The Gibbs Paradox and Indistinguishability

Derivation

The Gibbs Paradox and Indistinguishability

D-238 Home PU-302 Threads matter · chance · symmetry Depends on Boltzmann Entropy from Microstate Counting, The Partition Function and F = -kT ln Z

Statement

For a monatomic classical ideal gas, treating the particles as distinguishable gives an entropy that is not extensive: doubling \(N\) and \(V\) at fixed density adds a spurious \(Nk_B\ln 2\), so removing a partition between two identical samples appears to create entropy (the Gibbs paradox). Dividing the \(N\)-particle partition function by \(N!\) — the number of permutations of identical particles — removes exactly this overcount, yielding the extensive Sackur–Tetrode entropy \(S = Nk_B\!\left[\ln\!\frac{V}{N\lambda^3} + \frac{5}{2}\right]\). The ad hoc \(1/N!\) is the classical shadow of quantum indistinguishability, exact in the non-degenerate limit \(n\lambda^3 \ll 1\).

Why it matters

Extensivity is not a cosmetic nicety: thermodynamics is built on the assumption that \(S\), \(U\), \(F\) scale with system size while \(T\), \(P\), \(\mu\) do not. A non-extensive entropy makes the chemical potential ill-defined and breaks the additivity that underlies phase equilibria and the law of mass action. The Gibbs paradox is the sharpest signal that purely classical state-counting is incomplete.

The resolution reaches beyond bookkeeping. The factor \(N!\) is precisely the order of the permutation group of \(N\) identical objects, and its appearance says that permuting identical particles does not produce a new physical state. Classical mechanics has no principled reason to impose this; quantum mechanics does, through the symmetry of the many-body wavefunction. The Gibbs correction thus anticipates bosons and fermions decades before their discovery.

Assumptions

Non-interacting particles.If dropped, the \(N\)-body partition function no longer factorizes into single-particle factors; the configurational integral produces virial corrections and the clean \(Z_1^N\) structure is lost.

Semiclassical phase space with cell size \(h^{3}\).Without Planck's constant setting the measure \(d^{3}q\,d^{3}p/h^{3}\), the phase-space volume is dimensionally arbitrary and the entropy carries an undetermined additive constant — \(\lambda\) cannot be defined.

Identical particles of a single species.If the particles are distinguishable (different species), there is no permutation symmetry to quotient by, the \(1/N!\) is absent, and the mixing entropy is physically real rather than spurious.

Non-degenerate (dilute) limit \(n\lambda^3 \ll 1\).If dropped, the mean occupation of single-particle states is not small; Bose–Einstein or Fermi–Dirac counting is required and \(Z_1^N/N!\) is no longer the exact many-body partition function — it is only the leading term.

Large \(N\) (Stirling's approximation).If dropped, \(\ln N! \neq N\ln N - N\); finite-size corrections \(\tfrac12\ln(2\pi N)\) survive and the entropy per particle acquires \(O(N^{-1}\ln N)\) terms that spoil exact extensivity.

Derivation

\[ Z_1 = \frac{1}{h^3}\int d^3q\,d^3p\; e^{-\beta \mathbf{p}^2/2m} = \frac{V}{h^3}\left(2\pi m k_B T\right)^{3/2} \equiv \frac{V}{\lambda^3},\qquad \lambda \equiv \frac{h}{\sqrt{2\pi m k_B T}}. \]

Single-particle canonical partition function; the Gaussian momentum integral factorizes over the three Cartesian directions and the spatial integral gives \(V\). \(\lambda\) is the thermal de Broglie wavelength. B

\[ Z_N^{\text{dist}} = Z_1^{\,N} = \left(\frac{V}{\lambda^3}\right)^{\!N}. \]

For non-interacting particles the Boltzmann factor separates, \(e^{-\beta\sum_i \epsilon_i}=\prod_i e^{-\beta\epsilon_i}\), so the integral factorizes. Treating each particle as a labelled, distinguishable object gives one factor of \(Z_1\) per particle. A

\[ F = -k_BT\ln Z_N^{\text{dist}} = -Nk_BT\left[\ln V + \tfrac{3}{2}\ln T + \tfrac{3}{2}\ln\!\frac{2\pi m k_B}{h^2}\right]. \]

