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Select Problems From Prausnitz Thermodynamics. These problems are well explained
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Advanced Thermodynamics KP8108—
Solutions to Selected Problems
(a companion to J. M. Prausnitz, R. N. Lichtenthalerand E. G. de AzevedoMolecular Thermodynamics,
2nd ed.)
Prof. Bjørn Hafskjold, Dep. of Chemistry (NTNU)Assoc. Prof. Tore Haug-Warberg, Dep. of Chem. Eng. (NTNU)
May 16, 2007
ii
Contents
Problem 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Problem 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2Problem 2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Problem 2.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Detailed derivation of Eqs. (3-53) and (3-54) . . . . . . . . . . . .. . . . . . . . 5Problem 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Problem 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Problem 3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Problem 3.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11Problem 5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Iteration 1:y1 = 0.01 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Iteration 2:y1 = 3.4835× 10−3 . . . . . . . . . . . . . . . . . . . . . . . . 15
Problem 5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Problem 5.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Problem 5.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
Iteration 1:y3 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25Iteration 2:y3 = 1.04× 10−4 . . . . . . . . . . . . . . . . . . . . . . . . . 26
Problem 6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Problem 6.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Problem 7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Problem 7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38Problem 10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
Method 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41Method 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
Problem 10.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44Problem 10.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
Question a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45Question b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47Question c (method I) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50Question c (method II) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
Problem 11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53Problem 11.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54Problem 11.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55Problem 12.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57Problem 12.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58Problem 12.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
iii
iv
Chapter 2
Problem 2.2
For the given equation of state,P(
Vn − b
)
= RT , and by using the Maxwell relation∂2A/∂T∂V =∂2A/∂V∂T , we find:
(
∂S∂V
)
T
=
(
∂P∂T
)
V
=R
Vn − b
=PT.
For an isothermal change:
∆S =∫ (
∂S∂V
)
T
dV =∫
RVn − b
dV = nR ln
(
V2 − nbV1 − nb
)
= nR ln
(
P1
P2
)
.
Alternatively, from∂2G/∂T∂P = ∂2G/∂P∂T , we find:(
∂S∂P
)
T
= −(
∂V∂T
)
P
= −nRP.
This formula integrates to the same expression for∆S as above. ForU we know thedifferentialdU = TdS − pdV and hence (∂U/∂V)T = T (∂S/∂V)T − P. Using the first ofthe Maxwell rules from above we get
(
∂U∂V
)
T
= T
(
∂P∂T
)
V
− P = 0.
Consequently,∆U = 0. Alternatively, we can write (∂U/∂P)T = T (∂S/∂P)T − P(∂V/∂P)T
which gives:(
∂U∂P
)
T
= T
(
∂S∂P
)
T
− P
(
∂V∂P
)
T
= T
(
∂V∂T
)
P
− P
(
∂V∂P
)
T
= −RTP+
RTP
= 0.
Again we find that∆U = 0. For the next derivative we shall useH = U + PV and(
∂H∂P
)
T
=
(
∂U∂P
)
T
+ V + P
(
∂V∂P
)
T
= 0+nRT
P+ nb − P
nRTP2
= nb.
1
For an isothermal change:
∆H =∫ (
∂H∂P
)
T
dP =∫
nbdP = nb (P2 − P1) .
The changes in Gibbs energy (G) and Helmholtz energy (A) are found from the definitionsof G andA, and the results above:
∆G = ∆H − T∆S = nb (P2 − P1) − nRT ln
(
P1
P2
)
,
∆A = ∆U − T∆S = −nRT ln
(
P1
P2
)
.
