Sackur–Tetrode entropy and the Gibbs paradox

Derivation · the atom

Sackur–Tetrode entropy and the Gibbs paradox

D-017 Home PU-203 Threads chance · matter Depends on D-013 · D-019 verified

Statement

The absolute entropy of a classical ideal gas can be computed from first principles — but only if identical particles are counted as indistinguishable, and the correction factor is not classical in origin.

Why it matters

This is where classical statistical mechanics visibly fails and announces the need for quantum theory, decades before quantum theory existed. Planck's constant h appears in a purely classical calculation — a genuine surprise, and it should be presented as one.

Assumptions

Non-interacting point particles.Real gases deviate; the virial expansion corrects this.

Non-degenerate: λ_th³ ≪ V/N.The thermal wavelength is much smaller than the interparticle spacing. Violated at low temperature or high density, where quantum statistics take over — and the result then predicts a negative entropy, which is how you know the assumption has failed.

Particles are indistinguishable.Introduced by hand as the 1/N! factor. There is no classical justification for it; that is the entire point of the paradox.

Derivation

Z₁ = V / λ_th³ , λ_th = h / √(2πmk_BT)

Single-particle partition function from D-013, integrating e^−βp²/2m over phase space and dividing by h³ to make the phase-space cell count dimensionless. That h is the first quantum intrusion, and it is unavoidable — a classical calculation cannot fix the size of a phase-space cell. A

Z_N = Z₁^N / N!

The N-particle partition function for independent particles is the product, divided by N! because permutations of identical particles are the same state, not different ones. Step 6 shows what goes wrong without this. A

F = −k_BT ln Z_N = −k_BT ( N ln Z₁ − ln N! )

Free energy from the partition function. B

ln N! ≈ N ln N − N

Stirling's approximation from D-019. Legitimate because N ~ 10²³; the relative error is of order lnN/N, utterly negligible. B

S = − ( ∂F/∂T )_V,N

Standard thermodynamic route. Differentiate, remembering that λ_th ∝ T^−1/2, which supplies the 3/2 and hence the 5/2 in the final bracket. C

Result

S = Nk_B [ ln ( V / Nλ_th³ ) + 5/2 ]

Reading. Absolute entropy — not a difference — for a classical ideal gas, and it contains h. Classical physics has no way to produce h, so a purely classical entropy is undefined up to an additive constant. Nature is not.

Units check. V/Nλ³ is dimensionless, so S carries the units of Nk_B: J K⁻¹. ✓ Note also that S is extensive: doubling V and N together doubles S, because the ratio inside the logarithm is untouched.

The paradox, stated properly

Tier 3 · watch out

Omit the N! at step 2 and the result becomes S = Nk_B[ln(V/λ³) + 3/2], which is not extensive — doubling the system more than doubles the entropy. Concretely: remove a partition between two identical gases and this expression predicts an entropy increase of 2Nk_B ln 2, even though nothing has physically changed and the process is trivially reversible by replacing the partition.

The N! removes the spurious term exactly. But its justification is that permuting identical particles does not produce a new state — a quantum statement about indistinguishability with no classical counterpart. For different gases the mixing entropy is real and is correctly given by the same formula, which is the check that the resolution is not merely a fudge.

Limiting cases

Large V/N (dilute). The logarithm grows and the non-degeneracy condition λ_th³ ≪ V/N is safely met; the classical result is exact.
Identical vs. distinguishable gases. For identical gases the mixing term cancels via N!; for distinguishable gases the same formula returns the real, measurable mixing entropy.
High T. λ_th ∝ T^−1/2 shrinks, deepening the non-degenerate regime — the formula becomes more accurate, not less.

Breaks when

The gas becomes degenerate — the formula gives negative entropy, violating the third law, and must be replaced by the quantum treatment in PU-302.
Internal degrees of freedom are excited — rotational and vibrational contributions must be added, as for any molecular gas.
Interactions matter, as in a dense or nearly condensing gas.

Failure modes

Treating the N! as a mathematical convenience. It is a physical statement, and it is the whole lesson of the derivation.
Applying the formula outside the non-degenerate regime and reporting a negative entropy without noticing.
Believing the mixing entropy is always spurious. For distinguishable gases it is real, measurable, and given correctly.

