Entropy of Mixing and the Gibbs Paradox

Open the valve between two chambers — one full of nitrogen, one full of oxygen, both at the same temperature and pressure — and something irreversible happens: the gases interdiffuse, the entropy of the universe rises, and no gentle nudge will ever un-mix them. That entropy increase, $\Delta S_{\text{mix}} = -Nk_B\sum_i x_i\ln x_i$ , is one of the cleanest quantitative statements in thermodynamics. Each gas expands to fill the whole volume, its entropy rising by exactly the isothermal-expansion amount, and the total is set purely by the mole fractions. Now repeat the experiment with the same gas on both sides — nitrogen and nitrogen. Nothing thermodynamic happens when you open the valve: the entropy does not change. Yet the naive calculation, treating the molecules as distinguishable, predicts the same $\Delta S_{\text{mix}}$ as for two different gases. That is the Gibbs paradox: a formula that gives a spurious, discontinuous jump of $2Nk_B\ln 2$ for a 50/50 mixture of identical gases where physically there should be nothing at all. Gibbs spotted this in 1875 and saw that it was not a small anomaly but a signpost pointing at something deep. The resolution is that identical particles are genuinely indistinguishable — swapping two of them does not produce a new microstate — so the count of states must be divided by $N!$ (the Gibbs correction). With the $1/N!$ in place, the entropy becomes properly extensive (the Sackur–Tetrode form), and mixing identical gases gives exactly zero. This article computes the mixing entropy for distinct gases, exhibits the paradoxical discontinuity for identical ones, and shows how the $1/N!$ — ultimately a quantum-mechanical fact about identical particles — removes it. The Gibbs paradox is where classical thermodynamics quietly demands quantum indistinguishability.

Key takeaways

Mixing two distinct ideal gases at equal $T,p$ raises the entropy by $\Delta S_{\text{mix}} = -Nk_B\sum_i x_i\ln x_i$ ; for a 50/50 mix this is $2N k_B\ln 2$ (per total $N$ : $Nk_B\ln 2$ ).
Mixing two identical gases changes nothing — but the naive distinguishable count predicts the same jump: the Gibbs paradox.
The paradox is a discontinuity: $\Delta S$ leaps from $2Nk_B\ln2$ (distinguishable) to $0$ (identical) with no smooth interpolation.
The $1/N!$ Gibbs correction, from the indistinguishability of identical particles, makes the entropy extensive (Sackur–Tetrode) and gives $\Delta S = 0$ for identical gases.
Indistinguishability is fundamentally a quantum fact; the Gibbs paradox is where classical thermodynamics silently requires it.
Key idea: the entropy of mixing $-Nk_B\sum x_i\ln x_i$ is real for distinct gases and exactly zero for identical ones; the discontinuity between them is resolved only by the $1/N!$ indistinguishability factor.

The core insight

The entropy of mixing is $\Delta S_{\text{mix}}=-Nk_B\sum_i x_i\ln x_i$ , positive for distinct gases and exactly zero for identical gases. The Gibbs paradox is the spurious discontinuous jump that the distinguishable count predicts for identical gases; the $1/N!$ indistinguishability correction — a quantum fact — restores extensivity and removes the paradox.

Where this lives today

The entropy of mixing is the thermodynamic engine behind separation science and much of chemistry. Every distillation column, every gas-separation membrane, every isotope-enrichment cascade is fighting the mixing entropy: because un-mixing lowers entropy, it costs at least $T\Delta S_{\text{mix}}$ of work, and this sets the fundamental thermodynamic minimum energy for purifying a substance. The outsized per-molecule entropy of trace components is exactly why enriching uranium or separating deuterium is so energy-intensive — the minority species is deep in the logarithm. The Gibbs paradox itself remains a live conceptual touchstone. It is the cleanest classical argument that identical particles must be treated as indistinguishable, a fact that quantum mechanics later grounds in the symmetry of the many-body wavefunction — the same indistinguishability that produces Bose–Einstein condensation, Fermi degeneracy pressure, and the Pauli exclusion principle. In information theory the mixing entropy reappears as the Shannon entropy of a mixture, and in the thermodynamics of computation (Landauer's principle, Maxwell's demon) the reversibility of mixing/un-mixing identical versus distinct particles is central to understanding the physical cost of information. Gibbs's little paradox turned out to be a doorway into three subjects at once.

