The Clausius-Clapeyron Relation

Derivation

The Clausius-Clapeyron Relation

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Statement

Along a curve of coexistence between two phases of a single-component substance, the two phases share the same temperature T, pressure P, and molar (or specific) Gibbs free energy g. Requiring that the equality g₁(T,P) = g₂(T,P) be preserved as one moves along the boundary yields the exact slope of that boundary in the P–T plane, dP/dT = L / (T Δv), where L is the molar latent heat of the transition and Δv is the change in molar volume across it.

Why it matters

The Clausius–Clapeyron relation is the exact thermodynamic constraint that ties together three separately measurable quantities: the slope of a phase boundary, the latent heat absorbed on crossing it, and the volume change of the substance. It is not a model or an approximation — it follows rigorously from the equality of Gibbs energies — so it serves as a stringent consistency check on any equation of state or measured phase diagram.

It explains why ice melts under pressure (a negatively sloped solid–liquid line), how boiling points shift with altitude, and it underlies the design of refrigeration cycles, the reading of geological phase diagrams, and the interpretation of superconducting and magnetic transitions where P–V is replaced by a conjugate field pair.

Assumptions

A single-component substance in bulk equilibrium.If a second component is present, chemical potentials must be balanced species-by-species and the single scalar g no longer captures the state; one obtains the more general multi-component coexistence conditions and a shifted boundary (colligative effects).

The transition is first-order: the two phases are distinct, with a genuine discontinuity in molar entropy and molar volume.If Δs → 0 and Δv → 0 together (a critical or continuous transition) the right-hand side becomes 0/0 and Clausius–Clapeyron is replaced by the Ehrenfest relations built from second derivatives.

Mechanical, thermal, and chemical equilibrium hold across the interface so that P, T, and g are each common to both phases.If the interface is curved or the phases are not in true equilibrium (e.g. a metastable supercooled phase), a Laplace-pressure or supersaturation correction shifts the coexistence condition.

Surface and interfacial contributions are negligible compared with bulk free energies.For small droplets or thin films the interfacial free energy per unit area is comparable to bulk terms, adding a curvature (Kelvin) term to g and bending the coexistence curve away from the bulk prediction.

Derivation

g₁(T, P) = g₂(T, P)

Coexistence condition. Two phases can be in equilibrium only if their molar Gibbs free energies (chemical potentials, since μ = g for one component) are equal; otherwise matter flows to the lower-g phase. A

dg = −s dT + v dP

Molar Gibbs differential. From g = u − Ts + Pv and the first law du = T ds − P dv, the Legendre-transformed potential has natural variables (T, P), so (∂g/∂T)_P = −s and (∂g/∂P)_T = v. A

dg₁ = dg₂ along the curve

Differentiate the constraint of step 1. Because g₁ = g₂ holds at every point of the coexistence curve, the changes in the two Gibbs energies for a displacement (dT, dP) that stays on the curve must be equal. B

−s₁ dT + v₁ dP = −s₂ dT + v₂ dP

Insert step 2 for each phase into step 3. Legal because both phases obey the same Gibbs differential, evaluated with their own molar entropy and volume at the shared (T, P). B

(v₂ − v₁) dP = (s₂ − s₁) dT

Collect dP and dT terms. Pure algebra: move volume terms to one side, entropy terms to the other. A

dP/dT = Δs / Δv, Δs ≡ s₂ − s₁, Δv ≡ v₂ − v₁

Divide through by dT and by Δv. Legal provided Δv ≠ 0 (first-order transition). This is already the exact Clapeyron slope. A

L = T Δs ⇒ Δs = L / T

Latent heat at constant T and P. The transition is reversible and isothermal, so heat absorbed is L = T Δs (equivalently, at constant P the latent heat equals the enthalpy change, L = Δh = T Δs). B

dP/dT = L / (T Δv)

Substitute step 7 into step 6. This trades the (harder to measure) entropy jump for the directly measurable latent heat. A

Result

dP/dT = L / (T Δv) = Δs / Δv

Reading. The slope of a first-order phase boundary in the P–T plane equals the latent heat divided by the absolute temperature times the volume change on crossing. A large latent heat steepens the line; a large volume change flattens it. The sign of the slope is fixed by the sign of Δv (since L > 0 and T > 0 for absorption of heat by the higher-entropy phase): if the higher-entropy phase is also the more voluminous, the boundary slopes up; if it is denser (as liquid water is, relative to ice), the boundary slopes down.

