Quantum Corrections to the Classical Ideal Gas

The ideal gas law $PV=Nk_BT$ is the most famous equation in thermodynamics, and it is wrong — a little — even for a gas with no interactions at all. Real particles are identical, and identical quantum particles are haunted by their own indistinguishability: the many-body wavefunction must be symmetric under exchange (bosons) or antisymmetric (fermions), and that requirement alone, with not a single force between particles, nudges the pressure off the classical value. Bosons end up pushing slightly less hard than $Nk_BT$ ; fermions push slightly more. It is the cleanest demonstration in all of physics that statistics can masquerade as a force. The size of the effect is governed by one number, the degeneracy parameter $n\lambda^3$ , where $\lambda=h/\sqrt{2\pi mk_BT}$ is the thermal de Broglie wavelength — the number of particles crammed into a "quantum blur." When $n\lambda^3\ll1$ the blurs rarely overlap and the correction is tiny; as $n\lambda^3$ approaches unity the correction grows and the two species diverge. The systematic tool is the fugacity expansion of the grand potential, which turns the equation of state into a virial series $P/k_BT=\sum_\ell a_\ell n^\ell$ whose coefficients, for the ideal quantum gas, are exact pure numbers times powers of $\lambda^3$ . The leading correction is the second virial coefficient $B_2=\mp\lambda^3/2^{5/2}$ — negative for bosons (a statistical attraction, because identical bosons bunch), positive for fermions (a statistical repulsion, the exclusion hole). We recall the polylogarithm machinery by name from Equation of State of Ideal Quantum Gases, then extract $B_2$ , show it is a pure exchange effect with zero interaction, and read it as an effective statistical potential of range $\lambda$ . This is the high-temperature bridge from the classical ideal gas into the fully quantum world of degeneracy pressure and condensation.

Key takeaways

Even with no forces, identical quantum particles depart from the classical ideal gas through exchange correlations, controlled by the degeneracy parameter $n\lambda^3$ with $\lambda=h/\sqrt{2\pi mk_BT}$ .
The exact equation of state expands as a virial series; for the ideal quantum gas the cluster integrals are pure numbers, $b_\ell=(\pm1)^{\ell-1}\ell^{-5/2}/\lambda^3$ (upper sign bosons).
The leading correction is $\dfrac{PV}{Nk_BT}=1\mp\dfrac{n\lambda^3}{2^{5/2}}$ : bosons ( $-$ ) an effective attraction lowering the pressure, fermions ( $+$ ) an effective repulsion raising it.
The second virial coefficient is $B_2=\mp\lambda^3/2^{5/2}$ — a purely statistical contribution with no interaction potential ( $u=0$ ).
The same exchange term makes the pair correlation $g^{(2)}(0)=2$ for bosons (bunching) and $0$ for fermions (the exclusion hole), so statistics acts like an attractive well / repulsive bump of range $\lambda$ .
Key idea: quantum indistinguishability alone — with no dynamics — gives a second virial coefficient $B_2=\mp\lambda^3/2^{5/2}$ , an effective statistical attraction for bosons and repulsion for fermions that is the controlled first step from the classical ideal gas toward degeneracy and condensation.

The core insight

For an ideal quantum gas, symmetrization alone produces a second virial coefficient $B_2=\mp\lambda^3/2^{5/2}$ (bosons attract, fermions repel) with no interaction potential, so quantum statistics behaves like an effective force of range $\lambda$ — the leading correction to $PV=Nk_BT$ .

Where this lives today

The exchange virial coefficient is a workhorse of ultracold-atom thermometry. When a dilute Fermi gas of $^6$ Li or $^{40}$ K is cooled toward degeneracy, the first departure from the classical equation of state is exactly the $+n\lambda^3/2^{5/2}$ repulsion computed here, and measuring it — through the density profile of the trapped cloud — calibrates the temperature before the gas becomes fully degenerate. For Bose gases the corresponding $-n\lambda^3/2^{5/2}$ softening is the precursor to condensation, visible as an anomalous compressibility as the cloud approaches $T_c$ . The same pair-correlation physics drives the Hanbury Brown–Twiss experiments that are now routine with atoms: releasing a thermal cloud and measuring density–density correlations reproduces $g^{(2)}(0)=2$ for bosons and $g^{(2)}(0)=0$ for fermions, directly imaging the bunching and antibunching that this article traces back to exchange. What began as a small correction to the ideal-gas law has become a precision tool for reading the quantum statistics of matter.

