One-Component Phase Diagrams

Fill an ice cube tray with water and put it in the freezer. You are walking a path across a map you have never seen drawn, but that map exists, and it is called a phase diagram. Every point on it records which state of matter is stable at a given pressure and temperature. One-component phase diagrams matter because they compress an enormous amount of thermodynamic information into a single picture. Instead of memorizing melting points and boiling points at every pressure, you read them off curves. Engineers use these diagrams to design autoclaves and CO₂ fire extinguishers; geologists use them to infer the depth at which diamond forms. This article builds the p–T diagram for a pure substance from the ground up. You will learn what the regions mean. You will learn why the boundary lines have slopes. You will meet the triple point and the critical point, and see why one is fixed and the other can vanish into a single fluid phase.

Key takeaways

A one-component phase diagram plots pressure against temperature; each point specifies a unique equilibrium state.
Single-phase regions (solid, liquid, gas) are two-dimensional areas because the Gibbs Phase Rule allows two degrees of freedom there.
Boundary lines are one-dimensional because two phases coexist only along curves, leaving one degree of freedom.
The triple point is a single, invariant point where three phases coexist simultaneously.
The critical point marks the end of the liquid–vapor line, beyond which liquid and gas become indistinguishable.
Key idea: the number of degrees of freedom left over from the Phase Rule dictates whether you get an area, a line, or a point on the diagram.

The core insight

A one-component p–T phase diagram is a direct graphical translation of the Gibbs Phase Rule: regions where $F = 2$ become areas, coexistence conditions where $F = 1$ become curves, and the unique condition where $F = 0$ becomes the triple point.

The physical picture

Consider a sealed, rigid container holding a pure substance—say, carbon dioxide—at some fixed amount of matter. You control only two external variables: pressure $p$ and temperature $T$ . Every possible $(p,T)$ pair you could dial in corresponds to exactly one point on a two-dimensional map. The one-component phase diagram is simply that map, partitioned into regions according to which phase (or phases) is thermodynamically stable there. Recall the Gibbs Phase Rule, $F = C - P + 2$ , where $C$ is the number of components and $P$ the number of phases present at equilibrium. For a one-component system, $C = 1$ , so $F = 3 - P$ . When only one phase is present ( $P=1$ ), $F = 2$ : you can vary both $p$ and $T$ independently within a finite range and the single phase persists. This is why solid, liquid, and gas each occupy an open two-dimensional area on the diagram. When two phases coexist ( $P=2$ ), $F = 1$ : only one variable is free. Fix the temperature and the coexistence pressure is determined; this is precisely the vapor pressure at that temperature. As you sweep $T$ , the coexistence pressure traces a curve—the phase boundary. Three such curves generically appear: solid–liquid (fusion), liquid–gas (vaporization), and solid–gas (sublimation). When three phases coexist simultaneously ( $P=3$ ), $F = 0$ : there are zero degrees of freedom. Both $p$ and $T$ are pinned to unique values. This is the triple point, where the three boundary curves meet. It is not a region and not a line—it is a single, invariant point fixed by the substance's own molecular properties, immune to any external tuning by the experimenter. The liquid–gas boundary behaves differently from the other two: it terminates. As you move up that curve toward higher $p$ and $T$ , the liquid and gas become progressively more similar in density, until at the critical point ( $p_c$ , $T_c$ ) the distinction disappears entirely. Beyond this point, no boundary exists; the substance is a single supercritical fluid phase that fills space like a gas but can dissolve substances like a liquid.

Figure

Schematic p–T phase diagram of a pure substance, showing regions, boundary curves, the triple point, and the critical point

The core relations

F = C - P + 2

where: $F$ = degrees of freedom, $C$ = number of components, $P$ = number of phases in equilibrium.

F = 3 - P \quad (C = 1)

where: for a pure substance, only $P$ determines whether you get an area ( $P=1$ ), a curve ( $P=2$ ), or a point ( $P=3$ ).

\left(\frac{dp}{dT}\right)_{\text{coex}} = \frac{\Delta_{\text{trs}} H}{T\,\Delta_{\text{trs}} V}

where: this is the Clapeyron relation, giving the slope of any two-phase boundary; $\Delta_{\text{trs}}H$ and $\Delta_{\text{trs}}V$ are the transition enthalpy and volume change.

P_{\max} = C + 2

where: the maximum number of phases that can coexist simultaneously in a one-component system, occurring only when $F=0$ .

Derivation AWhy regions, lines, and points must have those exact dimensionalities

Goal. Show that the Phase Rule, applied to $C=1$ , forces single phases into areas, two-phase equilibria into curves, and three-phase equilibria into an isolated point—with no other possibility.

Step 1. Write the Phase Rule for one component. Starting from $F = C - P + 2$ and setting $C = 1$ :

F = 1 - P + 2 = 3 - P

This is an exact bookkeeping identity from Gibbs' derivation (each phase contributes $C+2$ intensive variables, mutual equilibrium imposes $C(P-1)$ independent equalities of chemical potential, plus $P-1$ equalities of $p$ and $T$ shared across phases—the algebra collapses to $F = C - P + 2$ ).

