Chapter 7:
Equilibrium and
Stability in OneComponent Systems
1
Important Notation
2
Learning objectives
Be able to:
•Identify the criterion for equilibrium subject to different
constraints (specifications);
•Use the concept of stability to predict when vapor-liquid
phase splitting occurs;
•Identify the conditions of phase equilibrium and of the
critical point;
•Compute the fugacity of a pure liquid or vapor;
•Use fugacity in calculations of vapor-liquid equilibrium.
3
The criteria for equilibrium
Consider, for simplicity, a simple thermodynamic system with
known amounts of substances (on a mass or molar basis) . To
define its equilibrium state it is necessary to specify two
additional independent variables. Examples:
P, T 
Common in the design of steady-state processes
T ,V 
Equilibrium in storage vessels of known T and V.
U ,V 
Tank charges and discharges.
4
The criteria for equilibrium
The criteria for equilibrium criteria for these specifications (and
many more) can be derived in an organized way, starting from
the equilibrium condition of a closed, isolated system:
M, U, V are specified and constant
5
Equilibrium in an isolated system
No mass in or out; neglect changes in kinetic and potential
energies. The energy and entropy balances are:
dU
dV
QP
dt
dt
dS Q
  S gen
dt T
with Sgen  0 because of the 2nd law of thermodynamics.
6
Equilibrium in an isolated system
0, no heat transfer to an isolated system
dU
dV
QP
 0  U  constant
dt
dt
0, no change in volume in an isolated system
0, no heat transfer to an isolated system
dS Q
  S gen  S gen  0
dt T
7
Equilibrium in an isolated system
dS
 S gen  0
dt
for systems in which M, U, V are constant.
Away from equilibrium, the system conditions change,
increasing its entropy.
After a long enough wait (from a fraction of a second to many
years, depending on the system), the system attains a
condition in which its state properties no longer change,
including the entropy. At this state, the equilibrium state, the
entropy has a maximum value.
8
Equilibrium in an isolated system
Consider an isolated system composed of two subsystems
containing a single chemical component:
N  N I  N II
U  U I  U II
V  V I  V II
S  S I  S II
9
Equilibrium in an isolated system
Consider an isolated system composed of two subsystems
containing a single chemical component:
N  N I  N II
U  U I  U II
V  V I  V II
S  S I  S II
Find the equilibrium condition if the internal wall is diathermal, rigid, and impermeable.
10
Equilibrium in an isolated system
Consider an isolated system composed of two subsystems
containing a single chemical component:
N  N I  N II
U  U I  U II
V  V I  V II
S  S I  S II
Find the equilibrium condition if the internal wall is diathermal, rigid, and impermeable.
N I  constant
N II  constant
V I  constant
V II  constant
11
Equilibrium in an isolated system
S  S I  S II
dS  dS I  dS II
II
 S I 



S
I
II
dS  
dU

dU

I 
II 

U

U

V I , N I

V II , N II
U  U I  U II  constant  dU II  dU I
 S I 

 S II 
I
dS  

dU


I 
II 
 U V I , N I  U V II , N II 
12
Equilibrium in an isolated system
 S I 

 S II 
I
dS  

dU


I 
II 
 U V I , N I  U V II , N II 
At equilibrium:
 S I 

 S II 
I
dS  

dU
0


I 
II 
 U V I , N I  U V II , N II 
 S I 

 S II 


0
I 
II 
 U V I , N I  U V II , N II 
13
Equilibrium in an isolated system
At equilibrium:
 S I 
 S II 


I 
II 
 U V I , N I  U V II , N II
This looks like a very abstract equilibrium condition.
Based on your daily experience, what state property would be
equal when equilibrium is reached in this problem?
14
Equilibrium in an isolated system
From Chapter 4:
dU  TdS  PdV  GdN
1
P
G
dS  dU  dV  dN
T
T
T
In each subsystem, V and N are constant:
I


