Van der Waals from the Virial Expansion

The van der Waals equation was born as an inspired guess: shrink the volume by $Nb$ for the space molecules occupy, and drop the pressure by $an^2$ for their mutual attraction. This article shows the guess is exactly right — to first order in density — and computes what $a$ and $b$ actually are in terms of the forces between molecules. That is the microscopic derivation the equation (aq-sm-van-der-waals-equation) takes for granted. The engine is the virial expansion (aq-sm-virial-expansion), whose leading correction is the second virial coefficient $B_2(T)=-\tfrac12\int f(\mathbf r)\,\mathrm{d}^3r$ , a single integral over the Mayer function $f=e^{-\beta u}-1$ (aq-sm-mayer-f-function). Feed it a realistic potential — an impenetrable hard core plus a shallow attractive tail — and the integral splits into precisely the two pieces van der Waals intended. The hard core contributes a temperature-independent excluded volume $b=\tfrac23\pi N_A\sigma^3$ ; the attractive tail, linearized for weak binding, contributes $-a/(k_BT)$ with $a$ an integral over the tail. The result is $B_2(T)=b-a/(k_BT)$ . The final move closes the loop: expand the van der Waals equation itself to order $n^2$ and read off its second virial coefficient. It is $b-a/(k_BT)$ — identical to the microscopic one. So matching the leading density correction identifies the phenomenological constants with definite integrals over the pair potential, and the folk recipe " $V\to V-Nb$ , $P\to P-an^2$ " is revealed as the exact low-density truncation dressed up as a closed equation. One must be honest about the status. Matching $B_2$ fixes $a$ and $b$ exactly, but the van der Waals equation then makes definite — and generally wrong — predictions for the higher coefficients $B_3,B_4,\dots$ . For hard spheres it gives $B_3=b^2$ , whereas the exact value is $\tfrac58 b^2$ , an overestimate by a factor $8/5$ . The van der Waals equation is therefore the simplest closed-form interpolation consistent with the exact low-density limit, not a systematic expansion — a distinction worth keeping sharp.

Key takeaways

The van der Waals constants $a$ and $b$ are not free parameters: they emerge from the second virial coefficient $B_2(T)$ of a gas of hard cores plus a weak long-range attraction.
Splitting $B_2(T)=-\tfrac12\int f\,\mathrm{d}^3r$ at the core radius gives a $T$ -independent excluded-volume piece $b$ and an attractive piece $-a/(k_BT)$ : exactly $B_2(T)=b-a/(k_BT)$ .
The hard core gives $b=\tfrac23\pi N_A\sigma^3$ — four times the molecular volume — while the attractive tail integrates to $a=2\pi N_A^2\int_\sigma^\infty|u_{\rm attr}(r)|\,r^2\,\mathrm{d}r>0$ .
Expanding the van der Waals equation to order $n^2$ reproduces this same $B_2$ , so the derivation is exact only to first order in density; the full equation is an approximate resummation of the excluded-volume effect.
The construction is the cleanest bridge from the microscopic Mayer cluster expansion to a macroscopic equation of state, and it underlies the law of corresponding states.
Key idea: compute $B_2(T)=-\tfrac12\int(e^{-\beta u}-1)\,\mathrm{d}^3r$ for a hard-core-plus-attraction potential; the repulsion gives a $T$ -independent excluded volume $b$ and the attraction gives $-a/(k_BT)$ , so $B_2=b-a/(k_BT)$ — precisely the second virial coefficient of the van der Waals equation.

Where this lives today

The split that produces $a$ and $b$ — a hard repulsive reference carrying the fluid's structure, plus a weak attractive tail treated as a perturbation — is the founding idea of modern liquid-state theory. Barker–Henderson and Weeks–Chandler–Andersen perturbation theories make this rigorous: the reference hard-sphere fluid is solved accurately (its structure factor is known nearly exactly), and the attraction is added as a controlled correction. These theories power the equations of state used to model everything from liquid metals to colloidal suspensions and protein solutions, where a short-ranged attraction on top of excluded volume governs phase behaviour. The derivation also anchors primary metrology. Second (and third) virial coefficients computed from ab initio pair potentials for helium — the same $B_2=-\tfrac12\int f$ evaluated for a quantum-accurate potential — underpin acoustic and dielectric-constant gas thermometry, part of how the kelvin is now realized and how the Boltzmann constant was pinned down. In every case the through-line is the one this article establishes: a macroscopic constant of an equation of state is, at bottom, an integral over the forces between two molecules.