The physical picture

Picture two identical particles at temperature $T$ , each a wavepacket of width $\lambda$ . If the packets are far apart compared with $\lambda$ , they are effectively distinguishable and the gas is classical. Bring them within $\lambda$ of each other and their symmetrized joint wavefunction changes the odds of finding them at a given separation. For bosons the symmetric combination enhances the amplitude to be at the same place — they bunch. For fermions the antisymmetric combination forbids it — an exclusion hole opens up around each particle. Neither effect involves any force; it is the algebra of the wavefunction. The first figure is the effective potential this produces. If you insisted on describing the exchange correlation as though it came from a potential $u_{\rm eff}(r)$ , you would find an attractive well for bosons (they behave as if drawn together) and a repulsive bump for fermions (as if pushed apart), both of range set by the thermal wavelength $\lambda$ . This is a fictitious potential — there is no real $u(r)$ — but it captures exactly how statistics shifts the pressure.

Figure

Effective statistical potential from exchange — bosons feel an attractive well (bunching), fermions a repulsive bump (exclusion hole), both of range λ; no real force is present

The second figure is the payoff and the punchline: the pair correlation function $g^{(2)}(r)$ , which measures the relative probability of finding a second particle a distance $r$ from a given one. For distinguishable classical particles it is flat at $1$ . For bosons it rises to $g^{(2)}(0)=2$ — twice as likely to find them together, the bunching that Hanbury Brown and Twiss measured in starlight. For fermions it drops to $g^{(2)}(0)=0$ — the exclusion hole, zero probability of coincidence. That factor-of-two split at $r=0$ , with no force present, is the quantum correction.

Figure

Pair correlation g²(r) — bosons rise to g²(0) = 2 (bunching), fermions fall to g²(0) = 0 (exclusion hole), classical flat at 1; the split at r = 0 is the exchange effect with no interaction

To turn this picture into a number we use the fugacity expansion: write the pressure and density as power series in $z=e^{\mu/k_BT}$ , invert to get pressure as a series in density, and read off the coefficients. The result is the virial series, whose first term past the ideal law is exactly the $B_2=\mp\lambda^3/2^{5/2}$ we anticipated from the correlation function.

The core equations

The starting point is the fugacity (cluster) expansion of the grand potential for an ideal quantum gas. Both the pressure and the density are single power series in $z$ , and their coefficients are the cluster integrals $b_\ell$ — for the ideal gas, exact pure numbers.

\frac{P\lambda^3}{k_BT}=\sum_{\ell\ge1} b_\ell\,z^\ell,\qquad n\lambda^3=\sum_{\ell\ge1}\ell\,b_\ell\,z^\ell,\qquad b_\ell=\frac{(\pm1)^{\ell-1}}{\ell^{5/2}} \qquad (1)

where: the upper sign is for bosons, the lower for fermions; $\lambda=h/\sqrt{2\pi mk_BT}$ is the thermal wavelength. Read it as: the $\ell$ -th term counts the contribution of an $\ell$ -particle exchange cluster, weighted by $\ell^{-5/2}$ . The first term ( $b_1=1$ ) is the ideal gas; the second ( $b_2=\pm2^{-5/2}$ ) is the leading quantum correction. Inverting the density series to express the pressure as a series in $n$ gives the virial expansion, whose second coefficient is the object of interest.

