Statistical Thermodynamics — Complete Notes for Competitive Exams
Thermodynamics tells you what happens — energy flows from hot to cold, equilibria shift, entropy increases. What it never explains is why at the molecular level. Statistical thermodynamics is exactly the bridge that fills that gap. It takes quantum-mechanical properties of individual molecules — their energy levels, degeneracies, masses — and converts them into the bulk thermodynamic quantities you measure in the lab: internal energy, entropy, heat capacity, equilibrium constants.
For GATE, CSIR-NET, IIT-JAM, BITSAT, TGT, and PGT exams, this chapter carries significant weight. The derivations look scary at first glance. Once you see the structure, though — and there really is a clean, logical structure — the whole chapter falls into place rather naturally.
13A — The Boltzmann Distribution
Configurations and Weight
Consider N molecules distributed across energy states ε₀, ε₁, ε₂, ... with N₀ molecules in state 0, N₁ in state 1, and so on. This particular arrangement — written as {N₀, N₁, N₂, ...} — is called a configuration. Different configurations can be realised in different numbers of ways. The number of distinct ways of achieving a given configuration is its weight, W.
W = N! / (N₀! × N₁! × N₂! × N₃! × ...)
The configuration where all N molecules sit in state 0 — written {N, 0, 0, ...} — has W = 1. Just one way to achieve it. The configuration {N−2, 2, 0, ...} can be achieved in ½N(N−1) ways, which is enormously larger. This combinatorial fact is why systems almost never pile all molecules into a single state — not because it's forbidden, but because statistically it almost never happens when N is large.
Work through a quick number to get the intuition. For N = 20 with the configuration {3, 6, 5, 4}, the weight is 18!/(3!×6!×5!×4!) ≈ 515 million. That's just one of many configurations, but already it dwarfs {20, 0, 0, 0} = 1.
Stirling's Approximation
Factorials of large numbers are unwieldy. The approximation that makes the maths tractable is:
Using this, the log-weight simplifies to:
ln W = N · ln N − Σᵢ Nᵢ · ln Nᵢ
The Most Probable Distribution
The system essentially always sits in its maximum-weight configuration, because that configuration is so overwhelmingly more probable than any other when N is large. To find it, we maximise ln W subject to two hard constraints:
- Number constraint: Σ Nᵢ = N (the total number of molecules is fixed)
- Energy constraint: Σ Nᵢεᵢ = E (total energy is fixed for an isolated system)
The mathematical tool for constrained maximisation is Lagrange's method of undetermined multipliers. Introducing constants α and −β (the negative sign on β is chosen for later convenience), the condition dln W = 0 gives:
ln(Nᵢ/N) = α − βεᵢ
After applying the number constraint to evaluate α, this becomes the celebrated Boltzmann distribution:
Nᵢ / N = e−βεᵢ / q where β = 1/kT
The denominator q = Σᵢ e−βεᵢ is the partition function. It's not just a normalisation constant — it contains all the thermodynamic information about the system, analogous to the role the wavefunction plays in quantum mechanics.
The Partition Function — Physical Meaning
At very low temperatures, β → ∞ and e−βεᵢ → 0 for every excited state. So q → g₀, the degeneracy of the ground state. At very high temperatures, every Boltzmann factor approaches 1 and q approaches the total number of states. In general:
Population Ratios — The Working Formula
When you only care about the relative population of two levels, the partition function cancels entirely:
This ratio falls exponentially as energy separation increases or temperature drops. At T = 0, everything is in the ground state. As T → ∞, all states approach equal population — regardless of their energies.
Worked Example: Rotational population ratio for HCl (B̃ = 10.591 cm⁻¹) at 25°C.
The J = 0 level is non-degenerate (g₀ = 1). The J = 1 level is triply degenerate (g₁ = 3). Energy separation: ε₁ − ε₀ = 2hcB̃. Using kT/hc = 207.225 cm⁻¹ at 298 K:
N₁/N₀ = 3 × e−2 × 10.591/207.225 = 3 × e−0.1022 ≈ 2.708
J = 1 is more populated than J = 0 despite being at higher energy — purely because of its threefold degeneracy. You'll see this pattern repeatedly in rotational spectroscopy problems.