Units check. L in J·mol⁻¹, T in K, Δv in m³·mol⁻¹. Then L/(T Δv) has units J·mol⁻¹ / (K · m³·mol⁻¹) = J / (K · m³) = (N·m)/(K·m³) = N/(K·m²) = Pa·K⁻¹, i.e. pressure per temperature, exactly dP/dT. Molar quantities may be replaced by specific (per-kg) quantities throughout without changing the units of the ratio.

Limiting cases

Liquid–vapour or solid–vapour, ideal dilute gas. Then v_gas ≫ v_cond so Δv ≈ v_gas = RT/P, giving dP/dT ≈ LP/(RT²), i.e. d(ln P)/dT = L/(RT²) — the integrated form ln P = −L/(RT) + const (the “Clausius–Clapeyron equation” proper).
Melting (solid–liquid). Δv is small and can have either sign, so dP/dT is large in magnitude and steep; the sign follows Δv (positive for most substances, negative for water, bismuth, gallium).
Near the critical point. Δv → 0 and L → 0 together; the slope tends to a finite limit set by the ratio, and the vapour-pressure curve terminates.
Latent heat weakly temperature-dependent. Treating L ≈ const over a modest range lets the integrated form be used directly for pressure–temperature interpolation.

Breaks when

Continuous (second-order or higher) transitions, e.g. the normal–superconducting transition in zero field, the λ-transition of liquid helium, or a ferromagnetic Curie point: here Δs = Δv = 0 so dP/dT = 0/0 is undefined; the slope is instead governed by the Ehrenfest relations involving jumps in c_P, α, and κ_T.
Approach to the critical point where mean-field vanishing of Δv and L is modified by critical fluctuations; the naïve extrapolation fails and coexistence terminates.
Non-equilibrium or metastable crossing (supercooling, superheating, glass formation): g₁ = g₂ no longer holds, so the derived slope does not describe the observed transition path.
Curved interfaces / finite systems (droplets, capillaries, nanopores): Laplace and Kelvin corrections add curvature-dependent terms to g, shifting coexistence away from the bulk line.

Failure modes

Using the integrated ln P = −L/RT + c for melting. That integrated form assumes an ideal-gas phase (Δv ≈ RT/P); for solid–liquid lines Δv is nearly constant and tiny, so the exact dP/dT = L/(TΔv) must be used, not the exponential.
Dropping the factor of T. Writing dP/dT = L/Δv conflates the latent heat L = TΔs with the entropy jump Δs; the 1/T is not optional.
Sign errors in Δv. Defining Δv = v_final − v_initial inconsistently with L (heat absorbed going the same direction) flips the predicted slope; both differences must run “phase 1 → phase 2” in the same sense.
Assuming L is constant over wide ranges. L itself varies with T (falling to zero at the critical point); using a fixed L far from the calibration temperature gives large errors.
Confusing molar and specific quantities. Mixing L per mole with Δv per kilogram gives a numerically wrong slope by a factor of the molar mass.
Forgetting T must be absolute (kelvin). Using °C in T Δv is a common and severe error.

Discussion

The relation is a statement about geometry in state space forced by a conservation-of-equilibrium condition. The equality g₁ = g₂ defines a one-dimensional curve in the two-dimensional (T,P) plane; demanding that a displacement stay on that curve is exactly the requirement dg₁ = dg₂, and the slope drops out with no further physics. Everything specific to the substance enters only through the two material quantities Δs and Δv. This is why the relation is exact and model-free: it is a corollary of the first and second laws packaged into the Gibbs potential, whose natural variables (T,P) are precisely the controlled variables of a phase diagram.

The water anomaly is the textbook payoff. For the ice–water line, melting absorbs heat (L > 0, liquid is higher entropy) but liquid water is denser than ice, so Δv = v_liquid − v_solid < 0 and dP/dT < 0: increasing pressure lowers the melting point. The slope is steep (about −13.5 MPa·K⁻¹) because Δv is small. The popular “ice skating melts by pressure” story is quantitatively too weak to explain skating, but the sign and magnitude of the boundary are precisely what Clausius–Clapeyron delivers.

The relation generalises far beyond P–V systems. Any first-order transition between phases distinguished by a pair of conjugate variables obeys an analogous slope equation: for a magnetic transition, dH/dT = −Δs/Δm (with H, m the field and magnetisation); for the superconducting transition in a field, one recovers the relation between the critical-field curve and the entropy and (via a second derivative) the specific-heat jump. The Gibbs-equality argument is the common root of all of them.