The physical picture

Watch two molecules approach. At short range they cannot interpenetrate — a hard wall — so configurations where their centres are closer than the diameter $\sigma$ are simply forbidden. That removal of configuration space is the excluded volume: it pushes the pressure up, and it is temperature-independent because a wall is a wall at any temperature. At longer range the molecules feel a gentle attraction, so configurations where they sit near each other are weighted more (the Boltzmann factor exceeds one). That extra weight lowers the energy and reduces the pressure — the $-a/(k_BT)$ term. The single second virial coefficient carries both effects, one positive and constant, one negative and temperature-dependent. The first figure is the potential and its split. A realistic pair potential is a steep repulsive core at $r<\sigma$ and a shallow attractive well beyond. The Mayer function $f=e^{-\beta u}-1$ turns this into something bounded and short-ranged: $f=-1$ inside the core (the Boltzmann factor vanishes), and $f\approx\beta|u_{\rm attr}|>0$ in the attractive region at not-too-low temperature. The integral $B_2=-\tfrac12\int f\,\mathrm{d}^3r$ therefore receives a positive contribution $b$ from the core (where $f=-1$ ) and a negative contribution $-a/(k_BT)$ from the well (where $f>0$ ).

Figure

The hard-core-plus-attraction potential u(r) (steep wall at σ, shallow well beyond) and its split into the two van der Waals pieces: the core gives the excluded volume b, the attractive tail gives a

The second figure is the payoff and the punchline. Because attraction carries the $1/T$ , it dominates at low temperature and fades at high temperature, so $B_2(T)$ rises from strongly negative values (attraction-dominated) at low $T$ , crosses zero at the Boyle temperature $T_B=a/(bk_B)$ , and saturates toward the hard-core value $b$ at high $T$ . The two van der Waals constants are literally the intercept and the slope of this curve: $b$ is the high- $T$ plateau, and $a$ sets where the zero-crossing sits.

Figure

Punchline: B₂(T) = b − a/(k_BT) rises from strongly negative at low T (attraction), crosses zero at the Boyle temperature T_B = a/(bk_B), and saturates at the excluded volume b at high T — the two constants are the plateau and the crossing

The van der Waals equation can even be read off without any integral, by two heuristic substitutions in the ideal gas law: replace $V$ by the free volume $V-Nb$ , and reduce the pressure by a mean-field attraction $\propto n^2$ . The virial derivation shows this folk recipe is exact to first order in density and tells you precisely what $a$ and $b$ mean — the whole point of this article.

The core equations

The starting point is the virial expansion in density, whose second coefficient is a single integral over the pair potential.

\frac{P}{k_BT}=n+B_2(T)\,n^2+B_3(T)\,n^3+\cdots,\qquad B_2(T)=-\frac12\int f(\mathbf r)\,\mathrm{d}^3r \qquad (1)

where: $n=N/V$ — number density; $B_\ell(T)$ — the $\ell$ -th virial coefficient ( $B_1\equiv1$ ); $f(\mathbf r)=e^{-\beta u(r)}-1$ — the Mayer function (owned by aq-sm-mayer-f-function); $\beta=1/(k_BT)$ . The identity $B_2=-b_2=-\tfrac12\int f$ is delivered by the cluster expansion; the inversion of the fugacity series that produces it is owned by aq-sm-virial-expansion. For a spherically symmetric potential the pair integral collapses to a one-dimensional radial integral.

B_2(T)=-2\pi\int_0^\infty\!\left(e^{-\beta u(r)}-1\right)r^2\,\mathrm{d}r \qquad (2)

where: the factor $2\pi$ is $\tfrac12\times4\pi$ (half from the pair symmetry, $4\pi r^2$ from the angular integration). This single integral is the whole microscopic content of the second virial coefficient; everything below is evaluating it for the van der Waals model potential. The model is the simplest realistic one: an impenetrable core and a weak attractive tail.

u(r)=\begin{cases}+\infty, & r<\sigma\\[3pt] u_{\rm attr}(r)<0, & r>\sigma,\ |u_{\rm attr}|\ll k_BT\end{cases} \qquad (3)

where: $\sigma$ — hard-core diameter (closest centre-to-centre approach); $u_{\rm attr}$ — the shallow attractive tail. The weakness condition $|u_{\rm attr}|\ll k_BT$ is what lets us linearize the Boltzmann factor in the tail region, and it is the assumption that makes the attractive contribution scale as $1/T$ . Evaluating (2) on this model splits it cleanly into the two van der Waals constants.

