A boiling point is a pressure, not a temperature

Nothing about water says one hundred degrees. What says it is the air standing on top of it — ten tonnes of atmosphere per square metre, 101 kilopascals — and take that air away and the number goes with it. On the summit of Everest water at a rolling boil is at 71 °C, and in a sealed pot at twice atmospheric pressure it is at 119.5 °C. Neither is a different substance.

Everything below is where that curve comes from — one equation with one input — and where reading it backwards turns a thermometer into an altimeter.

Where boiling actually happens #

Water does not wait for 100 °C to become vapour. It evaporates at every temperature, because the distribution of molecular speeds always has molecules in its tail energetic enough to leave the surface. Boiling is a stronger condition: vapour forming in the bulk of the liquid, in bubbles, rather than only at the free surface.

A bubble can exist only if the vapour inside it pushes out at least as hard as the liquid around it pushes in. So the condition for boiling is an equality between two quantities that have nothing to do with each other: the vapour pressure of the liquid, which depends on the liquid and its temperature and on nothing else, and the ambient pressure, which is set by the weight of whatever is above and knows nothing about what is in the pot. The boiling point is where the two cross, and is therefore a property of the pair rather than of either member — which is the whole claim of this page.

Vapour pressure itself is the exponential that runs through the whole of thermal physics: the population of a state falls as $e^{-E/kT}$, and the state here is “out of the liquid”, whose price is the latent heat.

The steepness of that exponential is what makes the boiling point sharp, and it is worth putting a number on. An energy of ten kT costs a factor of $10^{-4.3}$, twenty kT costs $10^{-8.7}$, and thirty kT costs $10^{-13.0}$; at 300 K one kT is 25.9 meV. Water’s vaporisation enthalpy of 40.65 kJ/mol is 0.42 eV per molecule, which at its own boiling point is 13.1 kT — far enough out along the exponential that the vapour pressure changes by about three and a half per cent for every kelvin. A pot does not drift gently through boiling; it arrives.

That last number is the essay in miniature. If a kelvin is worth 3.5 per cent of a pressure, then a boiling point measured to a kelvin fixes a pressure to 3.5 per cent, and the reverse. The curve reads in either direction, and both readings have been made into instruments.

Clausius–Clapeyron, in one derivation #

On a phase boundary the two phases coexist, and coexistence has one meaning: the two arrangements have the same free energy. Neither is preferred, which is why any mixture of them is stable and why the transition absorbs heat at no change of temperature.

Write $g$ for the Gibbs free energy per mole. The boundary is the locus where $g_1 = g_2$. Move a short distance along it, and both change by the same amount, because they remain equal. Each changes by $\mathrm{d}g = -s\,\mathrm{d}T + v\,\mathrm{d}p$, so

$$-s_1\,\mathrm{d}T + v_1\,\mathrm{d}p = -s_2\,\mathrm{d}T + v_2\,\mathrm{d}p \quad\Longrightarrow\quad \frac{\mathrm{d}p}{\mathrm{d}T} = \frac{\Delta s}{\Delta v} = \frac{L}{T\,\Delta v},$$

using $\Delta s = L/T$, which is the entropy jump the latent heat pays for. That is Clapeyron’s relation, and there is no approximation in it at all — only the second law and the definition of a latent heat. The slope of a phase line is a latent heat divided by a temperature and a volume change.

Two approximations turn Clapeyron’s relation into something integrable. Treat the vapour as an ideal gas, so that $v_{\text{gas}} = RT/p$, and neglect the volume of the liquid, which at 100 °C is 0.06 per cent of the vapour’s. Then $\Delta v \approx RT/p$ and

$$\frac{\mathrm{d}\ln p}{\mathrm{d}T} = \frac{L}{RT^2}, \qquad p = p_0\exp\left[-\frac{L}{R}\left(\frac1T - \frac1{T_0}\right)\right].$$

The second form is the whole boundary. One latent heat, one point on the curve, and every other point follows. Nothing about hydrogen bonds, molecular shape or intermolecular forces appears; all of that is compressed into the single number $L$.

The gap between those two figures is nine kelvin, and it is not a gap in the physics. It is the cost of one arbitrary choice about which single value to hold constant — a fair warning about how much a “constant” latent heat is carrying.

The straight line whose slope is the latent heat #

Take logarithms of the integrated form and the boundary becomes a straight line — not approximately, but exactly, for constant $L$:

$$\log_{10} p = \log_{10} p_0 - \frac{L}{R\ln 10}\left(\frac1T - \frac1{T_0}\right).$$

Plotted with $\log p$ up the axis and $1/T$ along it, the gradient is $-L/R$. That slope is not a summary of the curve, nor a parameter correlated with the latent heat. It is the latent heat, in different units.

