Chapter 7
Phase Behavior of Pure Fluids
1
„
„
In this chapter we examine conditions for the
existence of an equilibrium state in a singlecomponent system, which is a useful reference
to solve problem of phase equilibrium.
Equilibrium in multicomponent systems will be
considered in following chapters 8 and 9.
From four basic relations:
dG = VdP − SdT
EQ(1)
Integrate EQ(1) at constant T from P1 to P2 for IG:
∫
2
1
dG = dG
IG
2
1
P2
P2
p1
p1
= G − G = ∫ VdP = RTd ln( P )
2
1
When P1 =0: It means (ΔG
IG
→ −∞) and it is singularity we don't want to see this calculation.
Hence, a new definition of fugacity (potential of escape) was employed for the G calculation.
i.e.:
dG (T , P) = RT d ln f (T , P )
V
T
Why is G, not H, A, S?
2
7.1 The criteria for equilibrium
Balance equations for a closed system (at constand T and P)
3.1-4b
4.1-5b
dU
dV
= Q& − P
dt
dt
dS Q& &
= + Sgen
dt T
(7.1-1)
(7.1-2)
Second law of thermodynamics
S& ≥ 0 (the equality holding at equilibrium or for reversible processes)(7.1-3)
gen
For a constant-volume system exchanging no heat with its surroundings,
dU
=0
dt
so that
U = constant
dS &
= Sgen ≥ 0
dt
(7.1-4)
S = maximum at equilibrium in a closed system at constant U and V
(7.1-5)
( S ) N ,U ,V = maximum
The entropy function can only be increased during the approach to
equilibrium because the sign of entropy generation.
3
To illustrate the use of this equilibrium criterion, consider the very simple, initially
nonuniform system shown in Fig. 7.1-1.
There a single-phase, single-component fluid in an adiabatic, constant-volume container
has been divided into two subsystems by an imaginary boundary.
Each of these subsystems is assumed to contain the same chemical species of uniform
thermodynarnic properties. However, these subsystems are open to the flow of heat and
mass across the imaginary internal boundary, and their temperature and pressure need
not be the same.
Yes Mass/Energy
ndary
u
o
b
ary
n
i
g
a
Im
I
II
No M
ass
/Ene
rgy
Figure 7.1-1 An isolated non-equilibrium system
4
For the composite system, the total mass (moles), internal energy, volume, and entropy are:
N = N I + N II ⎫
I
II ⎪
U = U +U ⎬
V = V I + V II ⎪⎭
(7.1-6)
Now considering the entropy to be a function of internal energy, volume, and mole
number, we can compute the change in the entropy of system I due to changes in N I
U I and V I from Eq. 6.2-5c. ( dU = TdS − PdV + GdN ; dS =
⎛ ∂S I ⎞
⎛ ∂S I ⎞
⎛ ∂S I ⎞
I
I
dS = ⎜ I ⎟
dU + ⎜ I ⎟
dV + ⎜ I ⎟
dN I
⎝ ∂U ⎠V I ,N I
⎝ ∂V ⎠U I , N I
⎝ ∂N ⎠U I ,V I
I
P
G
1
dU + dV − dN )
T
T
T
Chain − rule
PI
GI
1
I
I
= I dU + I dV − I dN I
T
T
T
PII
GII
1
II
II
II
Similarly, dS = II dU + II dV − II dN II
T
T
T
PI
PII
GI
GII
1
1
I
II
I
II
I
II
I
dS = dS + dS = I dU + II dU + I dV + II dV − I dN − II dN II
T
T
T
T
T
T
EQ(f)
5
Total number of molecules, total internal erergy, and total volume are constant in the composite.
dN =dN I + dN II = 0 or dN I = −dN II
dU =dU I + dU II = 0 or dU I = −dU II
(7.1-7)
dV =dV I + dV II = 0 or dV I = − dV II
Consequently, EQ(f) becomes:
⎛ P I P II ⎞ I ⎛ G I G II ⎞ I
1 ⎞
⎛ 1
I
dS = ⎜ I − II ⎟ dU + ⎜ I − II ⎟ dV − ⎜ I − II ⎟ dN
(7.1-8) dS=0
T
T
T
T
⎝T T ⎠
⎝
⎠
⎝
⎠
Now since S = maximum (or dS = 0) for all system varations at constan N , U , and V
at Equilibrium
(here all variations of the non-zero inependent varables dU I , dV I and dN I at constant total
number of moles, total internal energy, and total volume), we conclude that
1
1
⎛ ∂S ⎞
= 0 so that I = II or T I = T II
⎜
I ⎟
T
T
⎝ ∂U ⎠V I , N I
(7.1-9a)
P I P II
⎛ ∂S ⎞
I
II
⎜ I ⎟ I I = 0 so that I = II or P = P
T
T
⎝ ∂V ⎠U , N
(7.1-9b)
}
Mechanical Equilibrium
G I G II
⎛ ∂S ⎞
I
II
Chemical Equilibrium
(7.1-9c)
⎜ I ⎟ I I = 0 so that I − II or G = G
T
T
⎝ ∂N ⎠U ,V
Therefore, the equilibrium condition for the system illustrated in Fig. 7.1-1 is satisfied
if both subsystems have the same temperature, the same pressure, and the same molar
Gibbs energy. For a single-component, single-phase system, this implies that the composite system should be uniform in T, P and G.
6
Equilibrium criteria and Stability condition
dS = 0 => a possible equilibrium state of the closed, isolated system (i.e. S is a maximum)
From calculus we know that dS = 0 is necessary but not sufficient condition
for S to achieve a maximum value.
Since at dS = 0, the following two cases can be frequently happened.
Smaximum :
dS = 0 and d 2 S < 0
(true equilibrium state)
Smimimun :
dS = 0 and d 2 S > 0
(unstable state)
Metastable inflection point is no good: d 2 S = 0
In − summary :
At Smaximum :
dS = 0
determined equilibrium criteria.
At Smaximum:
d 2S <0
determined stability condition.
Why
2
d S<0
determined stability condition?
please see stability criteria!
7
Alternative equilibrium criteria (G and A)
For a closed system at constant temperature and volume (no PV work)
the energy and entropy balances are
dU &
dS Q& &
dS
= Q and
= + Sgen ; Q& = T
− TS&gen
dt
dt T
dt
d (U − TS )
dU &
dS
= Q =T
− TS&gen ;
= −TS&gen ≤ 0
dt
dt
dt
A = minimum for equilibrium in a closed system at constant T and V
(7.1-10)
For a closed system at constant temperature and pressure (with PV work)
the energy and entropy balances are
dU &
dV
dS Q& &
dS
=Q−P
and
= + Sgen ; Q& = T
− TS&gen
dt
dt
dt T
dt
dU &
dV
dS
dV d (U + PV − TS )
=Q −P
=T
− TS&gen − P
;
= −TS&gen ≤ 0
dt
dt
dt
dt
dt
G = minimum for equilibrium in a closed system at constant T and P
(7.1-11)
It is a ls o th e c o n s e q u e n c e o f d
2
S < 0 (i.e . S&
gen
It is m o s t im p o rta n t o f th e e q u ilib riu m c rite ria !
= 0 ).
8
9
ILLUSTRATION 7.1-2
Proving the Equality of Gibbs Energies for Vapor-Liquid Equilibrium
Use the information in the steam tables of Appendix A.III to show that Eq. 7. l-9c is satisfied at
G I = G II
l00 o C and 0.10135 MPa.
SOLUTION
From the saturated steam temperature table, we have at l00 o C and 0.10135 MPa
Hˆ L = 419.04 kJ/kg
Hˆ V = 2676.1 kJ/kg
Sˆ L = 1.3069 kJ/ ( kg K )
Sˆ V = 7.3549 kJ/ ( kg K )
Since Gˆ = Hˆ − TSˆ , we have further that
Gˆ L = 419.04 − 373.15 × 1.3069 = −68.6 kJ/kg
and
Gˆ V = 2676.1 − 373.15 × 7.3549 = −68.4 kJ/kg
which, to the accuracy of the calculations here, confirms that Gˆ L = Gˆ V (or,
L
V
G = G ) at this vapor-liquid phase equilibrium state.