Definition of the Helmholtz free energy from the partition function. The \(T\)-dependence enters through \(\lambda^{-3}\propto T^{3/2}\). A

\[ S^{\text{dist}} = -\left(\frac{\partial F}{\partial T}\right)_{V,N} = Nk_B\left[\ln\!\frac{V}{\lambda^3} + \frac{3}{2}\right]. \]

Differentiate \(F\) at fixed \(V,N\); the \(\tfrac32\) comes from \(\partial(\,T\ln T^{3/2})/\partial T\) and equivalently equals \(U/T\) with \(U=\tfrac32 Nk_BT\). This is the distinguishable-particle entropy. B

\[ S^{\text{dist}}(\alpha N,\alpha V) = \alpha N k_B\left[\ln\!\frac{\alpha V}{\lambda^3}+\frac32\right] = \alpha\,S^{\text{dist}}(N,V) + \alpha N k_B\ln\alpha. \]

Scale the system by a factor \(\alpha\) at fixed intensive state (fixed \(T\), fixed density \(N/V\)). Extensivity demands \(S(\alpha N,\alpha V)=\alpha S(N,V)\), but a spurious \(\alpha N k_B\ln\alpha\) survives because \(\ln V\) is not intensive. This is the Gibbs paradox. C

\[ Z_N = \frac{1}{N!}\,Z_1^{\,N} = \frac{1}{N!}\left(\frac{V}{\lambda^3}\right)^{\!N}. \]

Configurations related by a permutation of the \(N\) identical particles are the same physical microstate, yet the labelled integral of Step 2 counts each of the \(N!\) permutations separately. Dividing by the order \(N!\) of the permutation group removes the overcount — the Gibbs correction. C

\[ \ln Z_N = N\ln\!\frac{V}{\lambda^3} - \ln N! \;\xrightarrow{\text{Stirling}}\; N\ln\!\frac{V}{\lambda^3} - N\ln N + N. \]

Take the logarithm and apply \(\ln N! \approx N\ln N - N\) (valid for large \(N\)). Note \(N!\) is \(T\)-independent, so \(U=k_BT^2\,\partial_T\ln Z_N = \tfrac32 Nk_BT\) is unchanged. B

\[ S = \frac{U-F}{T} = \frac32 Nk_B + k_B\!\left[N\ln\!\frac{V}{\lambda^3} - N\ln N + N\right] = Nk_B\!\left[\ln\!\frac{V}{N\lambda^3} + \frac52\right]. \]

Recompute the entropy with the corrected \(F=-k_BT\ln Z_N\). The \(-N\ln N\) combines with \(\ln V\) into the intensive ratio \(V/N\); the \(+1\) from Stirling lifts \(\tfrac32\to\tfrac52\). This is the Sackur–Tetrode equation. B

\[ S(\alpha N,\alpha V) = \alpha N k_B\!\left[\ln\!\frac{\alpha V}{\alpha N\lambda^3}+\frac52\right] = \alpha\,S(N,V). \]

Repeat the scaling test of Step 5. Because \(S\) now depends on \(V,N\) only through the intensive ratio \(V/N\), the spurious \(\ln\alpha\) cancels and extensivity is restored exactly. C

Result

\[ S = Nk_B\left[\ln\!\left(\frac{V}{N\lambda^3}\right) + \frac{5}{2}\right],\qquad \lambda=\frac{h}{\sqrt{2\pi m k_B T}} \]

Reading. The entropy per particle is set by the logarithm of the number of thermal cells available to one particle, \(V/N\) divided by the quantum volume \(\lambda^3\). The \(1/N!\) that produced the intensive argument \(V/N\) is the counting of indistinguishable permutations; it makes \(S\) extensive and quietly encodes that identical particles have no labels. The formula is trustworthy only while \(n\lambda^3\ll1\), i.e. while cells are sparsely occupied and Maxwell–Boltzmann statistics hold.

Units check. \(\lambda = h/\sqrt{2\pi m k_BT}\) has units \(\mathrm{J\,s}/\sqrt{\mathrm{kg\cdot J}} = \mathrm{J\,s}/\sqrt{\mathrm{kg^2 m^2 s^{-2}}} = \mathrm{J\,s}/(\mathrm{kg\,m\,s^{-1}}) = \mathrm{m}\), a length. Hence \(V/(N\lambda^3)\) is dimensionless, its logarithm is a pure number, and \(S=Nk_B\times(\text{number})\) has units of \(\mathrm{J\,K^{-1}}\), as an entropy must.

