Maxwell Construction and Coexistence

A mean-field equation of state — the van der Waals equation is the archetype — is analytic in volume, and analyticity is fatal: it cannot reproduce the sharp, non-analytic kink that a genuine first-order phase transition puts into the free energy. So below the critical temperature the van der Waals isotherm does something visibly wrong. As you compress the gas the pressure falls (gas branch), then rises along a middle segment where $(\partial P/\partial v)_T>0$ , then falls again (liquid branch). That rising segment has negative compressibility: squeeze it and it pushes back less, not more. No homogeneous material behaves that way, and the loop is telling you the single-phase description has broken down. The real isotherm replaces the loop with a flat plateau. Over a whole range of volumes the fluid does not stay homogeneous — it separates into liquid and gas coexisting at one shared pressure $P_{\text{sat}}(T)$ . James Clerk Maxwell showed in 1875 how to read that plateau straight off the analytic loop: draw the horizontal line at the pressure for which the two lobes of area it cuts from the loop are exactly equal. This is the equal-area rule, and it is the structural result this article owns and proves. The rule is not a fudge. Two phases coexist only when they agree on temperature, pressure, and chemical potential. Integrating $\mathrm{d}\mu=v\,\mathrm{d}P$ along the isotherm and demanding equal $\mu$ at the two endpoints yields the equal-area condition exactly — nothing added by hand. Read in the free energy instead of the pressure, the same statement is a convexification: the mean-field Helmholtz free energy $f(v)$ is non-convex over the loop, the physical free energy is its convex hull, and the double-tangent line that convexifies $f$ projects onto the Maxwell plateau. The construction also organizes the subcritical phase diagram — binodal, spinodal, and the metastable band of superheated liquid and supercooled vapour between them — but it stops there; the full critical behaviour and the phase diagram of real fluids belong to later chapters.

Key takeaways

Below $T_c$ a van der Waals isotherm develops an S-shaped loop with a middle segment where $(\partial P/\partial v)_T>0$ — mechanically unstable and unphysical.
The Maxwell construction replaces the loop by a horizontal coexistence plateau at $P_{\text{sat}}(T)$ , drawn so the two lobes of area between loop and plateau are equal — the equal-area rule.
The equal-area rule is derived: it is exactly the thermodynamic condition $\mu_{\text{liq}}=\mu_{\text{gas}}$ at common $T$ and $P$ , via $\mathrm{d}\mu=v\,\mathrm{d}P$ and one integration by parts.
Equivalently it is the convex-hull (double-tangent) replacement of the non-convex mean-field free energy; states on the plateau are phase mixtures whose proportions obey the lever rule.
Plateau endpoints trace the binodal (coexistence curve); the loop extrema $(\partial P/\partial v)_T=0$ trace the spinodal; between them lie metastable superheated liquid and supercooled vapour.
Key idea: replace the unphysical loop by the horizontal line whose pressure equalizes the two phases' chemical potentials — geometrically $\oint v\,\mathrm{d}P=0$ , i.e. the equal-area rule, i.e. the convexification of the free energy by a common tangent.

The core insight

Replace the unphysical loop of a mean-field isotherm by the horizontal line whose pressure equalizes the chemical potentials of the two phases; geometrically this is the equal-area rule $\oint v\,\mathrm{d}P=0$ across the loop, i.e. the convexification of the free energy by a double tangent.

Where this lives today

The Maxwell construction is the everyday tool that turns any analytic ("cubic") equation of state into a usable phase diagram. Modern engineering equations of state — Soave–Redlich–Kwong, Peng–Robinson, and their descendants — all produce van-der-Waals-like loops below the critical temperature, and process simulators apply exactly this equal-area rule (in the form $\mu_{\text{liq}}=\mu_{\text{gas}}$ ) to compute saturation pressures, vapour–liquid equilibria, and flash separations for the chemical, petroleum, and refrigeration industries. The metastable branches the construction bounds are not academic: superheated liquids drive the operation of bubble chambers and the hazard of boiling-liquid explosions, and supersaturated vapours are what cloud and bubble chambers image. Its deepest role is conceptual. The same convex-hull logic — replace a non-convex free energy by its convex envelope, producing a flat coexistence region — governs every mean-field first-order transition. In the mean-field Ising model and in Landau theory a non-convex free energy $f(m)$ in a field is convexified into a flat magnetization jump, exactly analogous to the liquid–gas plateau; the shared name "Maxwell construction" is used throughout statistical field theory. And the construction's one clear failure — its mean-field prediction $\Delta\rho\propto(T_c-T)^{1/2}$ for the shape of the coexistence curve near criticality is wrong, because fluctuations dominate there — is precisely the signpost that pointed toward the renormalization group and the correct, non-mean-field theory of critical phenomena taken up later in this track.