A subtle point of rigour concerns the direction of differentiation in step 3. One does not set dg₁ = dg₂ = 0 (the Gibbs energy is not stationary along the curve); rather one equates the two total differentials because the difference g₁ − g₂ is identically zero on the curve, so its directional derivative along the curve vanishes. Equivalently, the coexistence curve is a level set of the function Δg(T,P) = g₁ − g₂, and its slope is dP/dT = −(∂Δg/∂T)/(∂Δg/∂P) = −(−Δs)/(Δv) — the implicit-function theorem applied to a level set, which requires precisely Δv = (∂Δg/∂P) ≠ 0, the condition that fails at a continuous transition.

Common misconceptions. (i) The relation is not restricted to boiling or the ideal-gas approximation — that is only the integrated Clausius–Clapeyron equation, a special case. The exact Clapeyron slope dP/dT = L/(TΔv) holds for melting, sublimation, and solid–solid transitions alike. (ii) A positive latent heat does not imply a positively sloped boundary; the slope’s sign is carried by Δv. (iii) The relation says nothing about the rate of a transition (kinetics); it is purely an equilibrium statement about where the phases coexist.

Worked examples

Boiling point of water vs altitude (vapour line, ideal-gas approximation)

Estimate the boiling temperature of water where atmospheric pressure is P = 0.70 atm (≈ 3000 m elevation), given L_vap = 40.7 kJ·mol⁻¹ and boiling at T₀ = 373.15 K, P₀ = 1.00 atm. A

v_gas ≫ v_liq ⇒ Δv ≈ v_gas = RT/P ⇒ d(ln P)/dT = L/(RT²)

Vapour-line limit of the result; molar volume of the liquid is ≈ 1/1600 that of the vapour at 1 atm and is dropped. A

ln(P/P₀) = −(L/R)(1/T − 1/T₀)

Integrate assuming L constant over the small range. B

1/T = 1/T₀ − (R/L) ln(P/P₀)

Solve for T. Rearrangement only. A

1/T = 1/373.15 − (8.314/40700) ln(0.70) = 2.6799×10⁻³ + (2.0427×10⁻⁴)(0.35667)

Insert numbers. ln 0.70 = −0.35667, and −(R/L)ln(P/P₀) = −(2.0427×10⁻⁴)(−0.35667) = +7.286×10⁻⁵. A

1/T = 2.7528×10⁻³ K⁻¹ ⇒ T = 363.3 K

Reciprocate. A

T_boil ≈ 363 K ≈ 90 °C at 0.70 atm

Reading. Water boils about 10 °C cooler at 3000 m, consistent with everyday experience and with tabulated values (≈ 90 °C). Units check. R/L has units (J·mol⁻¹·K⁻¹)/(J·mol⁻¹) = K⁻¹, matching 1/T.

Pressure dependence of the melting point of ice (solid–liquid line, exact Clapeyron)

Find dT/dP for the ice–water boundary at T = 273.15 K. Data: L_fus = 6.01 kJ·mol⁻¹, molar volumes v_liq = 18.02×10⁻⁶ m³·mol⁻¹, v_ice = 19.66×10⁻⁶ m³·mol⁻¹. A

Δv = v_liq − v_ice = (18.02 − 19.66)×10⁻⁶ = −1.64×10⁻⁶ m³·mol⁻¹

Melting runs solid → liquid, so Δv and L both refer to that direction; liquid is denser, hence Δv < 0. A

dP/dT = L / (T Δv)

Exact result; the ideal-gas simplification is not applicable to a condensed–condensed transition. A

dP/dT = 6010 / [273.15 × (−1.64×10⁻⁶)] = 6010 / (−4.480×10⁻⁴)

Insert numbers; denominator T Δv = 273.15 × (−1.64×10⁻⁶) = −4.480×10⁻⁴ m³·K·mol⁻¹. A

dP/dT = −1.342×10⁷ Pa·K⁻¹ = −13.4 MPa·K⁻¹ ⇒ dT/dP = −7.45×10⁻⁸ K·Pa⁻¹

Reciprocate for the shift of melting point with pressure. A

dT/dP ≈ −7.4×10⁻⁸ K·Pa⁻¹ ≈ −0.0075 K per atmosphere

Reading. Raising the pressure by 100 atm lowers the melting point of ice by only ≈ 0.74 K — the negative slope reflects water’s anomalous density inversion, and the tiny magnitude reflects the small Δv. Units check. L/(TΔv) = J/(K·m³) = Pa·K⁻¹; its reciprocal is K·Pa⁻¹.

