer: A

• The Third Law defines the entropy reference: a perfect crystal has exactly one accessible microstate (W = 1) at 0 K, and S = k ln W = 0. This anchor makes absolute entropies measurable. Difficulty: easy

Q5. In the canonical ensemble, the probability of a microstate with energy E is proportional to:

A. exp(-E/kT) B. exp(E/kT) C. E squared D. 1/E

Answer: A

• The Boltzmann factor exp(-E/kT) weights states by energy; higher-energy states are exponentially less probable. The partition function normalises these weights to probabilities. Difficulty: medium

Q6. Which heat capacity is larger for an ideal gas at ordinary temperatures?

A. Cv, constant volume B. Cp, constant pressure C. They are equal D. The gas’s colour

Answer: B

• At constant pressure some absorbed heat does expansion work, so more heat is needed for the same temperature rise: Cp = Cv + R for one mole of ideal gas. The difference equals the gas constant R. Difficulty: easy

Q7. Maxwell-Boltzmann statistics apply when particles are:

A. distinguishable and low-density B. indistinguishable with antisymmetric wavefunctions C. indistinguishable with symmetric wavefunctions D. always superconducting

Answer: A

• Maxwell-Boltzmann counting assumes distinguishable classical particles at low density. Option B describes fermions (Fermi-Dirac); option C describes bosons (Bose-Einstein). Molecular gases at room temperature are well approximated by A. Difficulty: hard

See also

• Statistical Mechanics
• Chemical Kinetics — Practice Problems
• Physical Chemistry — Practice Problems