The part of the curve no fluid follows

Van der Waals’ equation is the simplest correction to the ideal gas that produces a liquid. Two changes: molecules occupy volume, so the space available is reduced; molecules attract each other, so the pressure on the walls is reduced. Written in units of each substance’s own critical values, every constant belonging to any particular fluid disappears and what is left is

$$p = \frac{8T}{3v-1} - \frac{3}{v^2}.$$

Plotting that below the critical temperature produces something that cannot be right.

Why the rising section is impossible #

A material whose volume increases with pressure has negative compressibility, and such a material cannot exist for a reason that needs no thermodynamics at all. Suppose a small region of it were momentarily compressed by a fluctuation. Its pressure would fall, so the surrounding material would push in harder, compressing it further, lowering its pressure again. Nothing stops the process. The same argument run the other way makes an expanded region expand without limit.

This is the fluid version of an unstable equilibrium — the same structure as a body balanced at the top of a hill, where a displacement produces a force in the direction of the displacement rather than against it. Stability requires $\partial p/\partial v < 0$, and the middle of the loop has the wrong sign.

So the equation is being asked a question about a state that does not occur, and it answers, because an algebraic expression will answer anything. What actually happens is that the substance stops being one thing.

The line, and the condition that fixes its height #

Below the critical temperature the substance separates into a dense phase and a dilute one, which coexist at a single pressure. On the diagram the state runs along a horizontal line, and the position along the line records the proportions rather than a density: at the left end everything is liquid, at the right end everything is vapour, and in the middle the two are present in the ratio the lever rule gives.

The height of that line cannot be chosen freely. Two phases in equilibrium must have equal temperature, equal pressure and equal Gibbs energy per particle — the last because otherwise particles would move from the phase of higher Gibbs energy to the phase of lower, which is what “in equilibrium” excludes.

Along an isotherm, $\mathrm{d}g = v\,\mathrm{d}p$. So the difference in Gibbs energy between the two ends of the tie line is $\int v\,\mathrm{d}p$ taken along the equation-of-state curve from one end to the other, and setting that integral to zero is exactly the statement that the areas above and below the line are equal. Maxwell’s equal-area rule is an equilibrium condition wearing a geometric disguise, and the same reasoning would place the tie line for any equation of state at all — including the one a gas cooling by being let go is worked out from.

For the isotherm drawn above, at nine-tenths of the critical temperature, the construction puts the line at 0.647 of the critical pressure and its ends at reduced volumes of 0.603 and 2.349 — a density ratio of 3.9 between the liquid and its vapour.

It is worth being explicit about how the construction is actually carried out, because the description “make the two areas equal” hides a search. The tie-line pressure is not given by any formula; it is found. For a chosen temperature, the isotherm’s two turning points are located, and they bracket the answer — a horizontal above the local maximum or below the local minimum cuts the curve once rather than three times, and there is nothing to make equal. Between those two pressures, every candidate line cuts the curve at three volumes, and the signed area it encloses varies smoothly from one sign at the bottom of the bracket to the other at the top. One bisection later, the areas agree to whatever precision is wanted.

That is worth saying because the bracket is easy to get wrong and the error is quiet. The obvious guess — search between zero and the pressure at the critical volume — fails above about 0.86 of the critical temperature, because the tie line sits above that pressure there, and the search then reports that no construction exists at a temperature where one plainly does. The figures here find the turning points first for exactly that reason.

The ground between the curves #

Sweeping the temperature and repeating the construction produces a map, and the map has three regions rather than two.

That middle band is the interesting part, and it is not a mathematical artefact. Both of its halves are producible.

Superheated liquid. Water in a clean, smooth container, heated gently and without dissolved gas, can be taken well past its boiling point without boiling — to about 280 °C at atmospheric pressure in careful work, and to well over 100 °C routinely in a microwave oven, where the heating is volumetric and there is nothing to nucleate on. It then flashes over all at once when disturbed, which is why the mug boils over on being lifted.

