Lee–Kesler method

: \ln P_{\rm r} = f^{(0)} + \omega \cdot f^{(1)} : 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 - \frac{15.6875}{T_{\rm r}}-13.4721 \cdot \ln T_{\rm r} + 0.43577 \cdot T_{\rm r}^6 with :P_{\rm r}=\frac{P}{P_{\rm c}} (reduced pressure) and 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 • 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==