The Phase Diagram of Water

Skate blades work because ice sits just below its melting point at 1 atm. Push a thin blade down, and the pressure under it rises enough to melt a whisker of ice into liquid water. This single fact — that squeezing ice can melt it — is written into the shape of water's phase diagram. Most substances get harder to melt as you squeeze them; water does the opposite. That single anomaly, traced to a negative-slope solid–liquid boundary, explains floating icebergs, frost heaving, and why lakes freeze from the top down. It also makes water's phase diagram the standard teaching example for the Clapeyron equation. This article maps out water's full $P$ – $T$ phase diagram: the triple point, the critical point, and the strange backward-leaning fusion line. You will see why the fusion slope is negative. Then you will compute real numbers from real data. Finally you will connect the diagram to the equation that governs every phase boundary you meet next.

Key takeaways

Water's phase diagram has three key points: the triple point (273.16 K, 611.657 Pa) and the critical point (647.10 K, 22.06 MPa).
The solid–liquid (fusion) boundary slopes backward — negative $dP/dT$ — because ice is less dense than liquid water.
This anomaly arises because $\Delta_{fus}V < 0$ : melting ice contracts the sample, unlike almost every other substance.
The triple point fixes the Kelvin scale by definition until 2019, and still anchors it via a fixed value today.
Above the critical point no liquid–gas interface exists; you have a single supercritical fluid.
Key idea: the slope of every phase boundary is set by the sign and size of $\Delta V$ and $\Delta S$ for that transition, and water's expansion on freezing flips the fusion line's sign.

The core insight

Every line in a one-component phase diagram is a Clapeyron line, $dP/dT = \Delta S/\Delta V$ , and water's fusion curve leans left instead of right because freezing water expands it, making $\Delta_{fus}V$ negative while $\Delta_{fus}S$ stays negative too — two negatives that combine into a negative overall slope.

The physical picture

A one-component phase diagram plots pressure against temperature, with regions of $P$ – $T$ space labeled by which single phase is stable and lines marking two-phase coexistence. This is the general framework from the study of one-component phase diagrams. Water follows that framework, but with one striking twist in the solid–liquid boundary. Follow the diagram from low temperature. At low $T$ and low $P$ , ice (specifically ordinary hexagonal ice, "ice Ih") sits in equilibrium with water vapor along the sublimation curve. This curve rises steeply, ending at the triple point, $T_3 = 273.16\text{ K}$ , $P_3 = 611.657\text{ Pa}$ (about 0.006 atm). Here solid, liquid, and vapor coexist simultaneously — the only point on the whole diagram where all three phases meet with zero degrees of freedom, exactly as the phase rule $F = C - P + 2$ demands for $C=1$ , $P=3$ . From the triple point, the vaporization curve climbs up and to the right, tracing the vapor pressure of liquid water as temperature rises. It terminates abruptly at the critical point, $T_c = 647.10\text{ K}$ (373.95 °C), $P_c = 22.06\text{ MPa}$ (about 217.7 atm). Beyond this point, the meniscus between liquid and gas vanishes; density differences between the two phases disappear and you have one continuous supercritical fluid. This is why supercritical water and supercritical CO₂ are used as tunable, meniscus-free industrial solvents. The real anomaly is the third curve, the fusion (melting) line, running from the triple point up and to the left — toward lower temperature as pressure increases. For almost every other pure substance (CO₂, benzene, most metals) this line tilts to the right: squeezing the solid raises its melting point, because the solid phase is denser and compressing favors the smaller-volume phase. Water inverts this because liquid water is denser than ice at 0 °C. Compressing the ice–water system favors the denser phase — the liquid — so increasing pressure lowers the melting point. Ice under enough pressure literally melts.

Figure

Schematic P–T phase diagram of water, not to scale

The negative fusion slope is exaggerated in the sketch — in reality it needs about 13.5 MPa (roughly 133 atm) to lower the melting point by just 1 K, so the line is nearly vertical but leans unmistakably left.