Step 2. Evaluate for $P = 1$ .

F = 3 - 1 = 2

Two independent intensive variables, $p$ and $T$ , may be chosen freely (within limits) while the system remains single-phase. Geometrically, the locus of all allowed $(p,T)$ pairs is an open two-dimensional set in the plane—a region with nonzero area.

Step 3. Evaluate for $P = 2$ .

F = 3 - 2 = 1

Only one intensive variable is independently choosable. If you fix $T$ , the coexistence pressure $p_{\text{coex}}(T)$ is uniquely determined by the equality of chemical potentials,

\mu_{\alpha}(p,T) = \mu_{\beta}(p,T)

This is one equation in two unknowns ( $p$ , $T$ ), so its solution set is generically a one-dimensional curve in the $p$ – $T$ plane—exactly matching $F=1$ .

Step 4. Evaluate for $P = 3$ .

F = 3 - 3 = 0

Now both variables are fixed. Demanding simultaneous equality of three chemical potentials,

\mu_{\alpha}(p,T) = \mu_{\beta}(p,T) = \mu_{\gamma}(p,T)

gives two independent equations in two unknowns. Generically this has a unique, isolated solution $(T_3, p_3)$ —the triple point. There is no freedom left: $F=0$ means zero dimensions, a single point.

Step 5. Rule out four-phase coexistence. Setting $P=4$ gives $F = 3-4 = -1$ , an unphysical negative number of degrees of freedom. This means four bulk phases of one pure substance cannot coexist at equilibrium under ordinary $p$ , $T$ control; three independent equalities of chemical potential cannot generically be satisfied by only two adjustable variables. So $P_{\max} = 3$ for $C=1$ , matching $P_{\max}=C+2$ .

Boxed result.

\boxed{F = 3-P \implies P=1 \to \text{area},\ P=2 \to \text{curve},\ P=3 \to \text{point},\ P\ge 4 \text{ forbidden}}

Worked examples

Example 1 — Reading a phase diagram foundationally. Water's triple point is at $T_3 = 273.16\ \text{K}$ (exactly, by definition of the former Kelvin scale) and $p_3 = 611.657\ \text{Pa}$ ( $6.117\times10^{-3}\ \text{atm}$ ). Its critical point is at $T_c = 647.10\ \text{K}$ ( $373.95\ ^\circ\text{C}$ ) and $p_c = 22.06\ \text{MPa}$ ( $217.7\ \text{atm}$ ). Question: At $T = 298\ \text{K}$ (room temperature) and $p = 1.00\times10^5\ \text{Pa}$ (1 atm, roughly), how many degrees of freedom does liquid water alone have, and is the state point above or below the triple point pressure? Step 1 — identify $P$ . At 298 K and 1 atm, water is a single liquid phase, so $P=1$ . Step 2 — apply the Phase Rule.

F = 3 - P = 3 - 1 = 2

Both $p$ and $T$ can be changed independently (within the liquid region) without a phase disappearing or appearing—consistent with liquid water occupying an area on the diagram. Step 3 — compare pressures.

p = 1.00\times10^5\ \text{Pa} \gg p_3 = 611.657\ \text{Pa}

The ratio is

\frac{p}{p_3} = \frac{1.00\times10^5}{611.657} \approx 163

So the state point sits about 163 times higher in pressure than the triple point—well inside the liquid region, far above the point where solid, liquid, and vapor could all coexist. Sanity check. This matches everyday experience: at 1 atm, liquid water is stable between 0 °C and 100 °C, comfortably including 298 K (25 °C). The Phase Rule correctly predicts $F=2$ , matching the fact that we can independently nudge the beaker's temperature and, in principle, apply modest extra pressure, without any phase transition happening.

Example 2 — Locating a state relative to the sublimation curve for CO₂ (exam-level). Solid carbon dioxide ("dry ice") sublimes directly to gas at ordinary atmospheric pressure. The CO₂ triple point is at $T_3 = 216.55\ \text{K}$ ( $-56.6\ ^\circ\text{C}$ ) and $p_3 = 5.185\times10^5\ \text{Pa}$ (5.11 atm). The critical point is at $T_c = 304.13\ \text{K}$ ( $30.98\ ^\circ\text{C}$ ) and $p_c = 7.38\times10^6\ \text{Pa}$ (72.8 atm). Question: Dry ice is placed in an open container at $p = 1.013\times10^5\ \text{Pa}$ (1 atm) and $T = 195\ \text{K}$ ( $-78\ ^\circ\text{C}$ , its normal sublimation temperature). Confirm using the Phase Rule why liquid CO₂ cannot exist here, and determine what must happen if the same solid is instead sealed in a rigid vessel and heated past 216.55 K while pressure is allowed to rise above 5.185×10⁵ Pa. Step 1 — compare 1 atm to the triple-point pressure.