The physical picture

Build the Helmholtz free energy per particle $f(v)$ by integrating the looped isotherm, since $P=-(\partial f/\partial v)_T$ . Where the loop rises, $f$ bulges upward into a non-convex "wiggle". A fluid whose homogeneous free energy sits on that bulge can do better: split into two phases whose average lies on the straight line drawn below the bulge. That line is the double tangent — it touches $f$ at two points, $v_{\text{liq}}$ and $v_{\text{gas}}$ , and its slope is a single number, $-P_{\text{sat}}$ . Between those two points the physical free energy is the line, not the curve. The bumpy $f$ is replaced by its convex hull, and that is the whole content of the construction, seen in the free energy. The same thing seen in the $P$ – $v$ plane is the equal-area rule. Chemical potential changes along the isotherm as $\Delta\mu=\int v\,\mathrm{d}P$ . For the liquid and gas endpoints — both at $P_{\text{sat}}$ — to share a chemical potential, the integral around the closed path (loop out, plateau back) must vanish: $\oint v\,\mathrm{d}P=0$ . That closed integral is exactly the signed area between loop and plateau, so "equal chemical potential" and "equal areas above and below the line" are the same statement. The first figure shows the loop with the equal-area tie-line and its two shaded lobes.

Figure

The vdW loop at T_r = 0.9 with the Maxwell equal-area tie-line at P_r = 0.647: the two shaded lobes (amber above, blue below) have equal area, and the plateau runs between the coexisting volumes v_l ≈ 0.60 and v_g ≈ 2.35

Picture the plateau as a tie-line. Any average volume between $v_{\text{liq}}$ and $v_{\text{gas}}$ is realized not by a homogeneous fluid at that density but by a mixture — some fraction liquid, some gas — and the fractions follow the lever rule, exactly as two masses balance on a beam about the average point. The system never sits at an intermediate density as a single phase. The loop's two extrema, where $(\partial P/\partial v)_T=0$ , are the spinodal points: inside them the homogeneous phase is absolutely unstable; between each spinodal point and the nearer plateau endpoint the phase is metastable — locally stable but globally beaten by phase separation, which is why superheated liquids and supercooled vapours can be coaxed into brief existence. The second figure shows the free energy and its convex hull, the picture behind the plateau.

Figure

The free energy f(v) is non-convex over the loop; the physical free energy is its convex hull — the common (double) tangent touching at v_l and v_g, whose constant slope −P_sat is the plateau pressure

The core equations

Everything starts from the van der Waals equation, written per particle. It is the mean-field equation whose loop we must repair; the article treats it as given (it is owned by aq-sm-van-der-waals-equation) and works out the coexistence.

\left(P+\frac{a}{v^2}\right)(v-b)=k_BT \qquad (1)

where: $P$ — pressure; $v=V/N$ — volume per particle; $T$ — temperature; $a$ — attraction (cohesion) parameter; $b$ — excluded volume per particle; $k_B$ — Boltzmann constant. Solving for pressure gives $P=k_BT/(v-b)-a/v^2$ , which loops for $T<T_c$ . Two phases coexist only when three intensive quantities agree. Temperature and pressure are shared by construction of the plateau; the substantive condition is equality of the chemical potential (the Gibbs free energy per particle).

\mu_{\text{liq}}(T,P_{\text{sat}})=\mu_{\text{gas}}(T,P_{\text{sat}}) \qquad (2)

where: $\mu_{\text{liq}},\mu_{\text{gas}}$ — chemical potentials of the coexisting phases; $P_{\text{sat}}(T)$ — the single saturation pressure they share. This equation, not any geometric averaging, fixes the plateau. Turning equation (2) into geometry uses $\mathrm{d}\mu=v\,\mathrm{d}P$ at fixed temperature and one integration by parts. The result is the equal-area rule.