Worked number

One mole of argon at 300 K and 1 atm: λ_th ≈ 1.6 × 10⁻¹¹ m and V/N ≈ 4.1 × 10⁻²⁶ m³, giving V/Nλ³ ≈ 1.0 × 10⁷ — comfortably non-degenerate, so the assumption holds. The formula returns S ≈ 155 J K⁻¹ mol⁻¹. The measured standard molar entropy of argon is 154.8 J K⁻¹ mol⁻¹. Agreement to three figures, from a calculation with no adjustable parameters, is the strongest possible evidence that the statistical interpretation of entropy is correct.

Run the check →

Discussion

The Sackur–Tetrode result is best read as a counting statement. The quantity V/(Nλ_th³) is the number of thermal phase-space cells available per particle: the box holds V/λ_th³ distinguishable-by-position cells of linear size λ_th, shared among N particles. Its logarithm is therefore the entropy per particle in units of k_B, and the whole formula says S = k_B ln(number of ways to arrange N particles among the cells). The additive 5/2 is not decorative: it splits cleanly as 3/2 + 1. The 3/2 is the translational kinetic contribution — three quadratic momentum degrees of freedom at ½k_BT each, surfacing through λ_th ∝ T^−1/2 when you differentiate. The +1 is the survivor of Stirling's −N term, i.e. it is the direct fingerprint of the 1/N! indistinguishability factor. Remove the N! and you lose exactly that +1 (5/2 → 3/2) and, more importantly, the N inside the logarithm.

That last point is the whole architecture of the result. Extensivity — the demand that S(λV, λN, T) = λS(V, N, T) — forces the logarithm's argument to depend on V and N only through the intensive ratio V/N (equivalently the density n = N/V). The N! supplies precisely the N in the denominator that converts the non-extensive ln V into the extensive ln(V/N). And because λ_th carries an h, the argument of the logarithm is a pure number with an absolute scale set by Planck's constant. This is why the formula gives an absolute entropy rather than an entropy up to a constant: h fixes the size of a phase-space cell, and hence fixes where the counting begins. A purely classical calculation cannot do this, which is the historical shock — a thermodynamic quantity measured with a calorimeter turns out to know about h.

The result is in fact the leading term of a quantum expansion, and reading it that way ties it to the rest of statistical mechanics. Differentiating the free energy with respect to particle number gives the chemical potential μ = k_BT ln(nλ_th³), so the single dimensionless combination nλ_th³ — the phase-space occupancy, or degeneracy parameter — governs everything. Non-degeneracy nλ_th³ ≪ 1 means μ is large and negative and the gas is dilute in phase space; the Sackur–Tetrode entropy is the first term of the fugacity (virial-in-nλ³) series that continues into the full Bose–Einstein and Fermi–Dirac results. As nλ_th³ → 1 the classical formula fails and, if pushed, returns negative entropy — the third law is the diagnostic that the expansion parameter has reached order unity and quantum statistics must take over (Bose condensation sets in near nλ_th³ ≈ 2.612). The 1/N! is the shadow this quantum indistinguishability casts on the classical limit: correct in leading order, unjustifiable within classical mechanics itself.

Common misconceptions.

The 1/N! is a bookkeeping trick to make the maths tidy. It is a physical statement about indistinguishability with no classical basis; it is the lesson, not the convenience.
Sackur–Tetrode gives an entropy change, like every other classical formula. It gives an absolute entropy, because h sets the zero — the reason it can be checked against a single tabulated number with no reference state.
The Gibbs mixing entropy is always fictitious. It is fictitious only for identical gases; for distinguishable species it is real, measurable, and given correctly by the same formula.
A negative entropy is just a sign the gas is very cold. It is a sign the formula is being used outside its domain (nλ_th³ ≪ 1 has failed) and must be replaced by the quantum treatment.

Worked examples

Example 1 — Absolute standard molar entropy of neon, from first principles. Compute the standard molar entropy of neon gas (m = 20.18 u) at T = 298.15 K and P = 1 bar = 10⁵ Pa, using no reference state and no adjustable constants, and compare with the calorimetric value 146.33 J K⁻¹ mol⁻¹.

λ_th = h√(2πmk_BT), VN = k_BTP

Symbols first. Per-particle volume from the ideal-gas law PV = Nk_BT, so only T and P are needed, not V and N separately.

m = 20.18 × 1.6605×10⁻²⁷ kg = 3.351×10⁻²⁶ kg

Convert the atomic mass to kg.

λ_th = 6.626×10⁻³⁴√(2π·3.351×10⁻²⁶·1.381×10⁻²³·298.15) = 2.251×10⁻¹¹ m

The thermal wavelength — about 0.23 Å, far below the interparticle spacing.

