Solution thermodynamics theory—Part IV Chapter 11

When we deal with mixtures of liquids or solids We define the ideal solution model Compare it to the ideal gas mixture, analyze its similarities and differences

Component i in a mixture of ideal gases This eqn. is obtained by combining Now we define Ideal solution model

Other thermodynamic properties for the ideal solution: partial molar volume

partial molar entropy in the ideal solution

partial molar enthalpy in the ideal solution

Chemical potential ideal solution Chemical potential component i in a Real solution Chemical potential Pure component i Subtracting: For the ideal solution

Lewis-Randall rule Lewis-Randall rule (Dividing by Pxi each side of the equation)

When is the ideal solution valid? Mixtures of molecules of similar size and similar chemical nature Mixtures of isomers Adjacent members of homologous series

Virial EOS applied to mixtures

How to obtain the cross coefficients Bij Mixing rules for Pcij, Tcij, wij, 11-70 to 11.73

Fugacity coefficient from virial EOS For a multicomponent mixture, see eqn. 11.64

problem For the system methane (1)/ethane (2)/propane (3) as a gas, estimate at T = 100oC, P = 35 bar, y1 =0.21, and y2 =0.43 Assume that the mixture is an ideal solution Obtain reduced pressures, reduced temperatures, and calculate

Results: methane (1) ethane (2) propane (3) Virial model Ideal solution

Now we want to define a new type of residual properties Instead of using the ideal gas as the reference, we use the ideal solution

Excess properties The most important excess function is the excess Gibbs free energy GE Excess entropy can be calculated from the derivative of GE wrt T Excess volume can be calculated from the derivative of GE wrt P And we also define partial molar excess properties

Definition of activity coefficient

Summary

Summary

Note that:

problem a) Find expressions for ln g1 and ln g2 at T and P The excess Gibbs energy of a binary liquid mixture at T and P is given by a) Find expressions for ln g1 and ln g2 at T and P Using x2 =1 – x1 GE/RT= x12 -1.8 x1 +0.8 x13

Since gi comes from We can use eqns 11.15 and 11.16

then And we obtain

If we apply the additivity rule and the Gibbs-Duhem equation At T and P (b and c) Show that the ln gi expressions satisfy these equations Note: to apply Gibbs-Duhem, divide the equation by dx1 first

Plot the functions and show their values GE/RT ln g1 ln g2