Supersaturated vapour. Vapour cooled below its condensation point without condensing is what a cloud chamber holds, and an ionising particle passing through leaves a trail of charged fragments that the vapour condenses on, drawing its own track.

Both states are stable against the small fluctuations that surround them and unstable against a large enough one. What separates “small” from “large” is a barrier, and the barrier has a size.

Why metastability survives at all #

A droplet forming in a supersaturated vapour has to pay for its surface before it is repaid by its bulk. The surface costs energy proportional to $r^2$ and the bulk repays energy proportional to $r^3$, so the free energy rises, peaks, and falls.

The pressure inside a curved surface exceeds the pressure outside by twice the surface tension over the radius, so a small droplet is at a higher pressure than a large one — and a droplet below a critical size evaporates even in a supersaturated vapour. That is the mechanical reason metastability survives: forming the new phase requires passing through sizes at which it is less stable, and nothing carries the system through them.

The free energy of a forming droplet has a barrier between the phase the substance is in and the phase it would rather be in — rising as the surface grows and falling as the volume does, with a maximum where the two cross. A fluctuation that clears the barrier grows without limit and one that does not collapses, so the transition waits for a fluctuation rather than proceeding when it becomes favourable.

The rate carries an exponential of minus the barrier over $kT$, which is what decides whether metastability lasts a microsecond or a year. That is the whole of the practical question: pure water in a clean container supercools to −40 °C because the barrier is high and the rate is negligible, and the same water with a speck of dust in it freezes at zero because the speck lowers the barrier to nothing.

A speck of dust, a scratch in the glass or a dissolved bubble reduces the barrier enormously, because a droplet forming on a surface has less of its own surface to pay for. That is the whole difference between a smooth microwaved mug and a kettle with limescale in it, and it is why bumping granules are added to laboratory flasks.

The point where the distinction stops existing #

As the temperature rises the two ends of the tie line approach each other; at the critical temperature they meet, and above it there is no line at all.

The existence of that end point is worth dwelling on. Solid and liquid are separated by a boundary with no end, because they differ by a symmetry — a crystal has one and a liquid does not, and a symmetry is either present or absent. Liquid and vapour differ only in density, which is a matter of degree, so their boundary can terminate. What can be gone round is what differs quantitatively; what cannot is what differs in kind.

From a kettle’s point of view, crossing the line is a plateau where heat goes in and the temperature does not move. That plateau is the coexistence region traversed at constant pressure — and the point where the distinction stops is where the plateau shrinks to nothing, which is the critical point. Above it there is no plateau, no latent heat, and no boiling to observe.

The lever rule, and what the horizontal line is a record of #

A point on the tie line is not a state of a homogeneous substance, and it is worth saying what it is a state of.

At a volume $v$ somewhere between the liquid volume $v_\ell$ and the vapour volume $v_g$, the sample is a mixture whose fractions follow from the volume adding up:

$$x_g = \frac{v - v_\ell}{v_g - v_\ell},$$

which is the lever rule, and is nothing more than the statement that total volume is the sum of the parts. Moving right along the line at fixed temperature and pressure converts liquid into vapour at a steady rate, and the heat that has to be supplied to do it is the latent heat.

So the horizontal line is a record of composition rather than of compression. Pushing a piston into a two-phase mixture does not raise the pressure; it condenses vapour, and the pressure stays exactly where the construction put it until the last of the vapour is gone. That is why a pressure gauge on a propane cylinder reads the same from full to nearly empty and then falls abruptly: the reading is the tie-line pressure at ambient temperature, which is a property of propane and the weather rather than of how much is left.

The two constants, and the gas that warms when it expands #

The equation has exactly two adjustable constants, and its most useful prediction is one in which the two fight each other and the winner depends on the temperature.

Let a gas expand through a valve or a porous plug, with no heat entering and no work extracted. The enthalpy is unchanged, and the question is what happens to the temperature. An ideal gas is unaffected — its energy depends on temperature alone, so nothing moves. A real gas changes temperature, and the sign is decided by the two corrections.