P_{\text{sat}}\,(v_g-v_l)=\int_{v_l}^{v_g}P(v)\,\mathrm{d}v,\qquad\text{equivalently}\qquad \oint v\,\mathrm{d}P=0 \qquad (3)

where: the left side is the rectangle under the plateau, the right the area under the analytic loop; their equality forces the two lobes between loop and plateau to have equal area. The closed integral $\oint v\,\mathrm{d}P$ (loop out, plateau back) vanishes because the plateau contributes nothing ( $\mathrm{d}P=0$ there), so the loop's signed area must cancel. On the plateau the fluid is a mixture, and the split between phases follows from simple volume additivity.

v=x_l\,v_l+x_g\,v_g,\qquad x_l+x_g=1\ \Rightarrow\ x_g=\frac{v-v_l}{v_g-v_l} \qquad (4)

where: $x_l,x_g$ — number fractions in liquid and gas; $v$ — overall volume per particle on the plateau. This is the lever rule: the average volume sits on the tie-line, and the phase fractions are the complementary lever arms. The plateau is bounded by two loci in the subcritical region. The binodal is the set of plateau endpoints; the spinodal is the set of loop extrema, and their merger defines the critical point.

\left(\frac{\partial P}{\partial v}\right)_T=0\ \text{(spinodal)},\qquad \left(\frac{\partial P}{\partial v}\right)_T=\left(\frac{\partial^2 P}{\partial v^2}\right)_T=0\ \text{(critical)} \qquad (5)

where: the spinodal marks the absolute limit of metastability (where the compressibility diverges); the critical point is where the two spinodal branches and the binodal all coincide at the isotherm's horizontal inflection. For van der Waals these conditions give $v_c=3b$ , $k_BT_c=8a/27b$ , $P_c=a/27b^2$ (derived in aq-sm-van-der-waals-equation) — the reference scales for the whole construction.

Derivation AMaxwell's equal-area rule from equality of chemical potentials

The goal is to show that the geometric equal-area prescription is a rigorous consequence of the thermodynamic coexistence conditions — equal $T$ , $P$ , and $\mu$ — and nothing more. Temperature and pressure are equal by the very construction of a horizontal plateau, so the entire content lies in forcing the chemical potentials of liquid and gas to match. The route is to integrate $\mathrm{d}\mu=v\,\mathrm{d}P$ along the analytic mean-field isotherm from the liquid endpoint to the gas endpoint, set the change to zero, and convert the $v\,\mathrm{d}P$ integral into a $P\,\mathrm{d}v$ integral by parts. The single subtlety to watch is that the integration path runs through the loop, where $P$ genuinely varies, even though both endpoints sit at the same plateau pressure — that is exactly why the loop's area, not zero, appears.

Step 1. Chemical potential along the isotherm. From the Gibbs–Duhem relation per particle, $\mathrm{d}\mu=-s\,\mathrm{d}T+v\,\mathrm{d}P$ , at fixed $T$ this is $\mathrm{d}\mu=v\,\mathrm{d}P$ . The analytic $P(v)$ defines $v(P)$ piecewise along the loop, so $\mu$ can be integrated along it. Why: coexistence is a statement about $\mu$ , and the isotherm gives us $\mu$ only through this differential — there is no other way to compare the two phases' free energies.

Step 2. Impose coexistence. Equality of chemical potentials means $\mu_g-\mu_l=\int_{l}^{g}v\,\mathrm{d}P=0$ , the integral taken along the loop from the liquid endpoint to the gas endpoint. Why: both endpoints are at $P=P_{\text{sat}}$ , but the path between them climbs and descends the loop, so the integral is not trivially zero — its vanishing is a real condition on $P_{\text{sat}}$ .

Step 3. Integrate by parts. $\int_l^g v\,\mathrm{d}P=\big[vP\big]_l^g-\int_{v_l}^{v_g}P\,\mathrm{d}v$ . The endpoint term is $v_g P_{\text{sat}}-v_l P_{\text{sat}}=P_{\text{sat}}(v_g-v_l)$ because both endpoints share the plateau pressure. Why: trading $v\,\mathrm{d}P$ for $P\,\mathrm{d}v$ turns an awkward pressure integral into the area under the loop, which is the quantity we can see in the $P$ – $v$ plane.

