Pure Component VLE in Terms of Fugacity
Purpose of this lecture:
To derive an expression for the calculation of fugacity of pure liquids
Highlights
Phase equilibrium expressed in terms of fugacities
The fugacity of a pure liquid at a given temperature can be calculated
through its vapour-phase fugacity coefficient and saturation pressure
The effect of pressure on liquid-phase fugacity is captured by the
Poynting factor
Reading assignment: Section 11.5 (pp. 396-401)
CHEE 311
Lecture 10
1
Pure Component VLE in Terms of Fugacity
Consider a pure component at its vapour pressure:
 Phase rule tells us, F=2-2+1 = 1 degree of freedom
 Therefore, at a given T, there can only be a single pressure,
Psat for which a vapour and a liquid are in equilibrium
P
liquid
gas
T
 Along the phase boundary, the chemical potentials are equal
 How do the fugacities of the liquid and gas relate?
CHEE 311
Lecture 10
2
Pure Component VLE in Terms of Fugacity
For the non-ideal, pure gas we can write:
ivap

Givap
 i (T)  RT ln fi
vap
11.38a
For a non-ideal liquid, we can define an analogous expression:
11.38b
liq
liq
liq
i  Gi  i (T)  RT ln fi
At equilibrium
liq
ivap  i (T)  RT ln fivap  liq


(
T
)

RT
ln
f
i
i
i
11.39
In terms of fugacity:
fivap  filiq  fisat
CHEE 311
Lecture 10
11.41
3
Review of Chemical Equilibrium Criteria
We have several different criteria for phase equilibrium. While they
stem from the same theory, they differ in practical applicability.
A system at equilibrium has the following properties:
 the total Gibbs energy of the system is minimized, meaning
that no change in the number of phases or their composition
could lower the Gibbs energy further
d(nG ) T,P  0
 the chemical potential of each component, i, is the same in
every phase within the system
in p phases


p
i  i  ...  i
 the fugacity of each component, i, is equal in every phase of
the system
in p phases
f   f   ...  f p
i
CHEE 311
i
i
Lecture 10
4
Calculating the Fugacity of Pure Liquids
The derivation of the fugacity of a pure liquid at a given T, P is
comprised of four steps:
Step 1. Calculate the fugacity of a vapour at Pisat
Pisat
ln ( fisat / P)  
0
( Z  1)
dP
P
Step 2. Calculate the change in Gibbs energy between Pisat and the
given pressure P using the fundamental equation:
dG = VdP - SdT
which after integration yields:
G i (T,P)  G i (T,Pi ) 
liq
liq
sat
(constant T)
P

Viliq dP
Pisat
Given that liquids are nearly incompressible (Viliq is not a strong
function of P) the integral is approximated as:
(A)
liq
sat
liq
sat
Gliq
(
T
,
P
)

G
(
T
,
P
)

V
(
P

P
)
i
i
i
i
i
CHEE 311
Lecture 10
5
Calculating the Fugacity of Pure Liquids
3. Using the definitions of fugacity:
liq
Gliq
(
T
,
P
)


(
T
)

RT
ln
f
i
i
i
sat
sat
Gliq
(
T
,
P
)


(
T
)

RT
ln
f
i
i
i
i
we can take the difference:
liq
sat
Gliq
(
T
,
P
)

G
(
T
,
P
)
i
i
i
 RT ln( filiq
(B)
/ fisat )
4. Substituting A into B:
RT ln( filiq / fisat )  Viliq (P  Pisat )
or
f
or
liq
i
 fi
sat
 Viliq (P  Pisat ) 
exp 

RT


filiq  isatPisat
CHEE 311
 Viliq (P  Pisat ) 
exp 

RT


Lecture 10
11.44
6
Calculating the Fugacity of Pure Liquids
We can now calculate the fugacity of any pure liquid using two
equations:
filiq

isatPisat
 Viliq (P  Pisat ) 
exp 

RT


11.44
and
isat
Pisat ( Z  1) 
 exp  
dP 
P
 0

11.35
The exponential within Equation 11.44 accounts for the change in
Gibbs energy as we compress the liquid from Pisat to the specified
pressure, P. This is known as the Poynting factor.
 Viliq (P  Pisat ) 
Poynting factor  exp 

RT


This contribution to fugacity is slight at all pressures near Pisat, and
is often assumed to be unity.
CHEE 311
Lecture 10
7
Example Problem
Calculate the fugacity of n-pentane at T= 25 oC and P=101 kPa. The
saturation pressure of n-pentane at 25 oC is Pisat =67.5 kPa.
CHEE 311
Lecture 10
8