B_2(T)=\underbrace{\frac{2}{3}\pi N_A\sigma^3}_{b}-\underbrace{\frac{2\pi N_A^2}{k_BT}\int_\sigma^\infty|u_{\rm attr}(r)|\,r^2\,\mathrm{d}r}_{a/(k_BT)}=b-\frac{a}{k_BT} \qquad (4)

where: $b=\tfrac23\pi N_A\sigma^3$ — molar excluded volume, four times the molecular volume; $a=2\pi N_A^2\int_\sigma^\infty|u_{\rm attr}|\,r^2\,\mathrm{d}r>0$ — molar cohesion, an integral over the attractive tail. (Per molecule, drop the $N_A$ factors.) This is the second virial coefficient predicted by the van der Waals equation, closing the identification.

Derivation AThe second virial coefficient of the hard-core-plus-tail gas gives b and a

The goal is to evaluate $B_2(T)=-2\pi\int_0^\infty(e^{-\beta u}-1)r^2\,\mathrm{d}r$ for the model potential (3) and show it equals $b-a/(k_BT)$ , with $b$ and $a$ definite integrals over the potential. The strategy is to split the radial integral at the core diameter $\sigma$ : inside the core the Boltzmann factor vanishes so $f=-1$ exactly, giving a temperature-independent volume; outside, the attraction is weak so we linearize the Boltzmann factor, giving a term proportional to $1/T$ . The subtle point to watch is the sign — the attractive tail has $u<0$ , so $e^{-\beta u}-1>0$ there, and the minus sign in front of $B_2$ turns it into the negative contribution $-a/(k_BT)$ .

Step 1. Split at the core. Write $B_2=-2\pi\left[\int_0^\sigma(e^{-\beta u}-1)r^2\,\mathrm{d}r+\int_\sigma^\infty(e^{-\beta u_{\rm attr}}-1)r^2\,\mathrm{d}r\right]$ . Why: the potential has two qualitatively different regions — an infinite wall and a weak well — so the integral naturally separates at $\sigma$ , and each region contributes one van der Waals constant.

Step 2. Core region. For $r<\sigma$ , $u=+\infty$ , so $e^{-\beta u}=0$ and $f=-1$ . The contribution is $-2\pi\int_0^\sigma(-1)r^2\,\mathrm{d}r=+\tfrac23\pi\sigma^3\equiv\tilde b$ (per molecule). Why: inside a hard wall no configuration is allowed, so $f$ takes its most negative value $-1$ everywhere; the minus sign out front flips it positive — excluded volume raises the pressure. Note there is no temperature here.

Step 3. Tail region. For $r>\sigma$ with $|u_{\rm attr}|\ll k_BT$ , expand $e^{-\beta u_{\rm attr}}-1\approx-\beta u_{\rm attr}=\beta|u_{\rm attr}|$ (since $u_{\rm attr}<0$ ). The contribution is $-2\pi\beta\int_\sigma^\infty|u_{\rm attr}|\,r^2\,\mathrm{d}r=-\tilde a/(k_BT)$ with $\tilde a=2\pi\int_\sigma^\infty|u_{\rm attr}|\,r^2\,\mathrm{d}r$ . Why: weak attraction lets the exponential be linearized, which is exactly what produces the $1/T$ dependence; the result is negative because attraction lowers the pressure.