13B — Molecular Partition Functions
Since molecular energy is a sum of contributions from different modes, the partition function factorises into a product. This is one of the most useful properties of q:
Translation, rotation, vibration, and electronic contributions can each be studied independently — which makes the whole framework enormously practical.
Translational Partition Function
Treating translational motion as a particle in a 3D box of volume V:
Λ is called the thermal de Broglie wavelength. It shrinks as mass or temperature increases, so qT grows with both. For an H₂ molecule in 100 cm³ at 25°C, Λ ≈ 71.2 pm and qT ≈ 2.77 × 10²⁶. An astronomical number of translational states are accessible at room temperature — which is why classical (equipartition) results work well for translation.
Rotational Partition Function
The characteristic rotational temperature is θR = hcB̃/k. When T >> θR, the high-temperature (classical) approximation applies:
Nonlinear molecule: qR = (1/σ)(kT/hc)3/2 (π/ÃB̃C̃)1/2
σ is the symmetry number — the number of indistinguishable orientations of the molecule produced by rotations.
| Molecule | σ | θR/K |
|---|---|---|
| Heteronuclear diatomic (CO, HCl) | 1 | — |
| Homonuclear diatomic (N₂, Cl₂, H₂) | 2 | H₂: 87.6, N₂: 2.88 |
| H₂O | 2 | — |
| NH₃ | 3 | — |
| CH₄ | 12 | — |
| C₆H₆ (benzene) | 12 | — |
H₂ has an unusually high θR = 87.6 K because its tiny moment of inertia means widely-spaced rotational levels. For most other molecules at room temperature, the high-T approximation is perfectly fine.
Vibrational Partition Function
Treating vibration as a harmonic oscillator — equally spaced levels separated by hcν̃ — gives:
where θV = hcν̃/k is the characteristic vibrational temperature.
- When βhcν̃ >> 1 (low T, high ν̃): qV ≈ 1 — only the ground state is significantly populated
- When T >> θV: qV ≈ kT/hcν̃ (classical limit)
For C–H stretches (ν̃ ≈ 3000 cm⁻¹), βhcν̃ ≈ 14 at room temperature. The partition function is essentially 1.000 — vibrations are frozen out entirely. For I₂ (ν̃ = 214.6 cm⁻¹) at 298 K, however, βhcν̃ = 1.035 and qV = 1.55 — both ground and first excited states are meaningfully populated.
For polyatomic molecules, each of the 3N−5 (linear) or 3N−6 (nonlinear) normal modes contributes its own qV. The overall vibrational partition function is the product of all mode contributions.
Example — H₂O at 1500 K (modes at 3656.7, 1594.8, 3755.8 cm⁻¹, kT/hc = 1042.6 cm⁻¹):
- qV(mode 1, βhcν̃ = 3.507) = 1.031
- qV(mode 2, βhcν̃ = 1.530) = 1.277
- qV(mode 3, βhcν̃ = 3.602) = 1.028
- Overall qV = 1.031 × 1.277 × 1.028 ≈ 1.353
Electronic Partition Function
Electronic energy gaps are typically tens of thousands of cm⁻¹ — enormous compared to kT (~200 cm⁻¹ at room temperature). So almost always qE = g₀, the ground-state degeneracy.
- Noble gas atoms: g₀ = 1, so qE = 1
- Alkali metal atoms (doublet ground state): g₀ = 2, so qE = 2
- NO molecule: doubly degenerate ground (²Π₁/₂) + doubly degenerate excited state (²Π₃/₂) at only 121 cm⁻¹. At 25°C, qE = 2 + 2e−121/207.2 ≈ 3.1
13C — Molecular Energies
The mean energy of a molecule follows directly from the partition function:
Energies are measured relative to the ground state, so εgs needs to be added back if you want the absolute energy.