Limiting cases

High \(T\) or low density (\(n\lambda^3\to0\)): \(V/(N\lambda^3)\gg1\), \(S>0\) and grows as \(\tfrac32 Nk_B\ln T\); the classical regime where Sackur–Tetrode is accurate.
Onset of degeneracy (\(n\lambda^3\to1\)): the argument of the log approaches unity, \(S/Nk_B\to\tfrac52\); quantum statistics take over and the formula ceases to be reliable.
\(T\to0\): Sackur–Tetrode gives \(S\to-\infty\), violating the third law — a diagnostic that the classical limit has broken down and the ground-state occupation is no longer sparse.
Identical-gas self-mixing: combining \((N,V)+(N,V)\to(2N,2V)\) at fixed \(T\) gives \(\Delta S=0\), as it must for a reversible partition removal.
Distinct-gas mixing: \(A\) and \(B\) each expand into the full volume, \(\Delta S = (N_A+N_B)k_B\ln2>0\); the entropy is real because there is no \(N!\) linking the two species.

Breaks when

Quantum degeneracy, \(n\lambda^3\gtrsim1\): at low \(T\) or high density (electrons in metals, liquid helium, neutron-star matter) the occupation numbers are \(O(1)\), \(1/N!\) is not the exact permutation count, and one must use the full Bose–Einstein or Fermi–Dirac partition functions. Sackur–Tetrode's divergence to \(-\infty\) is the warning flag.
Interacting particles: once \(U\) contains pair potentials, \(Z_N\neq Z_1^N/N!\); the configurational integral must be kept and yields virial/cluster corrections. Near condensation the factorized form fails entirely.
Internal structure or reactions: for molecules the single-particle \(Z_1\) must include rotational, vibrational and electronic factors; if the particle number itself changes (dissociation, chemical reaction) the fixed-\(N\) canonical treatment is inappropriate and one works in the grand canonical ensemble.
Small \(N\): for a handful of particles Stirling fails, the \(\tfrac12\ln(2\pi N)\) term matters, and "extensivity" is not even a meaningful demand.

Failure modes

Dividing \(U\) by \(N!\) as well. The \(1/N!\) is a state-count correction independent of \(T\); it drops out of \(U=k_BT^2\,\partial_T\ln Z\) entirely. Only \(F\) and \(S\) change, not the energy or heat capacity.
Applying \(1/N!\) to mixing of different gases. Students divide the combined system by \((N_A+N_B)!\) and wrongly find zero mixing entropy for distinct species. Each species carries its own \(N_A!\) and \(N_B!\); the cross-permutations do not exist.
Writing \(\ln(V/\lambda^3)\) instead of \(\ln(V/N\lambda^3)\). Forgetting the \(-N\ln N\) from Stirling leaves the non-extensive Step-4 entropy and re-introduces the paradox.
Claiming \(1/N!\) "derives" quantum mechanics. It is consistent with, and anticipated by, indistinguishability, but classically it is an ad hoc insertion; only quantum symmetrization justifies it from first principles.
Using \(\ln N!\approx N\ln N\) (dropping the \(-N\)). This loses the \(+1\) that turns \(\tfrac32\) into the correct \(\tfrac52\), giving a numerically wrong molar entropy.

Discussion

The Gibbs paradox is best read as classical statistical mechanics announcing its own incompleteness. Nothing in Hamiltonian mechanics forbids labelling particles: the phase-space trajectory of particle 1 and particle 2 are perfectly well-defined curves. Yet the thermodynamics only comes out right if we refuse to count relabelled configurations as distinct. The factor \(1/N!\) is the minimal patch that enforces this refusal, and its group-theoretic meaning — the order of \(S_N\), the symmetric group — is unmistakable in hindsight.

Quantum mechanics supplies the missing principle. A many-body state of identical particles must be symmetric (bosons) or antisymmetric (fermions) under exchange, so a permuted configuration is literally the same ray in Hilbert space, not a new one. In the non-degenerate limit the difference between symmetric and antisymmetric counting is negligible and both collapse to the Maxwell–Boltzmann result: the number of ways to distribute \(N\) particles among many sparsely-occupied states is \(1/N!\) times the labelled count. Thus \(Z_N=Z_1^N/N!\) is the leading term of a systematic expansion in the degeneracy parameter \(n\lambda^3\), whose next corrections are the exchange (statistics) contributions to the second virial coefficient.

