Gibbs Paradox

Take a box split by a partition into two equal halves of volume $V$ , each holding $N$ atoms of the same ideal gas at the same temperature and pressure. Slide the partition out. Nothing observable happens — the gas was uniform before and is uniform after, so a well-behaved entropy must not change. Yet the most natural classical calculation, counting single-particle phase-space volume without any correction for the interchangeability of atoms, insists the entropy jumps by $2Nk_B\ln2$ . That contradiction is the Gibbs paradox, and it lands squarely on the ideal gas and its absolute entropy, the Sackur-Tetrode equation. This article is the ideal-gas application of the paradox. The general concept — that permuting identical particles does not create a new microstate — is owned upstream by the foundations article on the Gibbs paradox, and the machinery of the $1/N!$ factor is owned by Correct Boltzmann Counting. Here the paradox becomes concrete and quantitative: we watch the spurious mixing entropy appear in the Sackur-Tetrode entropy, we watch the $1/N!$ turn the non-extensive $V^N$ into the extensive $(V/N)^N$ , and we watch the spurious $2Nk_B\ln2$ cancel exactly for identical gases while surviving intact for genuinely different ones. The sharp feature — the one this article proves and dwells on — is a discontinuity. The entropy of mixing of two different gases is $2Nk_B\ln2$ no matter how similar they are; the instant the two species become strictly identical it drops discontinuously to zero. There is no smooth interpolation. That step is the conceptual heart of the paradox and the reason it foreshadowed quantum statistics: indistinguishability is binary, not a matter of degree, and only a quantum origin for the $1/N!$ makes it exact.

Key takeaways

In the Sackur-Tetrode entropy of an ideal gas, naive classical counting (no $1/N!$ ) predicts a spurious entropy of mixing $2Nk_B\ln2$ even when two joined gases are identical — the Gibbs paradox.
Dividing the partition function by $N!$ (Correct Boltzmann Counting) turns the non-extensive $V^N$ into the extensive $(V/N)^N$ , and the spurious term cancels exactly: $\Delta S=0$ for identical gases.
For two genuinely distinct gases the entropy of mixing $\Delta S=2Nk_B\ln2=-N_{\text{tot}}k_B\sum_i x_i\ln x_i$ is real and survives.
The distinct-gas value equals the naive-identical value — the "paradox" number is the real mixing entropy mislabelled onto identical gases.
The mixing entropy is independent of how similar the species are, then drops discontinuously to zero when they become strictly identical — a genuine step, not a smooth limit.
The foundations article owns the general concept; Correct Boltzmann Counting owns the $1/N!$ ; this article owns the Sackur-Tetrode application and the discontinuity.
Key idea: in the ideal-gas entropy, the $1/N!$ replaces $\ln V$ with $\ln(V/N)$ , which makes the entropy extensive and cancels the spurious mixing entropy of identical gases exactly — while leaving the real mixing entropy of distinct gases untouched.

The core insight

In the Sackur-Tetrode entropy, the $1/N!$ factor replaces $\ln V$ with $\ln(V/N)$ : this single change makes the entropy an extensive (first-order homogeneous) function and cancels the spurious $2Nk_B\ln2$ mixing entropy of identical gases exactly, while the genuine mixing entropy of distinct gases survives intact.

Where this lives today

The Gibbs paradox is still load-bearing wherever entropy is computed for identical particles. In chemical thermodynamics, the mixing entropy $-R\sum x_i\ln x_i$ is the free-energy driver behind Dalton's law of partial pressures, ideal-solution behaviour, and the minimum work of separation — the thermodynamic floor on the cost of isotope enrichment, gas separation, and desalination. In statistical simulation, the $1/N!$ is what makes computed free energies of identical-particle systems extensive and comparable across system sizes; forgetting it is a classic bug that produces size-dependent chemical potentials. The paradox also remains a live conceptual arena: modern discussions of information, Landauer's principle, and the thermodynamics of small systems return to it to ask precisely when two subsystems count as "the same," and whether a Maxwell demon that could distinguish nominally identical particles would recover the mixing entropy. And its historical role endures in every quantum-statistics course, where the $1/N!$ is shown to be the dilute limit of Bose-Einstein and Fermi-Dirac counting — the classical shadow of wavefunction symmetry.