The attraction between molecules, which is the $a$ term, means that separating them costs energy. That energy comes out of the kinetic store, so expansion cools the gas. The finite size of the molecules, which is the $b$ term, works the other way: excluded volume makes the gas push harder than it otherwise would, so expanding it releases energy into the kinetic store and warms it.

Which wins depends on how much the molecules feel each other, and at high temperature they are moving too fast to be much deflected by an attraction. Carrying the calculation through gives a coefficient proportional to $2a/RT - b$, which changes sign at

$$T_{\text{inv}} = \frac{2a}{Rb} = \frac{27}{4}\,T_c.$$

The inversion temperature is six and three-quarter times the critical temperature, for every substance — another corresponding-states prediction, again wrong by tens of per cent and again right in structure and in order.

The consequences ran the whole of nineteenth-century cryogenics. Nitrogen’s critical temperature is 126 K, so its inversion temperature is well above room temperature and throttling compressed nitrogen at ordinary temperatures cools it. Repeat that in a counter-current heat exchanger, so each pass starts colder than the last, and the gas eventually liquefies. That is the Linde cycle, air was liquefied by it in 1895, and it works because of where nitrogen’s inversion temperature falls.

Hydrogen does not cooperate. Its critical temperature is 33 K, which puts its inversion temperature around 200 K — below room temperature. Compressed hydrogen throttled at room temperature gets hotter, and no amount of repetition liquefies it. Dewar’s solution in 1898 was to pre-cool hydrogen with liquid air first, taking it below its own inversion temperature before the cycle could begin to work.

Helium is worse again, with an inversion temperature near 40 K, and had to be pre-cooled with liquid hydrogen — which is why it was the last gas liquefied, in 1908, and why superconductivity was discovered three years later and not thirty years earlier. A two-constant correction to the ideal gas law predicted, in advance and correctly, which gases would need which others to get at them.

The inner curve is not really a line #

The map above draws two curves and calls the ground between them metastable, and the inner curve — the spinodal — is drawn as sharply as the outer. It should not be, and the reason is worth stating because it is the most important thing the mean-field picture gets wrong outside the critical region.

In the equation’s own terms the spinodal is unambiguous: it is where the compressibility changes sign, so inside it the homogeneous state has no barrier against separating and outside it there is one. That is a line, and crossing it is a qualitative change.

What actually happens is a crossover. As the spinodal is approached from outside, the nucleation barrier falls continuously toward zero, so the rate at which the metastable state decays rises continuously — through the eighty decades that make nucleation look like a threshold. Long before the barrier reaches zero it has fallen far enough that nothing survives long enough to be called metastable, so the practical edge of metastability is set by a rate rather than by the vanishing of a barrier.

There is a second failure alongside it, in the opposite direction. Near the spinodal the critical nucleus grows large and diffuse rather than small and sharp — it is no longer a droplet with a surface but a gentle undulation of density spread over many molecular diameters — so the droplet picture that gives the barrier its $r^2$ and $r^3$ terms stops applying exactly where the barrier is smallest.

Put together, the two mean that a sharp spinodal exists only in the mean-field limit, which is the limit of interactions with infinite range. For a real fluid with short-range forces, fluctuations smear the line into a band, and there is no experiment that locates it. What experiments locate instead is the kinetic limit of superheating: the temperature at which the decay rate reaches an observable value, which for water at atmospheric pressure is somewhere near 300 °C — comfortably short of where the equation puts its spinodal, and reproducible to a degree or two.

The general point is the one this essay keeps making about the equation. A mean-field theory draws sharp boundaries because it has averaged away the fluctuations that would blur them, and every such boundary should be read as the centre of a crossover rather than as a line. The coexistence curve survives that scrutiny, because it is an equilibrium condition; the spinodal does not, because it is a stability condition on an average.

Where the model stops #

Van der Waals’ equation is quantitatively poor and structurally right, and it is worth separating the two.

