Lecture #21

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Lecture 21 Objectives: 1. Be able to justify the Conformal Solutions Theory (CST) from both microscopic and macroscopic arguments. 2. Be able to compute properties of fluids based on two and three parameter CST. 3. Be able to analyze any general cubic EOS in terms of the parameters. 1. Conformal Solutions, corresponding states and pure fluids. (a) Define theory of corresponding states or conformal solution theory (CST): “Data for different fluids exhibit considerable uniformity when the thermodynamic coordinates are expressed in a suitable dimensionless form” van Ness & Abbot, p. 101. (b) Microscopic justification: Form of the potential is the same, i.e., the same function, but not the same parameters. Example Lennard-Jones fluid, write equations, draw picture of two different atoms (size, attractive well differences). Draw φ for each one, draw φ∗ for each one. (c) Macroscopic justification. Note that phase diagrams have a critical point. Mapping the critical point from one fluid to another may give the same phase envelope. NOTE that this is not a proof that such a scheme would work for other properties, whereas the microscopic justification is a proof. Use Figure 11.4 from Rowley (d) Reducing with respect to Tc and Pc is equivalent to reducing with respect to ² and σ, see partial proof in Rowley, or details in Reed and Gubbins. (e) Compressibility factor as a function of Tr , Pr . (f) Graph of compressibility factor as a function of Tr , Pr , Figure 11.3 from Rowley. (g) Three parameter corresponding states. Two issues: polarity and anisotropy. i. Scaling or shape factors, molecular parameters. If dipolar or quadrupolar interactions are present then two parameter CST cannot be used to accurately describe the fluid. Dipolar interactions can be made conformal by angle averaging the dipolar contributions, effectively introducing a third parameter, called the scaling or shape factor F . For a mixture a a and b molecules F is given by µ2a µ2b 3kT

+ µ2a αb + µ2b αa 4²0 σ 06 Note that F is a function of temperature. This gives the following reduced quantities: µ ¶ µ ¶5/2 P Fc Pr = Pc F µ ¶ µ ¶2 Fc T Tr = Tc F µ ¶ µ ¶−1/2 V Fc Vr = Vc F These reduced values can be used to compute fluid properties for weakly polar fluids from CST. F =1+

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ii. Acentric factor, compressibility factor expansion. Pitzer (1955) developed a way of accounting for the “non-central” or “acentric” (size-shape) interactions empirically. The acentric factor, ω is defined as ω = − log10 Prsat (at Tr = 0.7) − 1.0 Note that Prsat (at Tr = 0.7) is close to 0.1 for many simple fluids, so ω ≈ 0. Also note that Tr = 0.7 is close to the normal boiling point for many substances, so ω is relatively easy to come by. The first use of ω was in an expression for the compressibility factor Z(Tr , Pr , ω) = Z (0) (Tr , Pr ) + ωZ (1) (Tr , Pr ) Usually get Z (1) (Tr , Pr ) from tables or graphs. (h) Cubic EOS and two and three parameter corresponding states. Show how the van der Waals EOS is equivalent to two parameter conformal solution theory (CST). RT a − 2 V −b V

P = µ Ã

∂P ∂V

∂2P ∂V

¶

= − !T

=

2 T

RT 2a + 3 2 (V − b) V

2RT 6a − 4 3 (V − b) V

Can solve for a, b in terms of Tc , Pc , a=

27R2 Tc2 64Pc

b=

RTc 8Pc

Show that the Redlich-Kwong is a two-parameter CST EOS. P =

RT a − V − b T 1/2 V (V + b)

a = 0.42748

R2 Tc2.5 Pc

b = 0.086640

RTc Pc

Show that Soave-Redlich-Kwong is a three-parameter CST. P =

RT aα(T ) − V − b V (V + b)

a = 0.42747

R2 Tc2 Pc

b = 0.08664

RTc Pc

α(T ) = [1 + (1 − Tr1/2 )(0.480 + 1.574ω − 0.176ω 2 )]2

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hence, P = P (Tr , Pr , ω) 2. The generalized Cubic EOS. Cubic equations of state are pressure explicit functions that are cubic in the molar volume (or molar density). The general form of a cubic equation of state is P =

RT (V 2 + aV + b) V 3 + cV 2 + dV + e

Note that this gives any cubic equation with the appropriate values of a, b, c, d, e. This can be rearranged to P =

RT θ(V − η) − V − b (V − b)(V 2 + δV + ²)

Fill in the following table: EOS van der Waals Redlich-Kwong Soave-RK Peng-Robinson

θ

b

η

δ

²

3. van der Waals EOS. P =

a RT − 2 V −b V

From the above general cubic EOS, the van der Waals θ = a, η = b, and δ = ² = 0. Recall that b reduces the free volume of the fluid due to the repulsion between molecules, and a represents a mean-field attractive potential due to the other molecules. 4. Redlich-Kwong, empirical modification of vdW EOS. P =

a RT − V − b T 1/2 V (V + b)

From the general cubic EOS we see that θ = a/T 1/2 , η = b, δ = b, and ² = 0. Comparing with the van der Waals EOS what do we find? Class activity: Compare van der Waals and Redlich-Kwong EOS. 5. Soave EOS. This is a modification of the Redlich-Kwong EOS. P =

RT aα(T ) − V − b V (V + b)

What are the values? θ = aα(T ), η = b, δ = b, and ² = 0. α(T ) = [1 + (1 − Tr1/2 )(0.480 + 1.574ω − 0.176ω 2 )]2 where Tr = T /Tc and ω is the acentric factor, which is meant to describe the shape interactions between molecules. How many parameters are there?

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6. Peng-Robinson EOS was developed specifically for light hydrocarbons. Like most EOS, it was developed to improve performance or rectify defects of previous EOS. But, since there are only a limited number of parameters available all defects cannot be corrected. P =

aα(T, ω) RT − 2 V − b V + 2bV − b2

Where θ = a(T, ω), η = b, δ = 2b, and ² = −b2 . α(T, ω)1/2 = 1 + (1 − Tr1/2 )(0.37464 + 1.5422ω − 0.26992ω 2 ) How many parameters are there? Compare with the van der Waals EOS. What changes does it incorporate? 7. The Hard Sphere Equation of State All of the popular equations listed above use a very simple and very inaccurate approximation for the purely repulsive part of the pressure. This purely repulsive part is the hard sphere contribution. An empirical, but very accurate, equation of state for the hard sphere fluid was developed by Carnahan and Starling: 1 + η + η2 − η3 βP = ρ (1 − η)3 where η=

πρd3 6

and ρ is the number density, d is the hard sphere diameter.