VN = 1.381×10⁻²³·298.1510⁵ = 4.116×10⁻²⁶ m³, VNλ_th³ = 4.116×10⁻²⁶1.140×10⁻³² = 3.61×10⁶

Occupancy per particle ≫ 1: comfortably non-degenerate, so the formula applies.

S_m = R [ ln(3.61×10⁶) + 5/2 ] = 8.314 × (15.10 + 2.50) = 8.314 × 17.60

Per mole, Nk_B → R. Substitute numbers only now.

S_m = 146.3 J K⁻¹ mol⁻¹

Answer. The Sackur–Tetrode entropy of neon is 146.3 J K⁻¹ mol⁻¹, against the measured 146.33 J K⁻¹ mol⁻¹. Agreement to four figures, with no reference state and no fitted parameter, is direct experimental confirmation of the statistical definition of absolute entropy.

Example 2 — The Gibbs paradox, made numerical. Two rigid boxes, each of volume V holding one mole of the same monatomic gas at the same T and P, are joined by removing the partition. (a) What entropy change does the correct (with-N!) formula give? (b) What would the naive (no-N!) formula give? (c) Repeat (a) for the case where the two boxes hold different gases (0.500 mol argon, 0.500 mol neon, same T, P).

S = Nk_B[ ln(V/Nλ³) + 5/2 ] ⇒ S(λV,λN) = λS(V,N)

Correct formula is extensive because the argument depends only on V/N. Initial state: two identical systems, total 2S. Final: one system of (2V, 2N) at the same T, whose V/N is unchanged.

ΔS = S(2V,2N) − 2S(V,N) = 2S(V,N) − 2S(V,N) = 0

(a) No change — correct: removing a partition between identical gases is reversible (replace it and nothing has happened).

S_wrong = Nk_B[ ln(V/λ³) + 3/2 ] ⇒ ΔS = 2Nk_B ln 2

(b) Without the N!, ln V is not divided by N; the final 2N particles each "see" volume 2V instead of V, leaving a spurious 2Nk_B ln 2.

ΔS_spurious = 2(1 mol)R ln 2 = 2 × 8.314 × 0.6931 = 11.53 J K⁻¹

The magnitude of the paradox for one mole per box: a large, entirely fictitious entropy that no experiment measures.

ΔS_mix = n_ArR ln 2 + n_NeR ln 2 = (0.500+0.500)R ln 2 = 8.314 × 0.6931

(c) Different gases: each species really does expand from V to 2V at fixed T, and the same formula gives its nR ln 2. This mixing entropy is real.

ΔS_identical = 0 · ΔS_mix = 5.76 J K⁻¹

Answer. For identical gases the correct formula gives ΔS = 0; the naive formula gives a spurious 11.53 J K⁻¹. For argon–neon mixing it gives a real ΔS_mix = 5.76 J K⁻¹. That the same formula gives zero for identical gases and the measured value for different gases is the proof that the 1/N! is physics, not a fudge.

Problems

(Warm-up.) Compute the thermal de Broglie wavelength of helium (m = 4.003 u) at 300 K, and confirm the gas is non-degenerate at 1 atm by estimating V/Nλ³.
Solution

m = 4.003 × 1.6605×10⁻²⁷ = 6.647×10⁻²⁷ kg. Then 2πmk_BT = 6.283 × 6.647×10⁻²⁷ × 1.381×10⁻²³ × 300 = 1.730×10⁻⁴⁶; its square root is 1.315×10⁻²³. So λ_th = 6.626×10⁻³⁴ / 1.315×10⁻²³ = 5.04×10⁻¹¹ m (0.50 Å). With V/N = k_BT/P = 1.381×10⁻²³×300/1.013×10⁵ = 4.09×10⁻²⁶ m³ and λ³ = 1.28×10⁻³¹ m³, the occupancy is V/Nλ³ ≈ 3.2×10⁵ ≫ 1 — deeply non-degenerate, so Sackur–Tetrode applies. (Note helium, being light, has the largest λ_th of the noble gases and is the first to go quantum on cooling.)
(Direct application.) Compute the standard molar entropy of krypton (m = 83.80 u) at 298.15 K and 1 bar, and compare with the tabulated 164.08 J K⁻¹ mol⁻¹.
Solution

