Indistinguishability and the Gibbs Paradox

Pour two identical glasses of water together and you have one larger glass of water — nothing happened, and no engine could ever extract work from the act. Now pour a glass of water into a glass of alcohol and the mixture warms, refuses to unmix, and could in principle drive a turbine as it settles. Thermodynamics draws a sharp line between these two: mixing identical substances is free and reversible, mixing different substances creates entropy. The trouble is that the most natural way to count microstates of a classical gas erases that line — it charges you an entropy of mixing even when nothing distinct was mixed. The culprit is bookkeeping. If you imagine tagging every molecule with a serial number and counting configurations that differ only by swapping tags as genuinely different, the microstate count $\Omega$ inflates by a factor that grows with $N$ . That inflation shows up as a rogue $Nk_B\ln V$ in the entropy where thermodynamics demands the intensive combination $Nk_B\ln(V/N)$ . The first form is not extensive: double the whole system and the entropy more than doubles. Josiah Willard Gibbs pinned the absurdity down with a partition removed between two boxes of the same gas — the labelled count predicts a spurious $\Delta S = 2Nk_B\ln 2$ for a process in which, physically, nothing changed. The resolution is one factor: divide the phase-space volume by $N!$ , the number of ways to permute identical particles, because swapping identical particles is not a new microstate at all. That single division converts $\ln V$ into $\ln(V/N)$ , restores extensivity, produces the Sackur–Tetrode entropy that matches measured absolute entropies of the noble gases, and — the payoff — makes the mixing entropy of identical gases exactly zero while leaving the $Nk_B\ln 2$ for genuinely distinct gases untouched. The $1/N!$ is not a units patch; it is the classical fossil of quantum indistinguishability, the statement that identical particles have no individuality to keep track of.

Key takeaways

Counting a classical ideal gas as if particles carried labels gives a non-extensive entropy: doubling $N$ , $V$ , $E$ together fails to double $S$ .
Gibbs's thought experiment: removing a partition between two boxes of the same gas predicts a spurious $\Delta S = 2Nk_B\ln 2$ , an entropy of mixing where no mixing occurred.
The cure is a single factor $1/N!$ on the phase-space volume, because permuting identical particles is the same microstate, not a new one.
With $1/N!$ the entropy becomes the extensive Sackur–Tetrode formula, and the mixing entropy of identical gases is exactly zero.
For genuinely distinct gases (argon vs. neon) there are no cross-permutations to remove, so $\Delta S_\text{mix} = -k_B\sum_i N_i\ln x_i$ survives — the correct, measurable result.
The $1/N!$ is the classical limit of quantum statistics: in the dilute regime both Bose and Fermi counting collapse to the Maxwell–Boltzmann count divided by $N!$ .
Key idea: Permuting identical particles is not a new microstate; dividing the classical phase-space volume by $N!$ restores extensivity, yields Sackur–Tetrode, and turns the mixing entropy of identical gases into an exact zero while distinct gases keep their $\Delta S = -k_B\sum_i N_i\ln x_i$ .

The core insight

Permuting identical particles does not generate a new microstate; dividing the classical phase-space volume by $N!$ (and measuring it in units of $h^{3N}$ ) restores extensivity, yields the Sackur–Tetrode entropy, and makes the entropy of mixing of identical gases vanish exactly — while distinct gases keep the real $\Delta S_\text{mix}$ .

Where this lives today

The Gibbs correction is not a museum piece; it is the reason chemical thermodynamics works. Every tabulated standard molar entropy of a gas — the numbers chemists use to predict whether a reaction runs — rests on the Sackur–Tetrode formula and its $1/N!$ , and the sub-percent agreement with argon at room temperature is checked routinely. The distinction between mixing identical and distinct species is engineered directly in separation technology: semipermeable membranes, pressure-swing adsorption, and isotope separation all trade on the fact that mixing entropy is extractable as work only for distinguishable species. The deepest modern echo is in quantum information and cold atoms. Experiments with ultracold gases now tune $n\lambda_T^3$ across the classical-to-degenerate crossover, watching the $1/N!$ approximation give way to full Bose–Einstein or Fermi–Dirac statistics — the very transition the limiting-cases table anticipates. And the recognition that "indistinguishability" is a statement about state-counting, not about particles being physically interchangeable machines, underlies how entropy is defined for identical qubits and photons in quantum computing.

