to Apply the Lee-Kesler Correlation

The caution applies when your state is subcritical, near the vapor-liquid saturation curve. If your state is away from the vapor-liquid saturation curve, the following comments do not apply.

The Lee-Kesler correlation is a three-parameter corresponding states method for estimating thermodynamic properties of pure, nonpolar fluids [1]. For the compressibility factor Z, it takes the form

Z = Z_{0} + ω Z_{1}

where Z_{0} is the compressibility factor for fluids of nearly spherical molecules, ω is Pitzer's acentric factor, and Z_{1} corrects for nonspherical intermolecular forces.

Tables and charts provide values of Z_{0} and Z_{1}, from which Z and, hence, the molar volume can be computed. At subcritical temperatures, Z_{1} is typically negative (Z_{1} < 0), indicating that attractive forces dominate the nonspherical contribution to Z. At supercritical temperatures, Z_{1} is typically positive (Z_{1} > 0), indicating the dominance of repulsive forces that arise when molecules collide. Note that simple fluids have ω = 0.

Lee-Kesler tables and charts will locate the vapor-liquid saturation curve in their presentations for Z_{0}. But we emphasize that this is the saturation curve for simple fluids (ω = 0), and generally that curve does *not* correspond to the saturation curve for your fluid.

At subcritical temperatures, a positive value of ω combined with a negative value of Z_{1} means that, at a given reduced temperature, the vapor pressure of your substance will be smaller than that for simple fluids at the same Tr. This is illustrated in the Figure, which shows vapor pressure curves computed from the Lee-Kesler correlation for simple fluids (ω = 0), for fluids having ω = 0.15 (such as propane), and those having ω = 0.4 (such as octane).

Say you need the molar volume for propane at 240.4 K (Tr = 0.65) and 2.12 bar (Pr = 0.05). Propane has ω = 0.15. Tr = T/T_{c} and Pr = P/P_{c}.

Many Lee-Kesler tables have entries at Tr = 0.65 and Pr = 0.05, showing that simple fluids are vapors and having

Z_{0} = 0.9377 and Z_{1} = –0.0772.

The resulting value for the compressibility factor is

Z = 0.9377 + 0.15(–0.0772) = 0.926.

Hence, we find v = ZRT/P = 8730 cc/mol, which is a vapor volume.

But if you use the Lee-Kesler correlation to compute the vapor pressure for propane at 240.4 K, you find P_{sat} = 1.50 bar. Since the required state has P > P_{sat}, propane is actually liquid under these conditions. Directly calculating the Lee-Kesler parameters, we find

Z_{0} = 0.0089 and Z_{1} = –0.0039

Hence

Z = 0.0089 + 0.15(–0.0039) = 0.0083

From which we obtain v = 78.50 cc/mol, a liquid volume.

We may compare these two very different results (v = 8730 and 78.5 cc/mol) with experimental data for propane [2]. At 240 K and 2 bar, the data give

Z = 0.00775 and v = 77.3 cc/mol.

Propane is indeed liquid at the state of interest.

There is nothing fundamentally wrong with the Lee-Kesler correlation. The pitfall arises when a user misinterprets the saturation curves usually presented in tables and charts of Lee-Kesler parameters.

To avoid the pitfall, you must know the phase of your substance *before* you enter a Lee-Kesler chart or table. Charts and tables of Lee-Kesler parameters will not necessarily provide you with the correct phase. To identify the phase for a pure fluid from a small amount of data, we recommend the systematic procedure described in Lecture 5 of Lectures in Thermodynamics.

If you'd like to get numbers from the Lee-Kesler correlation, without recourse to a table or chart, use the Lee-Kesler Calculator on this website.

[ 1 ] B. I. Lee and M. G. Kesler, "A Generalized Thermodynamic Correlation Based on Three-Parameter Corresponding States," *AIChE J*, 21, 510 (1975).

[ 2 ] R. D. Goodwin and W. M. Haynes, *Thermophysical Properties of Propane from 85 to 700 K at Pressures to 70 MPa*, NBS Monograph 170, National Bureau of Standards, Washington, DC, 1982.