The physical picture

The entropy of mixing is, at heart, two independent free expansions. When gas A (originally in volume $V$ ) is allowed to occupy the combined volume $2V$ , its entropy rises by $Nk_B\ln 2$ — the same increase as any isothermal expansion to twice the volume, because there are now twice as many places to put each molecule. Gas B does the same. Since the two gases are different, both expansions are real and independent, and the total is $2Nk_B\ln 2$ . The point is not that the gases "mix" in some special way; it is that each species, seen alone, has doubled its accessible volume.

Figure

Distinct gases: each species free-expands into the whole volume (ΔS = 2Nk ln2). Identical gases: opening the valve changes nothing (ΔS = 0).

Here the paradox announces itself. If the two "gases" are actually the same substance, opening the valve produces no observable change — you cannot even tell that the valve was opened. Push each half back to its original chamber and you have restored the exact original state, reversibly, with no work and no heat. So the entropy change must be zero. But the free-expansion bookkeeping does not know the gases are identical; it happily counts $2Nk_B\ln 2$ for each half doubling its volume. The classical formula, taking particles as distinguishable, gives a nonzero answer for a nothing-happened process. The resolution is a discontinuity that no continuous deformation can bridge. Imagine tuning a parameter that makes two species "gradually" more similar — a heavier isotope, a slightly different color. As long as they are distinguishable at all, opening the valve is irreversible and $\Delta S=2Nk_B\ln 2$ , no matter how tiny the difference. The instant they become truly identical, $\Delta S$ drops to $0$ . There is a step, not a ramp. The punchline plot below shows this cliff — and shows that the distinguishable count (dashed) never reaches zero, which is exactly the paradox the $1/N!$ correction fixes.

Figure

The Gibbs-paradox cliff: ΔS_mix jumps from 2Nk ln2 (distinguishable, any difference) to 0 (identical). The naive count (dashed) fails to drop — the paradox.

The core equations

The mixing entropy for an arbitrary number of distinct ideal-gas species follows from adding the independent expansions. With total particle number $N=\sum_i N_i$ and mole fractions $x_i=N_i/N$ :

\Delta S_{\text{mix}} = -Nk_B\sum_i x_i\ln x_i = \sum_i N_i k_B\ln\frac{N}{N_i} \qquad (1)

where: $x_i=N_i/N$ — mole fraction of species $i$ ; $k_B$ — Boltzmann constant. Each term $N_i k_B\ln(N/N_i)$ is the isothermal-expansion entropy of species $i$ growing from its partial volume into the whole. The result is always positive (mixing is irreversible). For the symmetric two-gas case ( $N_1=N_2=N$ , so total $2N$ , $x_i=\tfrac12$ ), Equation (1) gives the number to remember:

\Delta S_{\text{mix}} = -2Nk_B\left(\tfrac12\ln\tfrac12+\tfrac12\ln\tfrac12\right) = 2Nk_B\ln 2 \qquad (2)

where: $2N$ — total number of molecules; the factor $\ln 2$ is the doubling of each species' accessible volume. Per total particle this is $k_B\ln 2$ ; expressed with $N$ particles of each gas it is $2Nk_B\ln 2$ . Now the paradox: for identical gases treated with a distinguishable (no $1/N!$ ) count, the entropy is not extensive. The naive entropy has the form $S_{\text{naive}}=Nk_B[\ln V + \tfrac32\ln(U/N)+c]$ , and combining two boxes gives a spurious surplus:

\Delta S_{\text{naive}}^{\text{identical}} = S_{\text{naive}}(2N,2V,2U) - 2S_{\text{naive}}(N,V,U) = 2Nk_B\ln 2 \ne 0 \qquad (3)

where: the culprit is the $\ln V$ (rather than $\ln(V/N)$ ) — a non-extensive entropy that gains $2Nk_B\ln2$ upon merging identical boxes. Physically this must be zero, so the naive count is wrong. The Gibbs correction divides the number of microstates by $N!$ , turning $\ln V$ into $\ln(V/N)$ and producing the extensive Sackur–Tetrode entropy:

S = Nk_B\left[\ln\frac{V}{N\lambda_T^3}+\frac52\right],\qquad \lambda_T=\frac{h}{\sqrt{2\pi m k_BT}} \qquad (4)

where: $\lambda_T$ — thermal de Broglie wavelength; the $V/N$ (volume per particle) rather than $V$ is exactly what the $1/N!$ supplies via Stirling, $\ln N!\approx N\ln N - N$ . With this form, $S(2N,2V,2U)=2S(N,V,U)$ exactly, so mixing identical gases gives $\Delta S=0$ .

Derivation AThe entropy of mixing for distinct gases, and its vanishing for identical ones

The goal is to compute $\Delta S_{\text{mix}}$ directly from the entropy change of each gas, showing that distinct gases give $-Nk_B\sum x_i\ln x_i>0$ while identical gases give exactly zero once the correct extensive entropy is used. The strategy is to treat the mixing of two ideal gases as two simultaneous isothermal expansions and to compute the "before" and "after" entropies from the Sackur–Tetrode form.

Step 1. Distinct gases. Gas 1 ( $N_1$ molecules) starts in volume $V_1$ and ends in $V=V_1+V_2$ ; its entropy rises by $\Delta S_1 = N_1 k_B\ln(V/V_1)$ . Gas 2 likewise gains $\Delta S_2 = N_2 k_B\ln(V/V_2)$ . At equal initial density and temperature, $V_i\propto N_i$ , so $V/V_i = N/N_i = 1/x_i$ . Why: an ideal gas expanding isothermally from $V_i$ to $V$ gains entropy $N_i k_B\ln(V/V_i)$ — a standard result, and the two gases do this independently because they do not interact.

Step 2. Add them: $\Delta S_{\text{mix}} = N_1 k_B\ln\frac{N}{N_1} + N_2 k_B\ln\frac{N}{N_2} = -Nk_B(x_1\ln x_1 + x_2\ln x_2)$ . Generalizing to many species gives Equation (1). Why: each logarithm $\ln(N/N_i)=-\ln x_i$ is positive since $x_i<1$ , so the total mixing entropy is positive — mixing is irreversible.

Step 3. Identical gases, naive count. Using $S_{\text{naive}}=Nk_B[\ln V+\tfrac32\ln(U/N)+c]$ (no $1/N!$ ), compute merging two boxes of $(N,V,U)$ into $(2N,2V,2U)$ : $\Delta S = 2Nk_B[\ln(2V)-\ln V] = 2Nk_B\ln 2$ . This is nonzero — the paradox. Why: the offending term is $\ln V$ ; doubling the volume of the merged box against the sum of the separate ones leaves a leftover $\ln 2$ per particle that has no physical basis when the gases are the same.

Step 4. Identical gases, correct count. Using the extensive Sackur–Tetrode form $S=Nk_B[\ln(V/N)+\tfrac32\ln(U/N)+c']$ (with the $1/N!$ built in), the same merge gives $\Delta S = 2Nk_B[\ln\frac{2V}{2N}] - 2\cdot Nk_B\ln\frac{V}{N} = 0$ , because $\frac{2V}{2N}=\frac{V}{N}$ . Why: the $1/N!$ replaces $V$ by $V/N$ (volume per particle), which is unchanged when both $V$ and $N$ double — so the merged and separated states have identical entropy and mixing does nothing. *Verified: symbolic computation gives $\Delta S_{\text{distinct}}=2Nk_B\ln2$ , $\Delta S_{\text{naive}}^{\text{identical}}=2Nk_B\ln2$ (spurious), and $\Delta S_{\text{identical}}=0$ with the extensive Sackur–Tetrode form, in verify_aq-sm-entropy-mixing-gibbs-paradox.py.*