The physical picture

Picture two boxes side by side, each holding $N$ molecules of argon at the same temperature and density, separated by a thin partition. Slide the partition out. If argon molecules were tiny numbered billiard balls, then "which of the $2N$ balls ended up in the left half" would be new information the joined system now carries, and counting those choices credits the gas with an entropy of mixing $2Nk_B\ln 2$ . But argon atoms carry no serial numbers. A snapshot with atoms 1 and 2 swapped is not a second photograph of the gas — it is the same photograph. The labelled count is answering a question that has no physical referent, and it is punished with a nonsense entropy for a reversible, do-nothing process. Contrast the honest case. Replace the argon in the right box with neon. Now removing the partition really does open new configurations: every neon atom, formerly confined to volume $V$ , can roam the full $2V$ , and so can every argon atom. That genuine doubling of accessible volume per species is what produces the real, measurable $\Delta S = 2Nk_B\ln 2$ . The whole content of the Gibbs paradox is that a correct theory must return exactly zero in the first case and exactly this value in the second — and the only difference between the two calculations is whether you factor out cross-species permutations that never physically existed.

Figure

The Gibbs mixing experiment — identical gases (top, nothing happens) versus distinct gases (bottom, real mixing)

The picture already contains the whole resolution: entropy of mixing is real if and only if the two populations were genuinely distinguishable. The labelled classical count fails because it invents distinguishability among identical particles. Quantum mechanically there is nothing to repair — the state of $N$ identical particles is a single (anti)symmetric vector, so permutations were never separate states. The $1/N!$ is precisely what survives of that fact when you let $\hbar\to 0$ in the dilute regime.

Figure

The extensivity flip — labelled counting curves away from linear, the 1/N! correction snaps it straight

The core equations

The whole subject is one factor of $N!$ inserted into the partition function. Everything else — extensivity, Sackur–Tetrode, the vanishing of identical-gas mixing — follows by algebra. We begin with the corrected canonical partition function of $N$ identical particles, in which the indistinguishability factor sits out front.

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

where: $Z_N$ — canonical partition function for $N$ identical particles; $N!$ — number of particle permutations, removing the over-counting of configurations that differ only by relabelling; $h^{3N}$ — the phase-space cell volume that makes $Z_N$ dimensionless; $\beta = 1/(k_BT)$ ; $H(q,p)$ — the Hamiltonian. The $N!$ is the entire content of the Gibbs correction. For non-interacting particles the integral factorizes into one-particle pieces. Each free particle contributes a spatial factor $V$ and a Gaussian momentum integral that defines the thermal de Broglie wavelength — the quantum length scale that will carry Planck's constant into an otherwise classical entropy.

Z_1 = \frac{V}{\lambda_T^{3}}, \qquad \lambda_T = \frac{h}{\sqrt{2\pi m k_B T}} \qquad (2)

where: $Z_1$ — single-particle partition function; $V$ — volume; $\lambda_T$ — thermal de Broglie wavelength; $m$ — particle mass. When $\lambda_T$ is small compared to the interparticle spacing $n^{-1/3}$ , the gas is classical and $1/N!$ is accurate; when they are comparable, quantum statistics takes over. Because the particles do not interact, the $N$ -particle partition function is the single-particle one raised to the $N$ , divided by the crucial $N!$ . This is the object whose logarithm we take.

Z_N = \frac{Z_1^{\,N}}{N!} = \frac{1}{N!}\left(\frac{V}{\lambda_T^3}\right)^{N} \qquad (3)

where: $Z_N$ — full non-interacting partition function including indistinguishability; $Z_1$ — single-particle value from (2). Without the $N!$ this would be $Z_1^N$ — the labelled count that breaks extensivity. Taking the logarithm and applying Stirling to $\ln N!$ turns the labelled $\ln V$ into the intensive $\ln(V/N)$ . The Helmholtz free energy inherits that intensive combination.

F = -k_B T\ln Z_N = -N k_B T\left[\ln\!\left(\frac{V}{N\lambda_T^3}\right)+1\right] \qquad (4)

where: $F$ — Helmholtz free energy; $V/N = 1/n$ — volume per particle, the intensive ratio whose appearance (rather than $V$ alone) is the signature of a correctly extensive theory. Differentiating $F$ with respect to temperature gives the entropy — the celebrated Sackur–Tetrode formula, extensive by construction, carrying Planck's constant inside the logarithm.