10
7.2 STABILITY OF THERMODYNAMIC SYSTEMS
What is the meaning of Stability Criterion of d 2 S < 0 at constant M , U , and V
Suppose that a small fluctuation in a fluid property, say temperature or pressure,
occurs in some region of a fluid that was initially at equilibrium; is the character
of the equilibrium state such that d 2 S < 0, and the fluctuation will dissipate, or
such that d 2 S > 0, in which case the fluctuation grows until the system evolves
to a new equilibrium state of higher entropy?
In fact, since we know that fluids exist in thermodynamically stable states , we will take
as an empirical fact that d 2 S < 0 for all real fluids at STABLE equilibrium, and instead establish
the restrictions placed on the equations of state of fluids by this stability condition.
System at Stable Equilibrium is better than at Unstable Equilibrium
11
Using the (imaginary)
subdivision of the system into two subsystems' and writing the extensive properties N ,
U , V , and S as sums of these properties for each subsystem, we were able to show in
Sec. 7.1 that the condition
dS = dS I + dS II = 0
(7.2-1)
for all system variations consistent with the constraints (i.e., all variations in dN I , dV I ,
and dU I ) led to the requirements that at equilibrium
T I = T II ; P I = P II ; G = G
I
II
Continuing, we write an expression for the stability requirement d 2 S < 0 for this
system and obtain
CV > 0
(7.2-12)
⎛ ∂V ⎞
⎜
⎟ <0
⎝ ∂P ⎠T
or KT = -
1 ⎛ ∂V ⎞
⎜
⎟ >0
V ⎝ ∂P ⎠T
(7.2-13)
12
„
Equations 7.2-12 and 7.2-13 may be thought of as
part of the philosophical content of thermodynamics.
In particular, thermodynamics alone does not give
information on the heat capacity or the equation of
state of any substance, such information can be
gotten only from statistical mechanics or experiment.
However, thermodynamics does provide
restrictions or consistency relations that must be
satisfied by such data; Eqs. 7.2-12 and 7.2-13 are
examples of this. C > 0
V
⎛ ∂V ⎞
⎜
⎟ <0
⎝ ∂P ⎠T
13
ILLUSTRATION 7.2-1
Using the Steam Tables to Show That the Stability Conditions Are Satisfied for Steam
i.e. show that Eqs. 7.2-12 and 7.2-13 are satisfied by superheated steam.
CV > 0
Solution
⎛ ∂V ⎞
⎜
⎟ <0
P
∂
⎝
⎠T
It is easiest to use Eq. 7.2-13 in the form (∂P / ∂V )T < 0, which requires that the volume
decrease as the pressure increase at constant temperature. This is seen to be true by using the
superheated steam table and obserbing that Vˆ decreases as P increases at fixed temperature.
For example, at 1000 o C
P ( MPa )
(
Vˆ m3／kg
0.50
)
0.8
1.0
1.4
1.8
2.5
1.1747 0.7340 0.5871 0.4192 0.3260 0.2346
Proving that CV > 0 or CˆV > 0 is a bit more difficult since
⎛ ∂Uˆ ⎞
ˆ
CV = ⎜
⎟
∂
T
⎝
⎠V
and the internal energy is not given at constant volume. Therefore, interpolation methods must
be used. As an example of how the calculation is done, we start with the following data from
the superheated vapor portion of the steam tables.
14
( C)
o
T
P = 1.8 MPa
P = 2.0 MPa
Vˆ m3 /kg
Uˆ ( kJ/kg ) Vˆ m3 /kg
Uˆ ( kJ/kg )
(
)
(
)
800
0.2742
3657.6
0.2467
3657.0
900
0.3001
3849.9
0.2700
3849.3
1000
0.3260
4048.5
0.2933
4048.0
To proceed, we need values of the internal energy at two different temperatures (pressures) and the same
specific volume. We will use P = 2.00 MPa and T = 1000 o C as one point; at these conditions
Vˆ = 0.2933 m3 /kg and Uˆ = 4048.0 kJ/kg. We now need to find the temperature at which
Vˆ = 0.2933 m3 /kg at P = 1.80 MPa. We use linear interpolation for this:
Vˆ (T ,1.80 MPa ) − Vˆ 800o C, 1.80 MPa
0.2933 − 0.2742
T − 800
=
=
900 − 800 Vˆ 900o C,1.80 MPa − Vˆ 800o C, 1.80 MPa
0.3001 − 0.2742
(
)
(
)
(
)
so that T = 873.75o C. Next we need the internal energy Uˆ at T = 873.75o C and P = 1.80 MPa
(since at these conditions Vˆ = 0.2933 m3 / kg). Again using linear interpolation,
(
)
(
o
o
873.75 − 800 Uˆ 873.75 C,1.80 MPa − Uˆ 800 C, 1.80 MPa
=
900 − 800
Uˆ 900o C,1.80 MPa − Uˆ 800o C, 1.80 MPa
=
(
)
(
(
)
)
)
Uˆ 873.75o C,1.80 MPa − 3657.6
3849.9 − 3657.6
we find that
Uˆ 873.75o C,1.80 MPa = Uˆ 873.75o C, 0.2933 m3 /kg = 3799.4 kJ/kg
(
)
(
)
15
For CˆV values between 1.8MPa < P < 2.0MPa:
Finally, replacing the derivative by a finite difference, and for the average temperature, i.e.,
(
)
T = 1000 + 873.75) / 2 = 936.9o C , we have
(
)
(
ˆ 1000o C,0.2933m3 /kg − Uˆ 873.75o C,0.2933m3 /kg
3
U
⎛
⎞
m
CˆV ⎜ T = 936.9o C,Vˆ = 0.2933
⎟≈
o
kg
(1000 − 873.75) C
⎝
⎠
)
4048.0 − 3799.4
kJ
= 1.969
>0
1000 − 873.75
kg K
Similarly, we would find that CˆV > 0 for all other conditions (e.g. 3 MPa < P < 4 MPa).
=
In Summary:
Steam Uˆ and Hˆ are weak functions of pressure when pressure is low.
16
7.3
PHASE EQUILIBRIA:
APPLICATION OF THE EQUILIBRIUM AND
STABILITY CRITERIA TO THE EQUATION OF STATE
Figure 7.3-1 Isotherm of the
Van der Waals equation in
The pressure-volume plane.
17
Maxwell’s rules/Level rules
In case : any intensive property θ = V
(
)
V = ω VV + ω LV = ω VV + 1 − ω V V
V
ωV
L
V
L
V −V
= V
= mole fraction of vapor phase
1 V −V L
L
L
ωV
V −V
1
=
=
1− ωV V V −V 2
(
Good for weight fraction as well
Vapor mole fraction Length of middle(Δ) to saturated liquid curve
)
=
Liquid mole fraction Length of middle(Δ) to saturated vapor curve
Theorem :
LHS =
N
N
=
= RHS
DN ± N
DN ± N
or
LHS =
N ± DN
N ± DN
=
= RHS
DN
DN
x 1 Δ 2o
18
ILLUSTRATION 7.3-1
Computing the Properties of a Two-Phase Mixture
Compute the total volume, total enthalpy, and total entropy of 1 kg of water at 100 o C, half by
weight of which is steam and the remainder liquid water.
SOLUTION
From the saturated steam temperature table at 100 o C, the equilibrium pressure is 0.10135 MPa
and
Vˆ L = 0.001004 m3 /kg Vˆ V = 1.6729 m3 /kg
Hˆ L = 419.04 kJ/kg Hˆ V = 2676.1 kJ/kg
Sˆ L = 1.3069 kJ/ ( kg K ) Sˆ V = 7.3549 kJ/ ( kg K )
Using Eq. 7.3-1a on a mass basis gives
Vˆ = 0.5 × 0.001004 + 0.5 ×1.6729 = 0.83645 m3 /kg
The analogous equation for enthalpy is
Hˆ = 0.5 × 419.04 + 0.5 × 2676.1 = 1547.6 kJ/kg
Sˆ = 0.5 ×1.3069 + 0.5 × 7.3549 = 4.3309 kJ/ ( kg K )
19
Phase equilibrium Solver is tried to find a better Pα !!!
x
For what’s purpose?
o
Figure 7.3-2 A low-temperature isotherm of the
Van der Waals equation.
20
Vapor pressure of a pure substance
Pa <P <Pb
G =G
dG = VdP − SdT
I
II
At constant T:
P2
ΔG = ∫ VdP
P1
Pa
Pb
Pα
Pα
Pa
Pb
G − G = 0 = ∫ VdP + ∫ VdP + ∫ VdP
V
L
I
(7.3-2)
II
In Figure 7.3-2:
I − LHS ( Area )below = II − RHS ( Area )above
21
Shadow equals to shadow
x
o
Figure 7.3-2 A low-temperature isotherm of the
Van der Waals equation.
