For benzene with
- Tc = 562.12 K
- Pc = 4898 kPa
- Tb = 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.
Read more about this topic: Lee-Kesler Method
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