Transcript No Slide Title

Calculating iv from the Virial EOS
We have used the virial equation of state to calculate the fugacity
and fugacity coefficient of pure, non-polar gases at moderate
pressures.
 Under these conditions, it represents non-ideal PVT
behaviour of pure gases quite accurately
 We will generalize the virial equation to allow the calculation
of mixture fugacities.
For mixtures, we use the truncated virial equation:
BP
Z  1
RT
(3.31)
where B is a function of temperature and composition according to:
n n
B    y i y jBij
1 1
(10.65)
Bij characterizes binary interactions between i and j; Bij=Bji
CHEE 311
J.S. Parent
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Calculating iv from the Virial EOS
Pure component coefficients (B11, B22, etc) are calculated as
previously (Equations 3.48, 3.50, 3.51), and cross coefficients are
found from:
Bij 
where,
ij 
RTcij
Pcij
i   j
2
and
Zcij 
Zci  Zcj
2
(Bo  ijB1 )
(10.70)
Tcij  TciTcj (1  k ij )
 Vci1/ 3  Vcj1/ 3 
Vcij  

2


3
Pcij 
ZcijRTcij
Vcij
(10.71-75)
Bo and B1 for the binary pairs are calculated using the standard
equations 3.50 and 3.51at Tr=T/Tcij.
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Calculating iv from the Virial EOS
We now have an equation of state that represents non-ideal PVT
behaviour of mixtures:
Z  1
or
BP
RT
(nZ  n)  n
BP
RT
We are equipped to calculate mixture fugacity coefficients from
equation 10.60
P  (nZ  n) 
dP
v
ln ˆ i   
ni  T,P,nj P
0 
1 P  (nB ) 

dP
 

RT 0  ni  T,nj

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P  (nB ) 
RT  ni  T,nj
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Calculating iv from the Virial EOS
The result of differentiation is:
ln ˆ kv
P 
1 in jn


B

y
y
(
2



)
  i j ik
kk
ij 
RT 
2 i1 j1

(10.69)
with the auxilliary functions defined as:
ik  2Bik  Bii  Bkk
ij  2Bij  Bii  Bkk
In the binary case, we have


(10.67)


(10.68)
P
B11  y 2212
RT
P
v
ˆ
ln 2 
B 22  y1212
RT
12  2B12  B11  B 22
ln ˆ 1v 
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6. Calculating iv from the Virial EOS
Method for calculating mixture fugacity coefficients:
1. For each component in the mixture, look up:
Tc, Pc, Vc, Zc, 
2. For each component, calculate the virial coefficient, B
RT
B  c (Bo  B1)
Pc
3. For each pair of components, calculate:
Tcij, Pcij, Vcij, Zcij, ij
and
o,B1
RTcij o
using
T
,
P
for
B
1
cij
cij
Bij 
(B  B )
Pcij
4. Calculate ik, ij and the fugacity coefficients from:
P 
1 in jn

v
ˆ
ln k 
B

y
y
(
2



)
  i j ik
kk
ij 

RT 
2 i1 j1

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6. Ideal Liquid Solutions
We have already developed a model for the chemical potential of
ideal solutions.
 Acknowledges the fact that molecules have finite volume and
strong interactions, but assumes that these interactions are
the same for all components of the mixture.
 This are the same assumptions used in Section 7.2 for ideal
mixtures of real gases.
The chemical potential of species i in an ideal solution is given by:
(10.26)
id
l
i (T,P, xi )  Gi (T,P)  RT ln xi
where Gil (T,P) represents the pure liquid Gibbs energy at T,P.
This reference state can be shifted to (T,unit pressure) using:
(10.37)
Gl (T,P)   (T)  RT ln f l
i
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i
i
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Ideal Liquid Solutions
Substituting for Gil (T,P) yields:
l
id
(
T
,
P
,
x
)


(
T
)

RT
ln
x
f
i
i
i i
i
To estimate the chemical potential of component i in an ideal liquid
solution, all we require is the composition (xi) and the pure liquid
fugacity (fil ).
The fugacity of a pure liquid can be calculated using:
fil

isatPisat
 Vl (P  Pisat ) 
exp 

RT


(10.41)
For those cases in which the ideal solution model applies, we
require only pure component data to estimate the chemical
potentials and total Gibbs energy of the liquid phase.
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Non-Ideal Liquid Solutions
Relatively few liquid systems meet the criteria required by ideal
solution theory. In most cases of practical interest, molecular
interactions are not uniform between components, resulting in
mixture behaviour that deviates significantly from the ideal case.
The approach for handling non-ideal liquid solutions is exactly the
same as that adopted for non-ideal gas mixtures. We define a
solution fugacity, fil as:
li (T,P)  i (T)  RT ln fˆil
(10.42)
 To use this approach, we require experimental data or
correlations pertaining to the specific mixture of interest
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Lewis-Randall Rule
The ideal solution model developed in Sections 7.2 and 7.4 is
known (in a slightly different form) as the Lewis-Randall equation:
fˆiid  xi fi
(10.84)
The solution fugacity of component i in an ideal solution (gas or
liquid) can be represented by the product of the pure component
fugacity and the mole fraction.
Whenever you apply an ideal solution model, you are using the
Lewis-Randall rule.
 This is an approximation that yields reasonable results for
similar compounds (benzene/toluene, ethanol/propanol)
 However, it is important that you appreciate the limitations of
this rule. When you cannot find mixture data, you may need
to use it (but I suggest you look harder).
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Liquid Phase Activity Coefficients
Based on our definition of solution fugacity:
li (T,P)  i (T)  RT ln fˆil
(10.42)
we could define a liquid phase solution fugacity coefficient:
fˆil