1

S
I
I
I
dS  I dU  
dU
I 
T

U

V I , N I
II


1

S
II
II
II
dS  II dU  
dU
II 
T

U

V II , N II
15
Equilibrium in an isolated system
At equilibrium:
 S I 
 S II 


I 
II 
 U V I , N I  U V II , N II
1
1
I
II


T

T
T I T II
Note, for example, that the pressures will in general be
different, because the internal wall is rigid.
16
Equilibrium in an isolated system
Consider an isolated system composed of two subsystems
containing a single chemical component:
N  N I  N II
U  U I  U II
V  V I  V II
S  S I  S II
Find the equilibrium condition if the internal wall is diathermal, moveable, and
impermeable.
17
Equilibrium in an isolated system
Consider an isolated system composed of two subsystems
containing a single chemical component:
N  N I  N II
U  U I  U II
V  V I  V II
S  S I  S II
Find the equilibrium condition if the internal wall is diathermal, moveable, and
permeable.
18
Equilibrium in an isolated system
The equilibrium state was found by setting:
dS  0
However, this mathematical condition is valid in a point of
maximum, in a point of minimum, and in an inflection point.
This is necessary condition for equilibrium, but is not sufficient.
Additionally, to characterize a point of maximum, the following
stability criterion must also be true:
d 2S  0
19
Equilibrium in systems with constant T and V
No mass in or out; neglect changes in kinetic and potential
energies. The energy and entropy balances are:
dU
Q
dt
dS 1 dU

 S gen
dt T dt
dS Q
  S gen
dt T
d U  TS 
 TS gen
dt
with Sgen  0 because of the 2nd law of thermodynamics.
20
Equilibrium in systems with constant T and V
d U  TS  dA

 TS gen  0
dt
dt
for systems in which M, T, V are constant.
Away from equilibrium, the system conditions change,
decreasing its Helmholtz energy.
After a long enough wait (from a fraction of a second to many
years, depending on the system), the system attains a
condition in which its state properties no longer change,
including its Helmholtz energy. At this state, the equilibrium
state, the Helmholtz energy has a minimum value.
21
Equilibrium in systems with constant T and P
No mass in or out; neglect changes in kinetic and potential
energies. The energy and entropy balances are:
dU
dV
QP
dt
dt
dS 1  dU d  PV  
 

  S gen
dt T  dt
dt 
dS Q
  S gen
dt T
d U  TS  PV 
 TS gen
dt
with Sgen  0 because of the 2nd law of thermodynamics.
22
Equilibrium in systems with constant T and P
d U  TS  PV  dG

 TS gen  0
dt
dt
for systems in which M, T, P are constant.
Away from equilibrium, the system conditions change,
decreasing its Gibbs energy.
After a long enough waiting (from a fraction of a second to
many years, depending on the system), the system attains a
condition in which its state properties no longer change,
including its Gibbs energy. At this state, the equilibrium state,
the Gibbs energy has a minimum value.
23
24
Example 1
Use the information available in the steam tables of Appendix
A.III of the textbook to show that the chemical potentials are
equal in the vapor and liquid phases at 100oC and 0.10135 MPa.
Solution:
The target is to show that:
L
V
ˆ
ˆ
G G G G
L
V
The Gibbs energy is not directly available in the steam table but
the enthalpy and entropy are:
Hˆ L  TSˆ L  Hˆ V  TSˆV
25
Example 1
At 100oC and 0.10135 MPa:
kJ
ˆ
H  419.04
kg
kJ
V
ˆ
H  2676.1
kg
kJ
L
ˆ
S  1.3069
kg.K
kJ
V
ˆ
S  7.3549
kg.K
L
Hˆ L  TSˆ L  Hˆ V  TSˆV
419.04  373.15 1.3069  2676.1  373.15  7.3549
kJ
kJ
68.6
 68.4
kg
kg
26
Example 2
Find the normal boiling temperature of n-hexane according to
the van der Waals and the Peng-Robinson equations of state,
using the XSEOS package.
Solution:
The normal boiling temperature is the temperature at which
there is vapor-liquid equilibrium (VLE) at the pressure of 1 atm
(1.0132 bar).
G G
L
G G
IG
G
R,L
,G
R ,V
R,L
V
 G G
IG
R ,V
Departure (residual) functions
27
Example 2
G G
IG
G
R,L
R,L
R,L
 G G
IG
G
R ,V
R ,V
R ,V
G
G

