1. CHEE 311J.S. Parent1 4. Chemical Potential in Mixtures When we add dn moles of a component to n moles of itself, we will observe (?) a change in Gibbs energy of: where  i,T,P represents the chemical potential of the pure component. In dealing with mixtures, we need to ask ourselves what the change in the Gibbs energy of a mixture will be if we add dn moles of a component.  Will the Gibbs energy change follow the above formula?  If so, why?  What, if any, differences need to be described?

2. CHEE 311J.S. Parent2 Chemical Potential in Mixtures Because the Gibbs energy is a function of composition, we expect the chemical potential to have a composition dependence as well. We are looking to derive  i as a function of T,P and y i  This requires an expression for the total Gibbs energy of a mixture:  to which the defining equation for  i is applied: 10.1 In lecture 5 we derived the dependence of (H, S) on (T,P) for a single component system.  What are the corresponding relationships for mixtures?

3. CHEE 311J.S. Parent3 Chemical Potential in Perfect Gas Mixtures The molecular conditions that produce a perfect gas mixture are the same as those of an ideal gas.  The gas must consist of freely moving particles of negligible volume and having negligible forces of interaction  Gibbs remarks “…every gas is a vacuum to every other gas (in the mixture)” If n moles of an perfect gas mixture occupy V at T, the pressure is: If n k moles of component k in this mixture were to occupy the same volume V at T: Which leads to the concept of partial pressure: (10.20)

4. CHEE 311J.S. Parent4 Properties of Perfect Gas Mixtures Gibbs Theorem: A total thermodynamic property of an ideal gas mixture is the sum of the total properties of the individual species, each at the mixture temperature, but at its own partial pressure, p i. In a mathematical form, a system property M varies according to: On a molar basis: = y A * + y B * G ig (T,P)G A ig (T,p A )G B ig (T,p B )

5. CHEE 311J.S. Parent5 Enthalpy of a Perfect Gas Mixture For a pure ideal gas at constant temperature: 6.20 6.22 Therefore, the molar enthalpy H ig (T,P) of a pure ideal gas is independent of pressure: H ig (T,p i ) = H ig (T,P) The total molar enthalpy of an ideal mixture of ideal gases is: H ig =  y i H i ig (T,p i ) or H ig =  y i H i ig (T,P) 10.23 0

6. CHEE 311J.S. Parent6 Entropy of a Perfect Gas Mixture We have seen that the entropy of a pure ideal gas is a function of T and P according to: 5.13, 6.23 Therefore, the entropy of a pure ideal gas at a partial pressure, p i relative to a total pressure P is for a given temperature: or The entropy of an ideal mixture of perfect gases is therefore, 10.25

7. CHEE 311J.S. Parent7 Gibbs Energy of an Ideal Gas The fundamental equation for a closed system gives us the dependence of Gibbs Energy on pressure and temperature: 6.11 For a pure ideal gas, i, at a given temperature, this reduces to: (constant T) which integrates to give the Gibbs Energy of this gas as a function of pressure: or This is the Gibbs energy of a pure, ideal gas at (T,p i ) relative to (T,P). This leads immediately to the Gibbs energy function for an ideal mixture of ideal gases (perfect gas mixture).

8. CHEE 311J.S. Parent8 Gibbs Energy of a Perfect Gas Mixture Using Gibbs Theorem to define the properties of a perfect gas mixture (slide 5), we have: In terms of the total Gibbs energy of a system of n moles: The total Gibbs energy is the sum of:  pure component Gibbs energies at reference state (T,P)  Gibbs energy term to account for component mixing

9. CHEE 311J.S. Parent9 Chemical Potential in a Perfect Gas Mixture Calculating the chemical potential for an perfect gas mixture begins with the definition of  i : which applied to our expression for nG ig gives: 10.26

10. CHEE 311J.S. Parent10 4. Chemical Potential in Ideal Solutions We previously developed an expression for the chemical potential of an perfect gas mixture:  Ideality requires freely moving particles of negligible volume and having negligible forces of interaction If we relax these criteria somewhat we arrive at the more general model of an ideal solution:  Require that all molecules are of the same size and all forces of interaction between molecules (like and unlike) are equal Clearly, perfect gas mixtures are a special case of the ideal solution model. The relaxed constraints allow us to characterize fluids where interactions are significant, such as liquids and non-ideal gases.