This is why a chemist measuring a latent heat boils a liquid at four or five pressures, plots the logarithm against the reciprocal temperature, and reads the answer off with a ruler rather than building a calorimeter. It is a measurement of an energy made entirely out of pressures and temperatures, and it works because the exponent is an energy over a temperature and nothing else — the same reason the atmosphere’s scale height is $kT/mg$.

The slope is a volume ratio, and water’s leans backwards #

On the vaporisation curve $\Delta v$ is enormous — a mole of water occupies 18 cm³ as a liquid and 30 litres as a vapour at its boiling point — and the latent heat dominates. On the melting curve $\Delta v$ is a difference between two condensed phases, a few per cent of either, and it decides everything.

Water’s solid is 917 kg/m³ against 1,000 for its liquid, so melting shrinks it: $\Delta v = M(1/\rho_\ell - 1/\rho_s) = -1.63 \times 10^{-6}$ m³/mol. With $L_{\text{fus}} = 6.01$ kJ/mol at 273.16 K, Clapeyron gives

$$\frac{\mathrm{d}p}{\mathrm{d}T} = \frac{6010}{273.16 \times (-1.63\times10^{-6})} = -13.5\ \text{MPa/K}.$$

The negative sign has one cause and one only: ice floats. Every other feature of water is irrelevant to it. Carbon dioxide’s solid is denser than its liquid, 1,562 against 1,178 kg/m³, and its melting line accordingly leans the other way at +4.5 MPa/K.

What that slope retires. Thirteen and a half megapascals per kelvin sounds steep, and it means the opposite: a whole atmosphere of extra pressure lowers the melting point of ice by 0.0075 K. Across the entire pressure axis of the hero figure — six and a half decades, up to 70 MPa, seven hundred atmospheres — the melting line moves 5.2 K. Skating happens at −5 or −10 °C, which would require 70 to 135 MPa held under the blade, and the melting line does not even reach that far: it terminates at 209.9 MPa and 251.2 K, where a denser crystal, ice III, takes over. Below −22 °C no pressure whatsoever melts ice.

So pressure melting cannot be why ice is slippery. The modern account has two parts, neither of them on the phase diagram: frictional heating, which melts a film at the contact and is why a slow-moving skate grips; and a disordered, liquid-like surface layer a few molecules thick that exists on ice at −10 °C at atmospheric pressure, because a molecule at a free surface has fewer neighbours to bond to. That second mechanism belongs to the physics of surfaces and the angles they make, not to Clapeyron.

The curve inverted, and what it costs #

Run the boundary the other way — pressure in, temperature out — and it becomes an instrument. What is needed first is a pressure profile, and the isothermal exponential is not good enough for it.

The isothermal column is a fine argument and a poor altimeter. Its scale height depends on the temperature it assumes — 6.4 km at 220 K, 8.4 km at 288 K, 11.7 km at 400 K — and the real atmosphere is not at any one of them, so by 20 km the 288 K model is out by seventy-one per cent. That matters here in a way it does not in the essay that derives it, because the error is being inverted: thirty per cent of a pressure is about three kelvin of boiling point, which is the difference between a usable instrument and a misleading one. What follows therefore uses a standard atmosphere with a 6.5 K/km lapse rate.

Cooking stops working. At 71 °C a pot on Everest is at a full, vigorous boil and cannot be made hotter by any amount of fuel, because every extra joule goes into vapour rather than into temperature. Starch gelatinises between about 70 and 80 °C and collagen breaks down near 90 °C, so at that pressure boiling is no longer a cooking method — however long the pot is left. Expeditions carry pressure cookers for that reason and not for speed.

The autoclave’s 121 °C is not a round number. Bacterial spores survive boiling water for hours; the temperature at which they are reliably killed in fifteen minutes is 121 °C, and the phase boundary says that reaching it needs a little over two atmospheres — the fifteen pounds per square inch of gauge pressure written on every autoclave. The figure above puts two atmospheres at 119.5 °C. The 19.5 kelvin between that and an open pot look small and are not: with an activation energy of about 300 kJ/mol, which is typical of the protein denaturation that kills a spore, the Boltzmann factor above makes those 19.5 kelvin a factor of 120 in rate. Fifteen minutes at 121 °C is thirty hours at 100 °C. Sterilisation is possible at all only because the boiling point can be moved.