0
RT
RT
Given an equation of state, this is, in general, a non-linear
equation in temperature that needs numerical solution.
Use XSEOS to solve it. How do the values you obtain
compare to the experimental value of 342 K?
28
Example 3
Plot the vapor pressure of n-hexane as function of temperature
according to the Peng-Robinson equation of state, using the
XSEOS package. Compute the vapor pressure from 300 K to 450
K with 10 K intervals. Also, plot ln P vs 1/T.
Solution:
The phase equilibrium equation derived in Example 2 is valid:
G
R,L
G
R ,V
It has to be solved to find the pressure at each specified
temperature along the interval.
29
Example 4
Plot the molar enthalpy of vaporization of n-hexane as function
of temperature according to the Peng-Robinson equation of
state, using the XSEOS package from 300 K to 450 K with 10 K
intervals.
Solution:
The molar enthalpy of vaporization is the difference between
the molar enthalpies of the vapor and liquid in equilibrium:
H
IG
H
R ,V
 H
IG
H
R,L
H
R ,V
H
R,L
Use the XSEOS solution of Example 3 as starting point.
30
Example 5
Plot the molar volume of the liquid and vapor phases in
equilibrium as function of temperature according to the PengRobinson equation of state, using the XSEOS package from 300 K
to 450 K with 10 K intervals.
Solution:
Use the XSEOS solution of Example 3 as starting point.
31
Stability of thermodynamic systems
In, the previous examples the assumption was that the
system had more than one phase.
In many applications, it is important to find out
whether a system splits into several phases. This is
called stability analysis.
There are two types of stability analysis: local and
global. This chapter covers local stability analysis.
32
Stability of thermodynamic systems
Consider an isolated system containing a single chemical
component. Will the system split in two phases?
N  N I  N II
?
U  U I  U II
V  V I  V II
33
Stability of thermodynamic systems
S
S I  S II
Entropy of the single-phase system
Entropy of the two-phase system
S   S  S
I
II
S
?
Key idea:
If, for every attempt to split the system,
S  0 , the original system is stable. In
other words, splitting the system does
not increase its entropy.
34
Stability of thermodynamic systems
The analysis begins by assuming the system will split into two phases, I
and II, with N I  N II .

N
 NS 
S  N S N  N S 
S  S   N S  S 
S  N S  N S
N
I
I
I
II
II
I
I
II
II
I
I
II
II
II
Now, obtain the molar entropy in phases I and II as a Taylor series
expansion around the conditions of the original system.
35
Stability of thermodynamic systems
 S 
 S 
I
I
S  S 

U


V




 U V
 V U
I
1  2 S 
I

U


2  U 2 V

1  2 S

2  V U

2
1  2 S
 
2  U V

I
I

U

V



 I
1  2 S 
I
I
 V U   2  V
2  V U



2
 ...
An analogous series expansion can be used for phase II.
36
Stability of thermodynamic systems




S  N I S  S  N II S  S 
I
II
  S 

 S 
I
I


 U  
 V 
 U V

 V U


2
2
2
1  S 
1  S 
I
I
I

N I   

U


U

V





2  U 2 V
2  U V 


 1  2 S  I

1  2 S 
I
I 2
 
 ...
 V U   2  V
2  V U
 2  V U 





  S 

 S 
II
II


 U  
 V 
 U V

 V U


2
2
2




1

S
1

S
II
II
II

N II   

U


U

V





2  U 2 V
2  U V 


 1   2 S  II

1  2 S 
II
II 2
 
 ...
 V U   2  V
2  V U
 2  V U 





37
Stability of thermodynamic systems
 S 
 S 
I
II
I
II
I
II
I
II




S  
N

U

N

U

N

V

N

V
 

 


 U V
 V U
1  2 S 

2 
2  U V

 N I U I


2
N
II
 U   
2
II
 2 S  I
I
I
II
II
II


N

U

V

N

U

V



 U V 
1  2 S   I
I 2
II 2 
II
 N V
 ...
 2   N V


2  V U 




Pink boxes are equal to zero because of energy and volume conservation.
38
Stability of thermodynamic systems
1  2 S 
S  
2 
2  U V

 N I U I


2
N
II
 U   
II
2
 2 S  I
I
I
II
II
II


N

U

V

N

U

V



 U V 
1  2 S   I
I 2
II 2 
II
 N V
 ...
 2   N V


2  V U 




II
N
II
I
II
N U  N U  0  U   I U
N
I
II
2
  N II
II 
II
 N I   I U   N II U
  N



2
I
  1

II
II 2
   I  N   N II  U

  N


2
N
II
 U 
II
39
2
Stability of thermodynamic systems
The analysis of the other second derivatives is similar. Then:
N II
S 
2
  2 S 
II

U

2 

U

V


2
 2 S
 2
 U V
 2 S 

II
II
II

U

V


V
 2


 V U


2



In matrix notation, we obtain what is called a quadratic form:
S 1
II



U
N II 2 
  2 S 
  2 S 



2 
II

U

V

U





U




V
II
V  
  II 
2
2
  S    S    V 
 U V   V 2  
U 

40
Stability of thermodynamic systems
If the quadratic form is negative definite, the original system is stable, i.e.:
 U II