There is a subtlety worth stating precisely: the \(1/N!\) does not act inside the phase-space integral, it multiplies it. The correct semiclassical prescription is that the physical phase space is the quotient of the labelled \(6N\)-dimensional space by the \(S_N\) action, and away from the coincidence set (where two particles share a phase point, of measure zero) the quotient volume is exactly \(1/N!\) of the covering volume. This is why the correction is a constant divisor and not a \(T\)- or \(V\)-dependent function — a point that also explains its silence in \(U\). The coincidence set becomes important only when wavepackets overlap, i.e. precisely when \(n\lambda^3\sim1\) and the semiclassical picture fails.

Common misconceptions. The paradox is not resolved by "the gases are really slightly different so mixing entropy is real" — Gibbs' point survives in the strict identical-particle limit. Nor is it a paradox about mixing at all; the mixing thought-experiment is merely the most vivid symptom of the deeper defect, non-extensivity, which is already visible in a single gas under rescaling (Step 5). And the \(1/N!\) is not a normalization of probability — the canonical distribution is already normalized by \(Z\); it is a correction to the measure that counts distinct states.

Worked examples

Example 1 — Standard molar entropy of argon. Compute \(S\) for one mole of argon gas at \(T=298.15\,\mathrm{K}\), \(P=1\,\mathrm{bar}=10^5\,\mathrm{Pa}\), and compare with the tabulated \(154.8\,\mathrm{J\,K^{-1}mol^{-1}}\).

\[ V = \frac{N k_B T}{P} = \frac{RT}{P} = \frac{(8.314)(298.15)}{10^5} = 2.479\times10^{-2}\ \mathrm{m^3}. \]

Ideal-gas law for one mole; symbols first, then numbers. A

\[ m = 39.95\times1.6605\times10^{-27} = 6.634\times10^{-26}\ \mathrm{kg}. \]

Argon atomic mass in kilograms (\(1\,\mathrm{u}=1.6605\times10^{-27}\,\mathrm{kg}\)). A

\[ \lambda = \frac{h}{\sqrt{2\pi m k_BT}} = \frac{6.626\times10^{-34}}{\sqrt{2\pi(6.634\times10^{-26})(1.381\times10^{-23})(298.15)}} = 1.599\times10^{-11}\ \mathrm{m}. \]

Thermal de Broglie wavelength; the radicand is \(1.716\times10^{-45}\,\mathrm{kg^2 m^2 s^{-2}}\). B

\[ \frac{V}{N\lambda^3} = \frac{2.479\times10^{-2}}{(6.022\times10^{23})(1.599\times10^{-11})^3} = \frac{2.479\times10^{-2}}{2.463\times10^{-9}} = 1.007\times10^{7}. \]

Number of thermal cells per particle; note \(n\lambda^3 = (V/N\lambda^3)^{-1}\approx10^{-7}\ll1\), so the classical formula is valid. B

\[ S = R\left[\ln(1.007\times10^7) + \tfrac52\right] = 8.314\,(16.12 + 2.50) = 8.314\times18.62. \]

Sackur–Tetrode with \(Nk_B=R\) for one mole. A

\[ S = 154.8\ \mathrm{J\,K^{-1}mol^{-1}} \]

Reading. The calorimetric standard entropy of argon is \(154.8\,\mathrm{J\,K^{-1}mol^{-1}}\) — agreement to four figures. This is the classic experimental confirmation that the \(1/N!\) correction (which supplies the \(-\ln N\) and the extra \(+1\)) is quantitatively correct, not merely a formal fix.

Units check. \(R\) in \(\mathrm{J\,K^{-1}mol^{-1}}\) times a dimensionless bracket gives \(\mathrm{J\,K^{-1}mol^{-1}}\).

Example 2 — The paradox made numerical: argon self-mixing. Two rigid boxes, each containing \(n_0=0.5\,\mathrm{mol}\) of argon in volume \(V\) at the same \(T\), are joined by removing the partition to give \(1\,\mathrm{mol}\) in \(2V\). Compute the mixing entropy predicted by the distinguishable formula and by Sackur–Tetrode.

\[ \Delta S = S(2N,2V) - 2\,S(N,V). \]

Entropy change on removing the partition, with \(N=n_0 N_A\) per box. A

\[ \Delta S^{\text{dist}} = 2Nk_B\!\left[\ln\tfrac{2V}{\lambda^3}+\tfrac32\right] - 2Nk_B\!\left[\ln\tfrac{V}{\lambda^3}+\tfrac32\right] = 2Nk_B\ln2. \]

