**Example Calculation**

For benzene with

- T
_{c}= 562.12 K - P
_{c}= 4898 kPa - T
_{b}= 353.15 K - ω = 0.2120

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}.

Read more about this topic: Lee-Kesler Method

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