\frac{P}{k_BT}=n+B_2(T)\,n^2+B_3(T)\,n^3+\cdots,\qquad B_2(T)=\mp\frac{\lambda^3}{2^{5/2}} \qquad (2)

where: $B_2$ is the second virial coefficient, carrying the units of volume. It is negative for bosons (statistical attraction, pressure below ideal) and positive for fermions (statistical repulsion, pressure above ideal). Crucially, $B_2$ relates to the cluster integral $b_2$ by $B_2=-b_2\lambda^3$ — the inversion flips the sign, a classic source of confusion. Dividing through, the equation of state takes the compact per-particle form that shows the correction directly.

\frac{PV}{Nk_BT}=1+B_2\,n=1\mp\frac{n\lambda^3}{2^{5/2}}+\cdots \qquad (3)

where: the correction $n\lambda^3/2^{5/2}$ has the exact coefficient $2^{-5/2}=0.1768$ and grows as the gas cools, since $\lambda^3\propto T^{-3/2}$ . A boson gas ( $-$ ) is softer than ideal; a fermion gas ( $+$ ) is stiffer. The whole effect vanishes as $T\to\infty$ , recovering the classical law. The microscopic origin is the exchange term in the two-particle density matrix, which fixes the pair correlation and hence $B_2$ by an elementary Gaussian integral.

g^{(2)}(r)=1\pm e^{-2\pi r^2/\lambda^2},\qquad B_2=-\tfrac12\!\int\!\big[g^{(2)}(r)-1\big]\,d^3r=\mp\frac{\lambda^3}{2^{5/2}} \qquad (4)

where: the $+$ sign gives boson bunching ( $g^{(2)}(0)=2$ ), the $-$ sign the fermion exclusion hole ( $g^{(2)}(0)=0$ ). The Gaussian range is the thermal wavelength $\lambda$ . Integrating the exchange bump reproduces $B_2$ exactly — proof that the virial coefficient is a pure statistics effect, built from free-particle wavefunctions with no interaction $u(r)$ anywhere.

Derivations

Derivation AThe second virial coefficient from the fugacity expansion

The goal is to derive $B_2=\mp\lambda^3/2^{5/2}$ from the cluster expansion of the ideal quantum gas, and to be scrupulous about the sign, which changes once between the cluster integral $b_2$ and the virial coefficient $B_2$ . The route is standard: write the pressure and density as series in the fugacity $z$ with the known cluster integrals $b_\ell=(\pm1)^{\ell-1}\ell^{-5/2}$ , invert the density series to get $z(n)$ to second order, and substitute into the pressure series. What to watch is the inversion step — it contributes a term of the same order as the direct $b_2$ term, and the two combine to give the final coefficient, with a net sign opposite to $b_2$ .

Step 1. Write the two series to second order in $z$ : $P\lambda^3/k_BT=z\pm\frac{z^2}{2^{5/2}}$ and $n\lambda^3=z\pm\frac{2z^2}{2^{5/2}}=z\pm\frac{z^2}{2^{3/2}}$ (the factor $\ell=2$ multiplies $b_2$ in the density). Why: from equation (1), $P$ carries $b_\ell$ and $n$ carries $\ell b_\ell$ ; the upper sign is bosons throughout.

Step 2. Invert the density series. Set $z=n\lambda^3+c\,(n\lambda^3)^2$ and match: substituting into Step 1 and cancelling the linear term forces $c=\mp2^{-3/2}$ , so $z=n\lambda^3\mp\frac{(n\lambda^3)^2}{2^{3/2}}$ . Why: the quadratic term of $n\lambda^3(z)$ must be undone; moving it across the equation reverses its sign, which is the first place the sign flips.

Step 3. Substitute into the pressure. $\frac{P\lambda^3}{k_BT}=z\pm\frac{z^2}{2^{5/2}}=\big(n\lambda^3\mp\frac{(n\lambda^3)^2}{2^{3/2}}\big)\pm\frac{(n\lambda^3)^2}{2^{5/2}}$ . Collect the quadratic terms: $\mp\frac{1}{2^{3/2}}\pm\frac{1}{2^{5/2}}=\pm\big(\frac{1}{2^{5/2}}-\frac{1}{2^{3/2}}\big)=\mp\frac{1}{2^{5/2}}$ (since $\frac{1}{2^{3/2}}=\frac{2}{2^{5/2}}$ ). Why: the inversion term ( $\mp2^{-3/2}$ ) dominates the direct term ( $\pm2^{-5/2}$ ) and flips the overall sign; the arithmetic $2^{-5/2}-2\cdot2^{-5/2}=-2^{-5/2}$ leaves exactly $\mp2^{-5/2}$ .

