Reading One-Component Phase Diagrams

Introduction

A pressure–temperature phase diagram is a map of which phase of a pure substance minimises chemical potential. Its regions, curves and special points encode different numbers of independent controls. Reading the diagram properly requires more than locating a labelled area: one must know what equilibrium means on a boundary, how the phase rule constrains it and why a metastable sample can appear outside its equilibrium region.

Core explanation

Inside a one-phase region of a one-component p–T diagram, one phase has the lowest chemical potential at that T,p. Two-phase coexistence curves satisfy μ α(T,p) = μ β(T,p). For C = 1, Gibbs phase rule F = C − P + 2 gives F = 2 in a one-phase region, so T and p can vary independently within it. On a two-phase curve, F = 1: choosing T fixes coexistence p, or choosing p fixes transition T. At a three-phase triple point, F = 0, so its T and p are fixed for that particular equilibrium assemblage.

The solid–vapour curve is a sublimation boundary, the solid–liquid curve a melting boundary and the liquid–vapour curve a boiling or condensation boundary. A curve's slope comes from Clapeyron dp/dT = ΔH/(TΔV) for a first-order transition. Most pure-substance solid–liquid lines slope upward, but water's familiar ice–liquid line slopes downward near ordinary pressure because melting reduces molar volume. The liquid–vapour curve ends at a critical point where the two fluid phases cease to be distinct. The solid–vapour curve does not turn into a liquid–vapour curve below the triple-point pressure; there is no stable liquid at those pressure–temperature conditions in the simple equilibrium map.

A phase diagram displays equilibrium stability, not the time required to reach it. Supercooled liquid can persist in a region where solid has lower μ, and supersaturated vapour can persist before droplets nucleate. Such metastable extensions may be observed experimentally but are not additional stable equilibrium regions. A spinodal is a more severe local-instability boundary within some models; it should not be confused with the ordinary coexistence curve, where a nucleation barrier can remain.

The diagram also does not state phase amounts. A point on a two-phase curve permits coexistence, but the quantity of each phase depends on total material and container constraints. A sample with too little material or inappropriate volume may occupy only one phase at the same external T,p despite the equality condition being possible.

Step-by-step reasoning

Read the axes and units, then locate the point relative to curves and special points. Name the stable phase or coexisting phases. Apply F = 3 − P for one component if asked about intensive freedom. For movement along a boundary, remember p and T are linked. If an observed phase disagrees with the equilibrium region, consider metastability and nucleation rather than redrawing the stable diagram.

Visual explanation

Draw a p–T map with solid at lower T, vapour at lower p and liquid between where stable. Meet three curves at a triple point and end the liquid–vapour curve at a critical point. Shade broad one-phase regions with F = 2, curves with F = 1 and the triple point with F = 0. Add a dashed supercooled-liquid extension behind the solid region.

Real-world analogy

A road map shows the officially shortest routes under given conditions, not every path a traveller might temporarily follow. A phase diagram similarly records thermodynamic destinations, while kinetic barriers may keep a material on a metastable route. The analogy helps separate equilibrium from observed history.

Real-world example

Solid carbon dioxide at ordinary atmospheric pressure sublimes rather than forming a stable liquid puddle. Its ambient-pressure path lies below the pressure required for liquid stability on the equilibrium diagram. Raising pressure can make a liquid region accessible. The conclusion comes from the location of boundaries, not from the name “dry ice.”

Why?

The phase with lowest μ minimises G at fixed T,p. Equal-μ conditions reduce the number of independent intensive choices by one per coexistence relation, producing curves and special points. Nucleation barriers explain why actual samples can retain a higher-μ phase for a while without invalidating the equilibrium diagram.

Common misconception

Two phases may coexist on a boundary, but they need not be present in equal amounts. The triple point is not a large three-phase region in p–T space; it is an invariant point for a pure substance. A supercooled liquid observed below the melting line does not make the stable liquid region larger.

Worked example

On a pure-substance diagram, suppose solid and liquid coexist at p = 1 bar and T = 280 K. At this pair μ s = μ l and F = 1 for the two-phase assemblage; choosing 280 K has already fixed the coexistence pressure to 1 bar in this imagined system. If the pressure is changed while T remains 280 K, the system generally leaves the two-phase boundary and one phase becomes stable. Which phase requires the sign of ΔV fus or the diagram's slope.

Quick check

1. How many independent intensive variables does a pure substance have at its triple point under the ordinary phase rule? Answer: Zero. C = 1 and P = 3 give F = 1 − 3 + 2 = 0, so triple-point temperature and pressure are fixed.

Exam focus

State axes and phase-rule values, and identify whether a point lies in a region, on a curve or at a triple point. Distinguish stable and metastable states. Do not extrapolate a liquid–vapour boundary beyond its critical endpoint, and do not infer amounts of coexisting phases from a p–T point alone.

Advanced insight

Polymorphism can add more than one solid phase and extra solid–solid coexistence curves. The basic chemical-potential and phase-rule analysis still applies, but a schematic three-phase map may omit these regions. At high pressure, even a familiar substance can exhibit unexpected solid structures, so diagrams are condition-specific.

Summary

One-component p–T diagrams map the phase with lowest μ. One-phase regions have F = 2, two-phase curves F = 1 and triple points F = 0. Curve slopes reflect entropy and volume changes, while the liquid–vapour line ends at a critical point. Metastable phases can persist outside their equilibrium stability regions because of kinetic barriers.

Practice questions

1. A pure material lies on its solid–vapour boundary. What phase change occurs on crossing toward the vapour region? Answer: Sublimation, conversion of solid to vapour, is favoured. On the boundary the two phases can coexist at equilibrium. 2. Why cannot both p and T be freely chosen along a one-component two-phase coexistence curve? Answer: Equal chemical potentials impose one relation between them, leaving F = 1. Choosing one fixes the other for coexistence. 3. A liquid remains present below its equilibrium freezing temperature. What term describes it and what may trigger transformation? Answer: It is supercooled and metastable. Formation of a crystal nucleus or seeding can trigger freezing to the lower-G phase.