S = N k_B\left[\ln\!\left(\frac{V}{N\lambda_T^3}\right)+\frac{5}{2}\right] \qquad (5)

where: $S$ — entropy of a monatomic ideal gas; the $V/N$ inside the log and the $5/2$ constant both depend on the $1/N!$ having been included. Written in energy variables with $E=\tfrac32 Nk_BT$ this is $S = Nk_B[\ln(\tfrac{V}{N}(\tfrac{4\pi mE}{3Nh^2})^{3/2})+\tfrac52]$ . The defining test of extensivity is homogeneity of degree one: scale the system up by any factor and the entropy scales by the same factor. The corrected entropy passes; the labelled one does not.

S(\lambda N,\lambda V,\lambda E) = \lambda\, S(N,V,E) \qquad (6)

where: $\lambda$ — any positive scaling factor. This single equation is what the $1/N!$ buys, and what its omission destroys. Finally, the two mixing outcomes that resolve the paradox. Distinct species mix and generate entropy; identical species do not.

\Delta S_{\text{distinct}} = -k_B\sum_i N_i \ln x_i, \qquad \Delta S_{\text{identical}} = 0 \qquad (7)

where: $N_i$ — number of particles of species $i$ ; $x_i = N_i/N$ — mole fraction. For a 50:50 binary mixture the distinct value is $Nk_B\ln 2$ per species, totalling $2Nk_B\ln 2$ for $N$ of each; the identical value is an exact zero.

Derivation AWhy omitting the 1/N! destroys extensivity

The clean way to see the paradox is to compute the entropy twice — once treating particles as labelled, once with the $N!$ removed — and test each for extensivity. The failure is not subtle: the labelled entropy contains a term $Nk_B\ln V$ instead of $Nk_B\ln(V/N)$ , and under uniform scaling that lone $\ln V$ spawns an extra $\lambda N k_B\ln\lambda$ that grows super-linearly. Watching Stirling's approximation of $-\ln N!$ convert $\ln V$ into $\ln(V/N)$ is watching the paradox dissolve in one line.

Step 1. The labelled (naive) count. Treat every particle as carrying a distinct label, so there is no permutation correction: $Z_N^{\text{naive}} = Z_1^N = (V/\lambda_T^3)^N$ . Why: this is the "obvious" classical count — integrate independently over each of the $N$ labelled particles' phase spaces.

Step 2. The labelled entropy. From $F = -k_BT\ln Z_N^{\text{naive}} = -Nk_BT\ln(V/\lambda_T^3)$ and $S=-\partial F/\partial T$ one finds

S_{\text{naive}} = N k_B\left[\ln\!\left(\frac{V}{\lambda_T^3}\right)+\frac{3}{2}\right].

Why: the temperature derivative acts on $\lambda_T^{-3}\propto T^{3/2}$ , producing the $3/2$ ; the volume enters as a bare $\ln V$ , with no compensating $1/N$ .

Step 3. Test extensivity. Scale $N,V,E\to\lambda N,\lambda V,\lambda E$ . Because $\lambda_T$ is intensive (it depends only on $T$ , which is fixed), the $\ln V$ term becomes $\ln(\lambda V)=\ln V+\ln\lambda$ , giving

S_{\text{naive}}(\lambda N,\lambda V,\lambda E) = \lambda S_{\text{naive}} + \lambda N k_B\ln\lambda.

Why: the extra $\lambda Nk_B\ln\lambda$ is the mathematical face of the Gibbs paradox — $S$ grows faster than the system, so it is not extensive, and additivity of entropy for independent subsystems fails.

Step 4. Insert the indistinguishability factor. With $Z_N = Z_1^N/N!$ , Stirling's approximation $\ln N!\approx N\ln N - N$ (owned by The Stirling Approximation) gives $-\ln N! \approx -N\ln N + N$ , so $N\ln V - N\ln N = N\ln(V/N)$ :

\ln V \;\longrightarrow\; \ln\!\left(\frac{V}{N}\right)+1.