Energy Contributions from Each Mode
At high enough temperatures (classical limit), the equipartition theorem gives ½kT per quadratic term in the energy. When this applies, you don't even need the partition function — just count the terms.
| Mode | Mean Energy | Condition |
|---|---|---|
| Translation (3D) | 3/2 kT | Always valid |
| Rotation — linear | kT | T >> θR |
| Rotation — nonlinear | 3/2 kT | T >> θR |
| Vibration (harmonic) | hcν̃/(eβhcν̃−1) | Exact; → kT when T >> θV |
| Electronic | ≈ 0 | Usually (unless thermally accessible) |
Molar Heat Capacity — Equipartition
The molar constant-volume heat capacity from equipartition:
where νR = 2 for linear molecules and 3 for nonlinear, and νV is the number of active vibrational modes (usually 0 at room temperature).
For the exact vibrational heat capacity (the Einstein formula):
This rises from zero at low T to R at high T (when T >> θV). The curve reaches within 10% of R by T ≈ θV.
Example — H₂O at 373 K: All three rotational modes (θR ≈ 13–40 K) are active. Vibrational temperatures (≈ 2300–5400 K) are far above 373 K, so vibrations are frozen. Predicted CV,m = 3/2 R + 3/2 R = 3R ≈ 25 J K⁻¹ mol⁻¹. Experimental value: 26.1 J K⁻¹ mol⁻¹. The small discrepancy reflects slight deviations from ideal gas behaviour.
13D — The Canonical Ensemble
The molecular partition function works beautifully for non-interacting molecules. Real gases interact. Liquids certainly do. Handling those requires the canonical ensemble.
The conceptual move is elegant. Take your actual system (with all its molecular interactions) and imagine replicating it Ñ times — all at the same temperature T, volume V, and particle number N, free to exchange energy with each other but isolated overall. This mental collection of replicas is the canonical ensemble.
The canonical partition function Q carries all the thermodynamic information for interacting systems:
where Eᵢ is the total energy of the entire system in microstate i. For independent molecules, Q reduces to:
- Distinguishable particles (e.g., molecules fixed at crystal lattice sites): Q = qN
- Indistinguishable particles (e.g., gas-phase molecules): Q = qN/N!
The N! factor corrects for overcounting — swapping two identical gas molecules doesn't create a physically new state. Molecules in a crystal, however, are identifiable by their positions, so no correction is needed.
• Microcanonical: constant V, E, N
• Canonical: constant V, T, N
• Grand canonical: constant V, T, μ
The mean energy of the system:
⟨E⟩ = −(∂ ln Q / ∂β)V
For a gas of N indistinguishable, independent molecules, Q = qN/N!, and this gives ⟨E⟩ = N⟨ε⟩ — exactly N times the single-molecule mean energy, as expected.
13E — Internal Energy and Entropy
Internal Energy from q
For N independent molecules, the internal energy relative to the ground state is:
U(0) is the internal energy when only the ground state is occupied — the T = 0 baseline including zero-point contributions. For a system of harmonic oscillators:
UmV(T) = UmV(0) + NAhcν̃ / (eβhcν̃ − 1)
The Boltzmann Formula for Entropy
The most famous equation in statistical mechanics — literally carved on Boltzmann's tombstone in Vienna:
W is the weight of the most probable configuration. As T → 0, every molecule collapses into the ground state, W = 1, ln W = 0, so S → 0. This is the statistical basis of the Third Law of thermodynamics — and it falls out naturally without being imposed as an external axiom.
Entropy in Terms of the Partition Function
For distinguishable independent molecules:
S = {U(T) − U(0)}/T + Nk ln q
For indistinguishable molecules (gas phase):
The extra term −Nk ln N + Nk = Nk(1 − ln N) from the N! correction is critical. Omitting it leads to the Gibbs paradox — the entropy of mixing of identical gases would be non-zero, which is unphysical.
The Sackur–Tetrode Equation
Applying the above to a monatomic perfect gas (where only translation matters) gives the remarkable Sackur–Tetrode equation:
This predicts the absolute entropy of a monatomic gas from first principles — nothing fitted, just mass and temperature as inputs. For argon at 25°C (Λ = 16.0 pm), this gives 154.8 J K⁻¹ mol⁻¹, in excellent agreement with experiment.
The physical interpretation writes itself. Heavier atoms → smaller Λ → larger qT → higher entropy. Larger volume → more translational states → higher entropy. Higher temperature → smaller Λ → again higher entropy. All consistent with thermodynamic intuition.