A thermometer is an altimeter. Read the same curve as a measurement of pressure and a thermometer in boiling water becomes a hypsometer, which nineteenth-century surveyors carried up mountains because it is lighter and less breakable than a mercury barometer.

It shares the barometric altimeter’s weakness exactly, and for the same reason: pressure at a fixed altitude moves by a few per cent with the weather, so a hypsometric height is only as good as the sea-level pressure assumed for it. Hence the second thermometer, at a known station.

Where the model stops #

The two neglects are tiny at the boiling point and fatal at the critical point. At 373 K the liquid’s volume is 0.06 per cent of the vapour’s and the vapour’s compressibility factor is about 0.985, so both cost well under two per cent. At the critical point the two densities are equal, $\Delta v$ is zero, and the constant-latent-heat curve reaches 37.4 MPa where the measured critical pressure is 22.1 — 70 per cent high. The cause is not the ideal gas but the latent heat: $L$ falls to zero there, and the integrated curve has never heard of that.

The size of that error depends on which single latent heat is chosen. Integrating with the 40.65 kJ/mol measured at the boiling point, as the straight-line figure does, reaches 26.0 MPa at the critical temperature instead — 18 per cent high rather than 70. Two defensible choices of one constant differ by a factor of four in the error, which measures how badly that constant is being asked to behave.

$L$ is not constant over the range at all. Water’s vaporisation enthalpy is 45.05 kJ/mol at the triple point and 40.65 at the boiling point, and the single value the two published endpoints imply, 43.32, is neither. The drift is measurable one atmosphere outside the fitted interval: at two atmospheres the curve gives 392.6 K against a measured 393.75, 1.15 K out.

The boundary is an equilibrium statement and says nothing about nucleation. A bubble has to pay for its own surface before it can grow, and a very small bubble has a very large internal pressure for that reason, so a liquid with nowhere for a bubble to start can be carried above its boiling point. Clean water in a smooth vessel in a microwave routinely reaches several degrees over 100 °C and then boils violently when disturbed. Bumping, superheating and the delay before a kettle actually boils all live outside this page, in the kinetics.

Nothing here is about mixtures. Dissolved salt lowers the vapour pressure and raises the boiling point; two miscible liquids boil over a range. Every line drawn here is for one pure substance.

The same relation, with a different Δv #

Clapeyron’s relation makes no reference to which phases are involved, so it applies wherever two phases meet.

Sublimation. At the triple point all three boundaries meet and the same molecule can take either route, so the latent heats add: $L_{\text{sub}} = L_{\text{fus}} + L_{\text{vap}} = 51.1$ kJ/mol for water. The sublimation curve in the hero figure is therefore not measured but predicted, and it reaches 103.2 Pa at 253.1 K against a measured 253.15 — agreement to within a twentieth of a kelvin, from an addition. Below the triple point’s 611 Pa there is no liquid water at any temperature, which is what freeze-drying exploits.

Helium, where the relation inverts and then breaks. Helium-4’s melting curve has a minimum, at about 0.775 K and 2.93 MPa. Clapeyron says immediately what that means: $\mathrm{d}p/\mathrm{d}T = 0$ requires $\Delta s = 0$, so at that point the solid and the liquid have the same entropy — and below it the solid is the more disordered phase, because the liquid has become an ordered quantum fluid. A little further along, the superfluid transition has $\Delta s = 0$ and $\Delta v = 0$ simultaneously, and Clapeyron’s relation degenerates to $0/0$: a second-order transition has no latent heat and no volume jump, and needs Ehrenfest’s replacement, which relates the slope to jumps in heat capacity and compressibility instead.

Chemical equilibrium. Replace the two phases by reactants and products and the identical algebra gives $\mathrm{d}\ln K/\mathrm{d}T = \Delta H/RT^2$ — the van 't Hoff equation, with a reaction enthalpy in place of a latent heat, and the reason a plot of $\ln K$ against $1/T$ yields $\Delta H$ from its gradient. With $\Delta v$ the volume change of the reaction instead, the same relation is Le Chatelier’s rule about pressure made quantitative: an equilibrium shifts towards the side of smaller volume, at a rate the volume change sets.

Clapeyron, Clausius, and a defined point on the kelvin #

Émile Clapeyron published the relation in 1834, in the memoir that rescued Carnot’s argument from obscurity by putting it into calculus and drawing the first indicator diagram of the Carnot cycle. His derivation was a cycle, not a free energy: vaporise a mole of liquid at temperature $T$, absorbing $L$; expand; condense at $T - \mathrm{d}T$; return.