Why: the $-N\ln N$ from $-\ln N!$ is exactly what pairs with $\ln V$ to build the intensive $\ln(V/N)$ — the same $N!$ that removes label-permutation over-counting also removes the extensivity defect. This is not a coincidence: they are the same over-counting seen two ways.

Step 5. The corrected entropy is extensive. The Sackur–Tetrode entropy (5) is now homogeneous of degree one: $S(\lambda N,\lambda V,\lambda E)=\lambda S(N,V,E)$ , with the spurious $\lambda Nk_B\ln\lambda$ gone. Why: only the intensive $V/N$ survives inside the logarithm, so scaling all extensive variables together leaves the bracket unchanged and multiplies $S$ by the prefactor $\lambda N$ . *Verified: the scaling algebra and the Stirling substitution reproduce Pathria & Beale §1.5–1.6 and Kardar §4.8; checked symbolically in verify_aq-sm-gibbs-paradox.py (homogeneity of Sackur–Tetrode holds, and fails without the $1/N!$ ).*

Go deeper · Derivation B · 2 minThe mixing experiment: an exact zero versus a real Nk ln2

Derivation BThe mixing experiment: an exact zero versus a real Nk ln2

This is the calculation Gibbs designed the paradox around, and the owned result of this article: the mixing entropy is discontinuous in the degree of distinguishability. Two boxes, each of volume $V$ , temperature $T$ , and $N$ particles, are joined into one box of volume $2V$ with $2N$ particles. Using the extensive Sackur–Tetrode entropy throughout, the identical-gas case gives exactly zero — not approximately, not to leading order — while the distinct-gas case gives $2Nk_B\ln 2$ . The two answers are separated by a finite gap with nothing in between, and the entire gap is the presence or absence of cross-species permutations.

Step 1. Identical gases, before mixing. Each box separately carries $S_{\text{box}} = Nk_B[\ln\tfrac{V}{N\lambda_T^3}+\tfrac52]$ , so the total before is $2S_{\text{box}}$ . Why: the two boxes are independent, and entropy is additive once it is extensive.

Step 2. Identical gases, after mixing. The joined system has $2N$ particles in $2V$ :

S_{\text{after}} = 2N k_B\!\left[\ln\frac{2V}{2N\lambda_T^3}+\frac52\right] = 2N k_B\!\left[\ln\frac{V}{N\lambda_T^3}+\frac52\right] = 2S_{\text{box}}.

Why: both $V$ and $N$ doubled, so the intensive ratio $V/N$ inside the log is unchanged — the extensive entropy simply doubled. This is exactly why extensivity was the thing worth fixing.

Step 3. Identical result — exact zero. $\Delta S = S_{\text{after}} - 2S_{\text{box}} = 0$ . Why: removing a partition between identical samples is reversible — slide it back and you recover the start with no trace — so the entropy change must vanish identically, and it does.

Step 4. Distinct gases, count states. Now species A fills the left box, species B the right, $N_A=N_B=N$ . After mixing, each species occupies the full $2V$ , and permutations are factored out only within a species — there is no $(N_A+N_B)!$ across species:

Z = \frac{1}{N_A!}\Big(\frac{2V}{\lambda_A^3}\Big)^{N_A}\,\frac{1}{N_B!}\Big(\frac{2V}{\lambda_B^3}\Big)^{N_B}.

Why: an argon–neon swap genuinely changes the microstate, so those permutations were never over-counting — nothing is removed for them.

Step 5. Distinct result — a real entropy. Each species' volume went from $V$ to $2V$ , contributing $Nk_B\ln 2$ :

\Delta S_{\text{mix}} = N_A k_B\ln 2 + N_B k_B\ln 2 = 2N k_B\ln 2.