Rotational Entropy (high-T limit)
Heavy molecules with large moments of inertia have small B̃, low θR, and large qR — so they have much higher rotational entropy than light molecules. Cl₂ has a rotational entropy of 58.6 J K⁻¹ mol⁻¹ at 25°C; H₂ has only 12.7 J K⁻¹ mol⁻¹.
Vibrational Entropy
This → 0 as T → 0, and rises indefinitely as T increases. For I₂ (θV = 309 K) at 25°C, SmV ≈ 8.38 J K⁻¹ mol⁻¹.
Residual Entropy
Some solids retain disorder even at T = 0. CO is the textbook example — the molecule looks nearly the same whether oriented as C≡O or O≡C in the crystal lattice, so molecules adopt random orientations. With two equally probable orientations for each of N molecules, W = 2N:
S(0) = Nk ln 2 = nR ln 2 ≈ 5.76 J K⁻¹ mol⁻¹
The measured residual entropy of CO is about 5 J K⁻¹ mol⁻¹ — very close. The small discrepancy comes from slight orientation preference.
Ice (H₂O) is another celebrated case. Each oxygen is surrounded by four hydrogens, two close (covalent O–H) and two far (hydrogen-bonded). Of the 2⁴ = 16 possible arrangements of H atoms around each O, only 6 satisfy the "two short, two long" condition. This gives W = (3/2)N and:
Sm(0) = R ln(3/2) ≈ 3.37 J K⁻¹ mol⁻¹ (experimental: 3.4 J K⁻¹ mol⁻¹) ✓
13F — Derived Functions: Helmholtz, Gibbs, and Equilibrium Constants
Key Thermodynamic Functions from Q
| Function | Expression |
|---|---|
| Helmholtz energy A | A(T) = A(0) − kT ln Q |
| Pressure p | p = kT (∂ ln Q / ∂V)T |
| Enthalpy H | H(T) = H(0) − (∂ ln Q / ∂β)V + kTV(∂ ln Q / ∂V)T |
| Gibbs energy G | G(T) = G(0) − nRT ln(qm/NA) for a perfect gas |
Substituting Q = qN/N! with qT = V/Λ³ into the pressure equation recovers the perfect gas law pV = nRT. This is the statistical proof that the parameter β in the Boltzmann distribution really equals 1/kT — it's not assumed, it's derived by matching to experiment.
Statistical Interpretation of Gibbs Energy
For a perfect gas of indistinguishable molecules: G(T) − G(0) = −nRT ln(qm/NA). As more states become thermally accessible — as qm/NA increases — the Gibbs energy decreases. The thermodynamic drive toward lower G is, at the molecular level, a drive toward maximising the number of accessible states. Entropy wins out.
Equilibrium Constants from Partition Functions
For the general gas-phase reaction aA + bB → cC + dD:
where νJ are the stoichiometric numbers (positive for products, negative for reactants), qJ,m⊖ is the standard molar partition function of species J, and ΔrE₀ is the difference in zero-point energies (from bond dissociation data).
This equation is remarkable. It says K is controlled by two things: the energy difference between ground states (the Boltzmann exponential, like enthalpic effects) and the relative density of accessible states (the partition function ratio, like entropic effects). A reaction can be endothermic and still have K > 1 at high temperature if the products have a much higher density of states — which is exactly the statistical explanation of why entropy-driven reactions are spontaneous.
Example — Na₂(g) → 2Na(g) at 1000 K: Using spectroscopic data (B̃ = 0.1547 cm⁻¹, ν̃ = 159.2 cm⁻¹, D₀ = 70.4 kJ mol⁻¹), qR(Na₂) = 2246, qV(Na₂) = 4.885, and the Na atom has a doublet ground state (g = 2). Substituting into eqn 13F.12 gives K ≈ 2.45 at 1000 K.
Solved Examples — Step by Step
Example 1 — Configuration Weight
Q: Calculate W for the configuration {1, 0, 3, 5, 10, 1} where N = 20.
Sol: W = 20! / (1! × 0! × 3! × 5! × 10! × 1!) = (2.432×10¹⁸) / (1×1×6×120×3628800×1) ≈ 9.31 × 10⁸
Example 2 — Two-Level Population
Q: Two conformations A and B of a molecule differ by 5.0 kJ mol⁻¹. What fraction is in B at 293 K?