The derivation is worth following because it needs nothing that was not already available in 1834. Run the loop between a liquid and its own vapour rather than between two gas states, and the area it encloses on pressure–volume axes is $\Delta v \,\mathrm{d}p$ — the change in volume on vaporising, times the small pressure difference between the two isotherms. That area is the net work. Meanwhile the Carnot ceiling fixes the work obtainable per unit of heat absorbed at $\mathrm{d}T/T$, and the heat absorbed is $L$. Equating the two expressions for the same work gives $\mathrm{d}p/\mathrm{d}T = L/(T\Delta v)$ directly, with no free energy anywhere in it.

The work round such a loop is the enclosed area, $\Delta v\,\mathrm{d}p$; the heat taken in is $L$; and Carnot’s ceiling fixes the ratio at $\mathrm{d}T/T$ for a reversible cycle. Setting $\Delta v\,\mathrm{d}p = L\,\mathrm{d}T/T$ gives the relation in a line. The slope of a phase line is thus a consequence of the second law, obtained sixteen years before Clausius named entropy. Clausius recast it in 1850 and supplied the ideal-vapour integration, which is why the integrated form carries both names while the exact relation carries only one.

The prediction that followed fastest was the melting line’s sign. James Thomson argued in 1849 that because ice expands on freezing, pressure must lower its melting point, and computed the size — about 0.0075 °C per atmosphere. His brother William measured it the following year and found 0.0074. A relation derived from a heat engine, applied to a block of ice, predicting seven thousandths of a degree that a mid-century laboratory could confirm, is as clean a test as thermodynamics offers.

The triple point earned a stranger distinction. Because it is a single point rather than a curve, it is reproducible in any laboratory with no calibration against anything else, and from 1954 until 2019 the kelvin was defined as 1/273.16 of the thermodynamic temperature of water’s triple point — a unit fixed by a phase boundary’s intersection. The 2019 redefinition fixed Boltzmann’s constant instead, at $1.380649\times10^{-23}$ J/K, and the triple point became a measured quantity again: 273.16 K, now with an uncertainty attached.

What the picture cannot show #

How much of each phase there is. A point on a phase line describes the two phases in any proportion — the free energies are equal, so the mixture is free. The whole boiling plateau in the heating figure, all 2,260 kJ of it, is one point on the vaporisation curve, and the fraction vaporised is a coordinate the p–T plane does not have.

The branches that lose. The diagram draws only the phase that wins. Supercooled liquid water has a higher vapour pressure than ice at the same temperature, and its curve is the vaporisation line extended below the triple point — invisible here, and the reason that in a cloud holding both ice crystals and supercooled droplets the ice grows while the droplets evaporate. Most rain in temperate latitudes begins that way, and a drawing of stable phases cannot show the metastable ones doing the work.

Time. There is no rate anywhere on the diagram. It says where a substance is heading and never how long the journey takes, which is the gap superheating and slow molecular transport live in.

Six and a half decades flatten everything familiar. The logarithmic pressure axis is what makes the whole boundary fit on one page, and it compresses the entire range a kitchen or a weather system ever sees into a sliver near 10⁵ Pa. On a linear axis the curve would hug the temperature axis and then turn vertical, and the exponential structure that is the whole content would be unreadable.

Why $L$ is 40 kJ/mol. Nothing in the p–T plane explains the one number the curve needs. That comes from hydrogen bonding — from the energy it takes to remove a molecule from its neighbours — and belongs to a molecular account, of the kind equipartition and the physics of surfaces both draw on. The phase boundary takes $L$ as given.

The ladder from here #

The rungs above this one: nucleation, and the barrier a bubble must pay before it can grow; the critical point, where the latent heat vanishes and the distinction between liquid and vapour ends; the Gibbs phase rule, which gets the triple point’s uniqueness from arithmetic; second-order transitions, where Clapeyron’s relation reads $0/0$ and Ehrenfest’s takes over; and the Maxwell construction, which locates the boundary on a van der Waals loop by making two areas equal.

The neighbouring ladders are close. The heat that changes no temperature is where the latent heat came from; what a system actually minimises is why equal free energies is the condition; the exponential that decides everything is the vapour pressure’s other face; and pressure as a rate of arrival is what the vapour does on the far side of the surface.

The claim to carry forward is the title. A phase boundary is not a table of measurements but one number’s consequence, and the temperature written on the side of a kettle is a fact about the sky.