Step 4. Read off $B_2$ . Since $\frac{P}{k_BT}=n+B_2n^2$ means $\frac{P\lambda^3}{k_BT}=n\lambda^3+B_2\frac{n^2\lambda^3}{1}$ , matching the quadratic coefficient gives $B_2=\mp\frac{\lambda^3}{2^{5/2}}$ : negative for bosons (attraction), positive for fermions (repulsion). Why: the sign that survives the inversion is opposite to the cluster integral $b_2=\pm2^{-5/2}$ , i.e. $B_2=-b_2\lambda^3$ ; bosons ( $b_2>0$ ) give an attractive $B_2<0$ . *Verified: the full symbolic inversion, the coefficient $2^{-5/2}=0.17678$ , the signs ( $-$ boson, $+$ fermion), the cluster integrals $b_\ell$ , and the relation $B_2=-b_2\lambda^3$ are checked in verify_quantum-corrections-classical-ideal-gas.py against Pathria §7.1 and Huang §8.*

Go deeper · Derivation B · 1 minThe exchange pair correlation and B₂ as a pure statistics effect

Derivation BThe exchange pair correlation and B₂ as a pure statistics effect

The goal is to derive the pair correlation $g^{(2)}(r)=1\pm e^{-2\pi r^2/\lambda^2}$ from the symmetry of the free two-particle wavefunction, and to show that integrating its exchange part reproduces exactly the $B_2=\mp\lambda^3/2^{5/2}$ of the previous card — with no interaction potential anywhere. The plan is to build the symmetrized/antisymmetrized two-particle density matrix from plane waves, extract the diagonal probability, and integrate. The point to hold onto is that every ingredient is a free-particle wavefunction of width $\lambda$ ; the only physics is the $\pm$ from exchange symmetry, so the resulting "force" is entirely statistical.

Step 1. Build the symmetrized density. For two free particles the (anti)symmetric two-body probability, thermally averaged over momenta, is $g^{(2)}(r)=1\pm|\langle e^{i\mathbf p\cdot\mathbf r/\hbar}\rangle_T|^2$ , where the thermal average of the plane-wave phase over a Maxwell distribution is a Gaussian. Why: symmetrizing (bosons, $+$ ) or antisymmetrizing (fermions, $-$ ) the product of single-particle states adds an exchange cross-term whose magnitude is the overlap of thermal wavepackets.

Step 2. Do the Gaussian average. $\langle e^{i\mathbf p\cdot\mathbf r/\hbar}\rangle_T=e^{-\pi r^2/\lambda^2}$ , so $g^{(2)}(r)=1\pm e^{-2\pi r^2/\lambda^2}$ . At $r=0$ , $g^{(2)}(0)=1\pm1$ : $2$ for bosons (bunching), $0$ for fermions (exclusion hole). Why: the Maxwell momentum distribution has width $\sqrt{mk_BT}$ , and its Fourier transform is a Gaussian in $r$ of width $\lambda$ ; squaring doubles the exponent to $2\pi r^2/\lambda^2$ .

Step 3. Relate to $B_2$ . The second virial coefficient is $B_2=-\tfrac12\int[g^{(2)}(r)-1]\,d^3r=\mp\tfrac12\int e^{-2\pi r^2/\lambda^2}\,d^3r$ . Why: $B_2$ measures the excess (or deficit) probability of pairs; the standard relation $B_2=-\tfrac12\int[g^{(2)}-1]d^3r$ turns the correlation directly into the virial coefficient.