Step 4. Assemble. Adding the two regions, $B_2(T)=\tilde b-\tilde a/(k_BT)$ per molecule; multiply $\tilde b$ by $N_A$ and $\tilde a$ by $N_A^2$ for molar constants. Why: the two regions contribute the two van der Waals constants; the temperature-independent $b$ from the wall and the $1/T$ attraction from the tail are precisely the ingredients van der Waals postulated. *Verified: the core integral $-2\pi\int_0^\sigma(-1)r^2\,\mathrm{d}r=\tfrac23\pi\sigma^3$ , the tail integral $a=\tfrac23\pi\varepsilon\sigma^3$ for a representative $-\varepsilon(\sigma/r)^6$ tail, and the assembled $B_2=b-a/(k_BT)$ are checked symbolically in verify_van-der-waals-derivation.py, matching Pathria & Beale §10.1 and Reif §10.4.*

Go deeper · Derivation B · 1 minExcluded volume: why b is four times the molecular volume

Derivation BExcluded volume: why b is four times the molecular volume

The goal is the famous geometric factor: the excluded volume per molecule $b=\tfrac23\pi\sigma^3$ is four times the volume of one molecule, not equal to it. The strategy is to compute the volume from which one sphere's centre is excluded by another, recognize it as eight molecular volumes, and then halve it because the exclusion is shared between the two partners of a pair. The point to watch is the two distinct factors of two — one from diameter-versus-radius in the forbidden ball, one from sharing the exclusion between the pair — which combine to the net factor of four.

Step 1. The forbidden ball. Two hard spheres of diameter $\sigma$ (radius $\sigma/2$ ) cannot have their centres closer than $\sigma$ . So the centre of sphere 2 is excluded from a ball of radius $\sigma$ about sphere 1, of volume $v_{\rm excl}^{\rm pair}=\tfrac43\pi\sigma^3$ . Why: the closest approach of centres is one full diameter, so the forbidden region has radius $\sigma$ , not $\sigma/2$ — this is the first factor of two.

Step 2. One molecular volume. A single molecule has radius $\sigma/2$ , so $v_{\rm mol}=\tfrac43\pi(\sigma/2)^3=\tfrac16\pi\sigma^3$ . Why: this is the actual physical volume a molecule occupies, the naive quantity one might guess for $b$ .

Step 3. The ratio is eight. $\dfrac{v_{\rm excl}^{\rm pair}}{v_{\rm mol}}=\dfrac{\tfrac43\pi\sigma^3}{\tfrac16\pi\sigma^3}=8$ . Why: the forbidden ball has twice the radius of a molecule, hence $2^3=8$ times the volume — a pair excludes eight molecular volumes.

Step 4. Share between the pair. The exclusion belongs jointly to the two members of the pair, so per molecule $\tilde b=\tfrac12 v_{\rm excl}^{\rm pair}=\tfrac23\pi\sigma^3=4\,v_{\rm mol}$ . Why: assigning half the pair's excluded volume to each partner turns the eight into four — the celebrated van der Waals factor, and it agrees exactly with the $B_2$ core integral of the previous card. *Verified: the pair-excluded-volume-to-molecular-volume ratio $=8$ and the per-molecule $\tilde b=\tfrac12 v_{\rm excl}=\tfrac23\pi\sigma^3=4v_{\rm mol}$ are checked symbolically in verify_van-der-waals-derivation.py, matching Kardar Ch. 5 and Reif §10.3.*

Limiting cases

The behaviour of $B_2(T)=b-a/(k_BT)$ across temperature regimes maps directly onto real-gas non-ideality.

Case	Result	Meaning
High temperature	$B_2\to b=\tfrac23\pi N_A\sigma^3>0$	Attraction negligible; only excluded volume survives, gas harder to compress than ideal.
Boyle temperature	$k_BT_B=a/b$ , so $B_2(T_B)=0$	Excluded volume and attraction cancel; the gas mimics ideal to order $n^2$ .
Low temperature	$B_2\to-a/(k_BT)<0$	Attraction dominates; pressure falls below ideal, foreshadowing condensation.
Dilute limit	$\dfrac{P}{k_BT}=n+B_2(T)n^2+O(n^3)$	Where the matching of $a,b$ is exact — vdW agrees with experiment here by construction.
Higher coefficients (hard spheres)	$B_3^{\rm vdW}=b^2$ vs exact $\tfrac58 b^2$	vdW overestimates $B_3$ by $8/5$ ; the resummation is exact only through $B_2$ .