p = 1.013\times10^5\ \text{Pa} < p_3 = 5.185\times10^5\ \text{Pa}

Since normal atmospheric pressure lies below the triple-point pressure, the horizontal line $p = 1\ \text{atm}$ never crosses the liquid region on the CO₂ diagram; it only crosses the solid–gas (sublimation) boundary. This is why dry ice sublimes rather than melts at atmospheric pressure — there is no accessible liquid phase along that isobar. Step 2 — locate the sublimation crossing. At $p=1$ atm the solid–gas boundary is crossed near $T \approx 195\ \text{K}$ , matching the well-known sublimation temperature of dry ice, $-78\ ^\circ\text{C}$ . Step 3 — sealed vessel, pressure allowed to rise. If the CO₂ is instead sealed in a rigid container, heating it drives $p$ upward along the solid (then liquid) saturation curve rather than staying fixed at 1 atm. Once both

T > T_3 = 216.55\ \text{K} \quad \text{and} \quad p > p_3 = 5.185\times10^5\ \text{Pa}

are satisfied simultaneously, the state point moves into the liquid region, and liquid CO₂ becomes thermodynamically stable — exactly how liquid CO₂ is produced commercially in pressurized cylinders. Step 4 — degrees of freedom check. While solid and gas coexist in the sealed vessel (heating along the sublimation curve before reaching $T_3$ ), $P=2$ and $F=3-2=1$ : fixing $T$ fixes $p$ uniquely along that curve. At the moment all three phases appear together at $(T_3,p_3)$ , $F=0$ —an instant, not a duration, on the heating path. Sanity check. $5.185\times10^5\ \text{Pa}/1.013\times10^5\ \text{Pa} \approx 5.12$ , matching the commonly quoted 5.11 atm triple-point pressure of CO₂—consistent with the reference values used.

Limiting cases & common traps

Situation	What the Phase Rule says	Common error
Single phase region	$F=2$ , an area	Thinking $p$ and $T$ are related inside a region — they are not; only at boundaries
On a boundary curve	$F=1$ , a line	Forgetting that fixing $T$ pins $p$ exactly; you cannot choose both freely on a curve
At the triple point	$F=0$ , one point	Believing the triple point can be shifted by adjusting external pressure — it cannot, it is fixed by the substance
Beyond the critical point	Single supercritical phase, no boundary	Assuming a liquid–gas boundary still exists above $T_c$ — it terminates at the critical point
Water's fusion curve	Negative slope ( $dp/dT<0$ ) because ice is less dense than liquid water	Assuming all solid–liquid boundaries tilt this way — water is the unusual exception

The chemist behind it

The systematic graphical language of phase diagrams traces to Josiah Willard Gibbs, whose 1876–78 memoir "On the Equilibrium of Heterogeneous Substances" derived the Phase Rule from first principles of chemical potential and entropy. Gibbs worked largely in isolation at Yale, publishing in a relatively obscure Connecticut journal; his results were popularized in Europe chiefly through Wilhelm Ostwald's German translation. It was the Dutch chemist Hendrik Willem Bakhuis Roozeboom who, starting in the 1880s, turned Gibbs' abstract rule into the concrete, testable p–T–composition diagrams used across chemistry and metallurgy today, mapping real triple points and critical loci for dozens of substances.

Origins · 1 minHendrik Willem Bakhuis Roozeboom and how this was discovered

Hendrik Willem Bakhuis Roozeboom

1854–1907 · Dutch chemist who translated Gibbs' Phase Rule into practical, experimentally verified phase diagrams

Ask the tutor: Want to see how the Clapeyron equation quantitatively derives the exact slope and curvature of these boundary lines, including why water's melting curve bends the "wrong" way?

Why are phase diagrams drawn with pressure and temperature rather than volume?

Pressure and temperature are the two intensive variables an experimenter directly controls with a barostat and thermostat, and per the Phase Rule they are exactly the right number ( $F=2$ maximum for $C=1$ ) to map single-phase regions as areas. Volume is extensive and changes discontinuously across a phase transition at fixed $p,T$ , so a $p$ – $V$ diagram, while useful for visualizing coexistence via tie-lines, does not give the same clean one-to-one region mapping that $p$ – $T$ provides.

Can the triple point pressure ever exceed the critical pressure?

No. The triple point always lies at lower temperature and pressure than the critical point along the same substance's coexistence curves, because the liquid–gas line must start at the triple point and rise continuously in $p$ and $T$ until it terminates at the critical point. If $p_3$ exceeded $p_c$ , the liquid phase would never have a continuous stability region connecting the two invariant points, contradicting the topology every real one-component diagram exhibits.

Does every substance show a solid–liquid–gas triple point, or can there be more than one?

A substance can have several solid–solid–liquid or solid–solid–gas triple points if it has multiple solid polymorphs (ice has at least fifteen known phases), but there is only one triple point involving the ordinary solid, liquid, and gas phases at their standard identities. Each additional pair of coexisting phases beyond the primary three requires its own invariant point, and the Phase Rule ( $F=0$ ) still governs each one individually.

Is the critical point the highest pressure or temperature on the diagram?

No—it is simply the point where the liquid–gas distinction disappears, not necessarily an overall pressure or temperature maximum for the substance's stability. Pressures far above $p_c$ are entirely accessible; you simply remain in the single supercritical fluid phase rather than crossing any liquid–gas boundary, since the boundary itself no longer exists past $(T_c,p_c)$ .