11. CHEE 311J.S. Parent11 Properties of Ideal Solutions Molar Volume: If two liquids of different molar volumes, V i, are mixed to generate an ideal solution, V id =  x i V i 10.82 This results from equivalent forces of interaction (A-A = B-B = A-B). In other words, the components must be virtually identical in every way except for chemical structure. Enthalpy: Given that the creation of an ideal solution results in no change in molecular interactions, we do not expect molecular energies to change upon mixing, and H id =  x i H i 10.83 where H i is the entropy of a pure component at the mixture T,P.

12. CHEE 311J.S. Parent12 Properties of Ideal Solutions Entropy: Although mixing of components to generate an ideal solution does not change intermolecular forces (hence U and H are constant), we expect an increase in entropy. In our development of the entropy of an ideal gas mixture, we applied Gibbs theorem and the pressure dependence of entropy.  A general treatment of the entropy of mixing can be derived from statistical mechanics (See me for references) The entropy of an ideal solution relative to its pure components is: S id =  x i S i - R  x i ln x i 10.81 where S i is the entropy of pure component i at the mixture T,P.

13. CHEE 311J.S. Parent13 Chemical Potential of an Ideal Solution Having defined H id and S id, the Gibbs energy and chemical potential immediately follow: G id = H id - T S id =  x i H i - T  x i S i + RT  x i ln x i =  x i G i + RT  x i ln x i 10.80 where G i represents the pure component Gibbs energy at the mixture T,P In terms of the total Gibbs energy of a system: nG id =  n i G i + RT  n i ln n i The chemical potential follows from this equation, leading to  i id = G i + RT ln x i 10.76 where our reference state is the pure component at T,P

14. CHEE 311J.S. Parent14 Origin of Raoult’s Law Ideal Vapour-Liquid Equilibrium (VLE) The simplest VLE condition exists if a perfect gas mixture is in equilibrium with an ideal liquid solution.  For the vapour phase:  i ig = G i ig + RT ln y i  For the liquid phase:  i il = G i l + RT ln x i At equilibrium, the chemical potential of each component must be equal in all phases:  i ig =  i id Using our model equations G i ig + RT ln y i = G i il + RT ln x i or RT ln y i / x i = G i il (T,P) - G i ig (T,P) (A) To proceed further, we need to consider the pure component Gibbs energies of both phases.

15. CHEE 311J.S. Parent15 Origin of Raoult’s Law How do the Gibbs energies of pure component i differ in the vapour and liquid states? 1. Assume G i il is not a strong function of pressure: G i il (T,P) = G i il (T, P i sat ) where P i sat is the vapour pressure of component i. 2. Calculate the change in Gibbs energy when the component as an ideal gas is compressed to the saturation pressure: G i ig (T,P) G i ig (T, P i sat ) Integrating from P to P i sat at constant T gives

16. CHEE 311J.S. Parent16 Origin of Raoult’s Law The change of Gibbs energy between a pure ideal gas at its saturation pressure and the system pressure, P, is: At P i sat, the pure component vapour and liquid are in equilibrium. Therefore, you should convince yourself that: G i il (T,P)  G i il (T, P i sat ) = G i ig (T, P i sat ) and (B) which is the expression we require to develop Raoult’s Law Substituting Equation B into Equation A: RT ln y i / x i = G i il (T,P) - G i ig (T,P) = RT ln P i sat / P

17. CHEE 311J.S. Parent17 Origin of Raoult’s Law Final expression for ideal VLE: y i / x i = P i sat / P or y i P = x i P i sat 12.19 partial pressure mole fraction vapour pressure of i in vapour of i in liquid of pure i How did we arrive at this expression?  Defined an ideal gas mixture  i ig = G i ig + RT ln y i  Defined an ideal solution  i il = G i l + RT ln x i  Applied the criterion for chemical equilibrium  i ig =  i il = *

18. CHEE 311J.S. Parent18 Ideal Phase Behaviour: P-xy Diagrams

19. CHEE 311J.S. Parent19 4. P,T-Flash Calculations If a stream consists of three components with widely differing volatility, substantial separation can be achieved using a simple flash unit. Questions often posed: Given P, T and z i, what are the equilibrium phase compositions? Given P, T and the overall composition of the system, how much of each phase will we collect? Feed z 1 z 2 z 3 =1-z 1 -z 2 T f, P f P,T Vapour y 1 y 2 y 3 =1-y 1 -y 2 Liquid x 1 x 2 x 3 =1-x 1 -x 2

20. CHEE 311J.S. Parent20 P-T Flash Calculations from a Phase Diagram For common binary systems, you can often find a phase diagram in the range of conditions needed. For example, a Pxy diagram for the furan/CCl 4 system at 30  C is illustrated to the right. Given T=30  C, P= 300 mmHg, z 1 = 0.5 Determine x 1, x 2, y 1, y 2 and the fraction of the system that exists as a vapour (V)

21. CHEE 311J.S. Parent21 Flash Calculations from a Phase Diagram Similarly, a Txy diagram can be used if available. Consider the ethanol/toluene system illustrated here at P = 1atm. Given T=90  C, P= 760 mmHg, z 1 = 0.25 Determine x 1, x 2, y 1, y 2 and the fraction of the system that exists as a liquid (L) How about: T=90  C, P= 760 mmHg, z 1 = 0.75?