The core relations

\frac{dP}{dT} = \frac{\Delta S}{\Delta V} = \frac{\Delta H}{T\Delta V}

where: $\Delta S$ , $\Delta H$ , $\Delta V$ are the entropy, enthalpy, and volume change of the transition at temperature $T$ (the Clapeyron equation, exact for any two-phase boundary).

\Delta_{fus}V = V_m(\text{liquid}) - V_m(\text{solid}) < 0 \quad \text{for water at } 273\text{ K}

where: $V_m$ is molar volume; ice has a larger molar volume (lower density, 0.9167 g/cm³) than water (0.99984 g/cm³) at 0 °C.

F = C - P + 2

where: $F$ is the number of degrees of freedom, $C=1$ component, $P$ the number of coexisting phases (Gibbs phase rule; $F=0$ at the triple point).

\left(\frac{dP}{dT}\right)_{fusion} = \frac{\Delta_{fus}H}{T\,\Delta_{fus}V} \approx -13.4\ \text{MPa K}^{-1}

where: using $\Delta_{fus}H = 6.01\text{ kJ mol}^{-1}$ and $\Delta_{fus}V = -1.63\times10^{-6}\text{ m}^3\text{ mol}^{-1}$ at 273.15 K, 1 atm.

The core relations — derivation

Derivation AWhy the fusion line bends backward

Goal. Show that the sign of $dP/dT$ along the solid–liquid boundary is fixed by the sign of $\Delta_{fus}V$ , and evaluate it numerically for water.

Step 1. Start from equilibrium of chemical potentials. Along any two-phase coexistence curve, the chemical potentials of the two phases are equal at every point:

\mu_{solid}(P,T) = \mu_{liquid}(P,T)

Moving along the curve by $(dP, dT)$ keeps both phases in equilibrium, so both chemical potentials must change by the same amount:

d\mu_{solid} = d\mu_{liquid}

Step 2. Use the fundamental relation for $\mu$ . For a pure substance, $d\mu = -S_m\,dT + V_m\,dP$ (from $dG = -S\,dT + V\,dP$ per mole). Substituting for each phase:

-S_{m,solid}\,dT + V_{m,solid}\,dP = -S_{m,liquid}\,dT + V_{m,liquid}\,dP

Step 3. Collect terms. Rearranging groups the entropy terms and volume terms separately:

(S_{m,liquid} - S_{m,solid})\,dT = (V_{m,liquid} - V_{m,solid})\,dP

\frac{dP}{dT} = \frac{\Delta_{fus}S}{\Delta_{fus}V}

This is the Clapeyron equation, exact and general — no ideal-gas or small- $\Delta V$ assumption used.

Step 4. Convert entropy to enthalpy at equilibrium. Since fusion at temperature $T$ occurs reversibly at constant $T$ and $P$ , $\Delta_{fus}S = \Delta_{fus}H / T$ . Substituting:

\frac{dP}{dT} = \frac{\Delta_{fus}H}{T\,\Delta_{fus}V}

Step 5. Insert the sign of each factor for water. Melting always absorbs heat, so $\Delta_{fus}H > 0$ for every substance, including water ( $+6.01\text{ kJ mol}^{-1}$ ). Temperature $T=273.15\text{ K}$ is always positive. The only place a substance's personality shows up is $\Delta_{fus}V$ . For water, liquid is denser than the solid, so

\Delta_{fus}V = V_{m,liquid} - V_{m,solid} < 0

Every other common factor is positive, so the whole right-hand side is negative.