m = 83.80 × 1.6605×10⁻²⁷ = 1.392×10⁻²⁵ kg. 2πmk_BT = 6.283 × 1.392×10⁻²⁵ × 1.381×10⁻²³ × 298.15 = 3.60×10⁻⁴⁵; root = 5.999×10⁻²³, so λ_th = 1.105×10⁻¹¹ m and λ³ = 1.347×10⁻³³ m³. With V/N = 4.116×10⁻²⁶ m³ (298.15 K, 10⁵ Pa), the occupancy is 4.116×10⁻²⁶/1.347×10⁻³³ = 3.06×10⁷. Hence S_m = R[ln(3.06×10⁷) + 5/2] = 8.314×(17.24 + 2.50) = 164.1 J K⁻¹ mol⁻¹, matching 164.08 to four figures. Krypton's larger mass (bigger denominator in λ) is exactly why its entropy exceeds neon's — heavier atoms pack more thermally-accessible states into the same box.
(Entropy change; watch what cancels.) Three moles of neon are taken from (250 K, 20.0 L) to (500 K, 10.0 L). Find ΔS, and explain why the value does not depend on h even though the absolute entropy does.
Solution

Because λ_th ∝ T^−1/2, ln(V/Nλ³) changes by ln(V₂/V₁) + (3/2)ln(T₂/T₁), so ΔS = nR[ln(V₂/V₁) + (3/2)ln(T₂/T₁)]. Substituting: ΔS = 3×8.314×[ln(10.0/20.0) + 1.5×ln(500/250)] = 3×8.314×[−0.6931 + 1.5×0.6931] = 3×8.314×0.3466 = +8.65 J K⁻¹. The heating (which alone would give +1.5nR ln 2) outweighs the compression (−nR ln 2). The additive constants inside the log — including every factor of h — are identical in the two states and subtract out; only ratios of V and T survive. This is precisely why classical thermodynamics, which only ever measures entropy differences, never needed h.
(Where the formula breaks.) Liquid helium has number density n ≈ 2.18×10²⁸ m⁻³. Treating it (wrongly) as a classical ideal gas, at what temperature would Sackur–Tetrode predict S = 0? Comment on the physical significance.
Solution

Setting the bracket to zero: ln(1/nλ³) + 5/2 = 0, i.e. nλ³ = e^5/2 = 12.18, so λ³ = 12.18/n = 12.18/2.18×10²⁸ = 5.59×10⁻²⁸ m³ and λ = 8.23×10⁻¹⁰ m. Inverting λ = h/√(2πmk_BT) gives T = h²/(2πmk_Bλ²) = (6.626×10⁻³⁴)² / (2π × 6.647×10⁻²⁷ × 1.381×10⁻²³ × (8.23×10⁻¹⁰)²) = 4.39×10⁻⁶⁷/3.91×10⁻⁶⁷ = ≈ 1.1 K. The prediction of zero (and, just below, negative) entropy occurs at nλ³ of order 10, i.e. deep in the degenerate regime where the classical formula is already invalid. Reassuringly, this ~1 K scale is right where real helium becomes a quantum fluid (the superfluid transition is 2.17 K): the third-law violation is the classical theory flagging its own breakdown at exactly the temperature quantum statistics take over.
(Chemical potential — connect to quantum statistics.) From the free energy F = −Nk_BT[ln(V/Nλ³) + 1], show that the chemical potential is μ = k_BT ln(nλ³). Evaluate it for argon at 300 K and 1 atm (where nλ³ ≈ 10⁻⁷), and explain what the sign tells you about the onset of quantum behaviour.
Solution

Write F = −Nk_BT[ln V − ln N − 3 ln λ + 1]. Then μ = (∂F/∂N)_T,V = −k_BT[ln(V/Nλ³) + 1] − Nk_BT·(−1/N). The two "1" terms cancel: μ = −k_BT ln(V/Nλ³) = k_BT ln(nλ³), with n = N/V. For argon: μ = 1.381×10⁻²³×300×ln(10⁻⁷) = 4.14×10⁻²¹×(−16.1) = −6.68×10⁻²⁰ J = −0.42 eV per atom (−40 kJ mol⁻¹). The large negative value is the signature of non-degeneracy: nλ³ ≪ 1 means each phase-space cell is almost never occupied, so quantum statistics (which care only about occupancy of the same state) are irrelevant and Bose/Fermi behaviour reduces to Boltzmann. As the gas is cooled or compressed, nλ³ → 1 and μ → 0⁻; a Bose gas condenses when μ reaches 0 (at nλ³ ≈ 2.612), while a Fermi gas crosses into μ > 0 (a positive Fermi energy). The chemical potential's approach to zero is thus the exact quantitative measure of how far the gas is from its quantum regime.