  2 S 
  2 S 



2 
II

U

V

U





U



V
II 
V  
  II   0
2
2
  S    S    V 
 U V   V 2  
U 

If the quadratic form is negative definite, all the eigenvalues of the matrix are
negative. Simplifying the notation:
 U II

II  S UU
V  
 S UV
S UV   U II 
 II   0

S VV   V 
Your textbook has a slightly different derivation, with the same conclusions section 41
7.2)
Stability of thermodynamic systems
Here, we will skip the formal solution using eigenvalues and use a simpler
property of negative definitive matrices that says determinants of their
submatrices along the diagonal are negative if of odd order, and positive if of
even order, i.e.:
S UU  0
S VV  0
S UU S VV   S UV   0
2
42
Stability of thermodynamic systems
S UU
  1 T  
 2 S     S  



 



2 
 U V  U  U V  V  U  V
1  T 
1
 2
  2 0
T  U  V
T CV
CV  0
43
Stability of thermodynamic systems
The analysis of the other derivatives involves more algebraic steps. A
summary of the implications of stability is:
CV  0
 P 

 0
 V T
CP  CV  0
44
Fugacity of a pure component
The fugacity coefficient of a pure component is defined as:
 G R T , P  
 G T , P   G IG T , P  
  exp 
  exp 

RT
 RT 


i.e., there is a simple and direct connection between the fugacity
coefficient of a pure component and the departure (residual) molar
Gibbs function at the same temperature and pressure:
G T , P   G T , P   G
R
IG
T , P   RT ln 
45
Fugacity of a pure component
RT ln   G T , P   G
IG
T , P 
For an ideal gas:
RT ln 
IG
G
IG
 T , P   G T , P   0
IG
 IG  1
The fugacity coefficient of a pure ideal gas is equal to 1.
46
Fugacity of a pure component
To find the fugacity coefficient, consider its change in an
isothermal process from zero pressure to the system pressure:
IG

G  T , P  
 G  T , P  
RTd ln   
 dP
 dP  
P
P

T

T
IG


G  T , P   
1  G  T , P  
d ln  

  dP
  
RT 
P
P

T

T 

ln 

0
IG


G  T , P   
 G  T , P  
1
d ln  

  dP
  

RT 0 
P
P

T

T  47

P
Fugacity of a pure component
P
1
IG


ln  
V
T
,
P

V
T
,
P
dP







RT 0
1 
RT 
ln  
V
dP



RT 0 
P 
P
48
Fugacity of a pure component
What are the units of the fugacity coefficient?
 G R T , P  
 G T , P   G IG T , P  
  exp 
  exp 

RT
 RT 


49
Fugacity of a pure component
Also, if you know how to evaluate
departure (residual) functions, you know
how to compute the fugacity coefficient.
 G R T , P  
 G T , P   G IG T , P  
  exp 
  exp 

RT
 RT 


50
Example 6
Find the expression of the fugacity coefficient of a pure
component according to the van der Waals equation of state.
51
Example 6
Find the expression of the fugacity coefficient of a pure
component according to the van der Waals equation of state.
Solution:
1 
RT 
ln  
V
dP



RT 0 
P 
P
RT
a
P
 2
V b V
A practical difficulty: expressing the molar volume as function of
pressure in the van der Waals equation of state is not easy.
52
Example 6
Way around: first, substitute P in the formula. To do that, it is
necesssary to express dP as function of dV.
RT
a
P
 2
V b V

RT
2a 
dP   
 3  dV
2
 V  b  V 
Then:



V 

1 
RT
RT
2
a
 
ln  
V

 3  dV
2

RT  
 RT
a    V  b  V 

V  b  2  
V  


53
Example 6
It is now a calculus problem… When studying these slides, do all
the steps to prove that:
2


b
2a
RTV
ln  

 ln 
2 
V  b RTV
 ab  aV  RTV 
For most of the equations of state commonly used for chemical
process design, the steps to derive expressions for fugacity
coefficients are analogous, but generally more complicated
because the equations of state are more complex.
54
Alternative expressions for
fugacity coefficients
v
f
1
ln