Problem 2.5
Because entropy is constant it would be natural to express the result as an explicit functionof S . However, this is not feasible because the equation of stateis expressed in terms ofTandV. We shall therefore use implicit differentiation and first write entropy as a functionof volume and temperature, differentiate, use some of the Maxwell relations, and then putthe differential ofS (V, T ) equal to zero:
dS =
(
∂S∂V
)
T
dV +
(
∂S∂T
)
V
dT =
(
∂P∂T
)
V
dV +CV
TdT = 0
Thermodynamically, the exhaust temperature does not depend on the gas flow rate throughthe turbine, and for convenience we choose one mole of gas as the basis for the calculation.With the van der Waal equation, we find for one mole of gas:
P =RT
v − b−
av2
⇒(
∂P∂T
)
V
=R
v − b
Inserted into the expression fordS = 0 this gives the following differential equation be-tweenv andT at constant entropy:
dvv − b
= −cV
RdTT
This is the adiabatic equation of state for a van der Waal gas.Integration at constantcv
gives
ln
(
v2 − bv1 − b
)
= −cV
Rln
(
T2
T1
)
whereT2 andv2 are unknown. It is given that the final pressure is atmospheric, and by usingthe equation of state, we could expressv2 by P2. This is, however, a little cumbersome, butsince the exhaust pressure is low, we try first by approximating the exhaust gas is an idealgas. This gives:
ln
(
T2
T1
)
= −RcV
ln
vig2− b
v1 − b
≈ −RcV
ln
(
RT2
P2(v1 − b)
)
2
where we have approximatedvig2 − b ≈ vig
2 = RT2/P2. Rearrangement gives
ln
(
T2
T1
)
=RcV
ln
(
T1
T2
)
−RcV
ln
(
RT1
P2(v1 − b)
)
= −R
cV
(
1+ RcV
) ln
(
RT1
P2(v1 − b)
)
= −RcP
ln
(
RT1
P2(v1 − b)
)
where we have made use ofcigP = cig
V + R. With the given numbers, we find
ln
(
T2
(350+ 273) K
)
= −8.314 J K-1 mol-1
33.5J K-1 mol-1ln
(
82.06cm3 atm mol-1 K-1(350+273) K
1 atm (600-45) cm3 mol-1
)
= −1.123
∴ T2 = 623 exp(−1.123)= 203 K
Checking the ideal-gas assumption:
V ig2 =
RT2
P2=
82.06 cm3 atm mol-1 K-1 203 K1 atm
= 16700 cm3 mol-1
Vvdw2 = 16400 cm3 mol-1
Note: The van der Waal result was found by an iterative solution of the cubic equation inV at 203 K and 1 atm. If we want to improve the ideal-gas assumption, we may insertVvdw
2 = 16400 cm3 mol-1 into the equation above:
ln
(
T2
T1
)
= −RcV
ln
vvdw2− b
v1 − b
= −8.314 J K-1mol-1
25.2 J K-1 mol-1ln
(
(16400− 45) cm3 mol-1
(600-45) cm3mol-1
)
= −1.117
∴ T2 = 623 exp(−1.117)= 204 K
Problem 2.6
The virial expansion of the compression factor is
z =PvRT= 1+
Bv+
Cv2+ · · ·
The compression factor for the van der Waal gas is
zvdw =v
RT
( RTv − b
−av2
)
=1
1− bv
−a
RT1v
We recognize the first term on the right-hand side as the sum ofa geometric series:
1
1− bv
= 1+bv+
(
bv
)2
+ · · ·
3
Ordering the van der Waal compression fcator in increasing powers of1v gives
zvdw = 1+(
b −a
RT
) 1v+
(
bv
)2
+ · · ·
and we conclude thatB = b −
aRT.
Problem 2.7
We start by recovering one result from Problem 2.2:(
∂U∂P
)
T
= −T
(
∂V∂T
)
P
− P
(
∂V∂P
)
T
.