Sol: ε = 5000 / (6.022×10²³) = 8.3×10⁻²¹ J. βε = 8.3×10⁻²¹ / (1.381×10⁻²³ × 293) = 2.05.
q = 1 + e−2.05 = 1 + 0.129 = 1.129. Fraction in B = e−2.05/1.129 = 11.4%
Example 3 — Rotational Partition Function (HCl)
Q: Calculate qR for HCl (B̃ = 10.591 cm⁻¹) at 25°C by (a) explicit summation and (b) the high-T approximation.
Sol (b): qR = kT/hcB̃ = 207.225/10.591 = 19.57. Exact summation gives 19.902 — agreement is excellent.
Example 4 — Vibrational Partition Function (I₂)
Q: Find qV for I₂ (ν̃ = 214.6 cm⁻¹) at 298 K.
Sol: βhcν̃ = 214.6/207.225 = 1.035. qV = 1/(1−e−1.035) = 1/(1−0.355) = 1.55
Only the ground and first excited vibrational states are significantly populated.
Example 5 — Mean Vibrational Energy (I₂)
Q: Calculate the molar vibrational internal energy of I₂ at 298 K (above zero-point).
Sol: ⟨εV⟩ = hcν̃/(eβhcν̃−1). hcν̃ = 6.626×10⁻³⁴ × 2.998×10¹⁰ × 214.6 = 4.26×10⁻²¹ J.
⟨εV⟩ = 4.26×10⁻²¹ / (e1.035−1) = 4.26×10⁻²¹ / 1.815 = 2.35×10⁻²¹ J
Molar: 2.35×10⁻²¹ × 6.022×10²³ = 1.41 kJ mol⁻¹
Example 6 — Electronic Partition Function (NO)
Q: Calculate qE for NO at 25°C (²Π₁/₂ ground state, g = 2; ²Π₃/₂ at 121 cm⁻¹, g = 2).
Sol: β × 121 cm⁻¹ = 121/207.225 = 0.5839. qE = 2 + 2e−0.5839 = 2 + 2(0.558) = 2 + 1.115 = 3.11
Example 7 — Residual Entropy of CO
Q: Estimate the residual molar entropy of CO.
Sol: Each molecule has 2 equally probable orientations (CO or OC). s = 2.
Sm(0) = R ln 2 = 8.314 × 0.693 = 5.76 J K⁻¹ mol⁻¹
Example 8 — Electronic Mean Energy
Q: An atom has a doubly degenerate ground level and a fourfold degenerate excited level at 600 cm⁻¹. Find its mean electronic energy at 25°C (kT/hc = 207.225 cm⁻¹).
Sol: qE = 2 + 4e−600/207.225 = 2 + 4e−2.895 = 2 + 4(0.0552) = 2.221
⟨εE⟩/hc = 4 × 600 × e−2.895 / 2.221 = 132.5/2.221 = 59.7 cm⁻¹
Multiple Choice Questions — Practice Set (25–30 Q)
Section A: Configurations and Boltzmann Distribution
Q1. The number of ways of achieving the configuration {N, 0, 0, ...} is:
(a) N! (b) N (c) 0 (d) 1
Ans: (d) 1
Q2. Stirling's approximation gives: ln(x!) ≈
(a) x ln x (b) x ln x − x (c) x ln x + x (d) (x−1) ln x
Ans: (b)
Q3. The Boltzmann distribution is obtained by maximising ln W subject to:
(a) Energy constraint only (b) Number constraint only (c) Both (d) Neither
Ans: (c)
Q4. At T → ∞ for a two-level system, the ratio N₁/N₀ approaches:
(a) 0 (b) 1 (c) g₁/g₀ (d) ∞
Ans: (c) g₁/g₀
Q5. The value of β in the Boltzmann distribution is:
(a) kT (b) 1/(kT)² (c) 1/kT (d) k/T
Ans: (c)
Q6. Two energy levels have the same energy but different degeneracies g₁ = 3, g₂ = 1. The ratio N₁/N₂ at any temperature T is:
(a) 1 (b) 3 (c) 1/3 (d) Temperature-dependent
Ans: (b) 3
Section B: Partition Functions
Q7. At T = 0, the molecular partition function q equals:
(a) 0 (b) 1 (c) g₀ (d) ∞
Ans: (c) g₀ (degeneracy of the ground level)
Q8. The thermal de Broglie wavelength Λ = h/(2πmkT)1/2. As mass increases, Λ:
(a) Increases (b) Decreases (c) Remains constant (d) Doubles
Ans: (b) Decreases
Q9. The translational partition function qT for a 3D ideal gas in volume V is:
(a) V/Λ (b) V/Λ² (c) V/Λ³ (d) VΛ³
Ans: (c)
Q10. The symmetry number of CO₂ is:
(a) 1 (b) 2 (c) 3 (d) 4
Ans: (b) 2
Q11. The symmetry number of benzene (C₆H₆) is:
(a) 6 (b) 24 (c) 12 (d) 3
Ans: (c) 12
Q12. For an equally spaced harmonic oscillator (spacing ε), qV = ?