Step 4. Read off the equal-area rule. Setting the Step-2 integral to zero, $P_{\text{sat}}(v_g-v_l)=\int_{v_l}^{v_g}P(v)\,\mathrm{d}v$ . The left side is the rectangle under the plateau; the right is the area under the loop; equality means the upper and lower lobes have equal area. Together with the two endpoint conditions $P(v_l)=P(v_g)=P_{\text{sat}}$ , this is three equations for the three unknowns $v_l$ , $v_g$ , $P_{\text{sat}}$ at each $T<T_c$ . Why: the equal-area statement is not an approximation or an averaging trick — it is precisely the equal- $\mu$ condition rewritten in coordinates you can plot. *Verified: the integration-by-parts identity $\int v\,\mathrm{d}P=P_{\text{sat}}(v_g-v_l)-\int P\,\mathrm{d}v$ and the equal-area ⇔ equal- $\mu$ equivalence are checked symbolically in verify_maxwell-construction.py; the construction matches Pathria & Beale §12.2 and Landau–Lifshitz §84.*

Go deeper · Derivation B · 2 minNumerically locating the coexistence plateau at T_r = 0.9

Derivation BNumerically locating the coexistence plateau at T_r = 0.9

The goal is to turn the equal-area rule into actual numbers for a definite subcritical isotherm, so the construction is concrete rather than schematic. Working in reduced units removes every substance constant (this is the corresponding-states form owned by aq-sm-law-of-corresponding-states), leaving the pure van der Waals problem $P_r=8T_r/(3v_r-1)-3/v_r^2$ . We fix $T_r=0.9$ , find the spinodal extrema that bracket the loop, then solve the three simultaneous coexistence equations for the plateau pressure and the two coexisting volumes. The point to watch is that the plateau pressure is fixed globally by the area balance — it is not the mean of the spinodal pressures, and it must fall strictly between them.

Step 1. The reduced isotherm and its slope. With $P_r=8T_r/(3v_r-1)-3/v_r^2$ , the slope is $\partial P_r/\partial v_r=-24T_r/(3v_r-1)^2+6/v_r^3$ . Setting it to zero at $T_r=0.9$ gives the two spinodal volumes $v_{r}^{\text{liq-spin}}$ and $v_{r}^{\text{gas-spin}}$ ; between them the slope is positive (the unstable middle branch). Why: the spinodal extrema are the boundary of mechanical instability, and the plateau pressure must lie between the pressures at these two extrema for a horizontal line to cut the loop in three places.

Step 2. The three coexistence equations. For a trial plateau pressure $P_r^{\text{sat}}$ , the equation $P_r(v_r)=P_r^{\text{sat}}$ is a cubic in $v_r$ with three real roots $v_l<v_m<v_g$ ; the outer two are the candidate coexisting volumes. The plateau is correct when the equal-area residual $P_r^{\text{sat}}(v_g-v_l)-\int_{v_l}^{v_g}P_r(v_r)\,\mathrm{d}v_r$ vanishes. Why: the two endpoint conditions are automatic once $v_l,v_g$ are roots at the same pressure; the equal-area condition is the third equation that pins $P_r^{\text{sat}}$ .

Step 3. Root-find the plateau. Bracketing $P_r^{\text{sat}}$ between the two spinodal pressures and driving the equal-area residual to zero gives $P_r^{\text{sat}}=0.647$ , with coexisting volumes $v_l=0.603$ and $v_g=2.349$ . The liquid is denser than critical ( $v_l<1$ ) and the gas rarer ( $v_g>1$ ), as it must be. Why: this is the honest output of the construction — a single computed number, not a guessed one; the liquid/gas volume ratio $v_g/v_l\approx3.9$ already shows the density contrast that widens as $T_r$ drops.

Step 4. Sanity against the spinodal. The spinodal pressures at $T_r=0.9$ are $P_r\approx0.420$ (gas-side, the loop minimum) and $P_r\approx0.724$ (liquid-side, the loop maximum). The equal-area plateau $0.647$ sits strictly between them — not at their midpoint $0.572$ — confirming that the coexistence pressure is set by the global area balance, not by averaging. Why: this is the sharpest check that the equal-area rule is doing real work; a naive "split the difference" prescription would give the wrong answer by more than ten percent. *Verified: the spinodal roots, the cubic coexistence solve, the equal-area residual → 0, the plateau $P_r^{\text{sat}}=0.647$ with $v_l=0.603$ , $v_g=2.349$ , and the plateau-between-spinodals check are all computed in verify_maxwell-construction.py (scipy brentq + quad).*

Limiting cases

The construction changes character across the subcritical region and switches off entirely above $T_c$ .