Using the non-extensive Step-4 entropy; the surviving term is the spurious contribution from Step 5. B

\[ 2N = 2n_0 N_A = N_A\ \Rightarrow\ \Delta S^{\text{dist}} = R\ln2 = 8.314\times0.6931 = 5.76\ \mathrm{J\,K^{-1}}. \]

One mole total, so \(2Nk_B=R\). A non-zero "entropy of un-mixing identical gases" — physically absurd. A

\[ \Delta S^{\text{ST}} = 2Nk_B\!\left[\ln\tfrac{2V}{2N\lambda^3}+\tfrac52\right] - 2Nk_B\!\left[\ln\tfrac{V}{N\lambda^3}+\tfrac52\right] = 0. \]

With the extensive Sackur–Tetrode entropy the argument \(V/N\) is unchanged under the doubling, so every term cancels. B

\[ \Delta S^{\text{dist}} = +5.76\ \mathrm{J\,K^{-1}} \quad(\text{spurious}),\qquad \Delta S^{\text{ST}} = 0 \quad(\text{correct}). \]

Reading. Removing a partition between two identical samples is reversible — reinsert it and nothing has changed — so the entropy change must vanish. The distinguishable formula wrongly manufactures \(R\ln2\); the \(1/N!\) correction is exactly what cancels it. By contrast, had the two boxes held argon and neon, the same expansion would give a real \(\Delta S=(n_A+n_B)R\ln2 = R\ln2 = 5.76\,\mathrm{J\,K^{-1}}\), because no \(N!\) links distinct species.

Units check. \(R\ln2\) has units of \(\mathrm{J\,K^{-1}}\); the logarithm's argument \(2\) is dimensionless.