Step 4. Evaluate the integral. $\int e^{-2\pi r^2/\lambda^2}d^3r=\big(\int e^{-2\pi x^2/\lambda^2}dx\big)^3=(\lambda/\sqrt2)^3=\lambda^3/2^{3/2}$ , so $B_2=\mp\tfrac12\cdot\lambda^3/2^{3/2}=\mp\lambda^3/2^{5/2}$ . Why: each Cartesian Gaussian gives $\lambda/\sqrt2$ ; the product of three and the factor $\tfrac12$ reproduce the exact $\mp2^{-5/2}\lambda^3$ — with $u(r)=0$ throughout, confirming the effect is pure statistics. *Verified: the Gaussian integral $\int e^{-2\pi r^2/\lambda^2}d^3r=\lambda^3/2^{3/2}$ , the resulting $B_2=\mp\lambda^3/2^{5/2}$ , and $g^{(2)}(0)=2$ (boson) / $0$ (fermion) are checked in verify_quantum-corrections-classical-ideal-gas.py; matches Pathria §5.5 and Feynman Statistical Mechanics ch. 1.*

Limiting cases

The correction is controlled by $n\lambda^3$ and the statistics; it vanishes in the classical corner and connects onto the fully quantum regimes.

Case	Result	Meaning
Classical, $n\lambda^3\ll1$	$PV/Nk_BT=1$	Wavepackets don't overlap; exchange negligible, ideal law recovered.
Bosons, leading order	$PV/Nk_BT=1-n\lambda^3/2^{5/2}$	Statistical attraction (bunching, $g^{(2)}(0)=2$ ); pressure below ideal.
Fermions, leading order	$PV/Nk_BT=1+n\lambda^3/2^{5/2}$	Statistical repulsion (exclusion hole, $g^{(2)}(0)=0$ ); pressure above ideal.
Full quantum, $n\lambda^3\gtrsim1$	series diverges	Push to all orders: fermions → degeneracy pressure; bosons → condensation.

Worked examples

1 · The quantum correction to helium gas at STP (confidence builder)

We compute how much the pressure of helium-4 gas at standard temperature and pressure differs from the ideal-gas value due to Bose statistics alone.

Step 1. The thermal wavelength at $T=273.15$ K, with $m=6.6465\times10^{-27}$ kg: $\lambda=h/\sqrt{2\pi mk_BT}=6.626\times10^{-34}/\sqrt{2\pi(6.6465\times10^{-27})(1.381\times10^{-23})(273.15)}$ . The radicand is $1.576\times10^{-46}$ , so $\lambda=6.626\times10^{-34}/1.255\times10^{-23}=\mathbf{5.28\times10^{-11}\ m}$ .

Step 2. The number density at STP is $n=P/k_BT=1.013\times10^{5}/(1.381\times10^{-23}\times273.15)=2.69\times10^{25}\ \mathrm{m^{-3}}$ .

Step 3. The degeneracy parameter is $n\lambda^3=2.69\times10^{25}\times(5.28\times10^{-11})^3=2.69\times10^{25}\times1.47\times10^{-31}=\mathbf{3.96\times10^{-6}}$ .

Step 4. Helium-4 is a boson, so $PV/Nk_BT=1-n\lambda^3/2^{5/2}=1-3.96\times10^{-6}/5.657=1-\mathbf{7.0\times10^{-7}}$ . The statistical attraction lowers the pressure by 0.7 parts per million — swamped by the real van der Waals attraction at this temperature, but a clean illustration that even a hypothetically interaction-free helium gas would not obey $PV=Nk_BT$ exactly.

2 · The exchange correction grows as the gas cools (exam level)

We show how the quantum correction for a fixed-density gas scales with temperature, and estimate the temperature at which it reaches 1% for a dense fermion gas.

Step 1. Since $\lambda\propto T^{-1/2}$ , the correction $n\lambda^3/2^{5/2}\propto T^{-3/2}$ at fixed $n$ . Cooling the gas by a factor of $4$ raises $\lambda$ by $2$ and the correction by $2^3=8$ .