xiP
that reflects deviations of the solution
fugacity from a perfect gas
ˆ li
mixture.
A more logical approach is to measure the deviations of the
solution fugacity from ideal solution behaviour. For this purpose,
we define the activity coefficient:
(10.89)
fˆil
i 
 this convenient parameter
xi is
fil used to correlate non-ideal
liquid solution data, just as i is used for gas mixtures
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Excess Properties of Non-Ideal Liquid Solutions
Most of the information needed to describe non-ideal liquid
solutions is published in the form of the excess Gibbs energy, GE.
Excess properties are defined as the difference between the actual
property value of a solution and the ideal solution value at the same
T, P, and composition.
ME(T,P, xn) = M(T,P, xn) - Mid(T,P, xn)
(10.86)
In defining excess properties, we use ideal solution behaviour as
our reference. Pure components cannot have excess properties.
Partial excess properties can also be defined:
MiE(T,P, xn) = Mi(T,P, xn) - Miid(T,P, xn)
where
E
Mi
CHEE 311
(10.87)
(nME )

ni T,P,nj
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Excess Properties of Non-Ideal Liquid Solutions
The partial excess Gibbs energy is of primary interest:
E
Gi

id
Gi  Gi
where the actual partial molar Gibbs energy is provided by
equation 10.42:
Gi  li  i (T )  RT ln fˆil
and the ideal solution chemical potential is:
id
l
Gi  id


(
T
)

RT
ln
x
f
i
i i
i
Leaving us with the partial excess Gibbs energy:
E
Gi  RT ln fˆil  RT ln x i fil
fˆil
 RT ln l
x i fi
 RT ln  i
CHEE 311
J.S. Parent
(10.90)
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Excess Properties of Non-Ideal Liquid Solutions
Why do we define excess properties for liquid solutions?
 They are more easily applied to experimental data
 Activity coefficients can be treated as partial molar properties
with respect to excess properties. Three important results
follow:
(nGE / RT )
ln  i 
ni
T,P,nj
The Gibbs-Duhem equation:
 xid ln  i  0
(10.94)
(10.98)
i
The summability relation, providing GE from lni data:
GE
  xi ln  i
RT i
CHEE 311
J.S. Parent
(10.97)
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Review of Thermodynamic Principles
Although thermodynamics applies to a great many problems of
engineering importance, CHEE 311 focuses on phase equilibrium
of multi-component systems.
In general, we are trying to describe
systems at equilibrium
 How many stable phases exist,
what are their compositions?
 What property of the system
determines its state?
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The Fundamental Equation
By combining the 1st and 2nd Laws, we derived the fundamental
equation, which relates changes in the internal energy of a system
to variations in volume, entropy and composition.
 In our calculations we are most interested not in changes of
V,S and composition, but P,T and composition
 We therefore defined the Gibbs energy, G
The fundamental
in 
terms
of 
Gibbs
dnGequation
 nVdP
nSdT
dni is:
 ienergy
This equation is useful because it relates changes of the pressure
(dP), temperature (dT) and composition (dni) of a system to
changes in Gibbs energy.
CHEE 311
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Chemical Equilibrium in terms of Gibbs
Energy
Our first definition of chemical equilibrium is based on the total Gibbs
energy (Section 4 notes, 14.1 text)
We considered a closed system (a vessel) in thermal and mechanical
equilibrium with its surroundings. We charged components to the system,
and watched it move towards a state of chemical equilibrium.
 What is the change in Gibbs energy as it does so?
Since these changes take place spontaneously, the second law tells us
that entropy must be created:
dSt + dSsurr  0
where dSt is the entropy change of the system and dSsurr of the
surroundings.
dSsurr is calculated from the heat transferred to maintain thermal and
mechanical equilibrium:
dSsurr = dQsurr / T = - dQ / T
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Chemical Equilibrium in terms of Gibbs
Energy
Substituting for dSsurr gives:
dSt  -dQ / T
or
TdSt  dQ
(A)
which describes the total entropy change of the system as it transfers heat
to/from its surroundings in an effort to reach equilibrium.
How much heat is governed by the first law:
dQ = dUt + P dVt
(B)
Substituting (A) into (B) yields,
TdSt  dUt + P dVt
or,
CHEE 311
dUt + P dVt - TdSt  0
J.S. Parent
(14.2)
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Chemical Equilibrium in terms of Gibbs
Energy
Equation 14.2 is a general equation which relates how the thermodynamic
properties of the system change as it moves towards chemical equilibrium:
dUt + P dVt - TdSt  0
(14.2)
The Gibbs energy is defined as:
Gt = Ut + PVt - TSt
from which changes to the Gibbs energy dG are:
dGt = dUt + P dVt + VtdP - TdSt - StdT
= (dUt + P dVt - TdSt)+ VtdP - StdT
If the system approaches equilibrium at constant T,P
dGt = dUt + P dVt - TdSt
which from equation 14.2 tells us:
dGtT,P  0
(14.3)
Spontaneous changes in the composition of the system that occur at
constant T,P must decrease the Gibbs energy.
CHEE 311
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Equilibrium in Terms of Chemical Potential
If the approach towards equilibrium decreases the Gibbs energy,
then the equilibrium state must reside at the minimum Gt. This
occurs when:
dGtT,P = 0
(14.4)
If the equilibrium state is comprised of two phases (a,b), we know
any change of the total Gibbs energy is:
d(nG) = d(naGa)+ d(n bGb)
Each phase is an open
system:
dnG a  nV a dP  nS a dT    ai dn ia
dnG b  nV bdP  nS bdT   bi dn bi
Therefore, changes of the total Gibbs energy at a given T,P are:
dG t    ai dn ia   bi dn bi  0
or,
a
b
a
 ( i  i )dn i  0
or,
 ai  bi
CHEE 311
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Equilibrium in Terms of Mixture Fugacity
We have developed equations to represent the chemical potential
of vapour and liquid phases.
For vapour mixtures:
 v (T,P)   (T)  RT ln fˆ v
i
i
i
i (T,P)  i (T)  RT ln fˆil
For liquid mixtures: l
Chemical equilibrium requires the equivalence of chemical potential
for all species in both phases. Therefore,
iv (T,P)  li (T,P)
or,
i (T )  RT ln fˆiv  i (T )  RT ln fˆil
or,
fˆiv  fˆil
Our working definition for equilibrium is now based on fugacity.
CHEE 311
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Raoult’s Law
We can apply our new definition of chemical equilibrium to derive
Raoult’s Law. In developing this simplified model, we assumed:
 The vapour behaves as a perfect gas mixture
 The liquid behaves as an ideal solution
v
The mixture fugacity
fˆiv  ˆfor
 yiP gas mixture is:
i ya
iPperfect
because iv= 1 for this simplified case.
l
sat


