Partition Functions: Worked Problems

Introduction

Partition function formulae are compact, but using them well requires care with units, degeneracies and zero-point conventions. This page works through three linked problems that draw on everything developed so far: vibrational populations and energies of iodine, the rotational distribution of hydrogen chloride, and the entropy of a two-level system. Each problem ends with a check against physical intuition, a habit that catches most numerical slips.

Core explanation

A useful conversion. Spectroscopic constants are given as wavenumbers in cm⁻¹. Multiplying by hc/k = 1.4388 cm K converts them into characteristic temperatures, so that every Boltzmann factor becomes exp(−θ/T). This avoids handling tiny joule values.

Problem 1: vibration of I₂ at 298 K. Iodine has ν̃ = 214.5 cm⁻¹, so θ V = 1.4388 × 214.5 ≈ 308.6 K and x = θ V/T = 1.036. Measuring energies from the zero-point level,

e^(−x) = 0.355, q V = 1/(1 − e^(−x)) = 1.550

Populations follow p v = (1 − e^(−x)) e^(−vx):

p₀ = 0.645, p₁ = 0.229, p₂ = 0.081

About a third of iodine molecules are vibrationally excited at room temperature, which is unusual; for N₂ (θ V ≈ 3390 K) the excited fraction is about 10⁻⁵. The molar vibrational energy above the zero-point level is

U V = Rθ V/(e^x − 1) = (8.314 × 308.6)/1.817 ≈ 1.41 kJ mol⁻¹

compared with the classical equipartition value RT = 2.48 kJ mol⁻¹. The vibration is partly, but not fully, active.

Problem 2: rotation of HCl at 298 K. With B = 10.59 cm⁻¹, θ R = 15.24 K and σ = 1, so the high-temperature form gives q R ≈ T/θ R = 19.6 (adding the small correction of 1/3 gives about 19.9). Relative populations are (2J + 1) exp(−J(J + 1)θ R/T):

J = 2: 5 × 0.736 = 3.68; J = 3: 7 × 0.541 = 3.79; J = 4: 9 × 0.360 = 3.24

The most populated level is J = 3, in agreement with J max ≈ (T/2θ R)^½ − ½ ≈ 2.6. The fraction in J = 3 is about 3.79/19.9 ≈ 0.19. This distribution explains the characteristic intensity envelope of the HCl infrared spectrum, with the strongest lines several steps from the band centre.

Problem 3: entropy of a two-level system. Take non-degenerate levels separated by ε, at a temperature where ε = kT. Then q = 1 + e^(−1) = 1.368 and the upper-level fraction is e^(−1)/q = 0.269. The molar internal energy is U = N Aε × 0.269 = 0.269RT. The entropy follows from

S = U/T + R ln q = R(0.269 + 0.313) = 0.582R ≈ 4.84 J K⁻¹ mol⁻¹

This lies below the limiting value R ln 2 = 5.76 J K⁻¹ mol⁻¹, reached only when both levels are equally populated at infinite temperature.

Formulae

θ = (hc/k)ν̃ with hc/k = 1.4388 cm K; q V = 1/(1 − e^(−θ V/T)); U V = Rθ V/(e^(θ V/T) − 1); q R ≈ T/(σθ R); S = U/T + R ln q for localised systems or per mode of internal motion.

Step-by-step reasoning

A reliable routine for any partition-function problem:

1. Convert every spectroscopic constant into a characteristic temperature. 2. Decide whether a closed form (harmonic oscillator, high-temperature rotor) is valid or a direct sum is needed. 3. Evaluate q, stating clearly where the energy zero lies. 4. Obtain populations as g i e^(−ε i/kT)/q. 5. Use U = −N(∂ ln q/∂β) and S = U/T + k ln Q. 6. Compare with the zero-temperature and high-temperature limits.

Visual explanation

Sketch a bar chart of vibrational populations for I₂: three bars falling geometrically, 0.65, 0.23, 0.08. Beside it, draw the HCl rotational chart: bars that rise from J = 0, peak at J = 3 and then decay. The contrast shows how degeneracy (2J + 1) reshapes a purely exponential Boltzmann distribution.

Real-world analogy

Estimating populations is like predicting seating in a stadium where lower tiers are cheaper. With a single seat per tier, every tier empties faster than the one below; when upper tiers are much wider, as for rotational levels, a middle tier can hold the largest crowd despite costing more.

Real-world example

Astronomers use the ratio of line intensities from different rotational levels of molecules such as CO in interstellar clouds to infer the gas temperature. The calculation is exactly the population analysis in Problem 2, run in reverse: measured populations give θ R/T and hence T.

Why?

Why does iodine have such a high vibrational population compared with nitrogen? Iodine atoms are heavy and the I–I bond is weak, so the vibrational wavenumber is low and θ V is similar to room temperature. Nitrogen's light atoms and triple bond give a θ V more than ten times larger.

Common misconception

"The most populated rotational level is always J = 0." That is true for each individual state, but a level with quantum number J contains 2J + 1 states, so the level population can peak well above J = 0.

Worked example

Question: Calculate the vibrational contribution to the molar entropy of I₂ at 298 K.

Reasoning: Use S V = R[x/(e^x − 1) − ln(1 − e^(−x))] with x = 1.036. First term: 1.036/1.817 = 0.570. Second term: −ln(0.645) = 0.439. Sum = 1.009, so S V = 8.314 × 1.009.

Answer: S V ≈ 8.4 J K⁻¹ mol⁻¹.

Quick check

1. What fraction of HCl molecules occupy the J = 0 level at 298 K, given q R ≈ 19.9? Answer: The J = 0 level has degeneracy 1 and zero energy, so its fraction is 1/19.9 ≈ 0.050.

Exam focus

State the energy zero in every answer, and remember the degeneracy factor 2J + 1 for rotational levels. Examiners reward a closing sanity check, such as comparing U with RT or S with R ln 2 for a two-level system.

Advanced insight

The harmonic formula overestimates populations of high vibrational levels for weakly bound molecules such as I₂, because anharmonicity brings levels closer together. Accurate thermochemical tables sum over measured anharmonic levels and include rotation–vibration coupling, corrections of a few per cent at high temperature.

Summary

Converting wavenumbers into characteristic temperatures makes partition-function problems straightforward. Iodine is significantly vibrationally excited at room temperature, with q V ≈ 1.55 and U V ≈ 1.41 kJ mol⁻¹. The HCl rotational distribution peaks at J = 3 because of level degeneracy. A two-level system at ε = kT has entropy 0.58R, below its limit of R ln 2.

Practice questions

1. Calculate θ V for a vibration of wavenumber 1000 cm⁻¹ and state whether it is significantly excited at 298 K. Answer: θ V ≈ 1439 K, so θ V/T ≈ 4.8 and only about 0.8% of molecules are excited; it is largely inactive. 2. What is the population ratio p₁/p₀ for I₂ at 298 K? Answer: p₁/p₀ = e^(−1.036) ≈ 0.355. 3. Estimate q R for HCl at 596 K using the high-temperature expression. Answer: q R ≈ 596/15.24 ≈ 39. 4. A two-level system has a doubly degenerate upper level. What is its entropy limit at very high temperature? Answer: Three states become equally populated, so S → R ln 3 ≈ 9.13 J K⁻¹ mol⁻¹.