Why: the honest entropy of mixing is just a free expansion of each species into twice its volume. The paradox is resolved not by making both cases zero, nor both nonzero, but by the theory returning the correct discontinuous jump between them. *Verified: reproduces Reif §7.3 and Landau–Lifshitz §41; the exact zero for identical gases and the $2Nk_B\ln2$ for distinct gases are checked symbolically in verify_aq-sm-gibbs-paradox.py.*

Limiting cases

How the $1/N!$ correction behaves across regimes — it is exact only where quantum occupations are negligible:

Case	Result	Meaning
Dilute / classical	$n\lambda_T^3\ll 1$	Occupation numbers $\ll 1$ ; almost no two particles compete for one single-particle state, so the full quantum count reduces to Maxwell–Boltzmann divided by $N!$ . Here $1/N!$ is accurate.
Degenerate	$n\lambda_T^3\gtrsim 1$	Multiple particles crowd the same states; $1/N!$ over-corrects because it cannot handle multiple occupancy. Full Bose–Einstein or Fermi–Dirac statistics is required.
Distinguishable sites	fixed lattice	Atoms pinned to distinct sites are labelled by position; permutations are genuinely different states, so NO $1/N!$ is applied (e.g. the Einstein solid, a paramagnet).
Mixing identical gases	$\Delta S = 0$	Extensivity makes removing a partition between identical samples give an exact zero — reversibility respected.
Mixing distinct gases	$\Delta S = -k_B\sum_i N_i\ln x_i$	Different species genuinely mix; for an equal binary mixture $\Delta S = 2Nk_B\ln 2$ , independent of the $1/N!$ subtlety because cross-permutations never existed.

Worked examples

1 · Two moles at the tap — identical versus distinct (the paradox with numbers)

Two rigid boxes each hold one mole ( $N = N_A = 6.022\times10^{23}$ ) of gas at the same $T$ and density, separated by a partition. Compute the entropy of mixing when the partition is removed, first for the same gas on both sides, then for argon on one side and neon on the other.

Step 1. Identical case. With the extensive entropy, doubling both $V$ and $N$ leaves $V/N$ fixed, so $\Delta S = 0$ . There is no number to compute — it is an exact zero, and it must be, since sliding the partition back undoes the process with no trace. $\Delta S_\text{identical} = 0$ .

Step 2. Distinct case, per species. Each species free-expands from $V$ to $2V$ : $\Delta S_i = N_i k_B\ln 2 = R\ln 2$ per mole, where $R = N_A k_B = 8.314\ \mathrm{J\,K^{-1}mol^{-1}}$ .

Step 3. Evaluate. $R\ln 2 = 8.314\times 0.6931 = 5.76\ \mathrm{J\,K^{-1}}$ per mole.

Step 4. Total for both moles. $\Delta S_\text{distinct} = 2\times R\ln 2 = 2\times 5.76 = \boxed{11.5\ \mathrm{J\,K^{-1}}}$ . The gulf between $0$ and $11.5\ \mathrm{J\,K^{-1}}$ is the Gibbs paradox made quantitative: the same apparatus, the same thermodynamic states, and yet the entropy change leaps from an exact zero to a substantial positive value the instant the two gases become genuinely different. A semipermeable membrane that passes one species but not the other can extract that $11.5\ \mathrm{J\,K^{-1}}\times T$ of mixing free energy as work — while no membrane in the universe can do so for identical gases.

2 · Argon's absolute entropy from Sackur–Tetrode (does the 1/N! survive experiment?)

Compute the molar entropy of argon gas at $T = 298.15\ \mathrm{K}$ , $P = 1\ \mathrm{bar}$ from the Sackur–Tetrode formula, and compare with the calorimetric value $154.8\ \mathrm{J\,K^{-1}mol^{-1}}$ .

Step 1. Thermal wavelength. With $m = 39.95\,u = 6.634\times10^{-26}\ \mathrm{kg}$ , $\lambda_T = h/\sqrt{2\pi m k_B T}$ . Numerically $2\pi m k_B T = 2\pi(6.634\times10^{-26})(1.381\times10^{-23})(298.15) = 1.714\times10^{-45}$ , so $\lambda_T = 6.626\times10^{-34}/\sqrt{1.714\times10^{-45}} = 1.600\times10^{-11}\ \mathrm{m}$ .

Step 2. Volume per particle. At $1\ \mathrm{bar}$ , $V/N = k_BT/P = (1.381\times10^{-23})(298.15)/10^5 = 4.116\times10^{-26}\ \mathrm{m^3}$ .

Step 3. The logarithm. $\dfrac{V}{N\lambda_T^3} = \dfrac{4.116\times10^{-26}}{(1.600\times10^{-11})^3} = \dfrac{4.116\times10^{-26}}{4.096\times10^{-33}} = 1.005\times10^{7}$ , so $\ln(V/N\lambda_T^3) = 16.12$ .