Common trapThe excluded volume equals the volume of the molecules

No — it is four times larger, and forgetting this is the classic error. The excluded volume $b=\tfrac23\pi N_A\sigma^3$ is four times the total molecular volume $N_A\cdot\tfrac43\pi(\sigma/2)^3=\tfrac16\pi N_A\sigma^3$ . Two factors conspire: a pair of spheres of diameter $\sigma$ excludes a ball of radius $\sigma$ between their centres (eight molecular volumes, from the diameter-versus-radius cube), and that exclusion is shared between the two partners (halving it to four). Dropping either factor — the pairwise $2^3$ or the per-particle $\tfrac12$ — gives the wrong $b$ . A related trap: treating the derived $B_2=b-a/(k_BT)$ as if the van der Waals equation were exact to all orders. It is exact only through $B_2$ ; its $B_3=b^2$ already overshoots the true hard-sphere $\tfrac58 b^2$ .

Worked examples

1 · Excluded volume of argon and its molecular size (confidence builder)

We turn a measured van der Waals constant into a molecular diameter, computing a real number from real data.

Step 1. Argon's critical data ( $T_c=150.7$ K, $P_c=4.86\times10^6$ Pa) invert to the molar $b=RT_c/(8P_c)=3.22\times10^{-5}$ m³/mol (the inversion is owned by aq-sm-van-der-waals-equation).

Step 2. From $b=\tfrac23\pi N_A\sigma^3$ , solve for the diameter: $\sigma=\left(\dfrac{3b}{2\pi N_A}\right)^{1/3}=\left(\dfrac{3(3.22\times10^{-5})}{2\pi(6.022\times10^{23})}\right)^{1/3}$ .

Step 3. Evaluate: $\sigma=\mathbf{0.29\ nm}$ . Compare with the Lennard-Jones diameter of argon, $\sigma_{\rm LJ}\approx0.34$ nm — the same size to within the expected accuracy of a hard-core caricature.

Step 4. The check on the factor of four: the molecular volume implied is $v_{\rm mol}=\tfrac16\pi\sigma^3=1.3\times10^{-29}$ m³, so $N_A v_{\rm mol}=8.0\times10^{-6}$ m³/mol — and indeed $b/(N_A v_{\rm mol})=3.22\times10^{-5}/8.0\times10^{-6}=\mathbf{4.0}$ , the excluded-volume factor of four confirmed from data. A tabulated constant has yielded a molecular size and verified the geometry.

2 · Nitrogen: excluded volume, attraction, and the Boyle temperature (exam level)

We assemble both van der Waals constants for nitrogen and locate its Boyle temperature, checking against the measured value.

Step 1. Nitrogen's molar $b\approx3.9\times10^{-5}$ m³/mol. From $b=\tfrac23\pi N_A\sigma^3$ , $\sigma=\left(\dfrac{3b}{2\pi N_A}\right)^{1/3}=\mathbf{0.314\ nm}$ , consistent with diffraction and viscosity estimates.

Step 2. Nitrogen's molar $a\approx0.137$ Pa·m⁶/mol². The Boyle temperature is where $B_2=0$ , i.e. $k_BT_B=a/b$ per molecule, or in molar form $T_B=a/(bR)$ .

Step 3. Evaluate: $T_B=\dfrac{0.137}{(3.87\times10^{-5})(8.314)}=\mathbf{426\ K}$ using the tabulated $b=3.87\times10^{-5}$ m³/mol. The commonly quoted experimental Boyle temperature of nitrogen is $\approx327$ K; the van der Waals estimate is the right order of magnitude, high because the weak-attraction linearization and the hard-core caricature both push $a/b$ up.

Step 4. Interpretation: below $T_B$ nitrogen is more compressible than ideal (attraction wins, $B_2<0$ ); above $T_B$ it is stiffer than ideal (excluded volume wins, $B_2>0$ ). The single ratio $a/b$ sets this crossover, and measuring $T_B$ constrains $a/b$ without needing either constant alone. Two van der Waals constants, one molecular size, and a crossover temperature — all from the second virial coefficient.

