Hong Life CalculatorsUnderstanding the Antoine equation vapour pressure calculator — concept and principle
The Antoine equation estimates the saturation vapour pressure of a pure liquid at a given temperature, that is, the pressure at which the liquid and its vapour are in equilibrium. Chemists and process engineers use it to size distillation columns, predict boiling points at reduced pressure and estimate evaporation losses.
How it is calculated
The equation is log₁₀P = A − B/(C + T), so P = 10^(A − B/(C + T)).
- A, B, C are empirical constants fitted to measured data for one substance over one temperature range.
- T is the temperature in °C (the constants entered must also be for °C).
- P comes out in the pressure unit the constants were fitted for. The defaults are the classic water constants (A = 8.07131, B = 1730.63, C = 233.426) for mmHg and °C, valid from about 1 to 100 °C. The calculator also converts to kPa using 1 kPa = 7.50062 mmHg.
The principle
Thermodynamics (the Clausius–Clapeyron relation) shows that, if the heat of vaporisation is roughly constant, ln P falls on a straight line against 1/T in kelvin. Real liquids bend slightly away from that line. In 1888 Louis Charles Antoine added a third constant C to the denominator, which shifts the temperature scale and absorbs most of that curvature. The result fits measured data closely within its range and is easy to invert.
Worked example
- Inputs: A = 8.07131, B = 1730.63, C = 233.426, T = 60 °C.
- C + T = 293.426, and B/(C + T) = 1730.63 / 293.426 ≈ 5.89801.
- log₁₀P = 8.07131 − 5.89801 ≈ 2.17330.
- P = 10^2.17330 ≈ 149.04 mmHg.
- In kPa: 149.04 / 7.50062 ≈ 19.87 kPa.
| A | Vapour pressure (mmHg) |
|---|---|
| 2.02 | 0.0001317705 |
| 4.04 | 0.013729295 |
| 6.05 | 1.430435354 |
| 8.07 | 149.0384192 |
| 10.1 | 15,527.0244 |
| 12.1 | 1,618,036.207 |
| 16.1 | 17,562,576,770 |
How to read the result
The vapour pressure tells you how strongly a liquid wants to evaporate. When it equals the surrounding pressure, the liquid boils: with the water defaults, T = 100 °C gives about 760 mmHg (one standard atmosphere), which is a good sanity check. At 20 °C the same constants give about 17.5 mmHg. A higher vapour pressure means faster evaporation and, for flammable liquids, a richer vapour above the surface. To find the boiling point at another pressure, solve T = B/(A − log₁₀P) − C.
Common mistakes
- Mixing constants fitted for kPa, bar or kelvin with inputs in mmHg and °C.
- Using natural log constants in a base-10 formula, or the reverse.
- Extrapolating far outside the fitted temperature range.
- Applying pure-component constants directly to a mixture (use Raoult's law with mole fractions instead).
Frequently asked questions
Why do different books list different Antoine constants for the same substance?
Each set is fitted to a particular data set, temperature range and unit system, so several equally valid sets exist. Always use constants together with the units and range quoted alongside them.
Can I use the equation above 100 °C for water?
Not with these constants. A separate set fitted for higher temperatures is needed, otherwise the error grows quickly.
Sources: NIST Chemistry WebBook, SRD 69 — phase change data and Antoine parameters · Poling, Prausnitz and O'Connell, The Properties of Gases and Liquids, 5th ed., McGraw-Hill