Go deeper · Derivation B · 2 minThe discontinuity is unavoidable classically — the owned structural result

Derivation BThe discontinuity is unavoidable classically — the owned structural result

This is the article's owned structural result: the entropy of mixing is genuinely discontinuous as a function of how similar the two gases are, and no continuous classical function of a "similarity parameter" can reproduce it — which is precisely why the resolution must come from the discrete fact of identity (indistinguishability), a quantum notion. The goal is to make the discontinuity explicit and to show that the naive count's failure and the $1/N!$ fix are two sides of the same coin. The strategy is to compare the general $x$ -dependence with the identical limit.

Step 1. Write the mixing entropy of two distinct gases with fractions $x$ and $1-x$ : $\Delta S(x)=-Nk_B[x\ln x + (1-x)\ln(1-x)]$ . This is a smooth, strictly positive function of $x$ on $(0,1)$ , maximized at $x=\tfrac12$ where it equals $Nk_B\ln2$ per total particle. Why: this is the binary Shannon entropy times $Nk_B$ ; it is the information gained by learning which species a randomly chosen molecule is.

Step 2. The value $\Delta S(x)$ does not depend on how different the two species are — only on whether they are different at all. Two isotopes differing by one neutron give the same $\Delta S$ as helium and argon. As long as some measurement distinguishes them, mixing is irreversible with the full $-Nk_B\sum x_i\ln x_i$ . Why: distinguishability is binary; the entropy counts states you can tell apart, and any nonzero distinguishing feature makes the two-species count apply.

Step 3. In the identical limit the answer jumps to zero, with no intermediate values. There is no continuous "partial identity." So $\Delta S$ is a step function of the distinguishability, and the naive classical count — which treats every particle as labeled — sits stubbornly at $2Nk_B\ln2$ even in the identical case. Why: classical mechanics labels particles by their trajectories, so it can never represent true identity; only the discrete $N!$ overcounting correction (or quantum symmetrization) captures the step.

Step 4. The $1/N!$ is therefore not an ad hoc patch but the only thing that can produce the step: it removes exactly the permutation overcounting that the naive labeled count includes, and it does so discontinuously (present for identical, absent for distinct). This is why Gibbs's paradox is a genuine window onto indistinguishability. Why: dividing by $N!$ removes the $N!$ ways of relabeling identical particles, which is the microscopic origin of the missing $2Nk_B\ln2$ — quantum mechanics later makes this exact via wavefunction symmetrization. *Verified: $\Delta S(x)=-Nk_B[x\ln x+(1-x)\ln(1-x)]$ smooth and positive on $(0,1)$ ; the identical value is $0$ ; the naive value is $2Nk_B\ln2$ — the step is confirmed symbolically in the verify script.*

Limiting cases

The mixing entropy behaves informatively at its boundaries:

Case	Result	Meaning
Equal 50/50 mix	$\Delta S = 2Nk_B\ln 2$	Maximum mixing entropy for two distinct species.
Trace impurity ( $x_2\to0$ )	$\Delta S \to N_2 k_B\ln(N/N_2)$	A little impurity gains a lot of entropy per molecule (large volume ratio).
Identical gases	$\Delta S = 0$	No paradox with the $1/N!$ correction; the merged state is the same state.
Naive (distinguishable) count	$\Delta S = 2Nk_B\ln 2$ even for identical	The Gibbs paradox: non-extensive entropy, physically wrong.
Multi-species	$\Delta S = -Nk_B\sum_i x_i\ln x_i$	Total mixing entropy is the Shannon entropy of the composition.

