Fugacity, Ideal Solutions,
Activity, Activity Coefficient
Chapter 7
Why fugacity?
 Equality of chemical potential is the
fundamental criterion for phase and
chemical equilibrium.
 It is difficult to use, because it
approaches -∞ as the concentration
approaches 0.
 Fugacity approaches 0 as the
concentration approaches 0.
i  g i  RT ln fi  Bi (T )
As chemical potential approaches -∞,
RT ln(f) must also approach -∞
Since RT is a constant, ln(f) must
approach -∞, and so f must approach 0
i  g i  RT ln fi  Bi (T )
The units of chemical potential are
kJ/mole – so RT was needed to make
the units work out, since natural logs
are dimensionless
Most practical calculations do not
include B(T)
i  g i  RT ln fi  Bi (T )
   (U  PV  TS ) 


 Bi (T ) 
 

n
i

T , P , n j


f i  exp 

RT






i  g i  RT ln fi  Bi (T )
We know that if phase 1 and phase 2 are in
equilibrium then:

(1)
i

( 2)
i
fi
(1)
 fi
( 2)
Our new criterion for
physical equilibrium
Pure substance fugacities
For pure substances…
gi  g
  g  RT ln f  B (T )
Take the derivative
  g  RT ln f  B (T )
dg  R(ln fdT  Td ln f )  dB
And recall that…
dg   sdT  vdP
At constant temperature….
Combine and rearrange
vdP  RTd ln f
v
  ln f 

 
 P T RT
But…
1

P
1
  ln P 

 
 P T P
And for an
ideal gas…
Which implies that for an ideal gas,
f=p
  ln P    ln f 

 

 P T  P T
For all gases, as the pressure approaches
zero, the gas approaches ideal behavior,
so…
f
lim    1 as P  0
P
You can check out the rest of the
math in Appendix C, however…

h h
  ln f 

 
2

T
RT

P
*

The * indicates an
ideal gas
f
 1 P

 P z - 1  
 exp 
 dP   exp  
dP 

0
0
P
RT
P




 RT
  1  RT
1  z 
  volume residual  
 v   v  1 
 P
 z  P
For an ideal gas, 0
 RT
  1  RT
1  z 
  volume residual  
 v   v  1 
 P
 z  P
Z=1 for an ideal gas, so 1-z=0
So what is the fugacity?
 For an ideal gas it is identical to the
pressure
 It has the dimensions of pressure
 Think of it as a corrected pressure,
and use it in place of pressure in
equilibrium calculations
 Calculations often include the ratio of
fugacity to pressure, f/P – called the
fugacity coefficient
Notice that f/P
approaches 1
as the value of
Pr goes down,
and the value
of Tr goes up—
in other
words, as the
gas
approaches
ideality
Fugacity of liquids and solids
 Follows the same equations as gases,
but in practice are calculated
differently
 The fugacity of gases and liquids is
usually much less than the pressure
The Fugacity of Pure gases
Illustrate this process with example 7.1,
page 134
 Estimate the fugacity of propane gas at
220 F and 500 psia
 The easiest way is to use figure 7.1
We need Pr and Tr
P
Pr 
PCR
T
Tr 
TCR
500 psi

 0.812
(41.9 *14.7) psi
680 R

 1.022
665.6 R
Pr  0.812
Tr  1.022
f  0.76
Now let’s calculate f, instead of
using a chart
 If you have reliable PvT measurements,
like the steam tables, you could use
them – which would give the most
reliable values
 For this problem, PvT measurements
were used to create Figure 7.2 – a plot
of z values. (Pv=zRT)
 The data are also presented in Table
7.A, so we don’t have to read a graph
Table 7.A – Volume residuals of
Propane at 220 F
Pressure, psia
Z
RT/P * (1-z)
0
1.0
3.8
100
0.9462
3.923
200
0.8873
4.110
300
0.8221
4.324
400
0.7787
4.582
500
0.6621
4.929
Recall that…
f
 1

 exp 

dP


P
 RT 0

P
I shamefully left out
units, in the interest
of space – see page
135
 changes with pressure, but not as strongly
as z, so we can approximate  as a constant
f
  P 
4.25*500



exp
 exp 

10.73*679.67  0.747
P
 RT 
We could also use an equation of
state (EOS) to find the fugacity
 The ideal gas law would tell us that
pressure and fugacity are equal –
which is obviously a really bad
estimate for this case
 Let’s try the “little EOS” from Chapter
2
 Equations 2.48 – 2.50
Little EOS
Pv=zRT
z  z  z
Pr 
0.422 
0
z  1   0.083  1.6 
Tr 
Tr 
Pr 
0.172 
1
z   0.139  4.2 
Tr 
Tr 
0
1
Eq 2.48
Eq 2.49
Eq 2.50
Substitute into Eq. 7.9
f
 P z - 1  
 exp  
dP 
P
 0 P

