Clapeyron and Clausius–Clapeyron Revisited

Introduction

The slope of a phase boundary is not just a feature to memorise from a diagram. It follows by differentiating equality of chemical potentials along the coexistence curve. The exact Clapeyron relation applies broadly to first-order transitions. The familiar straight-line vapour-pressure plot arises only after additional approximations about vapour volume, gas behaviour and enthalpy variation.

Core explanation

On the coexistence curve of pure phases α and β, μ α(T,p) = μ β(T,p). Move a small distance along that curve so equality remains true: dμ α = dμ β. Because dμ = −S m dT + V m dp for each phase, (V m,β − V m,α)dp = (S m,β − S m,α)dT. Thus dp/dT = ΔS m/ΔV m. At a reversible first-order transition, ΔS m = ΔH m/T, giving the exact Clapeyron equation dp/dT = ΔH m/(TΔV m), with Δ values defined consistently as β minus α.

For liquid–vapour coexistence, choose β = vapour and α = liquid. Usually ΔH vap > 0 and ΔV vap > 0, so the vapour-pressure curve slopes upward. If vapour behaves ideally and its molar volume greatly exceeds the liquid's, ΔV m ≈ V m,vap ≈ RT/p. Substitute into Clapeyron: dp/dT ≈ pΔH vap/(RT²), or dln p/dT ≈ ΔH vap/(RT²). This is the differential Clausius–Clapeyron approximation.

If ΔH vap is approximately constant from T₁ to T₂, integration gives ln(p₂/p₁) ≈ −(ΔH vap/R)(1/T₂ − 1/T₁). A plot of ln p against 1/T has approximate slope −ΔH vap/R. This line can curve when vaporisation enthalpy varies with T, vapour is nonideal or liquid volume is not negligible. Near a critical point, liquid and vapour become similar and ΔH vap and ΔV m both change substantially; the simple integrated form is especially unreliable.

For melting, do not use the ideal-vapour approximation. The exact Clapeyron relation still applies, and the sign of ΔV fus determines slope sign because ΔH fus is positive for ordinary melting. Water's ice-to-liquid ΔV fus is negative near ordinary conditions, giving a negative melting-line slope. The equation uses molar volumes of the actual phases and local T,p values along the boundary.

Step-by-step reasoning

Define the transition direction and compute ΔH m and ΔV m with consistent order. Start from μ α = μ β, differentiate along the boundary and derive dp/dT. Only for liquid–vapour equilibrium decide whether ideal vapour and negligible liquid volume are justified. For the integrated logarithmic formula, additionally state approximately constant ΔH vap and use kelvin and consistent units.

Visual explanation

Draw a p–T coexistence curve and a tangent labelled dp/dT. Beside it show separate μ α and μ β changes along the tangent, balancing −S m dT with V m dp. A second graph plots ln p against 1/T as an approximately straight descending line, with a slight curve added where assumptions fail.

Real-world analogy

Two runners can remain side by side only if a change in the track's slope offsets their different natural speeds. Along a coexistence line, changing T alters each phase μ differently because entropies differ; changing p offsets that difference according to their volumes. The analogy conveys compensation, not literal motion of phases.

Real-world example

Measuring vapour pressure at several temperatures permits an approximate ΔH vap to be estimated from the slope of ln p versus 1/T over a limited range. A careful analysis checks whether the line is sufficiently straight and whether nonideal gas behaviour or temperature-dependent enthalpy may bias the result.

Why?

Equality of phase chemical potentials must persist along an equilibrium boundary. Entropy differences make temperature changes favour one phase, while volume differences make pressure changes favour one phase. The coexistence slope is the ratio needed for these two effects to cancel. The vapour-pressure formula is this balance simplified for a voluminous ideal vapour.

Common misconception

The integrated Clausius–Clapeyron formula is not the exact Clapeyron equation and should not be applied to solid–liquid boundaries. A negative melting slope does not imply negative fusion enthalpy; it can result from a negative fusion volume. Also, a plot of ln p against T is not the standard approximately straight plot; the abscissa is 1/T.

Worked example

Take a liquid–vapour boundary near 300 K with ΔH vap = 40.0 kJ mol⁻¹. Under ideal-vapour and constant-enthalpy approximations, estimate p₂/p₁ for T₁ = 300 K and T₂ = 310 K. The integrated formula gives ln(p₂/p₁) = (40000/8.314)(1/300 − 1/310) ≈ 0.517. Thus p₂/p₁ ≈ e^0.517 ≈ 1.68. The pressure rises with temperature, as expected; the numerical ratio is only as good as the assumptions over the interval.

Quick check

1. Why is liquid molar volume often neglected in a low-pressure vaporisation calculation? Answer: Vapour molar volume is usually much larger than liquid molar volume, so ΔV m = V vap − V liquid ≈ V vap. The approximation can fail near critical conditions or high pressures.

Exam focus

Derive exact dp/dT = ΔH m/(TΔV m) before simplifying. State transition direction to fix signs. For liquid–vapour estimates, list ideal gas vapour, negligible liquid volume and nearly constant ΔH if integrating. Use natural logarithms and kelvin; evaluate whether a local slope or finite-temperature extrapolation is requested.

Advanced insight

Fugacity provides a route to nonideal vapour equilibrium, while measured heat capacities can update ΔH vap(T). A numerical integration of the exact equality of chemical potentials can outperform a linear ln p versus 1/T fit near conditions where simplified assumptions deteriorate.

Summary

Differentiating equal phase chemical potentials yields the exact Clapeyron slope dp/dT = ΔH m/(TΔV m). For a low-pressure ideal vapour with negligible liquid volume, dln p/dT ≈ ΔH vap/(RT²). Constant enthalpy then gives a linear relation in 1/T. Melting boundaries require the exact volume difference and can slope either way.

Practice questions

1. A fusion transition has ΔH fus > 0 and ΔV fus < 0. What is the sign of dp/dT? Answer: Negative, because T is positive and the denominator ΔV fus is negative. Pressure favours the smaller-volume liquid phase. 2. If ln p versus 1/T has slope −5000 K over a narrow range, estimate ΔH vap. Answer: Slope = −ΔH vap/R, so ΔH vap ≈ (5000 K)(8.314 J mol⁻¹ K⁻¹) = 41.6 kJ mol⁻¹, subject to the approximations. 3. Why should the integrated two-temperature form be used cautiously near a liquid–vapour critical point? Answer: Vapour nonideality increases, phase molar volumes converge and ΔH vap varies strongly toward zero. The assumptions of ideal vapour, negligible liquid volume and constant enthalpy fail.