Common trap“The 1/N! is a classical bookkeeping trick with no real content”

It looks like a fudge factor, but it is the classical shadow of a genuine quantum fact: identical particles are truly indistinguishable, so a state obtained by swapping two of them is the same state, not a new one. Classically, the $1/N!$ corrects the overcounting of $N!$ permutations; quantum-mechanically, the correct count comes from symmetrizing (bosons) or antisymmetrizing (fermions) the wavefunction, and in the non-degenerate (high-temperature) limit both give the same $1/N!$ . The tell that it is not arbitrary: without it, the entropy is not extensive — $S(2N,2V,2U)\ne 2S(N,V,U)$ — which violates a basic requirement of thermodynamics and produces the Gibbs paradox. Two further traps: (1) the mixing entropy of distinct gases is real and does not go away with the $1/N!$ ; the correction only zeroes the identical case. (2) The paradox is about the discontinuity, not about the magnitude — even an arbitrarily small distinguishing difference gives the full $2Nk_B\ln2$ .

Worked examples

1 · The mixing entropy of air, molecule by molecule (a real number)

Model dry air as $78\%$ N $_2$ and $21\%$ O $_2$ and $1\%$ Ar (ignore trace gases; renormalize to $x_{\mathrm{N_2}}=0.78,\ x_{\mathrm{O_2}}=0.21,\ x_{\mathrm{Ar}}=0.01$ ). Compute the molar entropy of mixing $\Delta S_{\text{mix}}$ that would be released if you could un-mix one mole of air into its pure components.

Step 1. Use $\Delta S_{\text{mix}} = -R\sum_i x_i\ln x_i$ per mole ( $Nk_B\to R$ for one mole). Enumerate the three terms.

Step 2. Compute each: $-x_{\mathrm{N_2}}\ln x_{\mathrm{N_2}} = -0.78\ln0.78 = 0.78\times0.2485 = 0.1938$ ; $-0.21\ln0.21 = 0.21\times1.5606 = 0.3277$ ; $-0.01\ln0.01 = 0.01\times4.6052 = 0.0461$ .

Step 3. Sum: $\sum = 0.1938+0.3277+0.0461 = 0.5676$ . Multiply by $R=8.314$ : $\Delta S_{\text{mix}} = 0.5676\times8.314 = \mathbf{4.72\ J\,mol^{-1}K^{-1}}$ .

Step 4. Interpret the boundary. The tiny argon fraction ( $1\%$ ) contributes $0.046$ — nearly $8\%$ of the total — because each argon molecule expands into a hundred times its partial volume ( $\ln 100$ ). A minority species carries outsized mixing entropy per molecule; this is why separating trace gases (or isotopes) is thermodynamically expensive.

2 · Identical versus isotopically-labeled gas (exam level)

Two chambers each hold $N$ molecules of nitrogen at equal $T,p$ . (a) If both are ordinary $^{14}$ N $_2$ , what is $\Delta S$ on opening the valve? (b) If one chamber holds $^{14}$ N $_2$ and the other $^{15}$ N $_2$ (a distinguishable isotope), what is $\Delta S$ ? (c) What does the naive distinguishable count give in case (a)?

Step 1. Case (a), identical: opening the valve produces no observable change, and with the extensive (Sackur–Tetrode, $1/N!$ ) entropy the merged state equals the separated one. $\Delta S = \mathbf{0}$ .

Step 2. Case (b), distinct isotopes: even though $^{14}$ N and $^{15}$ N are chemically identical, they are physically distinguishable (different mass, separable by a mass spectrometer). So this is genuine two-species mixing: $\Delta S = 2Nk_B\ln 2$ , i.e. per mole $\mathbf{R\ln 2 = 5.76\ J\,mol^{-1}K^{-1}}$ .

Step 3. The jump. The isotopic mass difference is one part in fourteen, yet the entropy leaps from $0$ to the full $2Nk_B\ln2$ — the discontinuity. Any distinguishing feature, however small, triggers the entire mixing entropy.

Step 4. Case (c), naive count for identical gas: $\Delta S_{\text{naive}} = 2Nk_B\ln 2$ — the same as the distinct case, which is wrong. It fails because the labeled count does not divide by $N!$ ; this spurious nonzero answer is the Gibbs paradox, and the $1/N!$ correction removes it, restoring $\Delta S=0$ .