From the given equation of state, the volume is
z =PvRT= 1+
BPRT
∴ v =RTP
(
1+BPRT
)
=RTP+ a −
bT 2
From this we find:(
∂V∂T
)
P
=RP+
2bT 3
(
∂V∂P
)
T
= −RTP2
(
∂U∂P
)
T
= −T
(
RP+
2bT 3
)
− P(
−RTP2
)
= −2bT 2
Integration at the constant temperatureτ then gives:
∆U(τ) =
π∫
0
(
∂U∂P
)
T=τ
dP = −2bτ2π
4
Chapter 3
Detailed derivation of Eqs. (3-53) and (3-54)
Start with this definition of the chemical potential:
µi =
(
∂A∂ni
)
V
Applied to Eq. (3-50) the definition leads to
µi =
∞∫
V
(
∂P∂ni
)
T,V,n j
−RTV
dV − RT∂
∂ni
∑
j
[
n j ln
(
P◦Vn jRT
)]
+ u◦i − T s◦i (1)
We note thatj = 1 is the only term that survives the differentiation of the sum. This term is
∂
∂ni
[
ni ln
(
P◦VniRT
)]
= ln
(
P◦VniRT
)
− 1 (2)
Furthermore, the standard state of componenti is pure ideal gas at temperatureT andpressureP◦ = 1 bar. The chemical potential ofi in this state is
µ◦i = g◦i = u◦i + RT − T s◦i (3)
Eqs. 2 and 3 into 1 gives
µi =
∞∫
V
(
∂P∂ni
)
T,V,n j
−RTV
dV − RT
[
ln
(
P◦VniRT
)
− 1
]
+ µ◦i − RT
µi − µ◦i =
∞∫
V
(
∂P∂ni
)
T,V,n j
−RTV
dV − RT ln
(
P◦VniRT
)
(4)
The relation between the chemical potential and fugacity is
µi − µ◦i = RT ln
(
fi
f ◦i
)
(5)
The fugacity of the standard state isP◦ = 1 bar. The fugacity is further expressed by thefugacity coefficient as
fi = ϕiPi = ϕiyiP
ϕi =fi
yiP(6)
5
Combination of Eqs. 4, 5, and 6 yields
RT ln
(
ϕiyiPP◦
)
=
∞∫
V
(
∂P∂ni
)
T,V,n j
−RTV
dV − RT ln
(
P◦VniRT
)
RT lnϕi =
∞∫
V
(
∂P∂ni
)
T,V,n j
−RTV
dV − RT ln
(
P◦VniRT
)
− RT ln
(
yiPP◦
)
=
∞∫
V
(
∂P∂ni
)
T,V,n j
−RTV
dV − RT ln( PVnRT
)
=
∞∫
V
(
∂P∂ni
)
T,V,n j
−RTV
dV − RT ln z (3-53)
For a pure component, the chemical potential is
µi =Gni= µ◦i + RT ln
(
fi
f ◦i
)
= µ◦i + RT ln
(
fi
P
)
+ RT ln
(
Pf ◦i
)
RT ln
(
fi
P
)
purei
= µi − µ◦i − RT ln
(
Pf ◦i
)
=Gni−
G◦
ni− RT ln
(
PP◦
)
(7)
The Gibbs energy is taken from Eq. (3-51) in the book:
Gni=
∞∫
V
(
Pni−
RTV
)
dV − RT ln
(
P◦VniRT
)
+PVni+ u◦i − T s◦i (8)
Note that
G◦
ni= h◦i − T s◦i = u◦i + RT − T s◦i (9)
and
PVni= RTz (10)
6
Eqs. 8, 9, and 10, inserted into 7, combine to:
RT ln
(
fi
P
)
purei
=Gni−
G◦
ni− RT ln
(
PP◦
)
=
∞∫
V
(
Pni−
RTV
)
dV − RT ln
(
P◦VniRT
)
+PVni+ u◦i − T s◦i
−(
u◦i + RT − T s◦i)
− RT ln
(
PP◦
)
=
∞∫
V
(
Pni−
RTV
)
dV − RT ln
(
PVniRT
)
+PVni− RT
=
∞∫
V
(
Pni−
RTV
)
dV − RT ln z + RTz − RT
=
∞∫
V
(
Pni−
RTV
)
dV − RT ln z + RT (z − 1) (3-54)
Note: There is probably an easier way to understand this equation. It is illogical to bringin expressions for bothA andG and pretend they are independent functions. The elegantsolution would be to bring in the homogeneous properties ofP = P(T,V,N) and developEq. 3-54 directly from Eq. 3-53. Suggestions are welcome.