Problems

De Broglie wavelength and degeneracy (level A). Compute \(\lambda\) for helium (\(m=4.0026\,\mathrm{u}\)) at \(T=300\,\mathrm{K}\), and evaluate \(n\lambda^3\) at standard number density \(n=2.69\times10^{25}\,\mathrm{m^{-3}}\). Is the classical limit justified?
Solution
\(m=4.0026\times1.6605\times10^{-27}=6.646\times10^{-27}\,\mathrm{kg}\). Radicand \(2\pi m k_BT = 2\pi(6.646\times10^{-27})(1.381\times10^{-23})(300)=1.730\times10^{-46}\), \(\sqrt{}=1.315\times10^{-23}\,\mathrm{kg\,m\,s^{-1}}\). \(\lambda = 6.626\times10^{-34}/1.315\times10^{-23}=5.04\times10^{-11}\,\mathrm{m}\ (\approx0.50\,\text{Å})\). Then \(\lambda^3=1.28\times10^{-31}\,\mathrm{m^3}\) and \(n\lambda^3=(2.69\times10^{25})(1.28\times10^{-31})=3.4\times10^{-6}\ll1\). The occupation of single-particle states is minuscule, so Maxwell–Boltzmann counting and the \(1/N!\) form are fully justified.
Why the energy is untouched (level B). Show explicitly that dividing \(Z_N\) by \(N!\) leaves \(U\) and \(C_V\) unchanged, and identify which thermodynamic potentials do change.
Solution
\(U=k_BT^2\,(\partial\ln Z_N/\partial T)_V\). Writing \(\ln Z_N=\ln(Z_1^N)-\ln N!\), the term \(\ln N!\) is independent of \(T\), so \(\partial_T\ln N!=0\) and \(U=k_BT^2\,\partial_T(N\ln Z_1)=\tfrac32 Nk_BT\), identical to the distinguishable case. Hence \(C_V=(\partial U/\partial T)_V=\tfrac32 Nk_B\) is also unchanged. The potentials that change are those built directly from \(\ln Z_N\): \(F=-k_BT\ln Z_N\) drops by \(+k_BT\ln N!\), and \(S=-(\partial F/\partial T)_V\) drops by \(k_B\ln N!\approx k_B(N\ln N-N)\). Because \(N!\) has no \(T\), \(V\) or \(U\) dependence, only the "counting" quantities \(F\), \(S\), \(\mu\) feel it.
Standard molar entropy of neon (level B). Compute the Sackur–Tetrode entropy of one mole of neon (\(m=20.18\,\mathrm{u}\)) at \(T=298.15\,\mathrm{K}\), \(P=1\,\mathrm{bar}\), and compare with the tabulated \(146.3\,\mathrm{J\,K^{-1}mol^{-1}}\).
Solution
\(V=RT/P=(8.314)(298.15)/10^5=2.479\times10^{-2}\,\mathrm{m^3}\). \(m=20.18\times1.6605\times10^{-27}=3.351\times10^{-26}\,\mathrm{kg}\). Radicand \(2\pi m k_BT=2\pi(3.351\times10^{-26})(1.381\times10^{-23})(298.15)=8.67\times10^{-46}\), \(\sqrt{}=2.945\times10^{-23}\). \(\lambda=6.626\times10^{-34}/2.945\times10^{-23}=2.250\times10^{-11}\,\mathrm{m}\), \(\lambda^3=1.139\times10^{-32}\,\mathrm{m^3}\). \(V/(N\lambda^3)=2.479\times10^{-2}/[(6.022\times10^{23})(1.139\times10^{-32})]=2.479\times10^{-2}/6.86\times10^{-9}=3.61\times10^{6}\). \(S=R[\ln(3.61\times10^6)+2.5]=8.314(15.10+2.50)=8.314\times17.60=146.3\,\mathrm{J\,K^{-1}mol^{-1}}\). Exact agreement with the calorimetric value.
Genuine mixing entropy (level B). One mole of gas \(A\) in volume \(V\) and one mole of a different gas \(B\) in an equal volume \(V\), at the same \(T\), are combined into \(2V\). Find \(\Delta S\) and contrast with the identical-gas case.
Solution
Each species is unaffected by the presence of the other (ideal gases), so each simply expands isothermally from \(V\) to \(2V\): \(\Delta S_A = n_A R\ln(2V/V)=R\ln2\), and likewise \(\Delta S_B=R\ln2\). Total \(\Delta S=(n_A+n_B)R\ln2 = 2R\ln2 = 2(8.314)(0.6931)=11.5\,\mathrm{J\,K^{-1}}\). This is physically real and irreversible: separating \(A\) from \(B\) again requires work. Had \(A\) and \(B\) been the same gas, the \(1/N!\) makes \(S\) depend only on \(V/N\), the density is unchanged by the merge (\(1\,\mathrm{mol}/V + 1\,\mathrm{mol}/V\to 2\,\mathrm{mol}/2V\)), and \(\Delta S=0\). The discontinuity between "infinitesimally different" and "identical" is the heart of the paradox and is resolved only by the quantum all-or-nothing nature of indistinguishability.
The \(1/N!\) as a quantum limit (level C). Starting from the grand-canonical result for an ideal gas, \(\ln\Xi = \pm\sum_{\mathbf{k}}\ln\!\left(1\pm e^{-\beta(\epsilon_{\mathbf{k}}-\mu)}\right)\) (upper sign fermions, lower bosons), show that in the dilute limit the canonical partition function reduces to \(Z_N=Z_1^N/N!\), and identify the leading correction.
Solution
Let \(z=e^{\beta\mu}\) be the fugacity. Expanding for small \(z\), \(\pm\ln(1\pm ze^{-\beta\epsilon_{\mathbf k}})=ze^{-\beta\epsilon_{\mathbf k}}\mp\tfrac12 z^2 e^{-2\beta\epsilon_{\mathbf k}}+\cdots\). Summing, \(\ln\Xi = zZ_1 \mp \tfrac12 z^2 Z_1(2) + \cdots\), where \(Z_1(2)=\sum_{\mathbf k}e^{-2\beta\epsilon_{\mathbf k}}\). To leading order \(\ln\Xi \approx zZ_1\) and \(\langle N\rangle=z\,\partial_z\ln\Xi \approx zZ_1\), so \(z\approx N/Z_1\). The canonical partition function is recovered from \(\Xi=\sum_N z^N Z_N\); matching the coefficient of \(z^N\) in \(\exp(zZ_1)=\sum_N (zZ_1)^N/N!\) gives \(Z_N=Z_1^N/N!\) — the Gibbs factor emerges as the leading term of the fugacity expansion. The first correction is the \(\mp\tfrac12 z^2 Z_1(2)\) term, the exchange contribution, which produces a statistics-dependent second virial coefficient \(B_2=\mp 2^{-5/2}\lambda^3\) (\(-\) for bosons, an effective attraction; \(+\) for fermions, an effective repulsion). It is of order \(n\lambda^3\) relative to the leading term, confirming that the classical counting is exact only as \(n\lambda^3\to0\).