/Trr  ff ((TTrr ))  Do the
 PP PPPcrr /T
 exp  
dP integration
P
 00

 Pr
 Now we can substitute in the
 exp  f (Tr )  values
=0.771
of Pr and Tr
 Tr



One more approach…
Use Thermodynamic tables
Recall that…
  g  RT ln f  B (T )
Rewrite this equation for two different states
g2  g1  RT ln f 2  ln f1 
f2
 RT ln
f1
(If you keep the temperature constant, and vary
the pressure)
f2
g 2  g1  RT ln
f1
f2
 g 2  g1 
 exp 

f1
 RT 
Divide both sides
by P1P2 and
rearrange
f 2 f1P1
 g 2  g1 

exp 

P2 P1P2
 RT 
At very low
pressures,
f and P are equal
f 2 P1
 g 2  g1 
 exp 

P2 P2
 RT 
But only at very
low pressures
Now plug in values from a table of
thermodynamic values – pg 137
f 2 P1
 g 2  g1 
 exp 

P2 P2
 RT 
Choose P1 as 1 psia
Btu
  1554.28  1735.54 lbm

f2
1 psia
lbm


exp 
*
44
.
062
lbmol

Btu
P2 500 psia
1
.
987
679
.
67
R
lbmol R


f2
 0.740
P2
There are obviously several ways to find
fugacity, depending on what you know and
how accurate you need to be
 Use a fugacity coefficient table like
Figure 7.1
 Use PvT data, to find z , then use
equation 7.9
 Use an equation of state
 Use thermodynamic tables and
equation 7.11
Fugacity of Pure Liquids and Solids
 We could compute the fugacity just
like we did in the previous example
 It is impractical for mathematical
reasons
 See examples 7.2 and 7.3
Fugacity of Solids and Liquids
 Does not change appreciably with
pressure
 Normally we approximate the fugacity
of pure solids and liquids as the pure
component vapor pressure
 We’ll return to this in Chapter 14
when we study cases where this
approximation is no longer valid
See Appendix C
Fugacities of species in mixtures
i  g i  RT ln fi  Bi (T )
vi
  ln f i 

 
 P T RT
 fi 
  1 as P  0
lim 
 Pxi 
See Appendix C
Fugacities of species in mixtures

hi  hi
  ln f i 

 
2
RT
 T  P
*

fi
 1 P

 exp 
 i dP 

Pxi
 RT 0

 
 P z -1

 exp  
dP 
0
P


These equations are the same as the pure
component equations, except that P has been
replaced with the partial pressure, xiP
Mixtures of Ideal Gases
nT RT
V
P
True only for ideal gases
  
nT RT

vi  
 ni T , P ,n j P
RT
vi 
P
 nT

 ni
But R, T and P
are constant,
so…
1


T , P , n j
and
RT
i 
 vi  0
P
All of which leads to…
f i  Pi  yi P
For mixtures of ideal gases, the fugacity of
each species is equal to the partial pressure of
that species
Can we extend this concept to ideal
solutions?
 Ideal solutions are like ideal gases
 Neither exist in nature – real gases and
solutions are much more complicated
 Many gases and solutions exhibit
practically ideal behavior
 It is often easier to work with deviations
from ideal behavior, rather than work
directly with the property of interest
 The compressibility factor z is an example
 Activity coefficient is similar for liquids
Ideal solutions
 The definition of an ideal solution is
that …
f i  xfi
0
Usually fi0 is defined as the pure
component fugacity – though this is not
the only choice
Any ideal solution has the following
properties
f i  xi f i
0
g i  g  RT ln xi
0
i
vi  v  0
0
i
s i  s   R ln xi
0
i
hi  h  0
0
i
True for gases,
liquids and
solids
Any ideal solution has the following
properties
f i  xi f i
0
g i  g  RT ln xi
0
i
vi  v  0
0
i
s i  s   R ln xi
0
i
hi  h  0
0
i
Tells us there is
no volume
change with
mixing
Any ideal solution has the following
properties
f i  xi f i
0
g i  g  RT ln xi
0
i
vi  v  0
0
i
s i  s   R ln xi
0
i
hi  h  0
0
i
Tells us that
there is no
heat of mixing
Any ideal solution has the following
properties
f i  xi f i
0
g i  g  RT ln xi
0
i
vi  v  0
0
i
s i  s   R ln xi
0
i
hi  h  0
0
i
Which means there
is a Gibbs free
energy change with
mixing
There is always
an entropy
change
associated with
mixing
Activity and Activity Coefficients
 Fugacity has dimensions of pressure
 This can cause problems if we don’t pay strict
attention to units
 Often we want a non-dimensional
representation of fugacity – which leads to
the activity
 When we deal with non-ideal solutions we’ll
want a measure of departure from ideal –
like z – which leads to the activity
coefficient
Activity
fi
ai  0
fi
Activity is defined as the ratio of the fugacity of
component i, to it’s pure component fugacity
Recall that for an
ideal gas…
fi  yi P
Recall that for an
ideal solution…
f i  xi f i
0
Usually – but
not
necessarily,
chosen as
the pure
component
fugacity
Activity
fi
ai  0
fi
Rearrange to give…
fi
xi  0
fi
 ai
For an ideal solution of
either gas or liquid,
activity is equivalent to
mole fraction
ai
i 
xi
Activity Coefficient
ai
i 
xi
fi

