ChemicalThermodynamics / VLEmediumCoursework

Vapour Pressure from the Antoine Equation

The Antoine equation for water, with $P$ in mmHg and $T$ in $^\circ\text{C}$ , is

$\log_{10}P = A - \frac{B}{C + T}$

with $A = 8.07131$ , $B = 1730.63$ , $C = 233.426$ .

Find the vapour pressure of water at $50^\circ\text{C}$ .

Given

T	50 degC	Temperature
A	8.07131 -	Antoine constant A
B	1730.63 -	Antoine constant B
C	233.426 -	Antoine constant C

Hint 1

These constants take temperature in °C, not kelvin — substitute directly.

Hint 2

The equation gives log₁₀P, not P.

Hint 3

Remember the final antilog: P = 10^(that value).

Worked solution — try the problem first

Substitute, keeping $T$ in °C as these constants require:

$C + T = 233.426 + 50 = 283.426$

$\frac{B}{C+T} = \frac{1730.63}{283.426} = 6.1061$

$\log_{10}P = 8.07131 - 6.1061 = 1.9652$

Take the antilog — this step is easily forgotten:

$P = 10^{1.9652} = 92.3\ \text{mmHg}$

$\boxed{P \approx 92\ \text{mmHg} \approx 12.3\ \text{kPa}}$

Sanity check. Water boils at 100 °C when its vapour pressure reaches 760 mmHg. At 50 °C it should be well below that, and 92 mmHg — about 12% of atmospheric — is right for the steam tables.

Two rules for Antoine. The constants are unit-specific: these give mmHg and take °C. A different tabulation might give kPa and take K, and the constants are not interchangeable. And each set is valid only over a stated temperature range — extrapolating beyond it is unreliable.

Concepts:Antoine equationVapour pressureCorrelation constants

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