It predicts a critical compressibility factor $p_c v_c/RT_c = 3/8 = 0.375$ for every substance. The measured values are 0.23 for water, 0.29 for argon and 0.27 for nitrogen — wrong by twenty to forty per cent, and wrong in the same direction for everything.

The failure is worst near the critical point, and the reason is instructive. Every analytic equation of state predicts that the difference between the two densities vanishes as the square root of the distance from the critical temperature. The measured exponent is about 0.33, not 0.5, and no analytic equation can produce it — because near the critical point the fluctuations become correlated over long distances and a description in terms of the average density alone stops being adequate. That is a subject of its own, and it belongs to whoever writes about criticality; what belongs here is the boundary itself, and the statement that the mean-field picture is qualitatively right everywhere and quantitatively right nowhere near the point where the two phases merge.

A cycle on pressure–volume axes encloses an area that is work, and one crossing the coexistence region does its work partly at constant temperature — which is what makes a steam plant possible and what this whole diagram is ultimately for. The metastable regions are excursions off that cycle, and an engine that entered one would not be running the cycle its designer drew.

Three numbers from one equation, and the check they provide #

The reduced form used throughout is not a cosmetic simplification. It is the assertion that all substances obey one equation once each is measured in its own units, which is the law of corresponding states, and it is testable.

The critical point is where the isotherm has an inflection with zero slope — both the first and second derivatives of pressure with respect to volume vanish — and imposing those two conditions on van der Waals’ equation fixes the critical point in terms of the two constants:

$$v_c = 3b, \qquad p_c = \frac{a}{27b^2}, \qquad T_c = \frac{8a}{27Rb}.$$

Three quantities from two constants, so one combination of them is a pure number with no adjustable content: the compressibility factor at the critical point comes out at three-eighths for every substance. That is the prediction quoted above as failing by twenty to forty per cent, and it fails in an informative way. Every measured value is below three-eighths, and the ordering across substances tracks how non-spherical the molecules are — argon at 0.29, nitrogen at 0.27, water at 0.23 — so the discrepancy is a measure of everything the equation left out rather than a random error.

The law of corresponding states itself survives the failure of the number. Plotting measured coexistence curves for a dozen substances in reduced units puts them nearly on top of one another, which is the useful half of the claim, and it holds for the simple fluids to a couple of per cent even though the constant does not.

What the pictures cannot show #

The isotherm figure draws the metastable branches as continuous curves, which suggests a substance can be moved smoothly along them. It can, but only slowly and only in one direction: the branch can be followed outward from the tie line and cannot be followed back, because the return crosses the barrier the other way and nucleation is not reversible.

The coexistence map shows two densities at each temperature and says nothing about where they are. A real two-phase sample is liquid at the bottom and vapour above, separated by a meniscus, and none of the thermodynamics above knows that gravity exists.

And the tie line is drawn as though the transition were sharp in every variable. It is sharp in density and not in the correlation length, which grows steadily as the critical point is approached and is what makes a fluid near it scatter light so strongly that it turns milky — critical opalescence, which is a visible property of a state these axes have no room for.

Where the ladder goes next #

The phase-change ladder began with a boiling point being a pressure rather than a temperature and continued through the latent heat that goes in without warming anything. This rung has added the equation of state and the construction that reads it. The rungs beyond are the triple point, where three phases meet and the degrees of freedom run out; the solid-liquid boundary’s peculiar negative slope for water; and the first-order-against-second-order distinction, which asks whether a transition has a latent heat at all.

The habit worth carrying away is what to do with an equation that answers a question about an impossible state. It is tempting to treat the loop as an error and to look for a better equation. The better move is to notice that an analytic expression covering two phases must have one, and to ask what condition selects the physical answer out of it. The equation supplies the candidates and the thermodynamics chooses, and that division of labour — a model that overproduces and a principle that selects — recurs wherever a single smooth description is stretched across a boundary it does not know about.