Step 6. Numeric evaluation. Molar mass of water is 18.015 g/mol. At 0 °C, ice density is 0.9167 g/cm³ giving $V_{m,solid} = 19.65\text{ cm}^3\text{/mol}$ ; liquid water density is 0.99984 g/cm³ giving $V_{m,liquid} = 18.018\text{ cm}^3\text{/mol}$ . Then

\Delta_{fus}V = 18.018 - 19.65 = -1.63\ \text{cm}^3\text{mol}^{-1} = -1.63\times10^{-6}\ \text{m}^3\text{mol}^{-1}

Plugging into Step 4 with $\Delta_{fus}H = 6.01\times10^3\text{ J mol}^{-1}$ and $T = 273.15\text{ K}$ :

\frac{dP}{dT} = \frac{6.01\times10^{3}}{273.15 \times (-1.63\times10^{-6})} = -1.34\times10^{7}\ \text{Pa K}^{-1} \approx -13.4\ \text{MPa K}^{-1}

Boxed result.

\boxed{\dfrac{dP}{dT}\bigg|_{fusion} \approx -13.4\ \text{MPa K}^{-1}, \text{ negative because } \Delta_{fus}V < 0}

Worked examples

Example 1 — How much pressure to melt ice by 1 °C? (foundational) Question: estimate the pressure increase needed to lower the melting point of ice by exactly 1.00 K, starting from the normal melting point (273.15 K, 1 atm = 101,325 Pa). Step 1. Because $\Delta_{fus}V$ and $\Delta_{fus}H$ change negligibly over such a small interval, treat $dP/dT$ as constant and equal to the value just derived:

\frac{dP}{dT} \approx -13.4\ \text{MPa K}^{-1}

Step 2. For $\Delta T = -1.00\text{ K}$ (cooling target, i.e. lowering the melting point by 1 K):

\Delta P = \frac{dP}{dT} \times \Delta T = (-13.4\times10^6\ \text{Pa K}^{-1})\times(-1.00\ \text{K}) = +1.34\times10^7\ \text{Pa}

Step 3. Add to the starting pressure:

P_{final} = 101{,}325\ \text{Pa} + 1.34\times10^7\ \text{Pa} \approx 1.35\times10^{7}\ \text{Pa} = 133\ \text{atm}

Sanity check: this matches the well-known textbook figure that it takes roughly 130–140 atm to depress water's freezing point by 1 K — consistent with the famous (if slightly overstated) claim that ice-skate blades melt ice under the skater's weight. A 70 kg skater on a blade contact area of ~10⁻⁴ m² produces only about 7×10⁶ Pa (≈70 atm), enough to lower the melting point by roughly 0.5 K — real ice skating relies mainly on frictional heating, but the pressure effect is genuine and in the right direction.

Example 2 — Locating a point relative to the phase diagram (applied) Question: liquid water is held at 300 °C (573.15 K) and 15.0 MPa. Using $T_c = 647.10\text{ K}$ , $P_c = 22.06\text{ MPa}$ , determine whether this is subcritical liquid, subcritical vapor, or supercritical fluid, and estimate whether it lies below or above the normal boiling curve. Step 1. Compare $T$ to $T_c$ :

573.15\text{ K} < 647.10\text{ K}

so the sample is below the critical temperature — a genuine liquid/vapor distinction is still possible, this is not supercritical fluid on temperature grounds alone. Step 2. Compare $P$ to $P_c$ :

15.0\text{ MPa} < 22.06\text{ MPa}

also below critical pressure, consistent with a subcritical two-phase system being possible at this $T$ . Step 3. Check against the real vaporization curve. The saturation (boiling) pressure of water at 300 °C is experimentally about 8.59 MPa (from steam tables). Since the actual pressure (15.0 MPa) exceeds the saturation pressure at that temperature,

P_{applied} = 15.0\text{ MPa} > P_{sat}(300^\circ\text{C}) = 8.59\text{ MPa}

the system sits on the liquid side of the vaporization curve — it is compressed liquid water, not vapor. Step 4. Sanity check using the diagram logic: at fixed $T$ , increasing $P$ above $P_{sat}$ always pushes you into the liquid region (higher pressure favors the denser, smaller-volume phase — ordinary behavior, unlike the fusion line). Since $573\text{ K} < 647\text{ K} = T_c$ , a phase boundary genuinely exists here, and the point sits just above it: stable liquid water, not supercritical fluid, at 300 °C and 15.0 MPa.

