(352p) A Recursive Formula Simplified n-Order Derivatives of Virial Coefficients from the Van Der Waals Theory of Cubic Equations of State
AIChE Annual Meeting
2020
2020 Virtual AIChE Annual Meeting
Engineering Sciences and Fundamentals
Poster Session: Thermodynamics and Transport Properties (Area 1A)
Wednesday, November 18, 2020 - 8:00am to 9:00am
We can state the nth virial coefficient Bn for hard-sphere equation of state as
Eq. 1 can be derived from the thermodynamic relations (see Hill, T. L., An Introduction to Statistical Thermodynamics, Dover Publications, Inc., New York, 1960; pp. 261) of universal law of equation of state for dilute gas and Bn(T) are virial coefficients which depend on intermolecular forces.
By analogy to Eqs.1-2, we can write the general nth virial coefficient Bn for cubic equations of state (see OâConnell-Haile: Thermodynamics (Fundamentals for Applications), pp. 154-155, Eq. 4.5.8) as
Or simply written as (where Cn is the nth virial coefficients):
We can then apply Eq. 4 to the Z-factor derived from the Lawal-Lake-Silberberg cubic equation of state as detailed below. Let us represent the Lawal-Lake-Silberberg (LLS) equation as
By transform to the Z-form
Or in Ï-form
Using partial fraction for further simplification and setting α = d + g and β = - dg, we have
or
By applying the thermodynamic criterion of Eq. 4 to Eq. 9, we arrives at the Recursive Formula for the virial coefficient Cn of nth order virial coefficients is derived from
This can be simplified into
We use binomial formula to expand any power of the summation (d + g) into a sum of the form
The binomial coefficients, which is denoted by is used to derive the coefficient of by the following formula
Then, we translate the result of the expansion into the individual pure substances molecular structure parameter, α and the molecular shape parameter, β; thus, α = (d + g) and β = - dg
For validation of the Recursive Formula, the solution of Cn in terms of parameters a = [a(T)], b = [b(T)], α and β
For n = 2, C2 = b (if n = 1, C1 = 1)
n = 3, C3 = b2 +
n = 4, C4 = b3
n = 5, C5 = b4 +
n = 6, C6 = b5 (