(a) 1 + e−βε (b) 1/(1 + e−βε) (c) 1/(1 − e−βε) (d) eβε − 1
Ans: (c)
Q13. At high temperature (T >> θV), qV approaches:
(a) 1 (b) hcν̃/kT (c) kT/hcν̃ (d) 0
Ans: (c)
Q14. The electronic partition function of a noble gas atom is:
(a) 0 (b) 2 (c) 1 (d) Temperature-dependent
Ans: (c) 1 (non-degenerate ground state, no low-lying excited states)
Section C: Molecular Energies
Q15. The mean translational energy of a molecule in 3D is:
(a) ½kT (b) kT (c) 3/2 kT (d) 5/2 kT
Ans: (c)
Q16. For a linear diatomic molecule at T >> θR, the mean rotational energy is:
(a) ½kT (b) kT (c) 3/2 kT (d) 2kT
Ans: (b) kT
Q17. CV,m for a gas of linear diatomic molecules at high temperature (all modes active) per mole is:
(a) 5/2 R (b) 3R (c) 7/2 R (d) 2R
Ans: (c) 7/2 R (3/2 R translation + R rotation + R vibration)
Q18. Which mode of molecular motion is always thermally active at any T > 0?
(a) Vibration (b) Rotation (c) Translation (d) Electronic
Ans: (c)
Section D: Entropy and Canonical Ensemble
Q19. For N indistinguishable, independent gas molecules, Q = ?
(a) qN (b) qN/N (c) N·q (d) qN/N!
Ans: (d)
Q20. The Boltzmann formula for entropy is:
(a) S = −k ln W (b) S = kW (c) S = k ln W (d) S = W/k
Ans: (c)
Q21. For a perfect crystal at T = 0 (W = 1), the statistical entropy is:
(a) k (b) Nk (c) ∞ (d) 0
Ans: (d) 0 — consistent with the Third Law
Q22. The Sackur–Tetrode equation applies to:
(a) Diatomic gases (b) Crystals (c) Monatomic perfect gases (d) Liquids
Ans: (c)
Q23. The residual molar entropy of a solid where each molecule can adopt s equally probable orientations is:
(a) R/s (b) R ln s (c) Rs (d) k ln s
Ans: (b)
Q24. The predicted residual entropy of ice arises from hydrogen-bond disorder and equals:
(a) R ln 2 (b) R ln(3/2) (c) R ln 4 (d) 0
Ans: (b) R ln(3/2) ≈ 3.4 J K⁻¹ mol⁻¹
Q25. In a canonical ensemble, every member has the same:
(a) Energy (b) V, T, and N (c) V, E, N (d) V, T, and μ
Ans: (b)
Section E: Derived Functions and Equilibrium
Q26. The Helmholtz energy A is related to the partition function Q by:
(a) A = A(0) + kT ln Q (b) A = A(0) − kT ln Q (c) A = kT/ln Q (d) A = −kT/Q
Ans: (b)
Q27. The pressure of a gas from statistical thermodynamics is:
(a) p = kT(∂ ln Q/∂V)T (b) p = −kT(∂ ln Q/∂V)T (c) p = kT·Q (d) p = Q/V
Ans: (a)
Q28. In an endothermic dissociation reaction, K can exceed 1 when:
(a) ΔrE₀ < 0 only (b) T = 0 (c) Products have a much higher density of states (d) Reactants have lower partition functions
Ans: (c)
Q29. Which ensemble allows matter to be exchanged between members?