The physical picture

Picture each atom wearing a tiny serial number. When you remove the partition between two identical gases, atom #7 drifts from the left half into the right. Naive counting rewards you with entropy for that migration — you "learned" that #7 crossed over. But identical atoms carry no serial numbers, even in principle: the configuration with #7 on the right and the configuration with #7 on the left are the same microstate. The $1/N!$ confiscates exactly the fictitious entropy that labelling would grant. There is no macrostate change, so there is no entropy change. Extensivity is the litmus test that exposes the error. Double the whole system at fixed density and the entropy should double. The naive ideal-gas entropy carries a term $Nk_B\ln V$ that scales faster than linearly once $N$ and $V$ grow together, breaking $S(\lambda N,\lambda V,\lambda E)=\lambda S$ . The $1/N!$ replaces $\ln V$ with $\ln(V/N)$ , a ratio, and homogeneity is restored. The paradox and the non-extensivity are two faces of the same missing factor.

Figure

Two equal boxes each with N atoms of the SAME gas at the same T,P. Remove the partition. With correct 1/N! counting the density V/N is unchanged, so the entropy change is exactly zero — an unobservable process makes no entropy.

Now imagine the same picture but with argon on the left and neon on the right. When the partition goes, each species independently spreads into the full doubled volume, and the real entropy $-N_{\text{tot}}k_B\sum_i x_i\ln x_i$ is liberated — the atoms genuinely have more room as a species. The punchline figure below plots the entropy of mixing against "how different the two gases are." It is a flat line at $2Nk_B\ln2$ for any distinction whatsoever, however faint, that plunges to a single point at zero the instant the gases become strictly the same. The discontinuity is the whole story.

Figure

Punchline: entropy of mixing versus how distinguishable the two gases are. For ANY distinction it is 2Nk ln2 (flat), then it drops DISCONTINUOUSLY to 0 at the identical point. Indistinguishability is binary, not gradual.

The core equations

The starting point is the uncorrected ideal-gas entropy — the one that produces the paradox. It follows from counting single-particle phase space with no permutation correction, and its flaw is that it is not extensive.

S_{\text{naive}} = Nk_B\left[\ln\!\left(\frac{V}{\lambda^3}\right)+\frac32\right] \qquad (1)

where: $N$ — number of atoms; $V$ — volume; $\lambda=h/\sqrt{2\pi mk_BT}$ — the thermal de Broglie wavelength (set by mass and temperature); $k_B$ — Boltzmann's constant. Reading it: the offending piece is $Nk_B\ln V$ . Under $N\to\lambda N,\ V\to\lambda V$ this does not scale linearly, because $\ln V$ is not a ratio — that is precisely the non-extensivity that breeds the paradox. Dividing the partition function by $N!$ and using Stirling's approximation turns equation (1) into the extensive Sackur-Tetrode form. The $1/N!$ is owned and derived in Correct Boltzmann Counting; here we use its consequence.

S = Nk_B\left[\ln\!\left(\frac{V}{N\lambda^3}\right)+\frac52\right] \qquad (2)

where: $V/N$ — the volume per atom, a density-dependent ratio; the $\tfrac52=\tfrac32+1$ , the extra $+1$ coming from the $-k_B\ln N!\approx-Nk_B(\ln N-1)$ correction. Reading it: replacing $\ln V$ with $\ln(V/N)$ is the single fix. Everything the paradox needs — extensivity, and the cancellation of the spurious mixing entropy — flows from that one swap. For two genuinely distinct gases the entropy of mixing is real. With correct counting each species independently fills the doubled volume, and the result depends only on mole fractions.

\Delta S_{\text{mix}} = -k_B\sum_i N_i\ln\frac{N_i}{N_{\text{tot}}} = -N_{\text{tot}}\,k_B\sum_i x_i\ln x_i \qquad (3)

where: $N_i$ — atoms of species $i$ ; $N_{\text{tot}}=\sum_iN_i$ ; $x_i=N_i/N_{\text{tot}}$ — mole fraction. Reading it: for two equal parts ( $x_1=x_2=\tfrac12$ ) and $N_{\text{tot}}=2N$ , this is $2Nk_B\ln2$ . It is independent of the atomic masses and of how different the species are — yet it vanishes discontinuously when they become identical. The corrected classical partition function carries both special ingredients — the $1/N!$ (indistinguishability) and the $h^{3N}$ (the phase-space cell size that makes $S$ absolute).