Case	Result	Meaning
Just below critical, $T\to T_c^-$	Loop amplitude $\to0$ ; binodal and spinodal merge	The plateau shrinks to a point at the horizontal inflection $(\partial P/\partial v)_T=(\partial^2 P/\partial v^2)_T=0$ ; the density gap closes and coexistence disappears.
At $T_c$	No loop, an inflection instead	The isotherm has $(\partial P/\partial v)_T=0$ at one point only; nothing to convexify — the construction is trivial here.
Deep coexistence, $T\ll T_c$	Wide plateau: $v_l\to b$ , $v_g\to\infty$	The liquid approaches close packing (the vdW excluded-volume limit), the gas is nearly ideal, and $P_{\text{sat}}\to0$ ; the coexistence line joins onto Clausius–Clapeyron.
Above critical, $T>T_c$	$(\partial P/\partial v)_T<0$ everywhere	No loop, no instability, no plateau — a single supercritical fluid; the construction has nothing to act on.
Infinite-range (Kac) limit	Convexification is exact	For a weak, infinitely long-ranged attraction the van der Waals equation becomes exact and the Maxwell rule is not an approximation but the literal convex hull of the free energy.

Common trapThe plateau is not the midpoint of the loop

A tempting mistake is to place the coexistence line by symmetry — through the loop's inflection, or halfway between the two spinodal pressures. That is wrong. The plateau pressure is fixed globally by the equal-area balance, which weights the whole loop, and it generically differs from any local or midpoint prescription. For van der Waals at $T_r=0.9$ the equal-area pressure is $P_r=0.647$ , while the midpoint of the two spinodal pressures ( $0.420$ and $0.724$ ) is $0.572$ — off by more than ten percent. Two further cautions: the unstable middle branch ( $\partial P/\partial v>0$ ) is genuinely unphysical, but the two branches between binodal and spinodal are real metastable states (superheated liquid, supercooled vapour), not errors; and the Maxwell rule is exact only where the free energy must be convexified (mean-field or infinite-range models), so for short-ranged real fluids it is a mean-field approximation to a transition whose true critical behaviour is not mean-field at all.

Worked examples

1 · The coexistence plateau of van der Waals at T_r = 0.9 (a real number from the rule)

We locate the liquid–gas plateau of the reduced van der Waals fluid at $T_r=0.9$ , computing the saturation pressure and coexisting volumes from the equal-area rule alone.

Step 1. In reduced units the isotherm is $P_r=\dfrac{8T_r}{3v_r-1}-\dfrac{3}{v_r^2}$ . At $T_r=0.9$ its slope $\partial P_r/\partial v_r$ vanishes at the two spinodal volumes; numerically these bracket a middle branch where the slope is positive. The spinodal pressures are $P_r\approx0.420$ (gas-side minimum) and $P_r\approx0.724$ (liquid-side maximum), so the plateau pressure must lie between them.

Step 2. A trial plateau at pressure $P_r^{\text{sat}}$ meets the isotherm where $P_r(v_r)=P_r^{\text{sat}}$ , a cubic $3P_r^{\text{sat}}v_r^3-(P_r^{\text{sat}}+8T_r)v_r^2+9v_r-3=0$ with three real roots; the outer two are the candidate liquid and gas volumes.

Step 3. Adjust $P_r^{\text{sat}}$ until the equal-area residual $P_r^{\text{sat}}(v_g-v_l)-\int_{v_l}^{v_g}P_r\,\mathrm{d}v_r$ is zero. Root-finding gives $\mathbf{P_r^{\text{sat}}=0.647}$ , with $\mathbf{v_l=0.603}$ and $\mathbf{v_g=2.349}$ . The liquid is compressed below the critical volume and the gas expanded above it, with a volume ratio $v_g/v_l\approx3.9$ .

Step 4. Check the number: the equal-area plateau $0.647$ sits between the spinodal pressures $0.420$ and $0.724$ but not at their midpoint $0.572$ . The construction has produced a definite coexistence pressure — $P_r=0.647$ at $T_r=0.9$ — that no local or averaging rule reproduces, which is precisely why the global equal-area balance is needed.

2 · The lever rule on the plateau (how much liquid at a given density)

We use the lever rule to find the phase fractions when a van der Waals fluid at $T_r=0.9$ is held at an average volume midway across the plateau.