22
x
o
Figure 7.3-3
A low-temperature isotherm of a real fluid, the horizontal
line also corresponded to data obtained from a V-L
experiment, which is the tie line.
23
x
o
o
x
x
o
Figure 7.3-4
The van der Waals fluid with the vapor-liquid coexistence region identified,
which all tie lines can be corresponded to data obtained from V-L experiments.
How to handle V-L equilibrium for the binary mixture?
[ANS]: X,O: experimental data pairs could be correlated by EOS to find the binary
interactional parameters (i.e. kij) in the a and b calculations for a binary mixture.
m
m
i
m
j
a ( x) = ∑∑ xi x j (aii a jj ) (1 − kij )
b( x ) = ∑
i
m
∑
j
⎛ bii + b jj
xi x j ⎜
⎝ 2
⎞
⎟ = Σxi bii
⎠
kij = 0 at i − molecule = j − molecule
i = 2 reduced to = Σxi bii
24
m
m
i
m
j
a ( x) = ∑∑ xi x j (aii a jj ) (1 − kij )
b( x ) = ∑
i
m
∑
j
⎛ bii + b jj
xi x j ⎜
⎝ 2
⎞
⎟ = Σxi bii
⎠
kij = 0 at i − molecule = j − molecule
i = 2 reduced to = Σxi bii
Figure 7.3-5
The PVT phase diagram for a substance with a single solid phase,
representing either by Equation of State or by experimental data.
25
C.P.:
Terminus of the vapor-liquid
coexistence curve in P-T plane
Figure 7.3-6
Phase diagram in the P-T plane
Figure 7.3-5.
projected from 3D diagram of
26
C.P. recognized by:
(1) Terminus of the vapor-liquid coexistence
curve in P-T plane.
(2) Peak of the vapor-liquid coexistence curve
in P-V plane.
PL
PH
Supposed G= f(T, P) at equilibrium condition,
G is minimum, which means LIQUID phase
(below phase transition T, TP) is equilibrium
with VAPOR phase (above TP) and both
phases are present at TP.
Then the angle (θ) decreased as pressure
(temperature) increased. At critical pressure,
θ=0 occurred at TC. It means:
⎛ ∂ GV
⎜
⎜ ∂T
⎝
⎞
⎛ ∂G L ⎞
⎟ =⎜
⎟ : GV (T , P ) = G L (T , P ); SV = S L
⎟
⎜ ∂T ⎟
C C
C C
⎠P ⎝
⎠P
7.1-9c
Figure 7.3-7 The molar Gibbs energy as a function of temperature
For the vapor and liquid phases of the same substance.
27
7.4 THE MOLAR GIBBS ENERGY AND
FUGACITY OF A PURE COMPONENT
For the vapor-liquid equilibrium in a pure fluid, the equality of molar
Gibbs energies in the coexisting phases is
G (T , P ) = G
L
V
(T , P )
dG = − SdT + VdP
⎛ ∂G ⎞
⎜
⎟ = −S
⎝ ∂T ⎠ P
⎛ ∂G ⎞
⎜
⎟ =V
∂
P
⎝
⎠T
(7.4-2)
Integration of Eq. 7.4-2 between any two pressures P1 and P2
(at constant temperature) yields
G (T1 , P2 ) − G (T1 , P1 ) = ∫ VdP
P2
P1
G
IG
(T1 , P2 ) − G
IG
(T1 , P1 ) = ∫
P2
P1
RT
dP
P
(7.4-3)
(7.4-4)
28
(
G (T1 , P2 ) − G
IG
(T1 , P2 ) ) − ( G (T1 , P1 ) − G IG (T1 , P1 ) ) = ∫P
P2
1
RT ⎞
⎛
V
−
⎜
⎟ dP (7.4-5a)
P
⎝
⎠
set P1 = 0
G (T , P ) − G
IG
(T , P ) = ∫0 ⎛⎜ V −
P
⎝
RT
P
⎞
⎟ dP
⎠
(7.4-5b)
dG
Define fugacity
⎧⎪ G (T , P ) − G IG (T , P ) ⎫⎪
⎧ 1
f = P exp ⎨
⎬ = P exp ⎨
RT
⎩ RT
⎪⎩
⎭⎪
∫
P
∫
P
0
RT ⎞ ⎫
⎛
V
−
⎜
⎟ ⎬ dP
P ⎠⎭
⎝
(7.4-6a)
RT ⎞ ⎫
⎛
V
−
⎜
⎟ dP ⎬
P
⎝
⎠ ⎭
(7.4-6b)
Define fugacity coefficient
⎧⎪ G (T , P ) − G IG (T , P ) ⎫⎪
⎧ 1
f
φ = = exp ⎨
⎬ = exp ⎨
P
RT
⎩ RT
⎪⎩
⎪⎭
f
= 1;
lim f = P
p →0
p →0 P
p →0
IG
f G (T , P ) − G (T , P )
1
=
ln =
P
RT
RT
IG
0
lim φ = lim
= RTd ln P
dG = RTd ln f
For a mixture at constant T:
∫
P
0
RT
⎛
V
−
⎜
P
⎝
⎞
⎟ dP
⎠
(dGi )T = RTd ln f i
f i : fugacity of i component in the mixture
Gi : partial Gibb energy of i component
29
Equilibrium criteria for pure compound
T I = T II
P I = P II
G =G
I
G
IG
II
(T , P ) + RT ln
f I (T , P )
P
=G =G =G
I
II
(T , P ) = f (T , P )
φ I (T , P ) = φ II (T , P )
f
I
II
IG
(T , P ) + RT ln
(7.4-7a)
(7.4-7b)
IG
(T , P ) =
P
Multi-component Phase Equation:
gave i component (igave )
Pxi I φi I = Pxi II φi II
(dGi )T = RTd ln fi
Re called :
f G (T , P ) − G
ln =
P
RT
f II (T , P )
1
RT
∫
P
0
RT
⎛
V
−
⎜
P
⎝
⎞
⎟ dP
⎠
⎛ fi
= RTd ln ⎜
⎝ xi
⎞
⎟ = RTd ln fi
⎠
30
Converting φ from (T, P) to (T, V)
IG
⎧ 1
f
⎪⎧ G (T , P ) − G (T , P ) ⎫⎪
φ = = exp ⎨
⎬ = exp ⎨
P
RT
⎩ RT
⎪⎩
⎪⎭
f
RT ⎞
1 P⎛
−
V
ln =
⎜
⎟ dP
P RT ∫0 ⎝
P ⎠
P
P
P
1
dP = d ( PV ) − dV = dZ − dV
V
V
Z
V
f
1
ln =
P RT
=
1
RT
∫
P
0
RT
⎛
−
V
⎜
P
⎝
1
⎞
=
d
P
⎟
RT
⎠
RT
⎛
−
V
∫V =∞ ⎜⎝ P
V
∫
P
0
RT ⎞ ⎫
⎛
−
V
⎜
⎟ dP ⎬
P
⎝
⎠ ⎭
(7.4-6b)
at constant T
⎞
P
⎞⎛ P
−
dZ
dV
⎟
⎟⎜
V
⎠⎝ Z
⎠
⎛ RT
⎞
P
−
∫V =∞ ⎜⎝ V ⎟⎠ dV − ln Z + ( Z − 1)
V
(7.4-8)
How to do it ???