0
xi f i
 i xi  ai
f i  ai f i
0
  i xi f i
The activity
coefficient is just
a correction
factor on x,
which converts it
to a
0
Why bother?
 Both are dimensionless
 They lead to useful correlations of
liquid-phase fugacities.
 The normal chemical equilbrium
statement – the law of mass action –
is given in terms of activities
 (The law of mass action is the definition
of the equilibrium constant, K)
For now…
 Activity is rarely used for phase
equilibria
 It will show up again in Chapters 12
and 13
 Activity coefficient is useful to us
now!!
For a pure species or for ideal
solutions
 Activity = mole fraction and..
 Activity coefficient,  =1
 We could redefine an ideal solution as
one where  equals 1
 This is like defining an ideal gas as
one where z=1
Activity coefficients
 For real solutions activity coefficients
can be either greater or less than 1
 Typically range between 0.1 and 10
From Appendix C (the general
case)
  ln  i 


 P T , xi

v v 

0
i
i
RT

h  hi
  ln  i 

 
2
RT
 T  P , x
0
i

0
Eq. 7.31
For an
ideal
solution
0
Eq. 7.32
Example 7.4
 At 1 atm pressure the ethanol-water
azeotrope has a composition of:
 10.57 mole% water
 89.43 mole% ethanol
 The composition is the same in the
vapor and liquid phases
 The temperature is 78.15 C
Example 7.4 cont
 At this temperature the pure
component vapor pressures are:
 Water 0.434 atm
 Ethanol 0.993 atm
 Estimate the fugacity and activity
coefficients in each phase
For an azeotrope…
xethanol=yethanol and xwater=ywater
 At one atm it is a reasonable
assumption that both the water and
ethanol behave as ideal gases
 Thus the fugacity in the gas phase is
equal to the partial pressure
gas
f ethanol
 0.8943atm
 yethanol *1atm
gas
f water
 0.1057atm
 ywater *1atm
Because we are at equilibrium the gas
phase fugacity equals the liquid phase
fugacity
liquid
0
f ethanol
 0.8943atm  aethanol f ethanol
0
f water  0.1057atm  awater f water
liquid
If this was an ideal solution, a would
be equal to x – but it’s not, so…
aethanol   ethanolxethanol
awater   water xwater
We use the pure component vapor
pressure as the standard fugacity
0
0
f ethanol  0.8943atm  aethanol fethanol
aethanolPethanol
0
0
f water  0.1057atm  awater f water
 awater Pwater
If this was an ideal solution, a would
be equal to x – but it’s not, so…
aethanol   ethanolxethanol
awater   water xwater
Remember, I’m using Pi0 to indicate
the pure component vapor pressure –
but Dr. deNevers is using p
Thus
 ethanol 
 water 
gas phase
ethanol
0
ethanol ethanol
f
x
P
gas phase
water
0
water water
f
x
P
0.8943atm
 1.007
0.8943 * 0.993atm
0.1057 atm
 2.31
0.1057 * 0.244atm
Thus
 ethanol 
 water 
gas phase
ethanol
0
ethanol ethanol
f
x
P
gas phase
water
0
water water
f
x
P
 1.007
 2.31
This
solution is
not ideal,
as shown
by the fact
that the
activity
coefficients
are not 1
Fugacity coefficient, Poynting
Factor and Alternative Notation
 The ratio of fugacity to pressure is called
the fugacity coefficient
 Often the symbol, n, is used
 Older literature
 Usually refers only to pure component fugacities
and pressures
 Sometimes the symbol, f, is used
 Newer literature
 Can refer to both pure components and mixtures
Older usage
f
fi   
* P n * P
 P  pure i
0
And since
Pi  xi i P
a, the activity
f i  n i * xi i P 
, the activity
coefficient
corresponds to
deviation from
ideality due to
mixing in the
gas or liquid
n, the fugacity
coefficient,
corresponds to
deviation from
ideal gas
behavior
Modern usage
 For pure species fi is exactly the same
as ni, and accounts only for the
departure from ideal gas behavior
 For mixtures fˆi accounts for not only
the deviation from ideal gas behavior,
but also for nonideal mixing behavior.
ˆ
fi   in i
Modern Usage
 In older usage, f i , is called the
fugacity of component i.
 In more modern usage it is often
represented as f i , and called the
partial fugacity – like partial pressure
 Obviously, it can’t be a partial molal
property, because fugacity is not an
extensive property
Poynting Factor
fi
 1 P