Step 1. From Example 1, $v_l=0.603$ and $v_g=2.349$ at $T_r=0.9$ . Take the overall volume per particle to be the arithmetic midpoint $v=\tfrac12(v_l+v_g)=1.476$ .

Step 2. The gas fraction is $x_g=\dfrac{v-v_l}{v_g-v_l}=\dfrac{1.476-0.603}{2.349-0.603}=\dfrac{0.873}{1.746}=0.500$ , and the liquid fraction is $x_l=1-x_g=0.500$ .

Step 3. So at the plateau's volume midpoint the system is half gas and half liquid by particle number — the average sits exactly on the balance point of the tie-line. Move the overall volume toward $v_l$ and the liquid fraction rises; toward $v_g$ and the gas fraction rises, linearly.

Step 4. Note the density picture is not symmetric: because $v_l$ and $v_g$ are not symmetric about the critical volume $v_r=1$ , an average density $\rho=1/v$ at the midpoint volume is $1/1.476=0.678$ , not the mean of the two phase densities. The lever rule is linear in volume, not in density — a distinction that matters whenever coexistence data are plotted against $\rho$ .

Origins · 1 mincritical point and how this was discovered

critical point

the merger of binodal and spinodal where liquid and gas become indistinguishable (van der Waals, 1873; Maxwell, 1875) · equal areas fix the coexistence line

The critical point is where the liquid–gas coexistence line ends: as temperature rises toward $T_c$ , the coexisting liquid and gas densities approach each other and merge, and above $T_c$ only a single supercritical fluid exists. In the van der Waals picture it is the point where the isotherm's loop has just disappeared into a horizontal inflection, so that the first and second volume derivatives of the pressure both vanish; there the binodal (coexistence curve) and the spinodal (limit of metastability) meet. Thomas Andrews established the existence of a critical temperature for carbon dioxide experimentally in 1869, and van der Waals's 1873 equation gave the first theory that produced a critical point from molecular assumptions, with the universal ratio $P_c v_c/k_BT_c=3/8$ . Maxwell's 1875 equal-area construction then supplied the missing piece — how to extract the physical coexistence line, and hence the whole subcritical phase diagram, from the analytic equation. The construction's mean-field prediction for the shape of the coexistence curve near the critical point ( $\Delta\rho\propto(T_c-T)^{1/2}$ ) is qualitatively right but quantitatively wrong, a discrepancy that a century later became the entry point to the renormalization group and the modern theory of critical phenomena.

History

1869

— Andrews measures the critical temperature of carbon dioxide, establishing the critical point experimentally.

1873

— van der Waals's equation of state produces a critical point and looped subcritical isotherms from molecular assumptions.

1875

— Maxwell states the equal-area construction in "On the Dynamical Evidence of the Molecular Constitution of Bodies" (Nature).

1893

— van der Waals and others recognize the metastable branches (superheating, supersaturation) bounded by the spinodal.

1900

— Gibbs's convex-hull view of coexistence (double-tangent construction) becomes the standard free-energy formulation.

Going further

Texts. Pathria & Beale, Statistical Mechanics, 3rd ed., §12.2 (van der Waals gas and the Maxwell construction). Kardar, Statistical Physics of Particles, Ch. 5 (interacting gas, phase coexistence). Reif, Fundamentals of Statistical and Thermal Physics, §8.6. Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 11 (abrupt phase transitions). Landau & Lifshitz, Statistical Physics Part 1, §84 (the critical point, van der Waals). J. C. Maxwell, "On the Dynamical Evidence of the Molecular Constitution of Bodies," Nature 11, 357 (1875).

Provenance. Base article: AllQFT track `sm`, "Maxwell Construction and Coexistence" (STAT.INTER.MAXCON.1252); source Pathria & Beale §12.2, Landau–Lifshitz §84. All symbolic and numeric claims re-verified here in verify_maxwell-construction.py: the van der Waals critical point and $Z_c=3/8$ , the reduced EOS, the equal-area ⇔ equal- $\mu$ equivalence via integration by parts, the middle-branch instability and the lever rule, and — computed fresh, not quoted — the $T_r=0.9$ coexistence numbers $P_r^{\text{sat}}=0.647$ , $v_l=0.603$ , $v_g=2.349$ (equal-area residual driven to zero; plateau strictly between the spinodal pressures $0.420$ and $0.724$ ). No base physics error was found; the base left the coexistence numbers un-computed, which we have now supplied. This article owns the equal-area construction; the full subcritical phase diagram and the critical exponents belong to later chapters (SM7).

