Heat Capacity from Partition Functions

Introduction

Heat capacity is one of the most revealing thermodynamic measurements: it shows directly which kinds of molecular motion are absorbing energy at a given temperature. Nineteenth-century physics predicted constant heat capacities from equipartition, but experiments showed that heat capacities fall at low temperatures — one of the first cracks in classical physics. Partition functions explain the data completely: each mode contributes only once kT is comparable with its level spacing.

Core explanation

Definition. C V = (∂U/∂T) V. Since U − U(0) = kT²(∂ln Q/∂T) V, the heat capacity is obtained by one further differentiation. Equivalently, C V = kβ²(∂²ln Q/∂β²) V, which shows that C V is proportional to energy fluctuations and so can never be negative.

Additivity. For a perfect gas, U is a sum of mode contributions, so C V is too:

C V = C V,T + C V,R + C V,V + C V,E

Translation. U T = (3/2)RT per mole, so C V,T = (3/2)R = 12.47 J K⁻¹ mol⁻¹ at all but the lowest temperatures, since translational levels are extremely closely spaced.

Rotation. Above θ R, C V,R = R for linear molecules and (3/2)R for non-linear ones. Most θ R values are only a few kelvin, so rotation is fully active at room temperature. H₂ is the exception, with θ R ≈ 88 K.

Vibration. Differentiating U V = Rθ V/(e^(θ V/T) − 1) gives, per mode,

C V,V = R x² eˣ/(eˣ − 1)², x = θ V/T

This is the Einstein function. For x ≪ 1 it tends to R (full classical value); for x ≫ 1 it collapses as R x² e^(−x). The switch-on happens over a fairly narrow range around T ≈ θ V/3 to θ V.

Equipartition as a limit. Classically, each quadratic term in the energy contributes ½R. Translation has three (kinetic), rotation of a linear molecule two, and a vibration two (kinetic and potential), giving R per vibrational mode. The partition function reproduces these values when T is far above the mode's characteristic temperature and shows how they fail below it.

Hydrogen. The heat capacity C V,m of H₂ is (3/2)R below about 50 K (only translation), rises towards (5/2)R by room temperature as rotation switches on, and climbs further above about 1000 K as vibration (θ V ≈ 6300 K) begins to contribute. This staircase is the classic demonstration of quantum freeze-out.

Solids. In a crystal each atom has three vibrational modes and no translation or rotation. Classical theory gives 3R ≈ 25 J K⁻¹ mol⁻¹ (the Dulong–Petit law). The Einstein model predicts the fall at low temperature; the Debye model, using a spread of frequencies, gives the observed C V ∝ T³ near absolute zero.

Formulae

C V = (∂U/∂T) V = kβ²(∂²ln Q/∂β²) V. Per mole: C V,T = (3/2)R; C V,R = R (linear) or (3/2)R (non-linear); C V,V = R x² eˣ/(eˣ − 1)² with x = θ V/T. For a perfect gas C p,m = C V,m + R.

Step-by-step reasoning

To predict C V,m of a gas:

1. Add (3/2)R for translation. 2. Add R (linear) or (3/2)R (non-linear) for rotation if T ≫ θ R. 3. For each vibrational mode calculate x = θ V/T and evaluate the Einstein function. 4. Add an electronic term only if there are low-lying excited levels. 5. Add R to convert to C p,m if required.

Visual explanation

Sketch C V,m/R against log T for H₂. There are plateaus at 1.5 and 2.5, joined by smooth S-shaped rises centred near θ R and θ V respectively. Each rise is the signature of one mode "unfreezing". A dashed line at 3.5 marks the full classical value that dissociation prevents H₂ from reaching.

Real-world analogy

A dimmer switch rather than an on–off switch: as temperature rises, each mode's contribution brightens smoothly from dark to full. Modes with small level spacings reach full brightness at very low temperatures; stiff vibrations need thousands of kelvin.

Real-world example

Diamond has an unusually low heat capacity at room temperature (about 6 J K⁻¹ mol⁻¹, far below 3R) because its stiff carbon–carbon bonds give very high vibrational frequencies. Einstein chose diamond to test his 1907 theory, and its success was early evidence for quantised energy in matter.

Why?

Why does a mode stop contributing to C V at low temperature? When kT is much smaller than the first excitation energy, almost every molecule sits in the ground level. Raising T slightly barely changes the population of excited levels, so almost no extra energy is absorbed.

Common misconception

"Heat capacity counts degrees of freedom." Only in the classical limit. The correct statement is that each mode contributes an amount between zero and its equipartition value, determined by the ratio of kT to its level spacing.

Worked example

Question: Estimate C V,m for Cl₂ at 298 K (θ V = 806 K, θ R < 1 K) and compare with the measured C p,m of 33.9 J K⁻¹ mol⁻¹.

Reasoning: Translation plus rotation give (5/2)R = 20.79 J K⁻¹ mol⁻¹. For vibration, x = 806/298 = 2.70, eˣ = 14.9, (eˣ − 1)² = 193, x²eˣ = 7.29 × 14.9 = 109. The Einstein function = 109/193 = 0.564, giving 0.564R = 4.69 J K⁻¹ mol⁻¹. Total C V,m ≈ 25.5 J K⁻¹ mol⁻¹, so C p,m = C V,m + R ≈ 33.8 J K⁻¹ mol⁻¹.

Answer: C V,m ≈ 25.5 J K⁻¹ mol⁻¹, giving C p,m ≈ 33.8 J K⁻¹ mol⁻¹, in excellent agreement with experiment.

Quick check

1. What is C V,m of a monatomic perfect gas, and why does it not depend on temperature? Answer: (3/2)R ≈ 12.5 J K⁻¹ mol⁻¹, because only translation contributes and its levels are so close that it is always fully active.

Exam focus

Be able to derive the Einstein function from q V, sketch C V against T for a diatomic gas, and explain the plateaus. Numerical questions typically combine (5/2)R with one or more vibrational terms; state which modes are active and why.

Advanced insight

A two-level system produces a very different heat-capacity curve: a peak (the Schottky anomaly) followed by a fall to zero at high temperature, because once both levels are equally populated no more energy can be absorbed. Such peaks appear in paramagnetic salts at low temperature and in NO from its spin–orbit levels, and they are used to measure small energy splittings.

Summary

C V follows from a second derivative of ln Q and, for a perfect gas, is a sum of mode contributions. Translation gives (3/2)R, rotation R or (3/2)R above θ R, and each vibration the Einstein function R x²eˣ/(eˣ − 1)², which tends to R only when T ≫ θ V. Modes freeze out when kT is small compared with their spacing, explaining the stepped heat capacity of H₂ and the low-temperature fall for solids.

Practice questions

1. Predict the high-temperature classical C V,m of CO₂, a linear triatomic. Answer: (3/2)R + R + 4R = (13/2)R ≈ 54.0 J K⁻¹ mol⁻¹. 2. What fraction of its classical value does a vibration contribute when T = θ V? Answer: With x = 1, e/(e − 1)² = 2.718/2.952 ≈ 0.92, so about 92 %. 3. Why is the room-temperature C V,m of N₂ almost exactly (5/2)R? Answer: Translation and rotation are fully active, but θ V ≈ 3400 K is so high that vibration contributes almost nothing at 298 K. 4. Explain why C V of a crystal falls towards zero as T → 0. Answer: All vibrational modes freeze out as kT becomes smaller than their level spacings, so almost no energy is absorbed per kelvin.