Statistical Entropy and the Sackur–Tetrode Equation

Introduction

Boltzmann's formula S = k ln W links entropy to the number of microstates of an isolated system. In practice we work at constant temperature, not constant energy, and we need an entropy formula expressed in terms of the canonical partition function. The result lets us calculate absolute entropies of gases from a handful of molecular constants — mass, bond length, vibrational frequency — with accuracy that rivals or beats careful low-temperature calorimetry. The Sackur–Tetrode equation for monatomic gases is the cleanest and most famous example.

Core explanation

Entropy from probabilities. For any ensemble, the Gibbs form of the entropy is S = −k Σ i P i ln P i. In the canonical ensemble P i = e^(−βE i)/Q, so ln P i = −βE i − ln Q. Substituting:

S = kβ Σ i P iE i + k ln Q Σ i P i = (U − U(0))/T + k ln Q

The first term reflects how energy is spread over levels; the second counts the accessible states.

Perfect gas. With Q = q^N/N! and Stirling's approximation, ln Q = N ln q − N ln N + N, giving

S = (U − U(0))/T + Nk ln(q/N) + Nk

Because ln q is a sum over modes, S is too: S = S T + S R + S V + S E. The N-dependent terms belong to translation, since only q T is proportional to V.

The Sackur–Tetrode equation. For a monatomic gas with q E = 1, only translation matters: q = V/Λ³ and U − U(0) = (3/2)NkT. Then

S = (3/2)Nk + Nk ln(V/(NΛ³)) + Nk = Nk ln(e^(5/2)V/(NΛ³))

For one mole of perfect gas, V/N = kT/p, so

S m = R ln(e^(5/2) kT/(pΛ³))

What it predicts. Since Λ ∝ (mT)^(−½), the equation shows that S m increases by (3/2)R ln M with molar mass (heavier atoms have more closely spaced translational levels), by (5/2)R ln T with temperature, and decreases by R ln p with pressure. The isothermal expansion result ΔS = nR ln(V f/V i) follows directly from the V dependence.

Comparison with experiment. Calorimetric entropies come from ∫C p/T dT from near 0 K, plus ΔH/T for each phase change, relying on the third law (S → 0 for a perfect crystal). For argon the two methods agree within experimental error. For CO, N₂O and H₂O ice, the statistical value is larger: their crystals trap orientational disorder, so S does not reach zero at 0 K. For CO, a random head-to-tail arrangement would give R ln 2 = 5.8 J K⁻¹ mol⁻¹; the observed discrepancy is about 4.6 J K⁻¹ mol⁻¹, showing that the disorder is substantial but not quite fully random.

Formulae

S = (U − U(0))/T + k ln Q. Perfect gas: S = (U − U(0))/T + Nk[ln(q/N) + 1]. Sackur–Tetrode: S m = R ln(e^(5/2) kT/(pΛ³)), with Λ = h/(2πmkT)^½. Residual entropy for two random orientations: R ln 2.

Step-by-step reasoning

To calculate S m for a monatomic gas:

1. Convert molar mass to the mass of one atom, m = M/N A. 2. Calculate Λ = h/(2πmkT)^½. 3. Calculate kT/p, the volume per atom. 4. Form kT/(pΛ³), take its natural logarithm, and add 2.5. 5. Multiply by R.

Visual explanation

Imagine a box divided into tiny cells of side Λ. The number of cells per atom, kT/(pΛ³), is typically around ten million at room conditions. The entropy is essentially R times the logarithm of this number of "places" available to each atom, plus a constant 5/2 from the energy spread.

Real-world analogy

A car park with many spaces per car offers enormous freedom in where each car stops; halve the car park and the choices shrink. Increasing pressure is like shrinking the car park, and the entropy falls logarithmically with the number of choices.

Real-world example

Thermodynamic databases for high-temperature chemistry — metallurgy, rocketry, atmospheric re-entry — take gas entropies almost entirely from statistical calculations. Measuring the entropy of monatomic sodium or iron vapour calorimetrically would be impractical; the Sackur–Tetrode equation plus the electronic partition function gives it reliably.