Prerequisites. Van der Waals Equation of State · Van der Waals from the Virial Expansion · Law of Corresponding States.

Leads to. Law of Corresponding States · Critical Exponents and Universality · Spinodal Decomposition.

Why does the coexistence line have to be horizontal at all?

Because coexisting phases share a pressure. If two phases are in mechanical equilibrium across an interface, the pressure must be equal on both sides (otherwise the interface accelerates). As you move along the plateau you are changing the proportion of liquid to gas at fixed intensive state — the temperature and pressure of both phases stay pinned at their coexistence values. So the physical isotherm is flat across the two-phase region, at height $P_{\text{sat}}(T)$ . The analytic van der Waals curve, being a single-phase equation, does not know about the interface and instead draws a loop; the Maxwell construction restores the flat plateau that mechanical equilibrium demands.

Is the equal-area rule exact, or is it an approximation?

Both, depending on the model. Given the van der Waals equation, the equal-area rule is an exact consequence of equilibrium thermodynamics — it is just $\mu_{\text{liq}}=\mu_{\text{gas}}$ rewritten, with no extra assumption. But the van der Waals equation itself is a mean-field approximation for a real short-ranged fluid, so the coexistence line it produces is only as good as that approximation. There is one clean case where everything is exact: a weak, infinitely long-ranged attraction (the Kac limit), for which van der Waals is the exact equation of state and the Maxwell rule is the literal convex hull of the free energy. For real fluids near the critical point, where fluctuations dominate, the mean-field construction gets the qualitative picture right but the critical exponents wrong.

What is the difference between the binodal and the spinodal?

The binodal (coexistence curve) is the locus of the Maxwell-plateau endpoints $v_l,v_g$ — where the global free energy is minimized by phase separation. The spinodal is the locus of the loop extrema $(\partial P/\partial v)_T=0$ — where a homogeneous phase loses even local stability. Between the binodal and the spinodal a single phase is metastable: locally stable, so it can persist (a superheated liquid, a supercooled vapour), but globally beaten, so a nucleation event will convert it. Inside the spinodal the phase is absolutely unstable and decomposes spontaneously without any nucleation barrier. The critical point is where binodal and spinodal touch.

How does the free-energy (convex-hull) picture connect to the equal-area picture?

They are the same statement in two coordinate systems. Pressure is the negative slope of the free energy, $P=-(\partial f/\partial v)_T$ . A straight segment of $f$ has constant slope, hence constant pressure — that is the plateau. The double tangent touches $f$ at $v_l$ and $v_g$ : equal slope at the two points means equal pressure, and equal intercept (which is $-\mu$ , since $\mu=f-vf'$ ) means equal chemical potential. So the common tangent encodes both coexistence conditions at once, and projecting its constant slope into the $P$ – $v$ plane gives the horizontal Maxwell line. Equal-area is the $P$ – $v$ face of the coin; convex-hull is the free-energy face.

Does the Maxwell construction give the full phase diagram?

No, and this article is deliberately scoped to the equal-area construction itself. What the construction gives you is, at each $T<T_c$ , the coexistence pressure and the two coexisting volumes — trace those over temperature and you get the binodal, the spinodal, and the metastable band between them. What it does not give is the correct critical behaviour: the shape of the coexistence curve as $T\to T_c$ comes out mean-field ( $\Delta\rho\propto(T_c-T)^{1/2}$ ), which is wrong for real fluids. The genuine critical exponents, the universality classes, and the renormalization-group treatment are the subject of later chapters (SM7 and beyond); the Maxwell construction is the classical, mean-field entry point to that story, not the whole of it.

Why can superheated liquids and supercooled vapours exist if the loop is unphysical?

Only the middle branch of the loop — the segment with $(\partial P/\partial v)_T>0$ between the two spinodal points — is genuinely unphysical, because its compressibility is negative. The two outer segments, between each spinodal point and the nearer binodal endpoint, still have $(\partial P/\partial v)_T<0$ , so a homogeneous phase there is locally stable: small perturbations decay. It is only globally unstable, meaning phase separation would lower the free energy if it could get started. Without nucleation sites — clean liquid, dust-free vapour — the phase can persist on these metastable branches, which is exactly how superheated liquids "bump" and supersaturated vapours condense onto tracks in a cloud chamber.