22. CHEE 311J.S. Parent22 Phase Rule for Intensive Variables For a system of  phases and N species, the degree of freedom is: F = 2 -  + N  # variables that must be specified to fix the intensive state of the system at equilibrium Phase Rule Variables: The system is characterized by T, P and (N-1) mole fractions for each phase  the masses of the phases are not phase-rule variables, because they do not affect the intensive state of the system  Requires knowledge of 2 + (N-1)  variables Phase Rule Equations: At equilibrium  i  =  i  =  i  for all N species  These relations provide (  -1)N equations The difference is F = [2 + (N-1)  ] - [(  -1)N] = 2-  +N

23. CHEE 311J.S. Parent23 Duhem’s Theorem: Extensive Properties SVNA12.2 Duhem’s Theorem: For any closed system of known composition, the equilibrium state is determined when any two independent variables are fixed. If the system is closed and formed from specified amounts of each species, then we can write: Equilibrium equations for chemical potentials (  -1)N Material balance for each species N  We have a total of  N equations The system is characterized by : T, P and (N-1) mole fractions for each phase 2 + (N-1)  Masses of each phase   Requires knowledge of 2 + N  variables To completely determined requires a knowledge of : [2 + N  ] - [  N] = 2 variables This is the appropriate “rule” for flash calculation purposes where the overall system composition is specified

24. CHEE 311J.S. Parent24 Ensuring you have a two-phase system Duhem’s theorem tells us that if we specify T,P and z i, then we have sufficient information to solve a flash calculation. However, before proceeding with a flash calc’n, we must be sure that two phases exist at this P,T and the given overall composition: z 1, z 2, z 3  At a given T, the maximum pressure for which two phases exist is the BUBL P, for which V = 0  At a given T, the minimum pressure for which two phases exist is the DEW P, for which L = 0 To ensure that two phases exist at this P, T, z i :  Perform a BUBL P using x i = z i  Perform a DEW P using y i = z i

25. CHEE 311J.S. Parent25 Ensuring you have a two-phase system If we revisit our furan /CCl 4 system at 30  C, we can illustrate this point. Given T=30  C, P= 300 mmHg, z 1 = 0.25 Is a flash calculation possible? BUBLP, x 1 = z 1 = 0.25 DEWP, y 1 = z 1 = 0.25 Given T=30  C, P= 300 mmHg, z 1 = 0.75 Is a flash calculation possible? BUBLP, x 1 = z 1 = 0.75 DEWP, y 1 = z 1 = 0.75

26. CHEE 311J.S. Parent26 Flash Calculations from Raoult’s Law Given P,T and z i, calculate the compositions of the vapour and liquid phases and the phase fractions without the use of a phase diagram. Step 1. Determine P i sat for each component at T (Antoine’s eq’n, handbook) Step 2. Ensure that, given the specifications, you have two phases by calculating DEWP and BUBLP at the composition, z i. Step 3. Write Raoult’s Law for each component: or (A) where K i = P i sat /P is the partition coefficient for component i.

27. CHEE 311J.S. Parent27 Flash Calculations from Raoult’s Law Step 4. Write overall and component material balances on a 1 mole basis Overall: (B) where L= liquid phase fraction, V= vapour phase fraction. Component: i=1,2,…,n(C) (B) into (C) gives which leads to: (D) Step 5. Substitute Raoult’s Law (A) into (D) and rearrange: (E)

28. CHEE 311J.S. Parent28 Flash Calculations from Raoult’s Law Step 6: Overall material balance on the vapour phase: into which (E) is substituted to give the general flash equation: 12.27 where, z i = overall mole fraction of component i V = vapour phase fraction K i = partition coefficient for component i Step 7: Solution procedures vary, but the simplest is direct trial and error variation of V to satisfy equation 12.27.  Calculate y i ’s using equation (E) and x i ’s using equation (A)