 exp 
 i dP 

Pxi
 RT 0

  1 0.9503atm

f i  Pxi exp 
 dP   pi

 RT 0

0
f i  pi * PF * f
0
sat
i
Estimating Fugacities of Individual
Species in Gas Mixtures
 If you don’t have an ideal gas – how
do you estimate the fugacity of a
gas?
 Use reliable PvT data
 Use an equation of state
Example 7.5
Example 7.5
 Table 7.F presents Volume residuals
() for a mixture of methane with nbutane, at 220 F
 Use this data to find the fugacity of
methane and the fugacity of n-butane
in a mixture that is:
 78.4 mole% methane
 21.6 mole% n-butane
This corresponds to
a mixture that is
50 wt% methane
 We’ll need the partial molal volume
residual,  methane
 Find its value at each of the pressures
in table 7. F, which will allow us to
integrate using the trapezoid rule
fi
 1 P

 exp 

dP

i

0
Pyi
RT


Volume Residual
Volume Residual, ft3/lbmol


At 220 F and 100 psia
10
9
8
6.6
ft3/lbmole
7
6
Use the method of tangent
intercepts
This is the
partial molal
volume residual
of methane
5
4
3
0.6
ft3/lbmole
Tangent Line, at
78.4% methane
2
1
0
0
20
40
60
Mole % Methane
80
100
Repeat the
process at
the other
pressures,
to create
Figure 7.10
avg0.290
Substitute
fi
 1 P

 exp 
 i dP 

Pyi
 RT 0



290
fi
 exp 
 10.73
Pyi


  0.961
 680 R 
psia ft 3
lbmole
psia ft 3
lbmoleR
f i  0.961Pyi  0.961*1000 psia * 0.784
f i  753 psia
Fugacities from an EOS for Gas
Mixtures
 PvT data are only available for a small
number of gas mixtures
 For other cases, using a reliable EOS
is a good (though not as accurate)
alternative
What equation to use?
fi
 1

 exp 

dP

i

Pyi
 RT 0

P
fi
  1 P z -1 
 exp 
dP


Pyi
 RT 0 P

This is easier for
the graphical
procedure, used
with PvT data
This is easier if we
are going to use an
EOS
But we have a problem
 The equations of state from chapter 2
are for singe pure species
 For mixtures we need a mixing rule
 Usually semi-empirical
 For a mixture of a and b, at some T and
P, we can find the pure component
values of z and use them in our mixing
rule to get a combination value of z
Any mixing rule will have the
form…
z mix  f ( z a , zb , ya , yb )T , P
One possibility is…

z mix  x z
1/ n
a a

1/ n n
b b
x z
Lewis and Randall fugacity rule
If n=1

zmix  xa za  xb zb

which is just a weighted average
It is equivalent to an ideal solution of non-ideal
gases
This is not rigorously correct, but is a useful
approximation, especially at pressures less
than a few atmospheres
Example 7.6
 Compares the Lewis Randall fugacity
rule results to those found with PvT
data
 I’ll let you work through it on your
own
Lewis Randall Rule
 Used because it is simple
 It is the next step in complexity after
the ideal gas law
 Unfortunately, sometimes gases exist
in states for which we can not
compute ni
Gases can exist as a mixture at conditions
where they would normally be liquids if
they were pure
 This makes finding the fugacity
coefficient hard
 You’ll need to use an equation of
state, because you can’t compare it
to some non-existent gas
 See Example 7.7
Other mixing rules
 The Lewis Randall rule is only one of
many possibilities
 They’ll be introduced in Chapters 9
and 10
Summary
1. Fugacity was invented because
chemical potential is awkward
2. For pure gases we correlate and
compute f/P based on either
measured PvT data, or an
appropriate EOS
Summary cont
3. For pure liquids and solids we usually
compute the fugacity from the vapor
pressure. The effect of increases
above the vapor pressure are small.
4. Ideal liquids are like gases, an
approximation
5. Activity and activity coefficient are
nondimensional representations of
fugacity
Summary Cont
6. Fugacity, activity, and activity
coefficient are calculated – they can
not be measured
7. For mixtures, we calculate fugacity
directly from PvT data, or from an
EOS
• When we use an EOS we need to use a
mixing rule
Summary cont.
8. The simplest mixing rule is the Lewis
and Randall rule – an ideal solution
of nonideal gases
9. For mixtures of liquids we’ll have to
wait for chapters 8 and 9 for
appropriate mixing rules