Clausius-Clapeyron Equation

How vapor pressure depends on temperature

Liquids in a closed container reach a dynamic equilibrium with their vapor: as fast as molecules leave the surface, others return. The pressure of that vapor is the substance's vapor pressure, and it depends only on temperature. Heat the liquid and more molecules have enough kinetic energy to break free; the vapor pressure climbs. Cool it and the climb reverses. Boiling is just the special case where vapor pressure rises to match ambient pressure.

The Clausius-Clapeyron equation puts this on a quantitative footing. Starting from the Clapeyron equation for any phase boundary (dP/dT = ΔS/ΔV) and applying two simplifications — the vapor is an ideal gas, and the liquid volume is negligible compared to the vapor — gives:

dP / dT = ΔH_vap · P / (R · T²)

Rearranging:  d(ln P) = (ΔH_vap / R) · dT / T²

Integrating between (T₁, P₁) and (T₂, P₂), assuming ΔH_vap constant:

  ln(P₂ / P₁) = −(ΔH_vap / R) · (1/T₂ − 1/T₁)

That last line is the workhorse two-point form. Three of the four state variables and ΔH_vap let you solve for the fourth.

Worked example — water on Mount Everest

Water boils at 100 °C (T₁ = 373.15 K, P₁ = 1.000 atm) at sea level. Atop Mount Everest the atmospheric pressure is roughly 0.333 atm (P₂). At what temperature does water boil there? Take ΔH_vap = 40,650 J/mol, R = 8.314 J/mol·K.

ln(P₂/P₁) = −ΔH_vap/R · (1/T₂ − 1/T₁)
ln(0.333/1.000) = −(40650 / 8.314) · (1/T₂ − 1/373.15)
−1.0996 = −4889 · (1/T₂ − 0.002680)

Solve:
  1/T₂ − 0.002680 = 1.0996 / 4889 = 0.0002249
  1/T₂ = 0.002680 + 0.0002249 = 0.002905
  T₂ = 1 / 0.002905 = 344.2 K
       = 71.1 °C

Reality: measured boiling point at 8,848 m altitude is approximately 71 °C — a near-perfect match. Climbers carry pressure-cooker stoves on long expeditions for exactly this reason: at 71 °C ordinary cooking takes hours, while 121 °C inside a sealed cooker behaves like sea-level boiling.

Vapor-pressure specs

Substance	Normal b.p.	ΔH_vap (kJ/mol)	P at 25 °C	Use of vapor pressure
Water	100 °C	40.7	23.8 mmHg	Humidity, weather, cooking
Ethanol	78.4 °C	38.6	59 mmHg	Distillation, breath testing
Mercury	356.7 °C	59.1	0.0017 mmHg	Thermometer hazard
Diethyl ether	34.6 °C	27.1	537 mmHg	Anaesthesia, lab solvent
Acetone	56.0 °C	31.3	231 mmHg	Nail polish remover
n-Octane	125.7 °C	41.5	14 mmHg	Gasoline volatility
Iodine (sublimes)	184 °C	62.4 (sub)	0.31 mmHg	Sublimation purification

The aerosol-can warning ("Do not store above 50 °C") is a Clausius-Clapeyron statement: the propellant's vapor pressure rises sharply with temperature, and at 50 °C may exceed the can's design pressure. A car's interior on a sunny day routinely hits 70 °C.

Linearised plot

ln P
  │                              ●
  │                       ●
  │                ●                Slope = −ΔH_vap / R
  │         ●
  │   ●
  │
  └─────────────────────────────→  1/T

A straight line on a log-pressure / inverse-temperature plot.
Steeper slope = larger ΔH_vap = stronger intermolecular forces.
Water (slope −4889 K) is steeper than ether (slope −3260 K).

Clausius-Clapeyron vs related relations

	Clapeyron	Clausius-Clapeyron	Antoine	Wagner
Form	dP/dT = ΔS/ΔV	ln(P₂/P₁) = −ΔH/R · Δ(1/T)	log P = A − B/(C+T)	ln(P/P_c) = polynomial in T_r
Phase boundaries	Any (sl, lv, sv)	Liquid–vapor, solid–vapor	Liquid–vapor	Liquid–vapor
Assumptions	None beyond equilibrium	Ideal vapor, V_liq ≪ V_gas	Empirical fit	Empirical fit
Range	All	Modest T range	Limited range per fit	Triple → critical point
Parameters	ΔH, ΔV measured	ΔH_vap	3 (A, B, C)	5–6
Typical accuracy	Exact (in principle)	±5% over 50 K window	±0.5% in fit window	±0.1% across full range
When to use	Phase diagrams generally	Quick calculations	Distillation design	Process simulators (Aspen)

Variants and refinements

• General Clapeyron equation. dP/dT = ΔS_trans / ΔV_trans applies to any first-order phase transition: melting, sublimation, allotropic changes. Ice melting under pressure is the famous example — ΔV is negative for water, so dP/dT is negative and skater blades partially melt the ice.
• Sublimation form. Replace ΔH_vap with ΔH_sub (= ΔH_vap + ΔH_fus). Used for solids like iodine, naphthalene, or freeze-drying coffee crystals. ΔH_sub is always larger because you pay both phase changes.
• Antoine equation. log₁₀ P = A − B / (C + T). Three empirical constants per substance, fitted to data in a specified range. Distillation column design uses Antoine constants from the DIPPR or NIST databases — accuracy ±0.5%, much better than constant-ΔH Clausius-Clapeyron over wide ranges.
• Wagner equation. Modern reference correlations (IAPWS for water, NIST REFPROP for refrigerants) use multi-term polynomial fits in reduced coordinates. Match data to a few parts per thousand from triple to critical point.
• Trouton's rule. ΔS_vap = ΔH_vap / T_b ≈ 85 J/mol·K for "normal" non-associating liquids. Combined with Clausius-Clapeyron, this gives a back-of-the-envelope way to estimate vapor pressure with no measured ΔH_vap.

Common pitfalls

• Treating ΔH_vap as constant. In reality ΔH_vap drops to zero at the critical point. The two-point form is fine for ±50 K windows; over wider ranges use Antoine or Wagner.
• Forgetting Kelvin. The 1/T term is ferociously sensitive to units — using °C produces 1/(T+273) magnitudes that are nonsense.
• Mixing pressure units inconsistently. Inside a logarithm only the ratio P₂/P₁ matters, so units cancel — but only if both pressures are in the same units. A torr/atm mismatch silently corrupts the answer.
• Applying it across a phase change. If your temperature range crosses the freezing point, you can't use a single ΔH_vap. Below the triple point the relevant quantity is ΔH_sub, and the slope changes.
• Mistaking it for a kinetic equation. Clausius-Clapeyron is purely thermodynamic — it tells you the equilibrium vapor pressure, not how fast evaporation happens. A puddle's drying rate depends on diffusion, wind, and humidity, not just vapor pressure.
• Plotting ln(P) vs T instead of 1/T. A surprisingly common student error. The relationship is linear in 1/T, not T. Plotting against T gives a curve and a useless "slope".