31
How to gat EQ(7.4-8) for pure component?
dP =
1
P
P
P
d ( PV ) − dV = dZ − dV
V
V
Z
V
f
1
ln =
P RT
∫
P
0
RT
⎛
−
V
⎜
P
⎝
1
⎞
=
dP
⎟
RT
⎠
RT
⎛
−
V
∫V =∞ ⎜⎝ P
V
P
⎞
⎞⎛ P
−
dZ
dV
⎟⎜
⎟
V
⎠⎝ Z
⎠
RT ⎞⎛ P ⎞ 1 V ⎛
RT ⎞ ⎛ P
⎞
⎛
−
−
−
V
dZ
V
dV
⎟
⎟⎜
⎟
∫V =∞ ⎜⎝ P ⎟⎜
∫
V =∞ ⎜
Z
RT
P
⎠⎝
⎠
⎝
⎠⎝ V
⎠
V ⎛ PV
RT ⎞
⎞ V ⎛ dZ ⎞ 1 V ⎛
=∫ ⎜
−
−
dZ ⎟ − ∫ ⎜
P
⎟
⎜
⎟ ( dV )
∫
=∞
V =∞ RTZ
V =∞
V
V ⎠
⎝
⎠
⎝ Z ⎠ RT
⎝
1
=
RT
V
1
= ∫ ( dZ ) − ∫ d ( ln Z ) −
V =∞
V =∞
RT
V
1
=
RT
V
RT ⎞
⎛
P
−
∫V =∞ ⎜⎝ V ⎟⎠ ( dV )
V
⎛ RT
⎞
P
−
⎟ dV − ln Z + ( Z − 1)
∫V =∞ ⎜⎝ V
⎠
V
(7.4-8)
Dependence of f on P and T(Gibbs-Helmholtz relation)
dG = RTd ln f
⎛ ∂G ⎞
⎛ ∂ ln f ⎞
=
=
V
RT
⎜
⎟
⎜
⎟
⎝ ∂P ⎠T
⎝ ∂P ⎠T
pressure dependence
(7.4-9a)
⎛ ∂G
⎞
⎛ ∂ ln f ⎞
RT ⎟ = V
⎜
=⎜
RT ⎝ ∂P ⎠⎟
⎜⎜ ∂P ⎟⎟
⎝
⎠T
For mixture:
⎛ f ⎞
H IG − H
V
d ⎜ ln i ⎟ = i 2 i dT + i dP
RT
RT
⎝ xi ⎠ x
f G (T , P ) − G (T , P )
From definition : ln =
P
RT
⎡ ⎛ f ⎞⎤
IG
⎢ ∂ ⎜ ln P ⎟ ⎥
∂ ⎪⎧ G (T , P ) − G (T , P ) ⎪⎫
⎝
⎠
⎢
⎥ =
⎬
P ⎨
∂T ⎩⎪
RT
⎢ ∂T ⎥
⎭⎪
⎢⎣
⎥⎦
P
IG
H iIG − H i
Vi IG − Vi
d (lnφi ) x =
dT −
dP
RT 2
RT
Through these DEMO equations, other
properties could be calculated by f.
⎧⎪ G (T , P ) − G IG (T , P ) ⎫⎪ It means Thermodynamics Kingdom
1 ∂
IG
=
G (T , P ) − G (T , P ) − ⎨
⎬ can be built up by Fugacity calculation!
2
RT ∂T
RT
⎪⎩
⎪⎭
⎧⎪ G (T , P ) − G IG (T , P ) ⎫⎪
(1) Adopting an EOS;
1
IG
=−
S (T , P ) − S (T , P ) − ⎨
⎬
(2) Calculating fugacity;
RT
RT 2
⎪⎩
⎪⎭
(3) Developing it using 7.4-9b.
1
IG
IG
⎡G T , P ) − T S (T , P ) ⎤⎦ − ⎡⎣G (T , P ) − T S (T , P ) ⎤⎦
=−
2 ⎣ (
RT
IG
H (T , P ) − H (T , P )
=−
temperaure dependence (7.4-9b)
RT 2
{
{
{
}
}
}
33
a. Fugacity of a pure gaseous species
a. Fugacity of a pure gaseous species
ln
f V (T , P )
P
1
=
RT
∫
V = Z V RT / P
V =∞
⎛ RT
⎞
V
V
−
P
⎜
⎟ dV − ln Z + Z − 1
⎝ V
⎠
(
)
(7.4-8)
Boundary condition :
At very low pressures, where a gas can be described by the ideal gas equation of state
PV = RT
or
ZV =1
we have
ln
f V (T , P )
P
=0
or
f V (T , P ) = P
(7.4-10)
34
ILLUSTRATlON 7.4-1 Computing Fugacity from Volumetric Data
Use the volumetric information in the steam tables of Appendix A.III to compute
the fugacity of superhealed steam at 300oC and 8 MPa.
SOLUTION
With tabulated volumetric data, as in the steam tables, it is most convenient to use Eq.7.4-6a:
⎡ 1 P⎛
RT ⎞ ⎤
f = P exp ⎢
V
−
⎜
⎟ dP ⎥
∫
0
RT
P
⎠ ⎦
⎝
⎣
From the superheated vapor steam tables at 300o C, we have the values in the next slide.
Numerically evaluating lhe integral using the data above, we find
∫
8 MPa
0
RT
⎛
−
V
⎜
P
⎝
3
⎞
−3 m MPa
and
⎟ dP ≈ −1.093 ×10
mol
⎠
3
⎡
⎤
−3 m MPa
−
×
1.093
10
⎢
⎥
mol
f = ( 8 MPa ) exp ⎢
⎥ = 8 exp ( −0.2367 ) MPa
3
m
MPa
⎢ 573.15 K × 8.314 ×10−6
⎥
⎢⎣
mol K ⎥⎦
= 8 × 0.7996 MPa = 6.397 MPa
f
Also, the fugacity coefficient, φ in this case is φ = = 0.7996
P
35
Steam Table
At 300 C
36
COMMENT
Had the same calculation been done at a much higher temperature,
the steam would be closer to an ideal vapor, and the fugacity coefficient would be closer to unity.
For example, the result of a similar calculation at 1000o C and 10 MPa yields:
f = 9.926 MPa and φ = f / P = 0.9926.
Conclusion :
High temperature and low pressure fluid approaches to ideal gas.
37
ILLUSTRATION 7.4-2 Alternative Fugacity Calculation
Use other data in the superheated vapor steam tables to calculate the fugacity of steam at 300o C
and 8 MPa, and check the answer obtained in illustration 7.4-1.
SOLUTION
o
At 300 C and 0.01 MPa we have from the steam tables
Hˆ = 3076.5 kJ/kg and Sˆ = 9.2813 kJ/(kg K). Therefore,
(
⎡ G (T , P ) − G IG (T , P ) ⎤
f (T , P ) = P exp ⎢
⎥
RT
⎢⎣
⎥⎦
)
kJ
Gˆ 300o C, 0.01 MPa = Hˆ − TSˆ = 3076.5 − 573.15 × 9.2813 = − 2243.1
kg
(
)
and G 300o C,0.01 MPa = − 2243.1 J/g × 18.015 g/mol = − 40 409 J/mol. Since the pressure
is so low (0.01 MPa) and well away from the saturation pressure of steam at 300o C (which is
8.581 MPa), we assume steam is an ideal gas at these conditions.
P
G
T
,
P
−
G
T
,
P
=
(
)
(
)
Then using Eq.7.4-3 for an ideal gas, we have
1
2
1 1
∫P VdP
2
(7.4-3)
1
G
IG
( 300
o
)
C, 8 MPa = G
IG
( 300
o
)
C,0.01 MPa + ∫
8 MPa
0.01 MPa
V IG dP
RT
8
dP = −40409 + RT ln
0.01 MPa P
0.01
J
= −40409 + 8.314 × 573.15ln 800 = −8555.7
mol
= −40409 + ∫
8 MPa
38
kJ
For real steam at 300o C and 8 MPa, we have, from the steam tables, Hˆ = 2785.0
and
kg
Sˆ = 5.7906 kJ/(kg K), so that
(
)
kJ
Gˆ 300o C, 8 MPa = 2785.0 − 573.15 × 5.7906 = −533.88
kg
J
and G 300o C, 8 MPa = −9617.9
P ↑ , then G ↓
mol
Now using Eq. 7.4-6a in the form
(
)
⎡ G (T , P ) − G IG (T , P ) ⎤
f (T , P ) = P exp ⎢
⎥
R
T
⎢⎣
⎦⎥
⎡ −9617.9 − ( −8555.7 ) ⎤
f 300 C, 8 MPa = 8 MPa exp ⎢
⎥
×
8.314
573.15
⎣
⎦
= 8 MPa exp [ −0.2229] = 6.402 MPa
(
o
)
which is in excellent agreement with the results obtained
in illustration 7.4-1 ( f = 6.397 MPa ) .
39
ILLUstration 7.4-3
Calculation of the Fugacity of Saturated Steam
i.e. Compute the fugacity of saturated steam at 300o C.
SOLUTION:
⎡ G (T , P ) − G IG (T , P ) ⎤
f (T , P ) = P exp ⎢
⎥
RT
⎥⎦
⎢⎣
The saturation pressure of steam at 300 o C is 8.581 MPa, and from Illustration 7.4-2 we have
V
Gˆ V = − 520.5 kJ/kg and G = − 9376.8 J/mol.
