Lee–Kesler method

The Lee–Kesler method ^[1] allows the estimation of the saturated vapor pressure at a given temperature for all components for which the critical pressure P_c, the critical temperature T_c, and the acentric factor ω are known.

Equations

${\displaystyle \ln P_{\rm {r}}=f^{(0)}+\omega \cdot f^{(1)}}$

${\displaystyle f^{(0)}=5.92714-{\frac {6.09648}{T_{\rm {r}}}}-1.28862\cdot \ln T_{\rm {r}}+0.169347\cdot T_{\rm {r}}^{6}}$

${\displaystyle f^{(1)}=15.2518-{\frac {15.6875}{T_{\rm {r}}}}-13.4721\cdot \ln T_{\rm {r}}+0.43577\cdot T_{\rm {r}}^{6}}$

with

${\displaystyle P_{\rm {r}}={\frac {P}{P_{\rm {c}}}}}$ (reduced pressure) and ${\displaystyle T_{\rm {r}}={\frac {T}{T_{\rm {c}}}}}$ (reduced temperature).

Typical errors

The prediction error can be up to 10% for polar components and small pressures and the calculated pressure is typically too low. For pressures above 1 bar, that means, above the normal boiling point, the typical errors are below 2%. ^[2]

Example calculation

For benzene with

• T_c = 562.12 K^[3]
• P_c = 4898 kPa^[3]
• T_b = 353.15 K^[4]
• ω = 0.2120^[5]

the following calculation for T = T_b results:

• T_r = 353.15 / 562.12 = 0.628247
• f⁽⁰⁾ = −3.167428
• f⁽¹⁾ = −3.429560
• P_r = exp( f⁽⁰⁾ + ω f⁽¹⁾ ) = 0.020354
• P = P_r · P_c = 99.69 kPa

The correct result would be P = 101.325 kPa, the normal (atmospheric) pressure. The deviation is −1.63 kPa or −1.61 %.

It is important to use the same absolute units for T and T_c as well as for P and P_c. The unit system used (K or R for T) is irrelevant because of the usage of the reduced values T_r and P_r.

See also

• Vapour pressure of water
• Antoine equation
• Tetens equation
• Arden Buck equation
• Goff–Gratch equation