P RT
l
f
1
ln

P RT
Z V RT
V
P

 RT

V
V


 P dV  ln Z  Z  1
 V

Z L RT
V
P

 RT

L
L


 P dV  ln Z  Z  1
 V







55
Fugacity of a pure component
We showed that the phase equilibrium condition for a pure
component is:
G G
L
G G
IG
G
R,L
R,L
V
 G G
G
IG
R ,V
R ,V
Using fugacity coefficients:
RT ln   RT ln 
L
V
56
Fugacity of a pure component
At equilibrium, the vapor and liquid phases have equal
temperature:
ln   ln 
L
V
ln   ln   0
L
V
 
L
V
These are just three different ways of expressing the VLE
condition in terms of fugacity coefficients.
57
Example 7
Find the normal boiling temperature of n-hexane according to
the Peng-Robinson equations of state, using the fugacity
coefficient function of the XSEOS package.
Solution:
The normal boiling temperature is the temperature at which
there is vapor-liquid equilibrium (VLE) at the pressure of 1 atm
(1.0132 bar). The phase equilibrium condition is:
ln  L  ln  V  0
Solve this non-linear equation to find the normal boiling
temperature.
58
Fugacity of a pure component
Starting from:
 L  V
Multiply both sides by pressure:
 L P  V P
The product in each side of this equation is the fugacity of the
pure component:
f L  fV
59
Fugacity of a pure component
f
f  P   
P
What are the units of fugacity in the S.I.?
60
Fugacity of a pure component
f
G  G  G  G  RT ln   G  RT ln
P
IG
R
G G
IG
IG
IG
 RT ln f  RT ln P
G G
ln f 
RT
IG
 ln P
61
Fugacity of a pure component
G G
ln f 
RT
IG
 ln P
Effect of changing pressure at constant temperature:

IG


G
RT
  ln f     G RT   
 

 
P
P
 P T 
T 
 
  ln P 


  P T
T
IG

1  G 
1 G 
1
  ln f 

 

 

 
 P T RT  P T RT  P T P
62
Fugacity of a pure component
Effect of changing pressure at constant temperature:
IG

1  G 
1 G 
1
  ln f 

 

 

 
 P T RT  P T RT  P T P
V V
1
  ln f 



 
 P T RT RT P
IG
V
  ln f 

 
 P T RT
63
Fugacity of a pure component
G G
ln f 
RT
IG
 ln P
Effect of changing temperature at constant pressure :

IG


G
RT
   G RT   
  ln f 
 

 
T
T
 T  P 
 P 
 
0
  ln P 


  T  P
P
64
Fugacity of a pure component
   G RT  
1

 
T

 P RT
G
 G 


 
2
 T  P RT
S H T S 



2
RT
RT
S
H
TS
H




2
2
RT RT
RT
RT 2
65
Fugacity of a pure component
Effect of changing temperature at constant pressure :

IG

   G RT     G RT
  ln f 
 

 
T
T
 T  P 
 P 
  ln f 


 T  P
H H 

H


IG
RT
2
 

P
R
RT
2
66
Fugacity of a pure liquid
Besides integrating formulas that depend on an equation of
state, there is another path to evaluating the fugacity of pure
liquids, especially useful for approximate calculations.
To prove it, consider the effect of pressure on the logarithm of
fugacity at constant temperature (shown on a previous slide):
V
  ln f 

 
 P T RT
Now compute the difference between ln f at the saturation
pressure and an arbitrary pressure, at the same temperature:
V
  ln f 
  
dP
 dP  
P T
RT
P sat 
P sat
P
ln f  ln f sat
P
67
Fugacity of a pure liquid
P
P




V
V
sat
sat sat
f  f exp  
dP    P exp  
dP 
 P sat RT

 P sat RT

Poynting correction
The formula above is rigorous – it involves no approximation.
In practice, it is common to use it with approximations.
•If the difference between the saturation pressure and the
pressure is small, the Poynting correction is nearly 1.
•If the saturation pressure is small, the saturated vapor can
be considered as an ideal gas and the fugacity coefficient is
approximately equal to 1.
68
Fugacity of a pure liquid
P
P




V
V
sat
sat sat
f  f exp  
dP    P exp  
dP 
 P sat RT

 P sat RT

Poynting correction
When making all these approximations:
f P
sat
69
Fugacity of a pure liquid
P
P




V
V
sat
sat sat
f  f exp  
dP    P exp  
dP 
 P sat RT

 P sat RT

Poynting correction
A less drastic approximation is to keep all terms but assume
the molar volume of the liquid is pressure-independent:
V  P  P sat  

f   sat P sat exp 
RT


70
Example 8
Compute the fugacity of pure liquid water at 300oC and 25 MPa
from steam table data.
Solution:
We will assume the molar volume is pressure-independent and
equal to the saturated liquid molar volume.
L
sat
V sat