Problems

Sublimation of dry ice. Solid CO₂ sublimes at T = 194.7 K at 1 atm, with L_sub = 25.2 kJ·mol⁻¹. Using the ideal-gas approximation for the vapour, estimate dP/dT at this point.
Solution
Δv ≈ v_gas = RT/P = (8.314)(194.7)/(1.013×10⁵) = 1.598×10⁻² m³·mol⁻¹. Then dP/dT = L/(TΔv) = 25200/[(194.7)(1.598×10⁻²)] = 25200/3.112 = 8.10×10³ Pa·K⁻¹ ≈ 8.1 kPa·K⁻¹ (≈ 0.080 atm·K⁻¹). Positive slope, as expected for a substance whose vapour is far more voluminous than the solid.
Latent heat from a vapour-pressure curve. The vapour pressure of a liquid doubles when the temperature rises from 300 K to 320 K. Assuming ideal-gas vapour and constant L, find L.
Solution
Integrated form: ln(P₂/P₁) = −(L/R)(1/T₂ − 1/T₁). Here P₂/P₁ = 2, so ln 2 = 0.6931. Compute 1/T₂ − 1/T₁ = 1/320 − 1/300 = 3.125×10⁻³ − 3.333×10⁻³ = −2.083×10⁻⁴ K⁻¹. Then L = −R ln(P₂/P₁)/(1/T₂−1/T₁) = −(8.314)(0.6931)/(−2.083×10⁻⁴) = 5.763/2.083×10⁻⁴ = 2.77×10⁴ J·mol⁻¹ ≈ 27.7 kJ·mol⁻¹.
Sign of a solid–solid boundary. A high-pressure solid phase B of a metal is denser than the low-pressure phase A, and the A→B transition on heating absorbs latent heat. Deduce the sign of dP/dT for the A–B boundary and explain physically.
Solution
Take the transition in the sense A→B. “Denser B” means v_B < v_A, so Δv = v_B − v_A < 0. “Absorbs heat on the A→B transition” means L > 0 for that direction, so Δs = L/T > 0. Then dP/dT = L/(TΔv) < 0 — negative slope. Physically: raising pressure favours the denser (smaller-volume) phase B, so the A→B boundary is reached at lower temperature as pressure rises, i.e. the boundary in the P–T plane runs down-and-to-the-right. (Note this is the less common case; usually the high-entropy phase is also more voluminous, giving positive slope.)
Pressure to melt ice at −2 °C. Using dP/dT = −13.4 MPa·K⁻¹ (from the worked example) as roughly constant, estimate the pressure needed to keep water liquid at T = −2.0 °C.
Solution
The melting curve shifts as ΔP ≈ (dP/dT) ΔT. To depress the melting point by ΔT = −2.0 K requires ΔP = (−13.4×10⁶ Pa·K⁻¹)(−2.0 K) = +2.68×10⁷ Pa ≈ 26.8 MPa, i.e. about 2.68×10⁷/1.013×10⁵ ≈ 265 atm above ambient. So roughly 270 atm total. (This large pressure is why the “pressure-melting” explanation of ice skating is quantitatively inadequate: skate loads give only a few tens of atmospheres.)
Ideal-gas limit consistency. Starting from the exact result dP/dT = L/(TΔv), derive the differential form d(ln P)/dT = L/(RT²) and state the two approximations used. Then, taking L = 40.7 kJ·mol⁻¹ for water at 373 K, evaluate d(ln P)/dT and the fractional change in vapour pressure per kelvin.
Solution
Approximation 1: neglect the condensed-phase molar volume, Δv ≈ v_gas. Approximation 2: treat the vapour as ideal, v_gas = RT/P. Substituting, dP/dT = L/(T·RT/P) = LP/(RT²), hence (1/P)dP/dT = d(ln P)/dT = L/(RT²). Numerically: L/(RT²) = 40700/[(8.314)(373)²] = 40700/(8.314×139129) = 40700/1.1569×10⁶ = 3.52×10⁻² K⁻¹. So the vapour pressure of water rises by about 3.5% per kelvin near its boiling point — a steep dependence familiar from pressure cookers.