Boltzmann's Entropy Formula

Introduction

Classical thermodynamics defines entropy through heat: dS = dq rev/T. This definition works perfectly but gives little insight into what entropy is . Ludwig Boltzmann provided the answer in a formula so important that it is carved on his tombstone in Vienna: S = k ln W . Entropy is a measure of the number of microscopic arrangements that are consistent with the macroscopic state of a system.

Core explanation

The formula. For an isolated system,

S = k ln W

where W is the number of microstates accessible to the system (for a macroscopic system, effectively the weight of the dominant configuration) and k = 1.381 × 10⁻²³ J K⁻¹. The Boltzmann constant gives entropy its units of J K⁻¹ and ensures agreement with the thermodynamic definition.

Why a logarithm? Consider two independent subsystems with W₁ and W₂ microstates. Each microstate of one can be combined with each microstate of the other, so the combined system has W = W₁W₂. Entropy, however, is extensive: the total should be S₁ + S₂. Only a logarithmic function turns products into sums: k ln(W₁W₂) = k ln W₁ + k ln W₂.

The second law. In an isolated system, a spontaneous change moves the system to macrostates with more microstates. Because W for the final state is overwhelmingly larger, ln W increases, and entropy increases. The second law is thus a statement of overwhelming probability rather than an absolute prohibition — but for 10²³ molecules the probabilities are so extreme that violations are never observed.

The third law. A perfect crystal at T = 0 has every molecule in its ground state, arranged in a single way. Then W = 1 and S = k ln 1 = 0, providing a molecular basis for the third law.

Residual entropy. Some crystals retain disorder at T = 0. In solid carbon monoxide, each molecule can point either way (CO or OC) with almost no energy difference, because the molecule has a very small dipole. If each of N molecules has two choices, W = 2^N and S = Nk ln 2. Per mole this is R ln 2 = 5.76 J K⁻¹ mol⁻¹; the experimental residual entropy is about 4.6 J K⁻¹ mol⁻¹, suggesting the disorder is not completely random. Ice has a residual entropy of about 3.4 J K⁻¹ mol⁻¹, close to Pauling's estimate R ln(3/2) = 3.37 J K⁻¹ mol⁻¹ based on the ways hydrogen atoms can be arranged among oxygen atoms.

Connecting with the Boltzmann distribution. Using Stirling's approximation for the dominant configuration gives ln W = −N Σ pᵢ ln pᵢ, where pᵢ = nᵢ/N. Hence S = −Nk Σ pᵢ ln pᵢ. When all accessible states are equally likely this reduces to Boltzmann's form; when the probabilities are unequal, it counts the effective number of states.

Formulae

S = k ln W; ΔS = k ln(W final/W initial); for isothermal ideal-gas expansion ΔS = nR ln(V₂/V₁); Gibbs form S = −k Σ pᵢ ln pᵢ (per system, pᵢ = probability of microstate i).

Step-by-step reasoning

To calculate an entropy change statistically:

1. Decide how the number of accessible microstates changes (for example, how many choices each molecule gains). 2. Write W final/W initial as a power of that factor, such as a^N. 3. Take the logarithm: ΔS = k ln(a^N) = Nk ln a. 4. Convert to molar quantities using Nk = nR.

Visual explanation

Draw a box divided into cells. With the gas confined to the left half, each molecule has a certain number of cells available; remove the partition and every molecule has twice as many. For N molecules, the number of arrangements multiplies by 2^N, and the entropy rises by Nk ln 2.

Real-world analogy

A combination lock with three dials of ten digits has 10³ settings; adding a fourth dial gives 10⁴. The "information" needed to specify the setting grows by the same amount with each dial — additive — even though the number of settings multiplies. Entropy behaves in exactly this logarithmic way.

Real-world example

When low-temperature heat-capacity measurements were used to find the entropy of CO, the value was lower than the entropy calculated from spectroscopy by about 4.6 J K⁻¹ mol⁻¹. The discrepancy was explained by S = k ln W as frozen-in orientational disorder, a striking confirmation of Boltzmann's interpretation.

Why?

Why does a gas expanding into a vacuum have increased entropy even though no heat flows? Entropy is a state function measuring accessible microstates. The expanded gas has more positions available to each molecule, so W and hence S are larger. The classical calculation via a reversible path gives the same answer, nR ln(V₂/V₁).

Common misconception

"Entropy is simply disorder." Disorder is a loose picture that can mislead. Entropy counts accessible microstates, including ways of distributing energy, not just visual messiness. For example, a crystallising solution can increase total entropy because the heat released spreads energy into the surroundings.

Worked example

Question: One mole of an ideal gas expands isothermally into a vacuum, doubling its volume. Use S = k ln W to find ΔS.

Reasoning: Each molecule has twice as many positions available, so W final/W initial = 2^N. Then ΔS = k ln 2^N = Nk ln 2. For one mole, Nk = R, so ΔS = 8.314 × ln 2 = 8.314 × 0.693 = 5.76 J K⁻¹.

Answer: ΔS = +5.76 J K⁻¹, identical to the classical result nR ln(V₂/V₁).

Quick check

1. Two independent systems have W₁ = 10²⁰ and W₂ = 10³⁰ microstates. What is the total entropy expressed in units of k? Answer: S/k = ln(10⁵⁰) = 50 ln 10 ≈ 115, the sum of the separate entropies.

Exam focus

Know S = k ln W, why the logarithm is needed for additivity, and how to derive ΔS = nR ln(V₂/V₁) from counting positions. Residual-entropy questions on CO or ice appear frequently: be prepared to calculate R ln 2 and explain why the measured value may be smaller.

Advanced insight

The Gibbs form S = −k Σ pᵢ ln pᵢ is formally identical to Shannon's measure of information in communication theory. This connection underlies Landauer's principle: erasing one bit of information in a device at temperature T must dissipate at least kT ln 2 of energy as heat, linking thermodynamics to the physical limits of computation.

Summary

Boltzmann's formula S = k ln W identifies entropy with the logarithm of the number of accessible microstates. The logarithm makes entropy additive for independent systems. The second law reflects the overwhelming probability of macrostates with more microstates, and the third law follows from W = 1 for a perfect crystal at absolute zero. Residual entropies of CO and ice confirm the statistical picture.

Practice questions

1. Calculate the molar residual entropy of a crystal in which each molecule can adopt any of three orientations of equal energy. Answer: S = R ln 3 = 8.314 × 1.099 = 9.13 J K⁻¹ mol⁻¹. 2. Calculate ΔS when 2.00 mol of ideal gas expands isothermally from 5.0 L to 20.0 L. Answer: ΔS = nR ln(V₂/V₁) = 2.00 × 8.314 × ln 4 = 23.1 J K⁻¹. 3. Explain why measured residual entropies are often smaller than the ideal value R ln a. Answer: Slight energy differences or correlations between neighbouring molecules make some arrangements less likely, so the disorder frozen in is only partial. 4. Why does S = k ln W imply S = 0 for a perfect crystal at T = 0? Answer: All molecules are in the ground state in a single arrangement, so W = 1 and ln 1 = 0.