G
IG
(
)
300o C, 8.581MPa = G
IG
(
)
300o C, 0.01 Pa + ∫
8.581 MPa
0.01 MPa
RT
dP
P
J
mol
= −40409 + 8.314 × 573.15ln 858.1 = −8221.6
Therefore,
⎡ −9376.8 − ( −8221.6 ) ⎤
f V 300o C, 8.581MPa = 8.581 MPa exp ⎢
⎥
8.314
×
573.15
⎣
⎦
= 8.581 MPa × 0.7847 = 6.7337 MPa
(
)
COMMENT:
(
)
(
)
At equilibrium, f L 300o C,8.581MPa = f V 300o C,8.581MPa = 6.7337 MPa
40
Fugacity coefficient: virial equation of state
At low to moderate pressures, the virial equation of state truncated
after the second virial coefficient may be used, if data for B(T) are available.
B (T )
PV
= Z = 1+
V
RT
ln
f v (T , P )
P
=
2B (T )
V
(7.4-11)
− ln Z =
2PB (T )
Z RT
− ln Z
(7.4-12)
Known T , P, B → get Z as follows → using (7.4-12) to find f v (T , P ) :
B (T )
⎡1 1
4 PB (T ) ⎤
⎥
where Z = 1 +
=⎢ +
1+
V
RT ⎥
⎢⎣ 2 2
⎦
V in terms of Z and using (7.4-11):
−b ± b 2 − 4ac
PB
= 0; Z=
Z - ZRT
2a
2
⎞
⎛ f ⎞ 1 V ⎛ RT
Adopting ln ⎜ ⎟ =
− P ⎟ d V − ln Z + ( Z − 1) (7.4-8)
⎜
∫
V
=∞
⎝ P ⎠ RT
⎝ V
⎠
referred to problem 7-14. Z and P in terms of V and substitute into 7.4-8.
41
ILLUSTRATION 7.4-4
Calculation of the Fugacity of a Gas Using the Virial Equation
Compute the fugacities of pure ethane and pure butane at 373.15 K and 1, 10, and 15 bar
assuming the virial equation of state can describe these gases at these conditions.
Data:
3
m3
−4 m
BET ( 373.15 K ) = −1.15 ×10
BBU ( 373.15 K ) = −4.22 ×10
;
mol
mol
SOLUTION: Using Eqs. 7.4-12, we find
−4
Ethane
Butane
P ( bar )
Z
f ( bar )
Z
f ( bar )
1
0.996
0.996
0.986
0.986
10
0.961
9.629
0.838
8.628
15
0.941
14.16
0.714
11.86
Since the pressures are not very high, these results should be reasonably accurate. However,
the virial equation with only the second virial coefficient will be less accurate as the pressure
increases. In fact, at slightly above 15 bar and 373.15 K, n-butane condenses to form a liquid.
In this case the virial equation description is inappropriate, as it does not show a phase change
or describe liquids.
42
Fugacity coefficient: van der Waals EOS
At higher pressure, a more complicated equation of state
must he used. By using the (not very accurate)
van der Waals equation, one obtains
⎛ f ⎞
1
ln ⎜ ⎟ =
⎝ p ⎠ RT
fv
V
a
⎛ PV
⎞
⎛ PV ⎞
− 1⎟ − ln ⎜
= ln
−
ln
+⎜
⎟
P
V − b RTV ⎝ RT
⎠
⎝ RT ⎠
A
V
V
= Z − 1 − ln Z − B − V
Z
(
)
(
)
where A =
aP
and B =
( RT )
2
⎛ RT
⎞
− P ⎟ d V − ln Z + ( Z − 1) (7.4-8)
⎜
V =∞
⎝ V
⎠
∫
V
(7.4-13)
bP
RT
43
Fugacity coefficient: Peng-Robinson EOS
For hydrocarbons and simple gases, the Peng-Robinson equation
(Eq. 6.4-2) provides a more accurate description. In this case we have
(
(
)
)
bP ⎤
⎡ V
+
+
Z
1
2
⎢
f
bP ⎞
a
⎛
RT ⎥
ln
= Z V − 1 − ln ⎜ Z V −
−
ln
⎢
⎥
⎟
bP
P
RT
V
bRT
2
2
⎝
⎠
⎢ Z + 1− 2
⎥
RT ⎦
⎣
⎡ Z V + 1+ 2 B ⎤
A
⎥
= Z V − 1 − ln Z V − B −
ln ⎢ V
(7.4-14a)
2 2B ⎢ Z + 1 − 2 B ⎥
⎣
⎦
aP
bP
where A =
an
d
B
=
2
RT
RT
( )
v
(
(
)
)
(
)
(
(
)
)
⎛ f ⎞ 1
ln ⎜ ⎟ =
⎝ p ⎠ RT
∫
V
V =∞
⎛ RT
⎞
− P ⎟ d V − ln Z + ( Z − 1) (7.4-8)
⎜
⎝ V
⎠
44
Calculation procedure for f from EOS
To use equations of state to calculate the fugacity of a gaseous species,
the following procedure is used:
(1) For a given value of T and P, use the chosen equation of state to
calculate the molar volume V or, equivalently, the compressibility
factor Z . When using cubic or more complicated equations of state,
it is the low-density (large V or Z ) solution that is used.
(2) This value of V or Z is then used in Eq. 7.4-12, 7.4-13, or 7.4-14,
as appropriate, to calculate the species fugacity coefficient, f / P,
and thus the fugacity (and enthalpy departure, G-H relation etc.). i.e. 7.4-9b.
Recalled that the Peng-Robinson equation-of-state programs
or MATHCAD worksheets in Appendix B can be used for this calculation.
45
ILLUSTRATION 7.4-5
Calculation of the Fugacity of a Gas Using the Peng-Robinson Equation of State
Compute the fugacities of pure ethane and pure butane at 373.15 K and 1, 10, and 15 bar,
assuming the Peng-Robinson equation describes these gases at these conditions.
SOLUTION
Using the Visual Basic computer program described in Appendix B.I-2, the DOS-based program
PR1 described in Appendix B.11-1, the MATHCAD worksheet described in Appendix B.III, or
the MATLAB program described in Appendix B.IV and the data in Table 4.6-1, we obtain the
following:
COMMENT
We see that these results are only slightly different from those computed with the virial equation of state. The differences would become larger as the pressure increases or the temperature
decreases.
46
Fugacity from corresponding-states chart
It is simple, but less accurate, to compute the fugacity of a pure
species using a specially prepared corresponding-states fugacity chart. To do this, we
note that since, for simple gases and hydrocarbons, the compressibility factor Z obeys
a corresponding-states relation, the fugacity coefficient f /P given by
Eq. 7.4-6 can also be written in corresponding-states form as follows:
fV
⎡ 1
= exp ⎢
P
⎣ RT
= exp
{∫ ( )
Pr
0
∫ (
P
0
V −V
IG
)
⎧ P ⎛ PV
⎤
⎞ dP ⎫
− 1⎟ ⎬
dP ⎥ = exp ⎨ ∫ ⎜
⎦
⎠ P⎭
⎩ 0 ⎝ RT
}
⎡ Vr Z C − 1⎤ d ln Pr = exp
⎣
⎦
{∫
Pr
0
Divided by Tc and Pc
⎡⎣ Z (Tr , Pr , ω ) − 1⎤⎦ d ln Pr
Pr
fV
= ∫ ⎣⎡ Z (Tr , Pr , ω ) − 1⎤⎦ d ln Pr
ln
0
P
PC VC
fV
= f (Tr , Pr , ω )
ln
=ZC
P
RTC
}
(7.4-15a)
(7.4-15b)
47
Figure 7.4-1 Generalized fugacity coefficients, Zc = 0.27
48
Fugacity is the KEY in phase equilibrium
Calculation procedure for phase equilibrium problem (e.g. VLE) at given T and P
V
L
(1) Obtain compressibility factor (Z , Z ) (Z
L ,V
i
V
=
7.4-14a,b
L ,V
PV
i
L
V
for mixture) using cubic Z (Z ) EOS;
RT
L
(2) Pick large root for Z and small root for Z ;
(3) Substitute Z ( Z ) into fugacity equation to compute f ( f );
V
L
V
L
(4) Check to see whether or not f = f (Px φ = Py φ for mixture) reached VLE;
L
V
L
i
i
V
i
i
If YES, then next; if NO, go looping using Bubble, Dew, or Flash method:
V ,L
(5) Printout P-T-x-y data as well as vapor and liquid molar volumes, V (V
V ,L
i
for mixture):
L ,V
Z RT
V =
P
Recalled at Z = 1, it is simplified to the ideal fluid.