Z_N = \frac{1}{N!}\,\frac{1}{h^{3N}}\int d^{3N}q\,d^{3N}p\;e^{-\beta H} = \frac{1}{N!}\left(\frac{V}{\lambda^3}\right)^{N} \qquad (4)

where: $h$ — Planck's constant, fixing the volume of a phase-space cell; $\beta=1/k_BT$ ; $H$ — the (purely kinetic, for an ideal gas) Hamiltonian. Reading it: the $1/N!$ makes $Z_N\propto(V/N)^N$ after Stirling, and $F=-k_BT\ln Z_N$ then yields the extensive entropy (2) and an intensive chemical potential $\mu=k_BT\ln(n\lambda^3)$ .

Derivation AThe spurious mixing entropy and its exact cancellation

The goal is to compute the entropy change when the partition between two identical ideal gases is removed, first with naive counting (which produces the paradox) and then with the $1/N!$ -corrected Sackur-Tetrode entropy (which resolves it), and to show the spurious $2Nk_B\ln2$ cancels exactly. The strategy is to write the before-and-after entropies for two boxes of $N$ atoms in $V$ merging into $2N$ atoms in $2V$ , using first equation (1) and then equation (2), and to track the single difference between $\ln V$ and $\ln(V/N)$ .

Step 1. Naive entropy before: two independent boxes, each contributing the uncorrected single-box entropy, $S_i=2Nk_B[\ln(V/\lambda^3)+\tfrac32]$ .

Step 2. Naive entropy after: $2N$ atoms in $2V$ with the same uncorrected formula, $S_f=2Nk_B[\ln(2V/\lambda^3)+\tfrac32]$ .

Step 3. Naive difference: $\Delta S_{\text{naive}}=S_f-S_i=2Nk_B\ln2>0$ . This is a positive entropy for an unobservable process — the Gibbs paradox in the flesh. Why the naive answer is wrong here but right for distinct gases. The number $2Nk_B\ln2$ is genuine — it is the entropy each species would gain by doubling its volume. The error is attributing it to identical gases, which have no species distinction to reward. The naive formula silently treats the two halves as distinguishable.

Step 4. Corrected entropy before (each side extensive): $S_i^{\text{corr}}=2Nk_B[\ln(V/N\lambda^3)+\tfrac52]$ .

Step 5. Corrected entropy after: $S_f^{\text{corr}}=2Nk_B[\ln(2V/2N\lambda^3)+\tfrac52]=2Nk_B[\ln(V/N\lambda^3)+\tfrac52]$ , because $2V/2N=V/N$ — the density per atom is unchanged.

Step 6. Result: $\Delta S=S_f^{\text{corr}}-S_i^{\text{corr}}=0$ . The $\ln(V/N)$ form makes the density the only thing that matters, and it did not change. Verified: the naive identical-gas mixing $\Delta S_{\text{naive}}=2Nk_B\ln2$ and the corrected $\Delta S=0$ , and the per-box difference $S_{\text{corr}}-S_{\text{naive}}=Nk_B(1-\ln N)=-k_B\ln N!$ (Stirling), are confirmed symbolically in verify_aq-sm-gibbs-paradox-sm3.py.

Go deeper · Derivation B · 1 minThe real mixing entropy of distinct gases and the discontinuity

Derivation BThe real mixing entropy of distinct gases and the discontinuity

The goal is to show that the same $2Nk_B\ln2$ is a real, physical entropy of mixing when the two gases genuinely differ, and that it drops discontinuously to zero the instant the species become identical — establishing the paradox's central feature. The strategy is to repeat the merge for two distinct species A and B, each with its own thermal wavelength, and to check that the wavelengths cancel so the result is independent of how similar the gases are.

Step 1. Initial entropy: species A ( $N$ atoms in $V$ ) plus species B ( $N$ atoms in $V$ ), each extensive with its own $\lambda$ : $S_i=Nk_B[\ln(V/N\lambda_A^3)+\tfrac52]+Nk_B[\ln(V/N\lambda_B^3)+\tfrac52]$ .

Step 2. Final entropy: A and B are distinguishable, so each independently fills $2V$ : $S_f=Nk_B[\ln(2V/N\lambda_A^3)+\tfrac52]+Nk_B[\ln(2V/N\lambda_B^3)+\tfrac52]$ .

