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:
d(nG ) T,P  m i T,P dni
where mi,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?
CHEE 311
J.S. Parent
1
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 mi as a function of T,P and yi
 This requires an expression for the total Gibbs energy of a
mixture:
nG  nH  TnS
 f (T,P, n1, n 2 ,...,nk )
 to which the defining equation for mi is applied:
 nG 

m i  
 ni  T,P,nj
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?
CHEE 311
J.S. Parent
2
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:
P
nRT
V
If nk moles of component k in this mixture were to occupy the same
n RT
volume V at T:
P  k
k
V
Which leads to the concept of partial pressure:
Pk  y k P
CHEE 311
J.S. Parent
(k  1,2,..., n)
(10.20)
3
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, pi.
In a mathematical form, a system property M varies according to:
nMig ( T, P)   ni Mig
i ( T, p i )
On a molar basis:
Mig ( T, P)   y i Mig
i ( T, p i )
Gig(T,P)
CHEE 311
= yA * G ig(T,p )
A
A
J.S. Parent
+ yB *
GBig(T,pB)
4
Enthalpy of a Perfect Gas Mixture
For a pure ideal gas at constant temperature:

V 
dH  C p dT   V  T
 dP
T P 
0 
dH ig
T
6.20
0
6.22
Therefore, the molar enthalpy Hig(T,P) of a pure ideal gas is
independent of pressure:
Hig(T,pi) = Hig(T,P)
The total molar enthalpy of an ideal mixture of ideal gases is:
Hig = S yi Hiig(T,pi)
or
Hig = S yi Hiig(T,P)
CHEE 311
J.S. Parent
10.23
5
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:
Cp
R
dS 
dT  dP
5.13, 6.23
T
P
Therefore, the entropy of a pure ideal gas at a partial pressure, pi
relative to a total pressure P is for a given temperature:
ig
S ig
(
T
,
p
)

S
i
i
i (T,P)
or
pi
R
   dP
P P
Sigi (T, p i )  Sigi (T,P)  R ln(p i / P)
The entropy of an ideal mixture of perfect gases is therefore,
Sig (T, P)   y i Sigi (T, pi )
  y i Sigi (T, P)   y i R ln(pi / P)
  y i Sigi (T, P)   y i R ln y i
CHEE 311
J.S. Parent
10.25
6
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
dG  VdP  SdT
For a pure ideal gas, i, at a given temperature, this reduces to:
ig
ig
(constant T)
dG  V dP
i
i
which integrates to give the Gibbs Energy of this gas as a function
pi
of pressure: ig
RT
ig
Gi (pi , T )  Gi (P, T ) 
or
 P dP
P
Gi (p i , T )  Gi (P, T )  RT ln
ig
ig
pi
P
This is the Gibbs energy of a pure, ideal gas at (T,pi) relative to
(T,P). This leads immediately to the Gibbs energy function for an
ideal mixture of ideal gases (perfect gas mixture).
CHEE 311
J.S. Parent
7
Gibbs Energy of a Perfect Gas Mixture
Using Gibbs Theorem to define the properties of a perfect gas
mixture (slide 5), we have:
Gig (T,P)   y i Gigi (T, p i )
  y i Gigi (T,P)   y i RT ln(p i / P)
  y i Gigi (T,P)  RT y i ln y i
In terms of the total Gibbs energy of a system of n moles:
nGig   niGig
i ( T, P)  RT  ni ln
ni
n
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
CHEE 311
J.S. Parent
8
Chemical Potential in a Perfect Gas Mixture
Calculating the chemical potential for an perfect gas mixture begins
with the definition of mi:
m ig
i
 nG ig 

 

 n i  T ,P,n j
which applied to our expression for nGig gives:
ni 

  niGig
(
T
,
P
)

RT
n
ln
 i
i
n 
ig

mi 
ni
T,P,n j
 Gig
i ( T, P )  RT ln y i
10.26
 m io  RT ln y i
CHEE 311
J.S. Parent
9
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.
CHEE 311
J.S. Parent
10
Properties of Ideal Solutions
Molar Volume:
If two liquids of different molar volumes, Vi, are mixed to generate
an ideal solution,
Vid = S xi Vi
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
Hid = S xi Hi
10.83
where Hi is the entropy of a pure component at the mixture T,P.
CHEE 311
J.S. Parent
11
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:
Sid = S xi Si - R S xi ln xi
10.81
where Si is the entropy of pure component i at the mixture T,P.
CHEE 311
J.S. Parent
12
Chemical Potential of an Ideal Solution
Having defined Hid and Sid, the Gibbs energy and chemical
potential immediately follow:
Gid = Hid - T Sid
= S xi Hi - TS xi Si + RTS xi ln xi
= S xi Gi + RT S xi ln xi
10.80
where Gi represents the pure component Gibbs energy at the
mixture T,P
In terms of the total Gibbs energy of a system:
nGid = S ni Gi + RT S ni ln ni
The chemical potential follows from this equation, leading to
miid = Gi + RT ln xi
10.76
where our reference state is the pure component at T,P
CHEE 311
J.S. Parent
13
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: miig = Giig + RT ln yi
 For the liquid phase:
miil = Gil + RT ln xi
At equilibrium, the chemical potential of each component must be
equal in all phases:
miig = miid
Using our model equations
Giig + RT ln yi = Giil + RT ln xi
or
RT ln yi / xi = Giil (T,P) - Giig (T,P)
(A)
To proceed further, we need to consider the pure component Gibbs
energies of both phases.
CHEE 311
J.S. Parent
14
Origin of Raoult’s Law
How do the Gibbs energies of pure component i differ in the vapour
and liquid states?
1. Assume Giil is not a strong function of pressure:
Giil (T,P) = Giil (T, Pisat)
where Pisat 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:
Giig (T,P)
Giig (T, Pisat)
Gig
i
P