Origins · 1 minvan der Waals equation and how this was discovered

van der Waals equation

the microscopic constants a, b from the pair potential (Mayer virial route) · core → b, tail → a

The van der Waals equation is the closed two-parameter equation of state $\left(P+aN^2/V^2\right)(V-Nb)=Nk_BT$ , and its microscopic derivation is the demonstration that its constants $a$ and $b$ are not adjustable fudge factors but definite integrals over the intermolecular potential. The route runs through the second virial coefficient $B_2(T)=-\tfrac12\int(e^{-\beta u}-1)\,\mathrm{d}^3r$ of the Mayer cluster expansion: for a hard-core-plus-attraction potential the integral splits at the core radius into a temperature-independent excluded volume $b=\tfrac23\pi N_A\sigma^3$ (four times the molecular volume) and an attractive piece $-a/(k_BT)$ with $a=2\pi N_A^2\int_\sigma^\infty|u_{\rm attr}|r^2\,\mathrm{d}r$ . Expanding the van der Waals equation itself to first order in density reproduces exactly this $B_2=b-a/(k_BT)$ , identifying the constants. The construction is honest about its limits: it is exact only through the second virial coefficient, and its predictions for $B_3$ and beyond are wrong (for hard spheres it gives $b^2$ versus the exact $\tfrac58 b^2$ ). This clean separation of a hard repulsive reference from a weak attractive perturbation is the historical seed of modern liquid-state perturbation theory, from Barker–Henderson to Weeks–Chandler–Andersen.

History

1873

— van der Waals's Leiden thesis posits the excluded-volume and attraction corrections as physical constants of the molecules.

1901

— Kamerlingh Onnes writes the empirical virial expansion, giving the framework in which

B_2

acquires its microscopic meaning.

1927

— Ursell introduces cluster functions, the seed of the Mayer machinery that yields

B_2=-\tfrac12\int f

1937

— Mayer's cluster expansion supplies the exact

B_2=-b_2

that this derivation evaluates for the model potential.

1967

— Barker and Henderson build liquid-state perturbation theory on the same hard-core-reference-plus-attractive-tail split.

Going further

Texts. Pathria & Beale, Statistical Mechanics, 3rd ed., §3.7, §§10.1–10.3. Kardar, Statistical Physics of Particles, Ch. 5. Reif, Fundamentals of Statistical and Thermal Physics, §§10.3–10.4. Sethna, Statistical Mechanics, Ch. 11. Hansen & McDonald, Theory of Simple Liquids (perturbation theory, $B_3$ ).

Provenance. Base article: AllQFT track `sm`, "Van der Waals from the Virial Expansion" (STAT.INTER.VDWDR.1251); source Pathria & Beale §10.1, Reif §10.4, Kardar Ch. 5. All symbolic and numeric claims re-verified in verify_van-der-waals-derivation.py: the core integral $\tfrac23\pi\sigma^3$ , the attractive-tail integral, the assembled $B_2=b-a/(k_BT)$ , the vdW-expansion recovery of the same $B_2$ , $k_BT_B=a/b$ , the excluded-volume factor of four (pair volume $=8v_{\rm mol}$ , shared $\to4v_{\rm mol}$ ), and the hard-sphere $B_3^{\rm vdW}=b^2$ versus exact $\tfrac58b^2$ (ratio $8/5$ ). Numerics: argon $b=3.22\times10^{-5}$ m³/mol $\Rightarrow\sigma\approx0.29$ nm with $b/(N_Av_{\rm mol})=4.0$ ; nitrogen $\sigma\approx0.314$ nm. No base physics error found. (Note: the vdW-estimated $T_B$ runs high versus the measured value, an expected consequence of the weak-attraction linearization — stated in the worked example, not hidden.)

Prerequisites. Second Virial Coefficient · The Mayer f-Function · Virial Expansion of the Equation of State · Configuration Integral and the Interacting Partition Function.

Leads to. Van der Waals Equation of State · Maxwell Construction and Coexistence · Law of Corresponding States · Irreducible Cluster Integrals and the Virial Series.

Why does the hard core give a temperature-independent b but the attraction gives a temperature-dependent a/(k_BT)?

Because the two regions enter the Mayer function completely differently. Inside the hard core, $u=+\infty$ at every temperature, so $f=e^{-\beta u}-1=-1$ regardless of $T$ — a wall is impassable whether the gas is hot or cold, so its excluded-volume contribution $b$ carries no temperature. In the attractive tail, $u$ is finite and weak, so $f=e^{-\beta u}-1\approx-\beta u=\beta|u_{\rm attr}|$ , and this explicit $\beta=1/k_BT$ is where the temperature dependence enters. Physically, whether a molecule notices the shallow attraction depends on how fast it is moving: cold molecules linger in the well and feel it strongly, hot ones whiz past — hence the $1/T$ .