Why?

Why does a heavier monatomic gas have higher entropy at the same T and p? The thermal wavelength Λ is smaller for heavier atoms, so translational energy levels are more closely spaced and more states are accessible within kT. More accessible states means greater entropy.

Common misconception

"The third law means every substance has zero entropy at 0 K." It applies to perfectly ordered crystals. Substances that freeze in disorder, such as CO or ice, keep residual entropy, and comparing statistical with calorimetric entropies is precisely how that residual entropy is detected.

Worked example

Question: Calculate the standard molar entropy of argon at 298 K and 1 bar. (m = 6.634 × 10⁻²⁶ kg.)

Reasoning: kT = 1.381 × 10⁻²³ × 298 = 4.115 × 10⁻²¹ J. Λ = 6.626 × 10⁻³⁴/(2π × 6.634 × 10⁻²⁶ × 4.115 × 10⁻²¹)^½ = 6.626 × 10⁻³⁴/4.142 × 10⁻²³ = 1.600 × 10⁻¹¹ m, so Λ³ = 4.10 × 10⁻³³ m³. kT/p = 4.115 × 10⁻²¹/10⁵ = 4.115 × 10⁻²⁶ m³. Ratio = 1.00 × 10⁷, ln = 16.12. S m = 8.314 × (16.12 + 2.50) = 8.314 × 18.62.

Answer: S m ≈ 155 J K⁻¹ mol⁻¹, matching the tabulated value of 154.8 J K⁻¹ mol⁻¹.

Quick check

1. By how much does the molar entropy of a perfect gas change when its pressure is increased tenfold at constant temperature? Answer: It decreases by R ln 10 ≈ 19.1 J K⁻¹ mol⁻¹, because S m contains the term −R ln p.

Exam focus

Derive S = (U − U(0))/T + k ln Q, then the Sackur–Tetrode equation, showing where the e^(5/2) comes from: (3/2) from energy and 1 from Stirling's approximation. Numerical calculations of S m for a noble gas are common; keep Λ in metres and pressure in pascals.

Advanced insight

The Sackur–Tetrode equation predicts S → −∞ as T → 0, violating the third law. This is not a flaw in statistical thermodynamics but a sign that the classical approximation V/(NΛ³) ≫ 1 has broken down. At very low temperature quantum statistics apply, and the entropy of an ideal Bose or Fermi gas correctly goes to zero.

Summary

In the canonical ensemble S = (U − U(0))/T + k ln Q. For a perfect gas S = (U − U(0))/T + Nk[ln(q/N) + 1], which separates into mode contributions. For a monatomic gas this gives the Sackur–Tetrode equation, S m = R ln(e^(5/2) kT/(pΛ³)), which reproduces measured entropies such as argon's 155 J K⁻¹ mol⁻¹. Differences between statistical and calorimetric entropies reveal residual entropy from frozen-in disorder.

Practice questions

1. How does S m of a monatomic perfect gas change when T doubles at constant pressure? Answer: S m contains (5/2)R ln T, so it increases by (5/2)R ln 2 ≈ 14.4 J K⁻¹ mol⁻¹. 2. Estimate the difference in standard molar entropy between krypton (M = 83.8 g mol⁻¹) and argon (M = 39.9 g mol⁻¹) at the same T and p. Answer: (3/2)R ln(83.8/39.9) = 12.47 × 0.742 ≈ 9.3 J K⁻¹ mol⁻¹, krypton higher. 3. Where does the "+1" inside Nk[ln(q/N) + 1] come from? Answer: From Stirling's approximation of ln N! = N ln N − N applied to the 1/N! for indistinguishable particles. 4. Why might the statistical entropy of solid-state CO exceed its calorimetric entropy? Answer: CO molecules freeze in with random CO/OC orientations, so the crystal retains residual entropy at 0 K that calorimetry, which assumes S(0) = 0, does not capture.