Limiting cases & common traps

Situation	Correct picture	Common error
Fusion line slope	Negative for water: $\Delta_{fus}V<0$	Assuming all melting curves tilt right like CO₂ or benzene
Above $T_c$ and $P_c$	One supercritical fluid, no meniscus	Calling it "compressed gas" or "dense liquid" — neither term applies
Triple point vs standard melting point	Triple point of water is 273.16 K at 611.657 Pa; standard melting point (1 atm) is 273.15 K	Treating these as the same point — they differ because the fusion line has finite (negative) slope
Ice skating	Pressure melting contributes, but frictional heating dominates	Believing blade pressure alone fully explains low-friction ice
Sublimation curve steepness	Governed by $\Delta_{sub}H = \Delta_{fus}H+\Delta_{vap}H \approx 51$ kJ/mol, much larger than $\Delta_{fus}H$	Confusing sublimation and fusion enthalpies when estimating slopes

The chemist behind it

The precision behind the modern triple-point value of water traces to decades of careful calorimetry and metrology, but the conceptual foundation — that a phase boundary's slope is fixed by $\Delta S$ and $\Delta V$ — belongs to Benoît Paul Émile Clapeyron, whose 1834 analysis of Carnot's cycle produced the equation used throughout this derivation. Water's own peculiar melting curve was already known qualitatively to 19th-century physicists (James Thomson, brother of Lord Kelvin, predicted the pressure-melting effect in 1849, confirmed experimentally by Kelvin himself soon after), making water's phase diagram one of the oldest quantitative tests of Clapeyron's relation.

Origins · 1 minBenoît Paul Émile Clapeyron and how this was discovered

Benoît Paul Émile Clapeyron

1799–1864 · French engineer-physicist who turned Carnot's abstract cycle into a quantitative equation of state relation, giving every phase boundary in this diagram its governing law

Ask the tutor: How does the Clapeyron equation simplify into the Clausius–Clapeyron equation for the vaporization and sublimation curves?

Why doesn't every liquid show this pressure-melting effect like water does?

Most solids are denser than their melts, so $\Delta_{fus}V = V_{liquid}-V_{solid} > 0$ , giving a normal positive fusion slope: squeezing favors the smaller-volume solid and raises the melting point. Water is unusual because hydrogen bonding builds an open, tetrahedral hexagonal lattice in ice that is less space-efficient than the more disordered, closer-packed liquid. Only a handful of substances share this property, including silicon, gallium, bismuth, and antimony — all of which also expand on freezing.

Is the triple point the same as water's normal melting point at 1 atm?

No, though they're close. The triple point is 273.16 K at 611.657 Pa, while the normal (1 atm) melting point is 273.15 K. The 0.01 K difference arises because the fusion curve has a small negative slope, so moving from 611.657 Pa up to 101,325 Pa (1 atm) shifts the equilibrium melting temperature down very slightly, by about 0.01 K, consistent with the ≈−7.4×10⁻⁸ K/Pa slope implied by −13.4 MPa/K.

What happens to water above the critical point exactly?

Above $T_c = 647.10$ K and $P_c = 22.06$ MPa, the distinction between liquid and gas disappears entirely: densities converge, the surface tension goes to zero, and no meniscus can form no matter how you adjust pressure. This single phase, called supercritical water, has solvent properties tunable continuously between gas-like and liquid-like by adjusting pressure and temperature, which is why it is used industrially for oxidizing waste and as a green reaction medium.

How does the phase rule apply to the different regions of water's diagram?

For one-component water, $F = C - P + 2 = 3 - P$ . In a single-phase region (ice alone, or liquid alone, or vapor alone), $P=1$ so $F=2$ : you can vary $T$ and $P$ independently. Along any coexistence curve, $P=2$ so $F=1$ : fixing $T$ fixes $P$ (or vice versa). At the triple point, $P=3$ so $F=0$ : no freedom remains, fixing the exact values 273.16 K and 611.657 Pa.