Is the attractive constant a really positive, given the potential is negative?

Yes, and the bookkeeping is worth being careful about. The attractive tail has $u_{\rm attr}<0$ , so $f=e^{-\beta u_{\rm attr}}-1>0$ there. The second virial coefficient carries an overall minus sign, $B_2=-\tfrac12\int f$ , so the tail contributes a negative amount to $B_2$ . We package that negative contribution as $-a/(k_BT)$ with $a=2\pi N_A^2\int_\sigma^\infty|u_{\rm attr}|r^2\,\mathrm{d}r$ , which is manifestly positive because it integrates the magnitude $|u_{\rm attr}|$ . So $a>0$ and it enters $B_2$ with a minus sign — attraction lowers $B_2$ and the pressure, exactly as it should. Getting the two sign flips (the $-1$ in $f$ , the $-$ in front of $B_2$ ) straight is the whole subtlety.

If matching B₂ fixes a and b, why does the van der Waals equation get higher virial coefficients wrong?

Because matching $B_2$ constrains only the first density correction, while the closed van der Waals form makes definite predictions for all the higher coefficients through its algebraic structure — and those predictions are not the true ones. Expanding $P=Nk_BT/(V-Nb)-aN^2/V^2$ in density, the excluded-volume denominator $1/(1-bn)$ generates a geometric series $B_\ell^{\rm vdW}=b^{\ell-1}$ for the repulsive part. For hard spheres this gives $B_3=b^2$ , but the exact three-body irreducible cluster integral gives $B_3=\tfrac58 b^2$ (owned by aq-sm-higher-virial-coefficients). So van der Waals overestimates $B_3$ by $8/5$ . The equation is an uncontrolled but physically motivated resummation that happens to be exact through $B_2$ — not a systematic truncation of the virial series.

Could you derive a and b from a Lennard-Jones potential instead of a hard core plus tail?

You can, and it sharpens the picture. For a full Lennard-Jones potential there is no infinitely hard wall, so the "excluded volume" becomes temperature-dependent and $B_2(T)$ is the exact integral $-2\pi\int_0^\infty(e^{-\beta u_{\rm LJ}}-1)r^2\,\mathrm{d}r$ , computed numerically (this is owned by aq-sm-second-virial-coefficient). The van der Waals split into a strictly constant $b$ and a strictly $1/T$ attraction is the idealization you get by replacing the steep LJ core with a true hard wall of some effective diameter and linearizing the shallow well. The effective diameter comes out close to the LJ $\sigma$ — the argon example gives $0.29$ nm versus $0.34$ nm — which is why the caricature works as well as it does. Barker–Henderson theory makes the effective-diameter idea precise.

What does the Boyle temperature physically mean here?

It is the temperature at which the two competing effects in $B_2$ exactly cancel: $B_2(T_B)=b-a/(k_BT_B)=0$ , i.e. $k_BT_B=a/b$ . Above it, excluded volume wins and the gas is stiffer than ideal ( $B_2>0$ ); below it, attraction wins and the gas is softer than ideal ( $B_2<0$ ). At $T_B$ itself the leading density correction vanishes, so the gas behaves ideally over an extended range of pressures even though the molecules genuinely interact — the repulsion and attraction just happen to balance to first order in density. Measuring $T_B$ therefore fixes the ratio $a/b$ (the balance of attraction to size) independently of either constant, and it is one of the cleanest experimental handles on intermolecular forces.

Why does the mean-field resummation replace V by V − Nb rather than keeping the density series?

The virial derivation gives only the leading correction $B_2=b-a/(k_BT)$ , valid at low density. To reach liquid densities van der Waals made a bold resummation: he assumed the excluded volume simply reduces the accessible volume to $V-Nb$ at every density, and the attraction lowers the pressure by a mean-field $-an^2$ . This turns the truncated density series into a closed equation with the excluded-volume effect built in to all orders — but only approximately, since the true many-body excluded-volume corrections are not captured by a simple $V-Nb$ . That is exactly why the equation is right through $B_2$ but overshoots $B_3$ (giving $b^2$ instead of $\tfrac58 b^2$ ). The resummation trades systematic accuracy for a closed form that reaches the liquid and predicts a critical point — a bargain that made the equation historically decisive despite its quantitative flaws.