L .V
L ,V
Similar to Fig 7.5-1
b. Fugacity of a pure liquid species
49
The fugacity of a pure liquid species can be computed in a number of ways, depending
on the data available. If the equation of state for the liquid is useful, we can again start
from Eq. 7.4-8, now written as
f L (T , P )
⎛ RT
⎞
(7.4-8b)
P
dV − ln Z L + Z L − 1
−
⎜
⎟
∫
V =∞
P
V
⎝
⎠
where the superscript L is used to indicate that the liquid-phase compressibility (high
density, small V and Z ) solution of the equation of state is to be used, and it is the
ln
1
=
RT
(
V = Z L RT / P
)
(
(
)
)
bP ⎤
⎡ L
1
2
Z
+
+
⎢
f
bP ⎞
a
⎛
RT ⎥
ln
ln
= Z L − 1 − ln ⎜ Z L −
−
⎢
⎥
⎟
P
RT ⎠ 2 2bRT ⎢ Z L + 1 − 2 bP ⎥
⎝
RT ⎦
⎣
⎡ Z L + 1+ 2 B ⎤
A
⎥
= Z L − 1 − ln Z L − B −
ln ⎢ L
(7.4-14b) ≈ (7.4-14a)
2 2B ⎢ Z + 1 − 2 B ⎥
⎣
⎦
To use this equation, we first, at the specified value of T and P, solve the PengL
(
(
)
)
(
)
(
(
)
)
Robinson equation of state for the liquid compressibility, Z L , and use this value to
compute f L (T , P ) .
50
ILLUSTRATION 7.4-6
Calculation of the Fugacity of a Liquid Using the Peng-Robinson Equation of State
Compute the fugacity of pure liquid n-pentane and pure liquid benzene at 373.15 K and 50 bar
using the Peng-Robinson equation of state.
SOLUTION
Using one of the Peng-Robinson equation-of-state programs in Appendix B with the liquid
(high-density) root, we obtain for n-pentane
Z PE ( 373.15 K, P = 50 bar ) = 0.2058
f PE ( 373.15 K, P = 50 bar ) = 6.22 bar
and for benzene
Z BZ ( 373.15 K, P = 50 bar ) = 0.1522
f BZ ( 373.15 K, P = 50 bar ) = 1.98 bar
51
Fugacity from liquid volume
If one has some liquid volume data, but not an equation of state, it is more convenient
to start with Eq. 7.4-6, which can be rearranged to
P⎡
fL
RT ⎤
(7.4-16a)
RT ln
= ∫ ⎢V −
dP
⎥
0
P
P
⎣
⎦
and perform the integration. However, one has to recognize that a phase change from a
vapor to a liquid occurs within the integration range as the pressure is increased from
zero pressure to the vapor pressure, and that the molar volume of a fluid is discontinuous at this phase transition. Thus, the result of the integration is
fL
IG
RT ln
= G (T , P ) − G (T , P )
P
P vap (T ) ⎡
P
RT ⎤
RT ⎤
⎡
⎛ f ⎞
+
−
−
Δ
=∫
+
ln
V
dP
RT
V
dP
⎟
⎜
∫ vap
⎢⎣
P =0
P ⎥⎦
P ⎥⎦
⎝ P ⎠phase change P (T ) ⎢⎣
(7.4-16b)
At constant T, what is meaning of the RHS three terms?
They are GL=GV and departures of Gibbs energies,
aren’t they?
52
⎛ f ⎞
Δ ⎜ ln ⎟
=0
⎝ P ⎠phase change
1st
∫
P vap (T )
0
No − pressure − change
RT ⎞
⎛
⎛ f ⎞
ln
V
dP
RT
−
=
⎜
⎟
⎜ ⎟
P ⎠
⎝
⎝ P ⎠sat,T
(7.4-17)
and
P
P
1
RT ⎞
⎛
V
dP
VdP
RT
−
=
−
∫Pvap (T ) ⎜⎝ P ⎟⎠ ∫Pvap (T )
∫Pvap (T ) P dP
P
P
= ∫ vap VdP − RT ln vap
P (T )
P (T )
3rd
P
Then substitute above to the previous 7.4-16b and delogrithmaticly :
⎡ 1 P
⎤
f L (T , P ) = fsat (T ) exp ⎢
V
dP
∫ vap
⎥
⎣ RT P (T )
⎦
⎛ f ⎞
⎡ 1 P
⎤
L
vap
f (T , P ) = P (T ) ⎜ ⎟ exp ⎢
VdP
∫ vap
⎥
⎝ P ⎠sat,T
⎣ RT P (T )
⎦
⎡ 1
Poynting pressure correction= exp ⎢
⎣ RT
⎤
VdP
∫Pvap (T ) ⎥⎦
P
(7.4-18)
represents discontinuity!
53
Equation 7.4-18 leads itself to several levels of approximation.
(1)The simplest approximation is to neglect the Poynting and (f /P )sat
terms and set the fugacity of the liquid equal to its vapor pressure
recalled real mixture :
at the same temperature, that is,
sat
f
L
(T , P ) = P (T )
vap
(7.4-19)
Rauolt ' s law : yi = xi Pi
The result is valid only when the vapor pressure is low and the liquid is at a low total
pressure. Although this equation applies to most fluids at low pressure, it is not correct
for fluids that associate, that is, form dimers in the vapor phase, such as acetic acid.
(2) A more accurate approximation is
⎛ f ⎞
f L (T , P ) = fsatL (T ) =f L T , P vap = P vap (T ) ⎜ ⎟
⎝ P ⎠sat,T
(3) Best approximation
(
)
(
vap
⎡
−
V
P
P
⎛ f ⎞
f L (T , P ) = P vap (T ) ⎜ ⎟ exp ⎢
RT
⎝ P ⎠sat,T
⎣⎢
) ⎥⎤
⎦⎥
(7.4-20)
(7.4-21)
54
ILLUSTRATION 7.4-7
Calculation of the Fugacity of Liquid Water from Density Data (steam table)
i.e. compute the fugacity of liquid water at 300 o C and 25 MPa.
SOLUTION:
From Eq. 7.4-18, we have
⎡ 1 P
⎤
(7.4-18)
VdP ⎥
f L (T , P ) = fsat (T ) exp ⎢
vap
∫
⎣ RT P (T )
⎦
Since liquids are not very compressible, we can assume (away from the critical point of steam)
that V ≈ V sat , so that
L
L
(
L
⎡ V sat
P − P vap (T )
f (T , P ) = fsat (T ) exp ⎢
RT
⎢⎣
L
(
) ⎤⎥
⎥⎦
)
From Illustration 7.4-3 we have fsat 300o C = 6.7337 MPa and from the steam tables, at 300 o C,
3
m
P vap = 8.581 MPa and Vˆ L = 0.001404
, so that
kg
V sat = 1.404 ×10−3 ×
L
g
m3 1 kg
m3
= 2.5293 ×10−5
× 3 ×18.015
mol
kg 10 g
mol
Consequently,
fL
(
3
⎡
⎤
−5 m
2.5293
×
10
25 − 8.581) MPa ⎥
(
⎢
mol
300o C, 25 MPa = 6.7337 MPa exp ⎢
⎥
3
⎢ 8.314 ×10−6 MPa m × 573.15 K ⎥
mol K
⎣⎢
⎦⎥
)
= 6.7337 exp [ 0.08715] MPa = 6.7337 ×1.0911 MPa = 7.347 MPa
55
c. Fugacity of a pure solid phase
The extension of the previous discussion to solids is relatively simple. In fact, if we
J
recognize that a solid phase may undergo several phase transitions, and let V be the
molar volume of the Jth phase and P J be the pressure above which this phase is stable
at the temperature T , we have
⎡ 1
⎛ f ⎞
f (T , P ) = P (T ) ⎜ ⎟ exp ⎢
⎝ P ⎠sat,T
⎣ RT
S
sat
∑∫
J =1
P J +1
P
J
⎤
J
V dP ⎥
⎦
(7.4-23)
Here P sat is generally equal to the sublimation pressure of the solid.