P

P


sat

f  f exp 
RT


71
Example 8
Saturated liquid molar volume at 300oC:
V
L
sat
3
m
1kg
g
3
 1.404 10

18.015

kg 1000 g
mol
3
m
2.5293 105
mol
The fugacity of the saturated liquid at 300oC is:
f sat  6.7337MPa  6.7337 106 Pa
Read illustration 7.4.3 in the book to see how to get this result.
72
Example 8
Saturation pressure at 300oC from steam table.
P sat  8.581MPa  8.581106 Pa
Then:
L
sat
V sat

P

P


sat

f  f exp 
RT


3


m
5
6
2.5293 10
  25  8.581 10 Pa 

mol
6.7337 106 Pa  exp 

J


8.314
 573.15 K


mol.K
7.347 106 Pa
73
Fugacity of a pure solid
At a given temperature, a solid may have different crystalline
structures (phases) at different pressures.
f  f
sat
exp 
j
P j 1

Pj
j
V
dP
RT
Assuming the molar volume of all solid phases is equal:
V S  P  P sat  

f  P sat exp 
RT


Here:
P sat : sublimation pressure
 sat =1: assuming small sublimation pressure
74
Example 9
Compute the fugacity of pure solid napthalene at 35oC and 50
bar.
Data:
3783
log10 P  bar   8.722 
T K 
sat
g
  1.025 3
cm
g
MW  128.19
mol
75
Example 9
Solution:
8.722 
3783
35 273.15
P sat  bar   10
 2.789 104 bar
g
128.19
3
cm
mol  125.06
VS 
g
mol
1.025 3
cm


cm3
4
125.06
  50  2.789  10  bar 

mol
f  2.789  104 bar exp 

3
 83.14 bar.cm  35  273.15  K 


mol.K
f  3.582 104 bar
76
Clayperon equation
Consider a pure component in vapor-liquid equilibrium. Then:
G G
L
V
Consider some perturbation is imposed, reaching a new state
of equilibrium. Then:
G
L
new
G
V
new
Subtracting these two expressions, we have:
G
L
new
G  G
L
V
new
 G  G  G
V
L
V
77
Clayperon equation
Assume the perturbation is arbitrarily small. Then:
dG  dG
L
V
But:
dG  SdT  VdP
 S dT  V dP
L
S
L
V
S
L
vap
  S dT  V dP
 dT  V
V
V
V
V
L
 dP
vap
vap
78
Clayperon equation




S S
 vap S
 P 
 V



L
 vap V
V V
 T G L GV
vap


V
L


S S
 vap H
 P 
 V



L
T  vap V
V V
 T G L GV
vap
V
L
The Clayperon equation is also valid for phase changes
between solid and liquid and between solid and vapor.
79
Clausius-Clayperon equation
Starting from the Clayperon equation:




S S
 vap H
 P 
 V



L
T  vap V
V V
 T G L GV
vap
V
L
Assume (this is why the Clausius-Clayperon equation is an
approximation):
RT
 vap V  V  V  V  vap
P
V
L
V
80
Clausius-Clayperon equation
Then:
vap
dP

dT
P vap  vap H
RT 2
1 dP vap d ln P vap  vap H


vap
2
P
dT
dT
RT
 vap H
d ln P vap

d 1 T 
R
81
Example 10
Estimate the molar enthalpy of vaporization of n-hexane at its
normal boiling point given the ln P vs. (1/T) plot. Use the
Clausius-Clayperon equation.
82
Antoine equation
Starting from the Clausius-Clayperon equation:
 vap H
d ln P vap

 B
d 1 T 
R
Integrating this equation, we obtain:
ln P
vap
B
 A
T
In this equation, A is an integration constant and B is defined as
shown above. This a simple relation between the saturation
pressure and temperature, but not accurate enough.
83
Antoine equation
The Antoine equation is an empirical modification:
ln P
vap
B
 A
T C
A, B, and C are obtained by fitting
experimental vapor pressure data.
Extensive parameter tables exist with
parameters for hundreds of components.
84
Riedel equation
Another empirical modification:
ln P
vap
B
 A   C ln T  DT 6
T
85
Harlecher-Braun equation
Yet another empirical modification:
ln P
vap
B
DP sat
 A   C ln T 
2
T
T
86
Recommendation
Read chapter 7 in the textbook and review all examples.
87