Viig
T
RT

P
Integrating from P to Pisat at constant T gives
ig
Gi (T, Pi
CHEE 311
sat
)
Gig
i
Pisat
dP
(T,P)  RT 
P P
J.S. Parent
15
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:
ig
Gi (T, Pi
sat
) - Gig
i (T, P)  RT ln (Pi
sat
/ P)
At Pisat, the pure component vapour and liquid are in equilibrium.
Therefore, you should convince yourself that:
Giil (T,P)  Giil (T, Pisat) = Giig (T, Pisat)
and
il
Gi (T, Pi
sat
) - Gig
i
(T, P)  RT ln (Pi
sat
(B)
/ P)
which is the expression we require to develop Raoult’s Law
Substituting Equation B into Equation A:
RT ln yi / xi = Giil (T,P) - Giig (T,P)
= RT ln Pisat / P
CHEE 311
J.S. Parent
16
Origin of Raoult’s Law
Final expression for ideal VLE:
yi / xi = Pisat / P
or
yi P = xi Pisat
partial pressure
of i in vapour
12.19
= mole fraction * vapour pressure
of i in liquid
of pure i
How did we arrive at this expression?
 Defined an ideal gas mixture
miig = Giig + RT ln yi
 Defined an ideal solution
miil = Gil + RT ln xi
 Applied the criterion for chemical equilibrium
miig = miil
CHEE 311
J.S. Parent
17
Ideal Phase Behaviour: P-xy Diagrams
CHEE 311
J.S. Parent
18
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.
Feed
z1
z2
z3=1-z1-z2
Tf, Pf
P,T
Vapour
y1
y2
y3=1-y1-y2
Liquid
x1
x2
x3=1-x1-x2
Questions often posed:
Given P, T and zi, what are the equilibrium phase compositions?
Given P, T and the overall composition of the system, how much of each
phase will we collect?
CHEE 311
J.S. Parent
19
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/CCl4 system at 30C is
illustrated to the right.
Given
T=30C, P= 300 mmHg, z1= 0.5
Determine
x1, x2, y1, y2 and the fraction of the
system that exists as a vapour (V)
CHEE 311
J.S. Parent
20
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, z1= 0.25
Determine
x1, x2, y1, y2 and the fraction of the
system that exists as a liquid (L)
How about:
T=90C, P= 760 mmHg, z1= 0.75?
CHEE 311
J.S. Parent
21
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
mi = mi  = mi 
 These relations provide (-1)N equations
The difference is
CHEE 311
for all N species
F = [2 + (N-1)] - [(-1)N]
= 2-  +N
J.S. Parent
22
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
Masses of each phase
 Requires knowledge of
2 + (N-1)

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
CHEE 311
J.S. Parent
23
Ensuring you have a two-phase system
Duhem’s theorem tells us that if we specify T,P and zi, 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, zi:
 Perform a BUBL P using xi = zi
 Perform a DEW P using yi = zi
CHEE 311
J.S. Parent
24
Ensuring you have a two-phase system
If we revisit our furan /CCl4 system at
30C, we can illustrate this point.
Given
T=30C, P= 300 mmHg, z1= 0.25
Is a flash calculation possible?
BUBLP, x1 = z1 = 0.25
DEWP, y1 = z1 = 0.25
Given
T=30C, P= 300 mmHg, z1= 0.75
Is a flash calculation possible?
BUBLP, x1 = z1 = 0.75
DEWP, y1 = z1 = 0.75
CHEE 311
J.S. Parent
25
Flash Calculations from Raoult’s Law
Given P,T and zi, calculate the compositions of the vapour and liquid
phases and the phase fractions without the use of a phase diagram.
Step 1.
Determine Pisat 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, zi.
Step 3.
Write Raoult’s Law for each component:
or
y iP  x iPisat
(A)
where Ki = Pisat/P is the partition
coefficient
for component i.
y K
x
i
CHEE 311
i
i
J.S. Parent
26
Flash Calculations from Raoult’s Law
Step 4.
Write overall and component material balances on a 1 mole basis
Overall:
(B)
L  V  F  1 mole
where L= liquid phase fraction, V= vapour phase fraction.
Component:
z i (1)  x iL  y i V
(B) into (C) gives
i=1,2,…,n
(C)
z i  x i (1  V )  y i V
which leads to:
yi 
z i  x i (1  V )
V
(D)
Step 5.
Substitute Raoult’s Law (A) into (D) and rearrange:
zi 
yi 
CHEE 311
yi
(1  V )
Ki
V
yi 
J.S. Parent
zi K i
1  V(K i  1)
(E)
27
Flash Calculations from Raoult’s Law
Step 6:
Overall material balance on the vapour phase:
i n
 yi  1
i 1
into which (E) is substituted to give the general flash equation:
i n
zi K i
1

i 1 1  V (K i  1)
12.27
where,
zi = overall mole fraction of component i
V = vapour phase fraction
Ki = 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 yi’s using equation (E) and xi’s using equation (A)
CHEE 311
J.S. Parent
28