Step 3. Difference: $\Delta S_{\text{mix}}=Nk_B\ln2+Nk_B\ln2=2Nk_B\ln2$ . The wavelengths $\lambda_A,\lambda_B$ cancel completely — the result is independent of the atomic masses and of how chemically alike A and B are. Why the answer does not depend on how different the gases are. Mixing entropy responds to the mere existence of an in-principle distinction, not its size. A helium/neon mixture and a mixture of two barely-different isotopes give the identical $2Nk_B\ln2$ per equal parts, because each species gains the same factor-of-two in accessible volume.

Step 4. The discontinuity: let B become identical to A. The two "distinct" sub-entropies now count the same atoms, and the correct treatment reverts to the previous derivation, giving $\Delta S=0$ . There is no path along which $2Nk_B\ln2$ shrinks smoothly to zero — the jump is a genuine step, because indistinguishability is a yes/no property. Verified: the distinct-gas mixing $\Delta S=2Nk_B\ln2$ (independent of $\lambda_A,\lambda_B$ ), its equality to the general $-N_{\text{tot}}k_B\sum x_i\ln x_i$ at $x=(\tfrac12,\tfrac12)$ , and its equality to the naive-identical value (the mislabelled point), are confirmed symbolically in the verify script.

Limiting cases

The entropy's behaviour in the relevant limits separates what is real from what is spurious.

Case	Result	Meaning
Identical gases, correct counting	$\Delta S=0$	Joining two equal volumes of the same gas at the same $T,P$ makes no entropy — the thermodynamic requirement.
Two distinct gases, equal parts	$\Delta S=2Nk_B\ln2$	Real, measurable mixing entropy; from $-N_{\text{tot}}k_B\sum x_i\ln x_i$ with $x_1=x_2=\tfrac12$ , $N_{\text{tot}}=2N$ .
General multi-species	$\Delta S=-N_{\text{tot}}k_B\sum_i x_i\ln x_i$	Independent of masses and of similarity — yet jumps to $0$ in the identical limit.
Naive counting, identical gases	$\Delta S=2Nk_B\ln2$ (spurious)	The paradox: the real mixing entropy mislabelled onto identical gases.
Classical validity	$n\lambda^3\ll1$	The $1/N!$ and the whole classical treatment hold only when $\lambda$ is small compared with the interatomic spacing.

Common trapThe entropy of mixing is unphysical, so just drop it

Two opposite over-corrections are both wrong. First, the $1/N!$ is not an optional or approximate fudge — it is exact, because identical atoms are exactly indistinguishable, and it must always be present in the classical partition function of identical particles. Second, do not "resolve" the paradox by declaring the entropy of mixing unphysical: for genuinely different species $\Delta S_{\text{mix}}=-N_{\text{tot}}k_B\sum x_i\ln x_i$ is real, measured in mixing calorimetry, and independent of how alike the species are. The correct statement is subtle: the mixing entropy is real for any distinction and drops discontinuously to zero only when the species become strictly identical. Deleting it wholesale would break Dalton's law and the thermodynamics of solutions.

Worked examples

Example 1 — Mixing one mole of argon with one mole of neon, then one mole of argon with one mole of argon (enumerate, apply, boundary number). Two flasks, each of volume $V$ at temperature $T$ : one mole of gas A on the left, one mole of gas B on the right. Open the valve so both gases fill the combined volume $2V$ .

Step 1. Enumerate. Total atoms $N_{\text{tot}}=2N_A$ ( $N_A$ = Avogadro's number), mole fractions $x_1=x_2=\tfrac12$ . Case (a): A = argon, B = neon (distinct). Case (b): A = argon, B = argon (identical).

Step 2. Apply equation (3). Case (a): $\Delta S=-2N_Ak_B(\tfrac12\ln\tfrac12+\tfrac12\ln\tfrac12)=2N_Ak_B\ln2=2R\ln2$ . Case (b): the gases are identical, so removing the valve is unobservable and $\Delta S=0$ .

Step 3. Boundary number. $2R\ln2=2(8.314)(0.6931)=\mathbf{11.53\,J\,K^{-1}}$ for the argon/neon case, and exactly $0$ for argon/argon. The same physical apparatus gives $11.53\,\mathrm{J/K}$ or $0$ depending only on whether the two gases are the same species — the discontinuity made numerical.