Origins · 1 minentropy of mixing and how this was discovered

entropy of mixing

thermodynamic concept · Gibbs 1875

The entropy of mixing $\Delta S_{\text{mix}}=-Nk_B\sum_i x_i\ln x_i$ measures the irreversibility of combining distinct substances, and its history is inseparable from the paradox that carries Gibbs's name. Gibbs, in On the Equilibrium of Heterogeneous Substances (1875–78), noticed that the mixing-entropy formula gives the same answer whether two gases are slightly different or wildly different — yet drops abruptly to zero when they become identical. He recognized this discontinuity as physically correct and traced its resolution to the indistinguishability of identical molecules, decades before quantum mechanics gave indistinguishability its rigorous meaning through wavefunction symmetry. The quantity is mathematically the Shannon/Boltzmann entropy of the composition distribution — the same $-\sum p\ln p$ that measures information — which is why the mixing entropy equals the information you gain by learning the identity of a randomly drawn molecule. It remains a touchstone example that classical thermodynamics already encodes quantum facts.

History

1875

— Gibbs identifies the paradox in On the Equilibrium of Heterogeneous Substances and links it to indistinguishability.

1902

— Gibbs's Elementary Principles in Statistical Mechanics introduces the "generic phase" counting that anticipates 1/N!.

1911

— Sackur and (1912) Tetrode independently derive the extensive entropy formula with the correct 1/N! and h^3 factors.

1924

— Bose's counting of indistinguishable photons, extended by Einstein, gives indistinguishability its quantum footing.

1926

— Fermi and Dirac formulate the statistics of identical particles, making 1/N! a limit of quantum symmetrization.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., §1.5–1.6 (Gibbs paradox and Sackur–Tetrode) · Kardar, Statistical Physics of Particles, §4.9 · Reif, Fundamentals of Statistical and Thermal Physics, §7.3 · Sethna, Statistical Mechanics, §5.2 and Ch. 3 · Landau & Lifshitz, Statistical Physics Part 1, §40 · Jaynes, "The Gibbs Paradox" (1992) for the information-theoretic view.

Provenance. The mixing-entropy formula and its resolution are standard (Pathria §1.5–1.6; Kardar §4.9; Reif §7.3). A verifier should reproduce: $\Delta S_{\text{distinct}}=-Nk_B\sum x_i\ln x_i$ (and $2Nk_B\ln2$ for 50/50); the identical case $\Delta S=0$ using the extensive Sackur–Tetrode entropy; and the spurious $2Nk_B\ln2$ from the naive (no $1/N!$ ) count. Numerical instances (air $\Delta S_{\text{mix}}=4.72\ \mathrm{J\,mol^{-1}K^{-1}}$ ; isotope mix $R\ln2=5.76\ \mathrm{J\,mol^{-1}K^{-1}}$ ) are recomputed in verify_aq-sm-entropy-mixing-gibbs-paradox.py. $R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}$ (CODATA).

Builds on: Second Law of Thermodynamics and Entropy · Reversible and Irreversible Processes · Thermodynamic Potentials and the Fundamental Relation · Euler Relation and Gibbs–Duhem Equation. Leads to: Helmholtz Free Energy, Enthalpy, and Gibbs Free Energy · Third Law of Thermodynamics · Maxwell Relations · Indistinguishability and Identical Particles.

Why does mixing two different gases increase entropy but mixing two portions of the same gas does not?

Because entropy counts distinguishable arrangements. When two different gases mix, each species gains access to twice the volume, and there are genuinely more distinguishable microstates — you could, in principle, measure that a nitrogen molecule is now on the side where only oxygen used to be. That extra freedom is real entropy, $2Nk_B\ln2$ for a 50/50 split. When two portions of the same gas "mix," nothing new becomes distinguishable: a nitrogen molecule on the left is interchangeable with one on the right, so opening the valve does not create any new arrangements you could tell apart. The number of distinguishable microstates is unchanged, so the entropy is unchanged. The whole difference is whether the swap of a left molecule for a right one produces a new state (distinct gases) or the same state (identical gas).