Step 2. Take a dense gas of $^3$ He (a fermion) at liquid density $n=1.6\times10^{28}\ \mathrm{m^{-3}}$ , $m=5.008\times10^{-27}$ kg. Set the correction to $0.01$ : $n\lambda^3=0.01\times2^{5/2}=0.0566$ , so $\lambda^3=0.0566/1.6\times10^{28}=3.54\times10^{-30}\ \mathrm{m^3}$ , giving $\lambda=1.52\times10^{-10}$ m.

Step 3. Invert $\lambda=h/\sqrt{2\pi mk_BT}$ for $T$ : $T=h^2/(2\pi mk_B\lambda^2)=(6.626\times10^{-34})^2/[2\pi(5.008\times10^{-27})(1.381\times10^{-23})(1.52\times10^{-10})^2]$ . The denominator is $2\pi\times5.008\times10^{-27}\times1.381\times10^{-23}\times2.31\times10^{-20}=1.004\times10^{-68}$ , so $T=4.39\times10^{-67}/1.004\times10^{-68}=\mathbf{44\ K}$ .

Step 4. So even at liquid-helium densities the exchange correction to $^3$ He reaches only 1% by about $44$ K, and grows rapidly below it — becoming order unity near $1$ K, exactly where $^3$ He's genuinely quantum behaviour (Fermi-liquid physics) takes over. The correction hits 1% at ≈44 K, confirming that quantum statistics is a low-temperature, high-density phenomenon.

Origins · 1 minexchange interaction and how this was discovered

exchange interaction

concept: W. Heisenberg & P. Dirac 1926 · The exchange interaction is the fictitious "force" born entirely of wavefunction symmetry, with no term in the Hamiltonian. Heisenberg and Dirac showed in 1926 that requiring symmetry or antisymmetry under particle exchange produces energy differences and correlations that mimic a real interaction — the same idea that underlies ferromagnetism, chemical bonding, and, here, the statistical attraction of bosons and repulsion of fermions. In the dilute ideal gas the exchange interaction shows up as the second virial coefficient

B_2=\mp\lambda^3/2^{5/2}

: a measurable shift in the equation of state produced by pure indistinguishability.

History

1924

— Bose and Einstein establish boson statistics; the enhanced pair probability is implicit.

1926

— Fermi and Dirac formulate the exclusion-based statistics for fermions.

1926

— Heisenberg and Dirac identify the exchange interaction as a consequence of wavefunction symmetry.

1927

— Uhlenbeck and Beth compute the quantum virial coefficients for ideal and interacting gases.

1938

— the ideal Bose gas virial series is understood as the approach to condensation.

1956

— Hanbury Brown and Twiss measure photon bunching, the direct observation of boson

g^{(2)}(0)=2

Going further

Standard treatments: Pathria & Beale, Statistical Mechanics, 3rd ed., §7.1 (quantum virial expansion) and §5.5 (exchange effects and the pair correlation); Huang, Statistical Mechanics, §8; Landau & Lifshitz, Statistical Physics Part 1, §56; Feynman, Statistical Mechanics, ch. 1. The Uhlenbeck–Beth cluster expansion is the classic reference for the interacting case. Provenance: adapted and upgraded from the AllQFT base article "Quantum Corrections to the Classical Ideal Gas" (ref STAT.IQGAS.QCORR.1240; Pathria & Beale, 3rd ed., §7.1 and §8.1), with the exchange virial $B_2=\mp\lambda^3/2^{5/2}$ , its sign, the cluster integrals $b_\ell$ , and the relation $B_2=-b_2\lambda^3$ independently python-verified (verify_quantum-corrections-classical-ideal-gas.py, 24 checks). Prerequisites: Quantum Statistics of Ideal Gases: Occupation Numbers · Density of States for Free Particles · Chemical Potential of a Fermi Gas vs Temperature. Leads to: Degenerate Fermi Gas at Zero Temperature · Bose-Einstein Condensation in the Ideal Gas · Degeneracy Pressure of Fermions.