Example 2 — Isotope mixing: are $^3$ He and $^4$ He "identical"? Repeat Example 1 with one mole of $^3$ He and one mole of $^4$ He.

Step 1. $^3$ He and $^4$ He are different species — different mass, different nuclear spin, and in principle separable (e.g. by diffusion or a superleak). They are distinguishable.

Step 2. Therefore the mixing is real: $\Delta S=2R\ln2=\mathbf{11.53\,J\,K^{-1}}$ , exactly as for argon/neon, and independent of the small mass difference.

Step 3. Boundary check. If instead you "mixed" $^4$ He with more $^4$ He, $\Delta S=0$ . The lesson: the mixing entropy does not care that $^3$ He and $^4$ He are chemically almost twins — any in-principle distinction gives the full $2R\ln2$ , while strict identity gives zero. This is why isotope separation carries a real thermodynamic cost set by $-R\sum x_i\ln x_i$ .

Origins · 1 minidentical particles and how this was discovered

identical particles

thermodynamic concept · Gibbs 1875–1878; paradox stated 1902

Treating identical particles as fundamentally interchangeable is the key that resolves the Gibbs paradox. The entropy of mixing quantifies the irreversibility of combining distinct substances, $\Delta S=-N_{\text{tot}}k_B\sum_i x_i\ln x_i$ . J. Willard Gibbs identified the paradox in his monumental memoir On the Equilibrium of Heterogeneous Substances (1875–1878): the entropy of mixing is independent of how different the gases are, yet must vanish for identical gases, implying a discontinuity that classical mechanics could not explain. Gibbs recognized that the resolution required treating identical particles as fundamentally interchangeable — an idea he formalized in his 1902 Elementary Principles in Statistical Mechanics with the "generic phase" (dividing by $N!$ ). The paradox thus anticipated quantum indistinguishability by more than two decades, since only quantum statistics rigorously justify the $1/N!$ .

History

1875

— Gibbs, in On the Equilibrium of Heterogeneous Substances, notes that the mixing entropy is independent of the difference between the gases yet must vanish for identical ones.

1902

— Gibbs, in Elementary Principles in Statistical Mechanics, introduces generic phases (the 1/N! counting) that resolve the paradox.

1912

— Sackur and Tetrode give the absolute, extensive entropy of a monatomic ideal gas, embedding the 1/N! and h.

1924

— Bose and Einstein develop quantum statistics that supply the microscopic origin of indistinguishability.

1926

— Fermi and Dirac complete the quantum statistics of identical particles, justifying the 1/N! as the dilute limit.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., §1.5 and §2.4 · Kardar, Statistical Physics of Particles, §4.8–4.9 · Reif, Fundamentals of Statistical and Thermal Physics, §7.3 · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, §3.2 and §5.2 · Landau & Lifshitz, Statistical Physics Part 1, §31.

Provenance. All formulae are standard. A verifier should confirm: (i) the naive identical-gas mixing $\Delta S_{\text{naive}}=2Nk_B\ln2$ and the corrected $\Delta S=0$ from the extensive Sackur-Tetrode form (Pathria §1.5, §2.4); (ii) the distinct-gas $\Delta S=2Nk_B\ln2$ , independent of $\lambda_A,\lambda_B$ , and its match to $-N_{\text{tot}}k_B\sum x_i\ln x_i$ at $x=(\tfrac12,\tfrac12)$ (Kardar §4.9, Reif §7.3); (iii) that the corrected $S(E,V,N)$ is first-order homogeneous while the naive one is not (Landau-Lifshitz §31); (iv) the numeric $2R\ln2=11.53\,\mathrm{J/K}$ . All checks are in verify_aq-sm-gibbs-paradox-sm3.py. The general concept is owned upstream by the foundations Gibbs-paradox article; the $1/N!$ is owned by Correct Boltzmann Counting.

Builds on: Classical Ideal Gas · Correct Boltzmann Counting · Sackur-Tetrode Equation Leads to: Classical Limit of Quantum Statistics · Thermal de Broglie Wavelength

Why does the entropy of mixing not depend on how different the two gases are?