Problem 3.2
The gas mixture is illustrated in Figure 1. If we consider thegas to be a pseudo pure
A, BxA = 0.25, xB = 0.75
P = 50 barT = 373 K
ϕA = 0.65,ϕB = 0.90
Figure 1: Properties of the gas mixture.
substance, the relation between the chemical potential andthe fugacity would be
µM = µ⊖M + RT ln
(
fM
f ⊖M
)
whereµ⊖M and f ⊖M are the chemical potential and the fugacity, respectively,in the standardstate. This would correspond to a molar Gibbs energy for the mixture,GM = µM. On the
7
other hand, the molar Gibbs energy for the mixture is given by
Gm =∑
i
µixi
where the chemical potential of each component is
µi = µ⊖i + RT ln
(
fi
f ⊖i
)
= µ⊖i + RT ln
(
ϕiPi
f ⊖i
)
Here,ϕi is the fugacity coefficient of componenti. EquatingGM andGm gives
µ⊖M + RT ln
(
fM
f ⊖M
)
=∑
i
xiµ⊖i + RT
∑
i
xi ln
(
ϕiPi
f ⊖i
)
(11)
which we can use to expressfM in terms of the component properties. We should, however,first make a remark on the relation between the standard states. Theµ⊖M is not the same asµ⊖i even if they refer to the same pressure, becauseµ⊖M contains the properties of an idealmixture:
µ⊖M =∑
i
xiµ⊖i + RT
∑
i
xi ln xi (12)
The standard fugacities, which are usually taken to be idealgas at 1 bar pressure, are thesame:f ⊖M = f ⊖i . Equations 11 and 12 give:
∑
i
xiµ⊖i + RT
∑
i
xi ln xi + RT ln
(
fM
f ⊖M
)
=∑
i
xiµ⊖i + RT
∑
i
xi ln
(
ϕiPi
f ⊖i
)
∴ ln
(
fM
1 bar
)
=∑
i
xi ln(
ϕiPi
1 bar
)
−∑
i
xi ln xi
With the given data, we find:
ln
(
fM
1 bar
)
= 0.25× ln (0.65× 0.25× 50) + 0.75× ln (0.90× 0.75× 50)
−0.25× ln 0.25− 0.75× ln 0.75= 3.725 3
fM = 1× exp(3.7253)= 41.5 bar
Problem 3.3
The two situations are shown schematically in Figure 2. Equilibrium between gas andliquid at 1 bar may be expressed as
µgasC2H6
(1bar) = µliqC2H6
(xC2H6 = 0.33× 10−4)
The same equilibrium at 35 bar may be expressed as
µgasC2H6
(35bar) = µliqC2H6
(xC2H6 = x)
8
gas gasH2O, C2H6 H2O, C2H6
P = 1 bar P = 35 barPH2O = 0.0316 bar PH2O = 0.0316 bar
PC2H6 ≈ 1 bar PC2H6 ≈ 35 barT = 298 K T = 298 K
liquid liquidH2O, C2H6 H2O, C2H6
xH2O ≈ 1.0 xH2O ≈ 1.0xC2H6 = 0.33× 10−4 xC2H6 =?
Figure 2: The two equilibrium situations.
