Internal Energy from the Partition Function

Introduction

With the canonical partition function in hand, the first thermodynamic property to extract is the internal energy. The result is compact: U is a derivative of ln Q with respect to temperature. From it, the familiar values (3/2)RT for a monatomic gas and (5/2)RT for a diatomic gas drop out naturally, along with the less familiar and very important fact that vibrations contribute much less than RT at room temperature. This page connects the counting of states to the energy measured by calorimetry.

Core explanation

Deriving the formula. In the canonical ensemble the mean energy is a probability-weighted average over system states:

⟨E⟩ = Σ i E i P i = (1/Q)Σ i E ie^(−βE i)

Notice that differentiating Q with respect to β brings down −E i from each exponential: (∂Q/∂β) V = −Σ i E ie^(−βE i). Therefore

⟨E⟩ = −(1/Q)(∂Q/∂β) V = −(∂ln Q/∂β) V

Volume is held constant because the energy levels depend on V. Energies in Q are measured from the lowest system state, so the thermodynamic internal energy is

U = U(0) − (∂ln Q/∂β) V

Using dβ = −dT/kT², this becomes the equivalent form U − U(0) = kT²(∂ln Q/∂T) V.

U(0). U(0) is the energy the system retains at absolute zero. For molecules it includes the zero-point vibrational energy and the electronic ground-state energy. It never affects heat capacities or the temperature variation of U, but it matters when comparing different substances, as in reaction energies.

Independent molecules. For a gas, ln Q = N ln q − ln N!. The N! term is independent of β, so

U − U(0) = −N(∂ln q/∂β) V

and because ln q is a sum over modes, U − U(0) is a sum of mode contributions.

Mode contributions (per mole).

- Translation: q T = V/Λ³ ∝ β^(−3/2), so −∂ln q T/∂β = 3/(2β), giving U T = (3/2)RT. - Rotation, linear molecule, T ≫ θ R: q R = 1/(σβhcB̃) ∝ β^(−1), giving U R = RT. A non-linear molecule, with q R ∝ β^(−3/2), gives (3/2)RT. - Vibration (energy from v = 0): q V = 1/(1 − e^(−βhcν̃)). Differentiation gives U V = N Ahcν̃/(e^(βhcν̃) − 1) = Rθ V/(e^(θ V/T) − 1) At T ≫ θ V this approaches RT; at T ≪ θ V it falls to nearly zero. - Electronic: a constant q E contributes nothing.

So a diatomic gas at room temperature has U − U(0) ≈ (5/2)RT, with vibration contributing little unless the bond is soft.

Formulae

U − U(0) = −(∂ln Q/∂β) V = kT²(∂ln Q/∂T) V. For a perfect gas: U − U(0) = −N(∂ln q/∂β) V. Molar mode contributions: U T = (3/2)RT; U R = RT (linear) or (3/2)RT (non-linear); U V = Rθ V/(e^(θ V/T) − 1).

Step-by-step reasoning

To find the contribution of any mode:

1. Write q for the mode as a function of β. 2. Take ln q. 3. Differentiate with respect to β at constant V. 4. Multiply by −N (or −N A per mole). 5. Check the high-temperature limit against equipartition.

Visual explanation

Plot U V/RT against T/θ V. The curve starts flat at zero, rises steeply through T ≈ θ V/3, and approaches 1 at high temperature. On the same axes, translation and rotation are horizontal lines at 1.5 and 1, because they are fully active at all ordinary temperatures.

Real-world analogy

A savings account pays interest only once the balance exceeds a threshold. Translational and rotational "accounts" have tiny thresholds and collect energy steadily as temperature rises; vibrational accounts have thresholds far above kT and stay nearly empty until the temperature approaches θ V.

Real-world example

Gas turbines and internal-combustion engines run hot enough to activate vibrations in N₂, O₂, CO₂ and H₂O. The extra energy stored in vibrations means more heat is needed per kelvin of temperature rise, reducing peak flame temperatures. Engineering models of combustion use exactly these partition function expressions.

Why?

Why does the derivative of ln Q give the mean energy? Q is a sum of Boltzmann factors; changing β reweights each state by an amount proportional to its energy. The fractional rate of change of Q with β is therefore the probability-weighted energy, which is precisely the average.

Common misconception

"Every molecule has (1/2)kT per degree of freedom, so a diatomic gas always has U = (7/2)RT." Equipartition holds only for modes whose level spacing is much less than kT. Vibrations of most small molecules are frozen at room temperature, and the partition function derivation shows exactly how much they contribute.

Worked example

Question: Calculate the molar vibrational contribution to U − U(0) for I₂ at 298 K (θ V = 309 K), and compare it with RT.

Reasoning: θ V/T = 309/298 = 1.037. e^1.037 = 2.82, so e^(θ V/T) − 1 = 1.82. U V = Rθ V/1.82 = 8.314 × 309/1.82 = 2569/1.82 ≈ 1410 J mol⁻¹. RT = 2478 J mol⁻¹.

Answer: U V ≈ 1.4 kJ mol⁻¹, about 57 % of the classical value RT, because I₂ vibrations are only partly active.

Quick check

1. What is U − U(0) for one mole of argon at 300 K, and which mode is responsible? Answer: (3/2)RT = 1.5 × 8.314 × 300 ≈ 3.74 kJ mol⁻¹, entirely from translation.

Exam focus

Examiners often ask for the derivation of ⟨E⟩ = −∂ln Q/∂β and the vibrational energy expression. Show the high- and low-temperature limits explicitly. Remember that U(0) cannot be obtained from Q, and that the N! plays no part in U.

Advanced insight

For interacting systems, Q does not factorise, but the same formula U − U(0) = −∂ln Q/∂β holds. In a classical liquid the kinetic part still gives (3/2)NkT, while intermolecular forces add a configurational energy that depends on the pair distribution function. This connects the energy of a liquid to its measured structure from X-ray or neutron scattering.

Summary

The internal energy follows from U − U(0) = −(∂ln Q/∂β) V = kT²(∂ln Q/∂T) V. For a perfect gas this reduces to −N(∂ln q/∂β) V, a sum of mode contributions. Translation gives (3/2)RT, linear rotation RT and non-linear rotation (3/2)RT, while vibration gives Rθ V/(e^(θ V/T) − 1), approaching RT only when T ≫ θ V. U(0) includes zero-point and ground-state energies.

Practice questions

1. Show that q ∝ β^(−3/2) leads to U − U(0) = (3/2)NkT. Answer: ln q = const − (3/2)ln β, so −∂ln q/∂β = 3/(2β) = (3/2)kT; multiplying by N gives (3/2)NkT. 2. Estimate U − U(0) for one mole of N₂ at 298 K. Answer: (5/2)RT = 2.5 × 8.314 × 298 ≈ 6.19 kJ mol⁻¹; vibration adds a negligible amount. 3. What is the high-temperature limit of U V, and why? Answer: RT, because e^(θ V/T) − 1 ≈ θ V/T when T ≫ θ V, so U V ≈ Rθ V/(θ V/T) = RT. 4. Why is the derivative taken at constant volume? Answer: The energy levels, especially translational ones, depend on V, so holding V fixed ensures only the populations change, not the levels themselves.