If there are no forces between the particles, how can there be a virial coefficient at all?

The virial coefficient measures deviations from $PV=Nk_BT$ , and those deviations can come from two sources: a real interaction potential $u(r)$ , or quantum exchange. For an ideal quantum gas $u(r)=0$ , but the particles are still identical, and the symmetry of the wavefunction correlates their positions — bosons bunch, fermions avoid each other. That correlation is exactly what a virial coefficient captures. The derivation uses only free-particle wavefunctions of width $\lambda$ ; the resulting $B_2=\mp\lambda^3/2^{5/2}$ is a genuine, measurable shift with no force behind it. Statistics does the job a potential would otherwise do.

Why is the boson correction negative and the fermion correction positive?

Bosons prefer to occupy the same states, so identical bosons are found close together more often than distinguishable particles — an effective attraction that lets the gas contract slightly, lowering the pressure below $Nk_BT$ (hence $B_2<0$ ). Fermions obey the exclusion principle, so an exclusion hole opens around each one, keeping them apart — an effective repulsion that props the pressure up above $Nk_BT$ (hence $B_2>0$ ). The signs come straight from the pair correlation: $g^{(2)}(0)=2$ (bunched, attractive) for bosons versus $0$ (anti-bunched, repulsive) for fermions.

Why does the cluster integral b₂ have the opposite sign to the virial coefficient B₂?

Because they live in different expansions. The cluster integral $b_2$ is the coefficient of $z^2$ in the fugacity series for the pressure; the virial coefficient $B_2$ is the coefficient of $n^2$ in the density series. Converting one series into the other requires inverting $n(z)$ , and that inversion contributes a term of the same order that flips the sign: $B_2=-b_2\lambda^3$ . For fermions $b_2=-2^{-5/2}$ (negative) but $B_2=+\lambda^3/2^{5/2}$ (positive). The safe habit is to trust the physical statement — fermions repel, so their $B_2$ is positive — and let that fix the sign of $b_2$ , not the other way round.

How big is this correction for an ordinary gas like air at room temperature?

Utterly negligible. For nitrogen or oxygen at room temperature $\lambda\approx0.2$ Å and $n\sim2.5\times10^{25}\ \mathrm{m^{-3}}$ , giving $n\lambda^3\sim10^{-7}$ and a quantum correction of order $10^{-8}$ — a hundred-millionth. The real second virial coefficient of air is thousands of times larger and comes entirely from the intermolecular van der Waals force, not from statistics. The quantum-statistical correction only becomes important for light particles (large $\lambda$ ) at low temperature and high density: helium below a few kelvin, electrons in metals, or ultracold atomic gases.

Is the "effective statistical potential" a real potential I could put in a Hamiltonian?

Not literally, but it is a useful fiction. There is no term $u_{\rm eff}(r)$ in the true Hamiltonian — the particles are genuinely free. What is real is the correlation $g^{(2)}(r)$ produced by symmetrization, and if you wanted to reproduce that same correlation with distinguishable classical particles, you would need an attractive well (bosons) or repulsive bump (fermions) of range $\lambda$ . People sometimes quote an "exchange potential" of order $k_BT$ at $r\sim\lambda$ for exactly this reason. It is a bookkeeping device: it gives the right $B_2$ , but it is temperature-dependent and vanishes classically, which no real potential does.

At what point does this leading correction stop being enough?

The virial series is a small- $n\lambda^3$ expansion, so the single $B_2$ term is trustworthy only while $n\lambda^3\ll1$ . Once $n\lambda^3$ approaches unity — a light gas cooled toward degeneracy — higher virial coefficients $B_3,B_4,\dots$ grow and the series eventually diverges. At that point you must abandon the expansion and use the full quantum equation of state: the Sommerfeld ( $z\to\infty$ ) limit for fermions, which gives degeneracy pressure, or the $z\to1$ saturation for bosons, which gives condensation. The exchange correction here is the first step of a road that leads all the way into those fully quantum regimes.