Hong Life CalculatorsUnderstanding the Clausius-Clapeyron vapour pressure calculator — concept and principle
The Clausius-Clapeyron equation predicts how the vapour pressure of a liquid changes with temperature. If you know the vapour pressure at one temperature (for example, the normal boiling point) and the enthalpy of vaporisation, you can estimate it at another temperature. It is used in distillation, drying and meteorology.
How it is calculated
ln(P₂ / P₁) = −(ΔH_vap / R) × (1/T₂ − 1/T₁), so P₂ = P₁ × exp[−(ΔH_vap / R)(1/T₂ − 1/T₁)]
- P₁: known vapour pressure (kPa) at temperature T₁.
- T₁, T₂: temperatures entered in °C and converted to kelvin.
- ΔH_vap: molar enthalpy of vaporisation in kJ/mol, converted to J/mol.
- R = 8.314462618 J/(mol·K). P₂ comes out in the same unit as P₁.
The principle
At equilibrium between liquid and vapour, the exact Clapeyron equation dP/dT = ΔH / (T ΔV) holds. Clausius simplified it with three assumptions: the liquid's volume is negligible compared with the vapour's, the vapour behaves as an ideal gas, and ΔH_vap is constant over the temperature range. The result is d(ln P) / d(1/T) = −ΔH_vap / R: a plot of ln P against 1/T is a straight line whose slope gives the enthalpy of vaporisation. Physically, vapour pressure grows roughly exponentially with temperature because the fraction of molecules with enough energy to escape the liquid follows a Boltzmann factor.
Worked example
- Inputs: P₁ = 101.325 kPa at T₁ = 100 °C (water's normal boiling point), T₂ = 80 °C, ΔH_vap = 40.66 kJ/mol.
- In kelvin: T₁ = 373.15 K, T₂ = 353.15 K.
- 1/T₂ − 1/T₁ = 0.00283166 − 0.00267989 = 0.000151770 K⁻¹.
- ΔH_vap / R = 40660 / 8.314462618 = 4890.27 K; exponent = −4890.27 × 0.000151770 = −0.742199.
- P₂ = 101.325 × e^−0.742199 = 101.325 × 0.476066 = 48.23736331 kPa.
| at temperature (°C) | Vapour pressure at T₂ (kPa) |
|---|---|
| 25 | 1,303.501437 |
| 50 | 366.4600055 |
| 75 | 123.6193366 |
| 100 | 48.23736331 |
| 125 | 21.1839015 |
| 150 | 10.25313269 |
| 200 | 3.023206508 |
How to read the result
Steam tables give about 47.4 kPa for water at 80 °C, so the estimate is within about 2%. The error grows with the size of the temperature step because ΔH_vap actually decreases as temperature rises. The result also means water would boil at 80 °C if the surrounding pressure were lowered to about 48 kPa, which is the principle of vacuum distillation and of lower boiling points at altitude.
Common mistakes
- Using °C in 1/T; this is the most frequent error and gives absurd results.
- Leaving ΔH_vap in kJ/mol while R is in J/(mol·K), which shrinks the exponent by 1000.
- Extrapolating far from the known point or close to the critical point, where the assumptions fail.
Frequently asked questions
How can I find ΔH_vap from two measurements?
Rearrange to ΔH_vap = −R ln(P₂/P₁) / (1/T₂ − 1/T₁).
Does this work for sublimation of solids?
Yes, using the enthalpy of sublimation instead of vaporisation.
Why is 40.66 kJ/mol used for water?
It is the enthalpy of vaporisation of water at its normal boiling point, 100 °C; at 25 °C it is higher, about 44 kJ/mol.
Sources: P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press · IAPWS, Revised Release on the IAPWS Industrial Formulation 1997 for the Thermodynamic Properties of Water and Steam