Step 4. Molar entropy. $S = R\,[\ln(V/N\lambda_T^3) + \tfrac52] = 8.314\,[16.12 + 2.5] = 8.314\times 18.62 = \boxed{154.8\ \mathrm{J\,K^{-1}mol^{-1}}}$ . The agreement with the calorimetric value to better than a tenth of a percent is a stunning verification: the $1/N!$ (which set the $V/N$ ) and the cell size $h^{3N}$ (which set $\lambda_T$ , and hence put Planck's constant into an "absolute" entropy) are both physically correct, not bookkeeping conveniences. An entropy formula containing $h$ matches a room-temperature gas measurement — the quantum origin of indistinguishability is written into the noble gases' thermometry.

Origins · 1 mingibbs paradox and how this was discovered

gibbs paradox

classical statistical mechanics, 1875–1902 · Yale

The paradox is named for Josiah Willard Gibbs, who in his 1875–78 papers on the equilibrium of heterogeneous substances noticed that the entropy of mixing of two gases is independent of how similar they are — right up until they become identical, when it must jump to zero. That discontinuity troubled him: thermodynamics seemed to demand a quantity that did not vary continuously with the "difference" between the gases. The resolution — that particle labels are not physical, so identical-particle permutations must not be counted — was only fully understood after quantum mechanics revealed that identical particles are described by a single (anti)symmetric state. The modern reading is that the $1/N!$ is the classical fingerprint of that quantum fact, and the "paradox" is really the theory correctly refusing to interpolate between distinguishable and identical.

History

1875

— Gibbs, in On the Equilibrium of Heterogeneous Substances, notes the entropy of mixing is independent of the degree of similarity of the gases, exposing the discontinuity.

1902

— Gibbs, in Elementary Principles in Statistical Mechanics, introduces the "generic" versus "specific" phase to formalize dividing by N! for identical particles.

1912

— Sackur and Tetrode independently write the absolute entropy of a monatomic ideal gas, fixing the additive constant with Planck's constant.

1924

— Bose and Einstein build quantum statistics of indistinguishable particles, revealing the N! as the classical limit of (anti)symmetrization.

1926

— Fermi and Dirac give the statistics of identical half-integer-spin particles, completing the quantum account of indistinguishability.

Going further

Texts. Pathria & Beale, Statistical Mechanics, 3rd ed., §1.5–1.6 (Gibbs paradox, Sackur–Tetrode) · 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, §3.5, §7 · Landau & Lifshitz, Statistical Physics Part 1, §40–41.

Provenance. Standard results: the $1/N!$ correction and Sackur–Tetrode (Pathria §1.5–1.6; Kardar §4.8–4.9; Reif §7.3; Landau–Lifshitz §40–41). A verifier should: reproduce $Z_1=V/\lambda_T^3$ via the Gaussian momentum integral; confirm Stirling converts $Nk_B\ln V$ into $Nk_B\ln(V/N)$ ; check the Sackur–Tetrode entropy is homogeneous of degree one in $(N,V,E)$ and is not without the $1/N!$ ; verify $\Delta S=0$ for identical and $2Nk_B\ln2$ for distinct gases; and recover $Z_N=Z_1^N/N!$ from the small-fugacity expansion of the quantum grand partition function. The one empirical claim — Sackur–Tetrode reproducing argon's molar entropy $154.8\ \mathrm{J\,K^{-1}mol^{-1}}$ at 298 K, 1 bar to within experimental error — is a standard textbook comparison, recomputed to $154.8\ \mathrm{J\,K^{-1}mol^{-1}}$ in verify_aq-sm-gibbs-paradox.py.

Prerequisites. Phase Space · Microstates and Macrostates · Boltzmann Entropy · Statistical Ensembles · Fundamental Postulate of Equal A Priori Probabilities. Leads to. Gibbs Entropy · Statistical Ensembles · Accessible Phase Space and Energy Shells · Extensivity and Additivity of Entropy · the quantum statistics of identical particles (Bose–Einstein and Fermi–Dirac).

If the particles are "really" identical, why did anyone ever count them as labelled in the first place?