(a) Microcanonical (b) Canonical (c) Grand canonical (d) Isothermal
Ans: (c)
Q30. The mean energy of a two-level system (0 and ε) at temperature T is:
(a) ε/(eβε − 1) (b) ε·e−βε/(1 + e−βε) (c) ε/2 (d) kT
Ans: (b)
Additional Practice Problems
- Calculate the weight of the configuration in which 16 objects are distributed as {0, 1, 2, 3, 8, 0, 0, 0, 0, 2}.
- At what temperature does a two-level system with energy separation 400 cm⁻¹ have the upper state population equal to one-third of the lower state?
- Calculate the translational partition function of a Ne atom in a 1.00 cm³ box at 300 K.
- H₂O has rotational constants 27.877, 14.509, and 9.287 cm⁻¹. Estimate qR at 298 K.
- Using equipartition, estimate CV,m for gaseous I₂, CH₄, and C₆H₆ at 25°C.
- Calculate qV for CS₂ at 500 K. Modes: 658 cm⁻¹ (stretch), 397 cm⁻¹ (bend, doubly degenerate), 1535 cm⁻¹ (stretch).
- An atom has a fourfold degenerate ground level, a non-degenerate excited level at 2500 cm⁻¹, and a doubly degenerate level at 3500 cm⁻¹. Find qE and the relative populations at 1900 K.
- Calculate the standard molar entropy of gaseous He and Xe at 298 K using the Sackur–Tetrode equation. Which is larger and why?
- Show that for a perfect gas of non-interacting atoms, (∂⟨E⟩/∂V)T = 0.
- Using the spectroscopic data for Na₂ (B̃ = 0.1547 cm⁻¹, ν̃ = 159.2 cm⁻¹, D₀ = 70.4 kJ mol⁻¹), evaluate K for Na₂(g) → 2Na(g) at 1000 K.
Quick Formula Summary
| Property | Formula | Notes |
|---|---|---|
| Weight W | N!/(N₀!N₁!...) | Maximise for most probable config |
| Stirling's approx. | ln(x!) ≈ x ln x − x | Valid for large x |
| Boltzmann distribution | Nᵢ/N = e−βεᵢ/q | β = 1/kT |
| Population ratio | Nᵢ/Nⱼ = (gᵢ/gⱼ)e−β(εᵢ−εⱼ) | Levels i and j |
| Partition function | q = Σ gᵢ e−βεᵢ | Sum over levels |
| qT | V/Λ³, Λ = h/(2πmkT)1/2 | Always ~10²⁶ for gases |
| qR (linear) | kT/(σhcB̃) | T >> θR |
| qR (nonlinear) | (1/σ)(kT/hc)3/2(π/ÃB̃C̃)1/2 | T >> θR |
| qV | 1/(1 − e−βhcν̃) | Harmonic oscillator |
| qE | Σ gᵢ e−βεᵢ (levels) | Usually = g₀ |
| Mean energy | ⟨ε⟩ = −(∂ ln q/∂β)V | From partition function |
| ⟨εT⟩ | 3/2 kT | Always active |
| ⟨εR⟩ linear | kT | T >> θR |
| ⟨εV⟩ | hcν̃/(eβhcν̃−1) | Harmonic; → kT at high T |
| Internal energy U | U(T) = U(0) − N(∂ ln q/∂β)V | Independent molecules |
| CV,m (Einstein) | R(θV/T)² eθV/T/(eθV/T−1)² | One vibrational mode |
| Entropy (S = k ln W) | Boltzmann formula | Cornerstone of stat. thermo |
| Sackur–Tetrode | Sm = R ln(Vme5/2/NAΛ³) | Monatomic perfect gas only |
| Residual entropy | Sm(0) = R ln s | s = number of orientations |
| Helmholtz energy | A = A(0) − kT ln Q | Q = canonical partition fn |
| Equilibrium constant | K = ∏(qJ,m/NA)νJ e−ΔE₀/RT | From spectroscopic data |