wherexC2H6 is the unknown denoted byx. In the gas phase, the effect of a change in thechemical potential at constant temperature is
dµi = v̄idP
wherev̄i is the partial molar volume of componenti. Since there is so little water in the gasphase, we assume it to be pure C2H6, in which case we can neglect the effect of a changein composition, andvi is the molar volume of C2H6, i.e. v̄i = vC2H6. The molar volume isgiven by the equation of state,
v =RTP
(
1− aP − bP2)
wherea = 7.63×10−3 bar−1 andb = 7.22×10−5 bar−2. Integration of the chemical potentialat constantT gives
µgasC2H6
(35bar) − µgasC2H6
(1bar) =
35 bar∫
1 bar
(
∂µC2H6
dP
)
T
dP = RT
35 bar∫
1 bar
(
1P− a − bP
)
dP
= RT
[
ln
(
35 bar1 bar
)
− a 34 bar−12
b(
352 − 12)
bar2]
In the liquid phase, the change in chemical potential is the sum of two contributions; onedue to the pressure change, and one due to the composition change. The effect of the changein pressure isdµC2H6 = v̄C2H6dP, where ¯vC2H6 is the partial molar volume of ethane in theliquid. Since the liquid is practically incompressible, itmay be assumed to be constantover the pressure change in question here, and in any case thecontribution is negligiblysmall for the liquid. The effect of the change in composition isdµC2H6 = RTd ln aC2H6,whereaC2H6 = fC2H6/ f ⊖C2H6
is the activity of ethane. We have to assume that the activity
9
coefficient of ethane, defined byγC2H6 = aC2H6/xC2H6, is constant over the concentrationrange considered here, which leads to
µliqC2H6
(x) − µliqC2H6
(x = 0.33× 10−4)
=
35 bar∫
1 bar
vC2H6dP + RT
xC2H6=x∫
xC2H6=0.33×10−4
d ln γC2H6 + RT
xC2H6=x∫
xC2H6=0.33×10−4
d ln xC2H6
≈ RT
xC2H6=x∫
xC2H6=0.33×10−4
d ln xC2H6 = RT ln( x0.33× 10−4
)
Equating the changes in the chemical potential of the gas andliquid phases give
RT ln( x0.33× 10−4
)
= RT
[
ln(35)− 7.63× 10−3 × 34−12× 7.22× 10−5
(
352 − 12)
]
ln( x0.33× 10−4
)
= ln(35)− 7.63× 10−3 × 34−12× 7.22× 10−5
(
352 − 12)
= 3.2517
∴ x = 0.33× 10−4 exp(3.2517)= 8.53× 10−4
Problem 3.7
Throttling is an isenthalpic process, and we shall therefore consider the conservation ofenthalpy in this problem. Since the equation of state is given explicitly in the volume asfunction ofP andT , we find the enthalpy from the following relation:
H =
P∫
0
[
V − T
(
∂V∂T
)
P,nT
]
dP +∑
i
nih0i
whereh0i is the molar enthalpy of purei at the actual temperature. The given volumetric
data for the mixture leads to[
V − T
(
∂V∂T
)
P,nT
]
= a +bT
wherea = 5 × 10−5 m3 mol−1 andb = −0.2 m3 K mol−1. Integration leads to the molarenthalpy
h =
(
a +bT
)
P +∑
i
xih0i
wherexi is the mole fraction of componenti. The conditionhin = hout gives(
a +b
Tin
)
Pin +∑
i
xih0i,in =
(
a +b
Tout
)
Pout +∑
i
xih0i,out
(
a +b
Tin
)
Pin =
(
a +b
Tout
)
Pout +∑
i
xi
(
h0i,out − h0
i,in
)
10
We haveh0i,out− h0
i,in = c0P,i (Tout − Tin) andc0
P =∑
ixic0
P,i = 33.5 J mol−1 K−1. Solving forPin
gives
Pin =
(
a + bTout
)
Pout + c0P (Tout − Tin)
(
a + bTin
)
=
(
5 · 10−5m3mol−1 − 0.2m3Kmol−1/200K)
105Pa+ 33.5Jmol−1K−1 (200K− 300K)(
5 · 10−5m3mol−1 − 0.2m3Kmol−1/300K)
= 55.9 · 105 Pa= 55.9 bar
Problem 3.9
The internal energy of mixing is (using one mole as basis):
∆mixu = uM − x1u1 − x2u2
Since the van der Waal equation is explicit inP:
P =RT
v − b−
av2,
