The Clapeyron Equation

Introduction

A phase diagram's boundary is not an arbitrary line. Along it, two phases have equal chemical potential, and their pressure and temperature responses determine the line's slope. The Clapeyron equation links this slope to the molar enthalpy and molar volume changes of the transition. It applies to solid–liquid, liquid–vapour and solid–vapour coexistence, making it a general bridge between measurable phase-change properties and diagram geometry.

Core explanation

For one component in phases α and β at equilibrium, μ α(T,P)=μ β(T,P). Move a small distance along the coexistence curve: the equality must remain true, so dμ α=dμ β. The thermodynamic differential for a pure phase is dμ=−S m dT+V m dP, where S m and V m are its molar entropy and volume. Substituting for both phases gives −S m,αdT+V m,αdP=−S m,βdT+V m,βdP. Rearranging yields dP/dT=(S m,β−S m,α)/(V m,β−V m,α)=ΔS m/ΔV m.

At a reversible phase transition in equilibrium, the molar enthalpy change is ΔH m=TΔS m. Therefore dP/dT=ΔH m/(TΔV m). The equation is local: ΔH and ΔV can vary with temperature and pressure, so a slope calculated at one point is not necessarily constant along the whole boundary. It presumes a genuine two-phase equilibrium and consistent definitions of initial and final phases. Reversing both ΔH and ΔV leaves the ratio unchanged.

For liquid vaporisation, ΔH vap>0 and vapour molar volume normally exceeds liquid molar volume, so ΔV>0 and dP/dT>0. Increasing pressure raises the boiling temperature along the ordinary boundary. Sublimation likewise usually has positive enthalpy and a large positive volume change, giving a positive slope. For melting, ΔH fus>0, but ΔV fus can be positive or negative. Most substances expand on melting and have positive solid–liquid slopes. Ice is less dense than liquid water near its normal melting point, so V liquid−V ice<0 and the water melting boundary slopes negatively.

Dimensional analysis is useful. ΔH has units J mol⁻¹, T is K and ΔV is m³ mol⁻¹. Since 1 J/m³=1 Pa, the slope has Pa K⁻¹. A common error is to insert ΔV in cm³ mol⁻¹ without converting to m³, creating a millionfold error. A negative slope is physically meaningful for a transition with a negative volume change; it is not evidence of a sign mistake if enthalpy and phase direction are defined consistently.

The equation also explains pressure's effect on phase stability. At fixed temperature, increasing pressure favours the phase with smaller molar volume because its chemical potential rises more slowly with pressure. Thus pressure can favour liquid over ice near melting, whereas it often favours solid over liquid for substances whose solid is denser. This argument follows directly from (∂μ/∂P) T=V m and agrees with the Clapeyron sign.

Clapeyron is exact within ordinary equilibrium thermodynamics, while the commonly used Clausius–Clapeyron integration adds approximations: vapour treated as ideal gas, condensed-phase volume neglected and enthalpy taken as roughly constant. Those approximations are helpful for vapour-pressure calculations but should not be silently applied to a melting line or near a critical point.

Step-by-step reasoning

Define β minus α for every transition property. Write μ α=μ β and differentiate along the boundary. Substitute dμ=−S m dT+V m dP, then obtain ΔS/ΔV and replace ΔS with ΔH/T. Check sign from ΔH and ΔV, convert volume units, and describe the result as a local slope unless an integration model is supplied.

Visual explanation

Draw a P-versus-T diagram with a two-phase line. Mark a short tangent whose slope is dP/dT and write ΔH/(TΔV) beside it. Sketch a usual positive melting line and a water-like negative melting line, labelling which phase has the smaller molar volume on each side.

Real-world analogy

Two runners staying level on different terrains must adjust their speeds as the route changes. Two phases staying equal in chemical potential similarly require linked changes of P and T. Their entropy and volume responses set the adjustment ratio. The analogy only conveys balancing rates; chemical potential equality supplies the actual equation.

Real-world example

Pressure cookers operate above atmospheric pressure, raising water's boiling temperature along the liquid–vapour boundary. In contrast, pressure applied to an ice–water coexistence system near ordinary conditions can favour the denser liquid and lower the melting temperature slightly. The effects have different signs because their phase-change volume differences differ.

Why?

Why does latent heat enter the boundary slope? Latent heat at equilibrium equals T times the entropy difference between phases. Entropy controls the temperature derivative of chemical potential, while volume controls its pressure derivative. The boundary slope balances those two tendencies so the phase chemical potentials remain equal.

Common misconception

Do not claim every melting line has a positive slope. Its sign depends on whether the liquid or solid has larger molar volume. Do not replace ΔV with vapour volume for a solid–liquid transition, where both volumes matter. The equation gives a slope, not by itself a full boundary at all temperatures.

Worked example

At an illustrative water melting point near 273 K, take ΔH fus=6.0 kJ mol⁻¹ and ΔV fus=−1.6 cm³ mol⁻¹. Convert ΔV to −1.6×10⁻⁶ m³ mol⁻¹. Then dP/dT=6000/[273(−1.6×10⁻⁶)]≈−1.37×10⁷ Pa K⁻¹, or about −13.7 MPa K⁻¹. The negative sign means a higher pressure corresponds locally to a lower melting temperature; the numerical inputs are rounded illustrative values.

Quick check

1. What is the sign of an ordinary liquid–vapour coexistence slope, and why can water's ice–liquid line slope negatively? Answer: The liquid–vapour slope is positive, because vaporisation has positive enthalpy and a large positive molar volume change. Melting ice decreases molar volume near ordinary conditions while requiring positive enthalpy, making ΔH/(TΔV) negative.

Exam focus

Derive from equal chemical potentials rather than quoting a formula without phase direction. State units and signs, especially for fusion. Distinguish the exact local Clapeyron relation from integrated vapour-pressure approximations.

Advanced insight

Near a critical point, the distinction between liquid and vapour disappears and both ΔV and ΔH tend toward zero, making simple latent-heat approximations unreliable. Clapeyron still describes coexistence where distinct phases exist, but critical behaviour requires careful limiting analysis rather than direct substitution of rounded small differences.

Summary

Differentiating μ α=μ β gives the Clapeyron slope dP/dT=ΔS m/ΔV m=ΔH m/(TΔV m). Enthalpy and volume changes control its sign and magnitude. Vapourisation normally slopes upward in P–T space; water's fusion line slopes downward because melting reduces volume. The relation is local and broadly applicable across first-order phase boundaries.

Practice questions

1. A substance expands on melting and has positive ΔH fus. What is the sign of its melting-line slope? Answer: Positive, because both ΔH fus and ΔV fus are positive in dP/dT=ΔH/(TΔV). 2. Why does increasing pressure usually favour a smaller-volume phase at fixed temperature? Answer: Chemical potential rises with pressure at a rate V m. The smaller-volume phase's μ rises more slowly, so it becomes relatively more stable. 3. A student uses ΔV=2.0 cm³ mol⁻¹ as 2.0 m³ mol⁻¹ in Clapeyron's formula. What error results? Answer: The slope magnitude becomes too small by a factor of one million, because 2.0 cm³=2.0×10⁻⁶ m³.