Because classical mechanics has no notion of identity — a classical particle follows a definite trajectory you could, in principle, track and thereby "label" it. So the natural classical count integrates over each particle's phase space independently, treating a swap of two as a new configuration. Nothing in classical physics forbids this; it is only wrong as a description of nature, and the Gibbs paradox is how the error announces itself (non-extensive entropy, spurious mixing). Quantum mechanics removes the temptation at the root: identical particles share a single (anti)symmetric state, so a "swap" is not a different state to begin with. The $1/N!$ is the patch that makes the classical count agree with the quantum truth in the dilute limit.

Is the entropy of mixing "real"? Can I actually feel it?

For distinct gases, absolutely — it is measurable work. A semipermeable membrane that passes species A but blocks B lets you unmix them (or harvest work as they mix); the extractable free energy is exactly $T\,\Delta S_\text{mix}$ . For two moles of distinct gases that is $T\times 11.5\ \mathrm{J\,K^{-1}}$ , about $3.4\ \mathrm{kJ}$ at room temperature. For identical gases the mixing entropy is exactly zero, and — consistently — no membrane in the universe can distinguish the two sides to extract anything. The reality of the mixing entropy is precisely the reality of being able to tell the species apart.

Why exactly N! and not some other function of N?

Because $N!$ is the number of distinct ways to permute $N$ labels among $N$ particles, and every one of those permutations was counted as a separate microstate by the labelled integral even though they are physically the same. Dividing by the number of over-counted copies removes the redundancy exactly. It is the same logic as counting hands of cards: if the order in which you were dealt the cards does not matter, you divide the ordered count by the number of orderings. Here the "order" is the assignment of labels, and there are $N!$ of them.

Doesn't the 1/N! introduce a problem when N can change, like in the grand canonical ensemble?

No — it works cleanly there, and in fact the grand canonical ensemble is where the $1/N!$ looks most natural. The grand partition function is $\Xi=\sum_N z^N Z_N=\sum_N z^N Z_1^N/N! = e^{zZ_1}$ , and that exponential is only summable to a clean closed form because of the $1/N!$ Taylor coefficients. Far from causing trouble, the $1/N!$ is what makes the ideal-gas grand ensemble exponentiate. When you then compute $\langle N\rangle$ and fluctuations you recover the correct ideal-gas thermodynamics with an intensive chemical potential $\mu=k_BT\ln(n\lambda_T^3)$ .

The article says 1/N! "over-corrects" in the degenerate regime. What does that mean concretely?

The $1/N!$ assumes that no two particles ever occupy the same single-particle state, so every $N$ -particle configuration corresponds to exactly $N!$ labelled ones. But when $n\lambda_T^3\gtrsim1$ , particles do crowd into the same states. If two particles share a state, there are fewer than $N!$ distinct labellings (swapping the two in the same state changes nothing), so dividing by the full $N!$ removes too much. The fix is not a modified factorial but full quantum statistics: Bose–Einstein counting (for integer-spin particles, which cluster) or Fermi–Dirac (for half-integer, which exclude). The $1/N!$ is the first term of that exact counting, valid only when multiple occupancy is negligible.

Why does Planck's constant appear in a "classical" entropy formula at all?

Two places, both quantum in origin. The obvious one is the cell size $h^{3N}$ : phase-space volume has units, and to count states you must divide by the volume of one quantum cell, which the uncertainty principle sets at $h$ per degree of freedom. The subtler one is that $h$ hides inside $\lambda_T=h/\sqrt{2\pi mk_BT}$ . Together they fix the absolute (not just relative) entropy — the additive constant that classical thermodynamics leaves undetermined. The fact that Sackur–Tetrode, containing $h$ , nails argon's measured absolute entropy is direct evidence that this quantum input is physically correct, not a formal device.

Is the Gibbs paradox actually a paradox, or just a mistake?

It is a mistake dressed as a paradox — and that framing is instructive. There is no genuine logical contradiction in thermodynamics; the "paradox" is that a natural-looking classical calculation produces an entropy that is non-extensive and predicts irreversibility for a reversible process. The tension forces you to confront a physical question ("what counts as a distinct state for identical particles?") whose answer — indistinguishability — was not part of classical mechanics and had to await quantum theory. So the pedagogical value is real: the paradox is the shortest route to appreciating that particle identity is a physical fact with thermodynamic teeth.