Since the sublimation (or vapor) pressure of a solid is generally small,
so that the fugacity coefficient can be taken equal to unity, it is usually
satisfactory to approximate Eq. 7.4-23 at low total pressures by
f S (T , P ) = P sub (T ) = P sat (T )
(7.4-24a)
or for a solid subject to moderate or high total pressures, by
(
S
⎡
V
P − P sat (T )
f
⎛ ⎞
S
sat
f (T , P ) = P (T ) ⎜ ⎟ exp ⎢
RT
⎝ P ⎠sat,T
⎢⎣
) ⎤⎥
⎥⎦
Low T
High T
(7.4-24b)
S
V
56
ILLUSTRATION 7.4-8
Calculation of the Fugacity of Ice from Density Data
Compute the fugacity of ice at -5o C and pressures of 0.1 MPa, 10 MPa, and 100 MPa.
Data: At -5o C, the sublimation pressure of ice is 0.4 kPa and its specific gravity is 0.915.
SOLUTION
We assume that ice is incompressible over the range from 0.4 kPa to 100 MPa,so that its molar
volume over this pressure range is
g
18.015
3
⎡V S ( P − P sat (T ) ) ⎤
cc
⎛ f ⎞
−5 m
S
sat
mol
V=
= 19.69
= 1.969 × 10
⎥
f (T , P ) = P (T ) ⎜ ⎟ exp ⎢
g
mol
mol
P
RT
⎝
⎠
⎢
⎥⎦
sat,T
0.915
⎣
cc
From Eq. 7.4-24b, we then have
3
⎡
⎤
−5 m
1.969
×
10
× ( 0.1 − 0.0004 ) MPa ⎥
⎢
mol
fice −5o C, 0.1MPa = 0.4 kPa exp ⎢
⎥
3
⎢ 268.15 K × 8.314 × 10−6 MPa m ⎥
⎢⎣
mol K ⎥⎦
= 0.4exp ⎡⎣8.797 × 10−4 ⎤⎦ kPa=0.4 × 1.00088 kPa=0.4004 kPa
(
)
Similarly,
3
⎡
⎤
−5 m
1.969
10
×
× (10 − 0.0004 ) MPa ⎥
⎢
mol
fice −5o C, 10 MPa = 0.4 kPa exp ⎢
⎥ = 0.4369 kPa
3
⎢ 268.15 K × 8.314 × 10−6 MPa m ⎥
⎢⎣
mol K ⎥⎦
3
⎡
⎤
−5 m
1.969
×
10
× (100 − 0.0004 ) MPa ⎥
⎢
mol
and fice −5o C, 100 MPa = 0.4 kPa exp ⎢
⎥ = 0.9674 kPa
3
⎢ 268.15 K × 8.314 × 10−6 MPa m
⎥
mol K
⎣⎢
⎦⎥
(
)
(
)
57
COMMENT
Generally, the change in fugacity of a condensed species (liquid or solid) with small pressure
changes is small. Here we find that the fugacity of ice increases by 10 percent for an increase in
pressure from the sublimation pressure to 100 bar, and by a factor of 2.5 as the pressure on ice
increases to 1000 bar.
58
ILLUSTRATION 7.4-9
Calculation of the Fugacity of Naphthalene from Density Data
The saturation pressure of solid naphthalene can be represented by
3783
log10 P sat ( bar ) = 8.722 −
(T in K )
T
The density of the solid is 1.025 g/cm3 and the molecular weight of naphthalene is 128.19.
Estimate the fugachy of solid naphthalene at 35 o C at its vapor pressure, and also at 1, 20, 30,
40, 50, and 60 bar.
SOLUTION
We start with Eq. 7.4-23 for a single solid phase with the assumption that the solid is incompressible:
(
sat
⎡
V
P
−
P
(T )
f
⎛ ⎞
S
sat
f (T , P ) = P (T ) ⎜ ⎟ exp ⎢
RT
⎝ P ⎠sat,T
⎢⎣
Using the data in the problem statement,
(
)
⎡8.722 − ( 3783 / ( 273.15 + 35 ) ) ⎤⎦
P sat 35o C = 10 ⎣
) ⎤⎥
⎥⎦
= 2.789 × 10−4 bar
59
Since the sublimation pressure is so low, we can use (see Fig. 7.4-1)
⎛ f ⎞
=1
⎜ ⎟
P
⎝ ⎠sat,T
Therefore,
(
Figure 7.4-1 Generalized fugacity coefficients
)
f S T = 35o C, P sat = 2.789 ×10−4 bar
At any higher pressure, we have
f
S
(T = 35 C, P ) = 2.789 ×10
o
−4
3
⎡ 128.19 cm3
−4
−6 m ⎤
bar ×10
⎢ 1.025 mol × P − 2.789 ×10
cm3 ⎥⎥
bar × exp ⎢
3
−5 bar m
⎢
⎥
× 308.15 K
8.314 ×10
⎢⎣
⎥⎦
K mol
(
)
The result is
P ( bar )
(
f s 35o C,P
) ( bar )
2.789 ×10−4
1
10
20
30
2.789 ×10−4
2.803 ×10−4
2.933 ×10−4
3.083 ×10−4
3.241×10−4
40
50
60
3.408 ×10−4
3.582 ×10−4
3.766 ×10−4
What will be this solid fugacity in supercritical
carbon dioxide extraction (500bar) in a mixture?
Does the Ponyting correction be UNITY still?
60
7.6 SPECIFICATION OF THE EQUILIBRIUM THERMODYNAMIC
STATE OF A SYSTEM OF SEVERAL PHASES: THE GIBBS PHASE
RULE FOR A ONE-COMPONENT SYSTEM
In this section we are interested in determining the amount of information, and its
type, that must be specified to completely fix the thermodynamic state of an equilibrium single-component, multiphase system. That is, we are interested in obtaining
answers to the following questions:
1. How many state variables must be specified to completely fix the thermodynamic
state of each phase when several phases are in equilibrium (i.e., how many degrees
of freedom are there in a single-component, multiphase system)?
2. How many additional variables need be specified, and what type of variable
should they be, to fix the distribution of mass (or number of moles) between the
phases, and thereby fix the overall molar properties of the composite, multiphase
system?
3. What additional information is needed to fix the total size of the multiphase system?
61
For a system of π -phases (C = 1)
The number of unknows to fix the thermodynamic state of π phases
T I and P I for phase I
T II and P II for phase II
M
The number of unknows = 2π
Equilibrium criteria
(1) T I = T II = L = T π
The number of independent equations = π -1
(2) P I = P II = L = Pπ The number of independent equations = π -1
(3) G I = G = L = G
II
Then
π
The number of independent equations = π -1
# of intensive variables
# of independent EQs
⎛ restrictions on
⎛ Number of state
⎞ ⎜
⎜
⎟ ⎜ these state variables
variables
needed
to
⎟ − ⎜ as a rsult of each of the
ℑ=Number of degrees of freedom=⎜
⎜ fix the state of each ⎟ ⎜
⎜
⎟ ⎜ phases being in
⎝ of the π phases
⎠ ⎜ equilibrium
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎠
= 2π − (π − 1) − (π − 1) − (π − 1) = 2π − 3π + 3 = 3 − π
(7.6-1)
For a V-L phase of the H2O system, What is Y ?
It means to fix T or P to settle down the system.
62
Two phases system (only good for pure component)
π
ω + ω +L = ∑ωi = 1
I
II
(7.6-2)
i =1
π
i ˆi
ω
∑ V = Vˆ
(7.6-3a)
Property change upon mixing for i
component in a mixture
i =1
π
∑ ω Hˆ
i
i
= Hˆ
(7.6-3a)
i =1
π
∑ ω θˆ
i
i
= θˆ
(7.6-3b)
i =1
63
ILLUSTRATION 7.6-1
Use of the Gibbs Phase Rule
a. Show, using the steam tables, that fixing the equilibrium pressure of a steam-water mixture
at 1.0135 bar is sufficient to completely fix the thermodynamic states of each phase. (This
is an experimental verification of the fact that a one-component, two-phase system has
only one degree of freedom.)
m3
ˆ
b. Show that additionally fixing the specific volume of the two-phase system V at 1
kg
is sufficient to determine the distribution of mass between the two phases.
c. What is the total enthalpy of 3.2 m3 of this steam-water mixture?
