The Clausius–Clapeyron Equation

Introduction

The exact Clapeyron equation relates a liquid–vapour boundary's slope to enthalpy and volume differences. Vapour usually occupies far more volume than the liquid, and at modest pressures it can often be approximated as an ideal gas. Those assumptions turn the boundary slope into a simple relation between vapour pressure and temperature. The resulting Clausius–Clapeyron equation is useful for interpolation, but its straight-line form has a limited temperature range because vaporisation enthalpy is not perfectly constant.

Core explanation

Along liquid–vapour coexistence, dP/dT=ΔH vap/[T(V g−V l)]. If V g≫V l, take ΔV≈V g. If the vapour behaves ideally, its molar volume is V g≈RT/P. Substitution gives dP/dT≈PΔH vap/(RT²), or dlnP/dT≈ΔH vap/(RT²). This differential relation predicts that equilibrium vapour pressure increases with temperature when ΔH vap>0. It is an approximation derived from the exact Clapeyron relation, not a separate unrelated law.

If ΔH vap is approximately constant between T₁ and T₂, integrate to obtain ln(P₂/P₁)=−(ΔH vap/R)(1/T₂−1/T₁). Equivalently lnP≈−ΔH vap/(RT)+constant over that interval. A plot of lnP versus 1/T should then have approximate slope −ΔH vap/R. Temperatures must be in kelvin; replacing reciprocal kelvin temperatures with Celsius values has no thermodynamic basis. The log is natural logarithm unless a factor of 2.303 is included for base-10 logarithms.

The saturation vapour pressure is a property of liquid–vapour equilibrium at a particular temperature. In a closed container with both phases present, evaporation and condensation continue microscopically while their rates balance. The pressure does not depend on how much vapour space is present once equilibrium and sufficient liquid are maintained under the simple model. If all liquid evaporates or a noncondensable gas is added, one must distinguish the vapour's partial pressure from the total pressure and check whether coexistence still holds.

Boiling occurs when a liquid's vapour pressure can match the surrounding pressure at a suitable nucleation condition. Lower external pressure therefore gives a lower boiling temperature, consistent with the positive liquid–vapour boundary slope. Vapour pressure itself is not the same as the rate of evaporation into open air: air movement, surface area and diffusion affect rate while equilibrium saturation pressure is a thermodynamic property.

The constant-enthalpy line is only an approximation. ΔH vap generally decreases as temperature approaches the critical point, where liquid and vapour cease to be distinct and latent heat tends toward zero. Vapour nonideality grows at higher pressure, and liquid molar volume becomes less negligible near critical conditions. A wide-range fit therefore needs a more complete vapour-pressure correlation or direct thermodynamic data.

In using two measurements, solve ΔH vap=−R ln(P₂/P₁)/(1/T₂−1/T₁). If P₂>P₁ for T₂>T₁, then ln(P₂/P₁)>0 while 1/T₂−1/T₁<0, giving a positive enthalpy as expected. This sign check is useful before pressing a calculator. Pressure units cancel inside a ratio but must be consistent between the two values.

The equation is often applied to sublimation as well, with ΔH sub and a solid–vapour boundary, under analogous ideal-vapour and negligible-solid-volume assumptions. It should not be applied with ΔH vap to a solid–liquid melting line, where no large ideal-gas volume dominates. The phase transition and its enthalpy must match the boundary being analysed.

Step-by-step reasoning

Identify the two phases and start from Clapeyron. Justify V g≫V l and ideal vapour before using the logarithmic form. Convert temperatures to kelvin, write the integrated equation and check the signs. If the interval is broad or near a critical point, report the result as an estimate rather than an exact pressure.

Visual explanation

Draw a curved vapour-pressure P versus T plot that rises with temperature. Beside it draw lnP against 1/T with a nearly straight descending line and mark slope −ΔH vap/R. Place a small droplet-and-vapour sketch on the coexistence line to show that the pressure is an equilibrium partial pressure.

Real-world analogy

A slope on a map can be approximated as constant over a short stretch but changes over a long winding route. Constant ΔH vap makes the lnP versus 1/T line locally straight; across a broad temperature range its slope changes. The analogy helps with approximation range but not the molecular reason for vaporisation enthalpy.

Real-world example

Water boils at a lower temperature at high altitude because atmospheric pressure is lower. A Clausius–Clapeyron estimate can approximate how its equilibrium vapour pressure changes between nearby temperatures, though accurate cooking or engineering calculations should use measured vapour-pressure tables over wider ranges.

Why?

Why does lnP appear rather than P directly? Replacing vapour molar volume by RT/P introduces P into the numerator of dP/dT. Dividing by P yields dP/P=dlnP, which integrates naturally. The logarithm is therefore a consequence of the ideal-gas volume approximation.

Common misconception

Vapour pressure increasing with temperature does not mean ΔH vap increases; the enthalpy often decreases near criticality. Another error is interpreting the straight-line equation as exact at all temperatures. It assumes nearly constant ΔH and ideal vapour. Celsius temperatures and mixed pressure units also invalidate routine calculations.

Worked example

Take water's vapour pressure as 1.00 atm at 373 K and use an illustrative constant ΔH vap=40.7 kJ mol⁻¹ to estimate P at 353 K. Compute ln(P₂/1.00)=−40700/8.314(1/353−1/373)≈−0.744. Thus P₂≈e⁻⁰·⁷⁴⁴ atm≈0.475 atm. The estimate is reasonable over this modest interval but depends on the constant-enthalpy and ideal-vapour assumptions.

Quick check

1. What is the sign of the slope of lnP versus 1/T for positive vaporisation enthalpy, and does P mean total gas pressure if air is present above the liquid? Answer: The slope is negative, approximately −ΔH vap/R. P is not the total pressure: it refers to the equilibrium partial pressure of that substance's vapour.

Exam focus

Derive the logarithmic relation from Clapeyron and list its assumptions. Use kelvin and natural logarithms, check pressure ratios and signs, and identify whether the process is vaporisation or sublimation before choosing its enthalpy.

Advanced insight

The exact temperature variation of ΔH vap is related to the heat-capacity difference between vapour and liquid. If that difference is significant over an interval, integrating with variable enthalpy improves the pressure prediction. Near a critical point, both phase distinction and latent heat vanish, so a simple constant-slope line cannot capture the limiting behaviour.

Summary

With vapour volume dominant and approximately ideal, Clapeyron becomes dlnP/dT≈ΔH vap/(RT²). Constant enthalpy then gives ln(P₂/P₁)=−ΔH vap/R(1/T₂−1/T₁). The formula predicts rising saturation vapour pressure and lower boiling temperatures at lower external pressure, but its assumptions limit wide-range accuracy.

Practice questions

1. If two vapour pressures are measured at T₂>T₁, which should be larger for an ordinary liquid with positive ΔH vap? Answer: P₂ should be larger. The integrated equation gives a positive ln(P₂/P₁) because 1/T₂−1/T₁ is negative. 2. Why is replacing T=80 °C by the number 80 wrong in a reciprocal-temperature calculation? Answer: Thermodynamic temperature must use the absolute kelvin scale; 80 °C is about 353 K, and reciprocals on the two scales are not related by simple substitution. 3. Can the same form estimate sublimation pressure? Answer: Yes, with ΔH sub and the solid–vapour boundary, if vapour is approximately ideal, solid volume is negligible and enthalpy is nearly constant over the interval.