Because mixing entropy responds to the existence of a distinction, not its magnitude. If the two species are distinguishable at all — by mass, spin, chemistry, anything in principle — then each species independently gains access to the full doubled volume, and each contributes $Nk_B\ln2$ , for a total $2Nk_B\ln2$ . The thermal wavelengths $\lambda_A,\lambda_B$ cancel in the calculation, so the atomic masses drop out entirely. A helium/neon mixture and a mixture of two nearly-identical isotopes give the same mixing entropy per equal parts. Only when the species become strictly identical — no distinction even in principle — does the term collapse to zero.

Is the 1/N! a quantum correction or a classical one?

It cannot be derived within classical mechanics, where every particle carries a definite trajectory and is in principle labelable. It is the classical shadow of quantum indistinguishability: in the dilute regime $n\lambda^3\ll1$ , the ratio of correct Bose or Fermi counting to naive distinguishable counting is exactly $1/N!$ , because with at most one particle per single-particle state the $N$ occupied states can be assigned to identical particles in only one way, while distinguishable counting overcounts by the $N!$ orderings. So the $1/N!$ is a genuinely quantum ingredient that must be imported into classical statistical mechanics — which is exactly why the paradox foreshadowed quantum statistics.

If atoms actually cross the partition, how can the entropy be zero?

Atoms do physically migrate — the point is that their migration creates no new microstate. Two configurations related by swapping identical atoms are the same microstate, so shuffling atoms across the (removed) partition does not increase the number of accessible states. Since entropy counts distinguishable microstates, and none are gained, $\Delta S=0$ . This matches the thermodynamic requirement that a process with no observable consequence produces no entropy. The naive calculation errs by implicitly treating the migrating atoms as tagged, which they are not.

How is this different from the foundations Gibbs-paradox article?

The foundations article owns the general concept: that permuting identical particles yields the same microstate, and the resulting $1/N!$ and extensivity argument in the abstract. This article is the ideal-gas application: it plants the paradox in the Sackur-Tetrode entropy specifically, shows the spurious $2Nk_B\ln2$ appear and cancel in that concrete formula, contrasts it with the real distinct-gas mixing entropy, and dwells on the discontinuity as a numerical, worked-out feature. The $1/N!$ machinery itself is owned by Correct Boltzmann Counting. Together the three form the extensivity cluster: concept, counting factor, and ideal-gas consequence.

What goes wrong with the chemical potential if you omit the 1/N!?

Without the factor, the free energy contains $Nk_BT\ln V$ instead of $Nk_BT\ln(V/N)$ , so the chemical potential $\mu=(\partial F/\partial N)_{T,V}$ becomes explicitly and non-intensively $N$ -dependent, violating the Gibbs-Duhem relation. Equilibrium conditions between subsystems and the law of mass action would then fail — a gas in contact with a copy of itself would spuriously exchange particles to lower a fictitious free energy. The $1/N!$ restores the intensive, density-only $\mu=k_BT\ln(n\lambda^3)$ , which is what underlies real chemical equilibrium.

Could a demon that distinguishes "identical" atoms recover the mixing entropy?

This is the sharp modern question the paradox poses. If a demon possessed a way to distinguish the atoms on the two sides — some hidden label we normally cannot access — then from its viewpoint the gases are distinct, and mixing them would carry $2Nk_B\ln2$ of entropy that the demon could in principle harness. The resolution, via Landauer's principle, is that the demon's measurement and memory carry their own entropy cost, so no net free lunch results. For truly identical quantum particles, though, there is no hidden label even in principle — the symmetry of the wavefunction forbids it — so the mixing entropy is genuinely zero, not merely inaccessible. The paradox thus sits at the crossroads of thermodynamics, information, and quantum identity.

Why does the 5/2 in Sackur-Tetrode become the relevant constant here?

The $\tfrac52$ is the fingerprint of correct counting. Starting from the naive $\tfrac32$ (from the $E^{3/2}$ density of states and the momentum integral), subtracting $k_B\ln N!\approx Nk_B(\ln N-1)$ via Stirling contributes an extra $+1$ , turning $\tfrac32$ into $\tfrac52$ and simultaneously turning $\ln V$ into $\ln(V/N)$ . So the appearance of $\tfrac52$ rather than $\tfrac32$ in the entropy is itself a signature that the $1/N!$ has been applied. If you ever see $\tfrac32$ with a bare $\ln V$ , the counting is naive and the Gibbs paradox is lurking; the $\tfrac52$ with $\ln(V/N)$ is the paradox-free, extensive form.