SOLUTION
a Using the saturation steam tables of Appendix A .III, we see that fixing the pressure at
1.0135 bar is sufficient to determine the temperature and the thermodynamic properties of
each phase:
3
m
Vˆ V = 1.6729
kg
T = 100 C
o
3
m
Vˆ L = 0.001044
kg
kJ
Hˆ L = 419.04
kg
kJ
Sˆ L = 1.3069
kg K
kJ
Hˆ V = 2676.1
kg
kJ
Sˆ V = 7.3549
kg K
64
(
)
b. Vˆ = ω VVˆ V + 1 − ω V Vˆ L
ωV + ωL = 1
m3
m3 ⎞
m3
V⎛
V
= ω ⎜1.6729
1
⎟ + 1 − ω 0.001044
kg
kg ⎠
kg
⎝
(
ω V = 0.5975;
)
ω L = 0.4025
V 3.2 m3
= 3.2 kg
c. M = =
m3
Vˆ
1.0
kg
Hˆ = ω V Hˆ V + 1 − ω V Hˆ L
(
)
kJ
kJ
kJ
+ 0.4025 × 419.04
= 1767.7
kg
kg
kg
kJ
H = MHˆ = 3.2 × 1767.7
= 5303 kJ
kg
= 0.5975 × 2676.1
65
7.7 THERMODYNAMIC PROPERTIES OF PHASE TRANSITIONS
I
II
G (T , P ) = G (T , P )
(7.7-1)
Supposed for a V-L equilibrium system of
dG = d G
I
II
(7.7-2)
H2O, What is the G vs T curve at P=C?
What is the
meaning of θ ?
V dP − S dT = V dP − S dT
I
I
II
II
⎛ S I − S II ⎞ Δ S
⎛ ∂P ⎞
=⎜ I
=
⎜
⎟
II ⎟
⎝ ∂T ⎠G I =G II ⎝ V − V ⎠ ΔV
Since :
G = H −T S = H −T S = G
I
ΔS
I
I − II
II
I
II
(7.7-3)
I
L −V
II
= −S1
L −V
L −V
(ΔT1L −V )
L −V
= − S 2 (ΔT2 L −V )
•
L −V
Pi ↑ , then S i
↓
L −V
= ΔH L −V / Ti = 0
C.P.: S i
ΔG 2
H −H
ΔH I − II
= S −S =
=
T
T
I
Δ G1
II
II
Then :
⎛ ∂P sat ⎞
ΔS ΔH
=
=
⎜
⎟
∂
Δ
T
V
T ΔV
⎝
⎠G I =G II
"Clapeyron equation" (7.7-4)
What is the meaning of LHS of the Clapeyron EQ?
2
1
There are slopes of three equilibrium lines for any substance.
Why the L-S slope is large?
Does the L-S slope be always positive?
ANS: θ is entropy difference between liquid and vapor lines in phase transition?
66
For the vapor-liquid coexistence curve with V >> V ,
Δ vap H
Δ vap H Δ vap H Δ vap H vap
dP vap Δ H
=
=
=
=
P
V
2
RT
dT
T ΔV T V V − V L
RT
TV
T vap
P
d ln P vap Δ vap H
=
dT
RT 2
V
(
ln
P vap (T2 )
P vap (T1 )
T2
Δ vap H
T1
RT 2
=∫
L
)
dT
"Clausius-Clapeyron equation"
(7.7-5a)
(7.7-5b)
For Δ vap H ≠ f (T )
P vap (T2 )
Δ vap H ⎛ 1 1 ⎞
=−
ln vap
⎜ − ⎟
P (T1 )
R ⎝ T2 T1 ⎠
(7.7-6)
67
ILLUSTRATION 7.7-1
Use of the Clausius-Clapeyron Equation
The vapor pressure of liquid 2,2,4-trimethyl pentane at various temperatures is given below.
Estimate the heat of vaporization of this compound at 25o C.
SOLUTION
Δ vap H
R
− ln ⎡⎣ P vap (T2 ) / P vap (T1 ) ⎤⎦
ln ( 8.00 / 5.33)
=
=−
= 4287.8 K
1 1
1
1
−
−
T2 T1
302.5 293.85
so that Δ vap H = 35.649
kJ
mol
68
Figure 7.7-1 The vapor pressure of 2,2,4-trimethyl
Pentane as a function of temperature.
69
Vapor pressure equations
(a)
(b)
(c)
ln P vap (T ) = −
Δ vap H
RT
B
ln P vap (T ) = A −
T
Antoine equation
ln P vap (T ) = A −
+C
(7.7-5a)
(7.7-7)
B
T +C
(7.7-8)
(d)
Riedel equation
(e)
B
+ C ln T + DT 2
T
Harlecher-Braun equation
ln P vap (T ) = A +
DP (T )
B
ln P vap (T ) = A + + C ln T +
T
T2
(7.7-9)
vap
(7.7-10)
70
ILLUSTRATION 7.7-2
Interrelations the Thermodynamic Properties of Phase Changes
The following vapor pressure data are available
Estimate each of the following:
a. Heat of sublimation of ice
b. Heat of vaporization of water
c. Heat of fusion of ice
d. The triple point of water
SOLUTION
a. Here we use Eq. 7.7-6 in the form
ln ⎡⎣ P
Δsub H
=−
R
sub
(T2 ) / Psub (T1 )⎤⎦
1 1
−
T2 T1
=−
ln [3.88 / 3.28]
= 6130 K
1
1
−
271.15 269.15
so that
Δ sub H = 6130 K × 8.314
J
J
kJ
= 50965
= 50.97
mol K
mol
mol
71
ILLUSTRATION 7.7-2(continued)
b. Similarly, here we have
Δ vap H
R
=−
ln ⎡⎣ P vap (T2 ) / P vap (T1 ) ⎤⎦
1 1
−
T2 T1
=−
ln [ 6.101/ 5.294]
= 5410 K
1
1
−
277.15 275.15
or
Δ vap H = 5410 K × 8.314
J
J
kJ
= 44979
= 44.98
mol K
mol
mol
c. Since
Δsub H = H ( vapor ) − H ( solid )
Δ vap H = H ( vapor ) − H ( liquid )
Δ fus H = H ( liquid ) − H ( solid ) = Δsub H − Δ vap H = 50.97 − 44.98 = 5.99
kJ
mol
72
ILLUSTRATION 7.7-2(continued)
d. At the triple-point temperature Tt , the sublimation pressure of the solid and the vapor pressure of the liquid are equal; we denote this triple-point pressure as Pt , Using Eq. 7.7-6 for
both the solid and liquid phases gives
ln ⎣⎡ Pt / P sub (T ) ⎤⎦
ln [ Pt / 3.880]
Δ sub H
= 6130 = −
=−
1 1
1
1
R
−
−
Tt T
Tt 271.15
and
Δ vap H
R
= 5410 = −
ln ⎡⎣ Pt / P vap (T ) ⎤⎦
1 1
−
Tt T
=−
ln [ Pt / 5.294]
1
1
−
Tt 275.15
The solution to this pair of equations is Tt = 273.279 K. and Pt = 4.627 mm Hg. The
reported triple point is 273.16 K and 4.579 mm Hg, so our estimate is quite good.
73
COMMENT
This example illustrates the value of thermodynamics in interrelating properties in that from two
sublimation pressure and two vapor pressure data points, we were able to estimate the heat of
sublimation, the heat of vaporization, the heat of fusion, and the triple point. Further, we can
now use the information we have obtained and write the equations
⎛ Psub ⎞
1 ⎞
⎛1
ln ⎜
⎟ = −6130 ⎜ −
⎟
⎝ T 271.15 ⎠
⎝ 3.880 ⎠
and
7.7-6
C-C equation
⎛ P vap ⎞
1 ⎞
⎛1
ln ⎜
⎟ = −5410 ⎜ −
⎟
⎝ T 275.15 ⎠
⎝ 5.294 ⎠
Consequently, we can also calculate the sublimation pressure and vapor pressure of ice and
water, respectively, at other temperatures. Using these equations, we find
(
)
P sub −10o C = 1.951 mmHg
which compares favorably with the measured value of 1.950 mm Hg. Also,
(
)
P vap +10o C = 9.227 mmHg
compared with the measured value of 9.209 mm Hg.
74
Homework
7.11, 7.12, 7.15b, 7.16a-2MPa,
7.25, 7.36, 7.44, 7.46
75