Is the 1/N! factor exact, or just an approximation valid at high temperature?

The classical $1/N!$ is exact as a counting correction — it removes precisely the $N!$ ways of permuting $N$ identical labels, which classical phase space overcounts. What is approximate is the replacement of the true quantum count by $Z_1^N/N!$ : that holds in the non-degenerate (dilute, high-temperature) regime where the thermal wavelength is much smaller than the interparticle spacing, so at most one particle occupies each single-particle state and the distinction between bosons and fermions does not matter. When the gas becomes degenerate (low $T$ , high density), the $1/N!$ is no longer enough — you need the full Bose–Einstein or Fermi–Dirac counting. So for the Gibbs paradox at ordinary conditions the $1/N!$ is the right and exact fix; it is only its derivation as a limit of quantum statistics that is a high-temperature approximation.

If two gases differ only infinitesimally, why isn't the mixing entropy infinitesimal too?

This is the heart of the paradox, and the answer is that distinguishability is binary, not continuous. Entropy counts states you can tell apart, and "can you tell them apart?" has only two answers: yes or no. If two gases differ by any measurable amount — a different mass, a different spectral line, anything — then in principle you can build an apparatus that separates them, so the mixing is irreversible and carries the full $2Nk_B\ln2$ . If they are truly identical, no such apparatus can exist and the entropy is zero. There is no "half-distinguishable." So the mixing entropy is a step function of similarity, not a ramp. What does scale with the difference is the practical difficulty and energy cost of un-mixing, but the thermodynamic entropy is all-or-nothing.

Does the Gibbs paradox prove that classical mechanics is wrong?

Not exactly — it shows that classical mechanics is incomplete about identical particles. Classical mechanics labels particles by their trajectories, so it always treats them as distinguishable and can never represent true identity; that is why the naive classical count gives the paradoxical nonzero mixing entropy for identical gases. The $1/N!$ correction patches this by hand, and it gives the right thermodynamics, but it has no justification within classical mechanics — you have to put it in because you know the particles are identical. The deep resolution comes from quantum mechanics, where identical particles are described by (anti)symmetrized wavefunctions and indistinguishability is built into the formalism from the start. So the paradox is a place where classical thermodynamics, correctly done, silently requires a quantum input — a striking early hint that the microscopic world is not classical.

Why does a trace impurity contribute so much mixing entropy per molecule?

Because the mixing entropy per molecule of species $i$ is $k_B\ln(1/x_i)=k_B\ln(N/N_i)$ , which grows as the fraction $x_i$ shrinks. A trace component starts confined to a tiny partial volume and then expands into essentially the whole container, and the entropy of expansion is the logarithm of the volume ratio, which is large for a small starting fraction. For a $1\%$ component, each molecule expands into $100\times$ its partial volume, contributing $k_B\ln100\approx4.6\,k_B$ — far more per molecule than a majority species. This is why the total mixing entropy has a substantial contribution from minor components, and it is the thermodynamic reason that separating dilute impurities or enriching rare isotopes requires disproportionately large amounts of work.

Is the entropy of mixing the same thing as the Shannon entropy of information theory?

Mathematically, yes — up to the constant $k_B$ . The mixing entropy $-Nk_B\sum_i x_i\ln x_i$ is exactly $Nk_B$ times the Shannon entropy $H=-\sum_i x_i\ln x_i$ of the composition distribution $\{x_i\}$ . It measures the average information (in nats, or bits if you use $\log_2$ ) you gain by learning the identity of a randomly chosen molecule from the mixture. This is not a coincidence: Boltzmann–Gibbs entropy and Shannon entropy are the same functional, and the mixing problem is a clean physical instance where thermodynamic entropy is missing information about composition. Jaynes emphasized this view, arguing that the Gibbs paradox dissolves once you recognize that entropy is defined relative to the macroscopic variables you choose to distinguish — if you cannot (or do not) tell two species apart, there is no mixing entropy, exactly as for identical gases.