Clausius-Clapeyron Vapor Pressure

Clausius-Clapeyron: ln(P₂/P₁) = −(ΔH_vap/R)·(1/T₂ − 1/T₁)

The Clausius-Clapeyron equation ties a liquid's vapor pressure to temperature by way of its enthalpy of vaporization: ln(P₂/P₁) = −(ΔH_vap/R)·(1/T₂ − 1/T₁), where R = 8.314 J/(mol·K) and temperatures are in kelvin. Once you have a reference point (P₁, T₁) and ΔH_vap, you can work out the vapor pressure at some other temperature T₂. Take water: ΔH_vap ≈ 40.7 kJ/mol, and at the boiling point T = 373 K, P = 101325 Pa. Drop T = 298 K (25°C) into the equation and you get P_vap ≈ 3.17 kPa, which lines up well with what's actually measured. Because vapor pressure falls off exponentially as temperature drops, a puddle that needs days to evaporate in winter clears in minutes during the summer.

Applications

It shows up in meteorology (relative humidity, dew point, the formation of clouds, fog and rain), in CFD simulations involving phase change, and in freeze-drying (lyophilization) for pharma and food, where vacuum and cold work together to sublimate water. Engineers also lean on it when designing distillation columns (stills, refineries), when adjusting cooking for altitude (water boils at 93°C in Bogotá and around 88°C at Everest base camp), and when picking refrigerants for HVAC systems.

FAQ

Does ΔH_vap change with temperature? A little, yes. The equation treats ΔH_vap as constant, which holds up fine over short temperature ranges. Once you stray far from the reference point, you're better off with more accurate forms like Antoine or Wagner.

Why use kelvin? The 1/T term only makes sense on an absolute scale. Plug in Celsius and you'd hit divisions by zero or negative reciprocals, and the formula falls apart.

Can I use it for solids (sublimation)? Yes. Just swap ΔH_vap for ΔH_sub. The same form covers dry ice, iodine, and naphthalene mothballs.