The solution fugacity
for
an
ideal
liquid
solution
V
(
P

P
)is:
i
i
l
sat sat
fˆi   i xii Pi
exp 

RT
  xiPi

sat
given that i=1, isat= 1, and we ignore the pressure dependence of
the Gibbs energy (Poynting factor = 1)
CHEE 311
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Raoult’s Law
fˆi  fˆi equilibrium in terms of fugacity:
Given our definition of chemical
v
l
 Vderive
)  Law:
i (P  P
we can apply our simplified
expressions
to
Raoult’s
i
sat sat
ˆv
l
yii P   i xii Pi
or
y i Pisat

xi
P
exp 

sat
RT


for systems where a perfect gas mixture is in equilibrium with an
ideal liquid solution.
l
sat expression:
When this is not the case, we must V
use
the full
)
i (P  P
 iisatPisat exp 
yi


xi
ˆ ivP
CHEE 311
J.S. Parent
i
RT


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Thermodynamic Calculations - Raoult’s Law
Bubble Point Pressure:
Given a liquid composition at a specified temperature, find the
composition of the vapour in equilibrium and the pressure
Given T, x1, x2, ... xn find P, y1, y2,... yn
Bubble Point Temperature:
Given a vapour composition at a specified pressure, find the
composition of the liquid in equilibrium and the temperature
Given P, x1, x2, ... xn find T, y1, y2,... yn
Governing Equation - Bubble Line
Equation:
sat
P   x iPi
i
then
CHEE 311
to find P or T
x iPisat
yi 
P
to find yi,…yn
J.S. Parent
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Thermodynamic Calculations - Raoult’s Law
Dew Point Pressure:
Given a vapour composition at a specified temperature, find the
composition of the liquid in equilibrium and the pressure
Given T, y1, y2,... yn find P, x1, x2, ... xn,
Dew Point Temperature:
Given a vapour composition at a specified pressure, find the
composition of the liquid in equilibrium and the temperature
Given P, y1, y2,... yn find T, x1, x2, ... xn
Governing Equation - Dew Line
1 Equation:
P
y
 i P sat
to find P or T
i
i
then
CHEE 311
xi 
y iP
Pisat
to find xi,…xn
J.S. Parent
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Thermodynamic Calculations
“Classic” P-T Flash Calculation:
Given an overall composition at a specified temperature and pressure, find
the composition of the liquid and vapour phases as equilibrium and the
relative amounts of vapour and liquid
Given T,P, z1, z2,... zn find x1... xn , y1,... yn, V, L
Governing Equation - Flash equation:

where Ki =
Pisat/P,
i
zi
1 0
1  V(K i  1)
to find V0
V=vapour phase fraction
Then
and
CHEE 311
zi
xi 
1  V(K i  1)
to find xi,…xn
to find yi,…yn
x iPisat
yi 
P
J.S. Parent
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