The Lee–Kesler method
allows the estimation of the saturated vapor pressure at a given temperature for all components for which the critical pressure Pc, the critical temperature Tc, and the acentric factor ω are known.
Equations
ln P r = f ( 0 ) + ω ⋅ f ( 1 ) {\displaystyle \ln P_{\rm {r}}=f^{(0)}+\omega \cdot f^{(1)}}
f ( 0 ) = 5.92714 − 6.09648 T r − 1.28862 ⋅ ln T r + 0.169347 ⋅ T r 6 {\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}}
f ( 1 ) = 15.2518 − 15.6875 T r − 13.4721 ⋅ ln T r + 0.43577 ⋅ T 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
P r = P P c {\displaystyle P_{\rm {r}}={\frac {P}{P_{\rm {c}}}}} (reduced pressure) and T r = T T c {\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%.
Example calculation For benzene with
Tc = 562.12 K Pc = 4898 kPa Tboiling = 353.15 K ω = 0.2120 the following calculation for T = Tb results:
Tr = 353.15 / 562.12 = 0.628247 f(0) = −3.167428 f(1) = −3.429560 Pr = exp( f(0) + ω f(1) ) = 0.020354 P = Pr · Pc = 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 Tc as well as for P and Pc. The unit system used (K or R for T) is irrelevant because of the usage of the reduced values Tr and Pr.
See also Vapour pressure of water Antoine equation Tetens equation Arden Buck equation Goff–Gratch equation
References