it is convenient to find the internal energy from
u =
∞∫
v
[
P − T
(
∂P∂T
)
V,nT
]
dv +∑
i
xiu0i
where
P − T
(
∂P∂T
)
V,nT
=RT
v − b−
av2−
RTv − b
= −av2
Note that the mixing must be assumed to occur at constant pressure,not at constant volume,which means that we must use different volumes for the lower integration limit for themixture and the two pure components. The molar internal energy for the mixture and eachof the pure components are:
uM = −aM
∞∫
vM
dvv2+
∑
i
xiu0i = −
aM
vM+
∑
i
xiu0i
u1 = −a1
v1+ u0
1
u2 = −a2
v2+ u0
2
The change in internal energy for the mixture is
∆mixu = −aM
vM+
∑
i
xiu0i + x1
(
a1
v1− u0
1
)
+ x2
(
a2
v2− u0
2
)
= −aM
vM+
x1a1
v1+
x2a2
v2
11
With the given data, we find
aM =∑
i
∑
j
xix j√
aia j(1− ki j)
= 0.52 × 106[
1.04+ 2×√
1.04× 4.17(1.0− 0.1)+ 4.17]
bar cm6 mol−2
= 2.24× 106 bar cm6 mol−2
vM =∑
i
xivi = 0.5(32.2+ 48.5) cm3 mol−1= 40.35 cm3 mol−1
∆mixu =
(
−2.2440.35
+0.5× 1.04
32.2+
0.5× 4.1748.5
)
106bar cm6 mol−2
cm3 mol−1= 363 J mol−1
12
Chapter 5
Problem 5.2
The question here is whether the fugacity of CO2 in the compressed gas is higher thanthe fugacity at saturation conditions. The actual temperature is lower than the triple-pointtemperature, and the condensed phase in equilibrium with the vapour, if it exists, is solid.To find the saturation pressure, we consider how it changes from the vapour pressure ofpure CO2 (component 1) to its new value after injection of H2 (component 2) until the totalpressure is 60 bar. We therefore consider a charging processas illustrated in Figure 3. TheH2 is inert in this case, in the sense that it does not dissolve inthe solid phase. The change
CO2 (g) CO2 (g) + H2 (g)P∗CO2
= 0.1392 bar ∆µ (g) PCO2 + PH2 = 60 barT = 173 K −→ T = 173 K
CO2 (s) ∆µ (s) CO2 (s)T = 173 K −→ T = 173 K
Figure 3: Available data for the state changes.
in chemical potential of CO2 in the gas phase is
∆µ1(g) = RT
c∫
⋆
d ln f1 = RT[
ln f1]c⋆ (13)
where f1 is the fugacity of CO2. The fugacity of pure CO2 vapour can be assumed to beequal to its vapour pressure (the saturation pressure of 0.1392 bar is quite low), but pleasevalidate this assumption by numerical calculations. The fugacity of CO2 in the compressedstate must be calculated somehow, and we shall be using
f1 = ϕ1y1P
whereϕ1 andy1 are the the fugacity coefficient and the mole fraction of CO2, respectively,andP is the total pressure (equal to 60 bar). The fugacity coefficient is estimated from van
13
der Waal’s equation:
z =PvRT=
vv − b
−a
RTv(14)
The mixing rules are
a =∑
i
∑
j
yiy j√
aia j
b =∑
i
yibi
The fugacity coefficient of component 1 in the mixture is calculated from Eq. 3-70 in Praus-nitz et al.:
lnϕ1 = lnv
v − b+
b1
v − b−
2√
a1∑
j y j√
a j
RTv− ln z (15)
Numerical values for the van der Waals parameters may be found in several textbooks, e.g.in P. W. Atkins,Physical Chemistry, 6th edition, Oxford, 1998:
a/atm cm6 mol−2 b/cm3 mol−1
CO2 (1) 3.640×106 42.67H2 (2) 0.2476×106 26.61
Alternatively we may use the corresponding state principleand calculatea = 27(RTc)2/64Pc
andb = RTc/8Pc from critical data. Do this on your own and compare the results with thetable. Because the mole fraction of CO2 is low, at most 0.01, we make a small error ifwe choose the van der Waal parameters to be those of pure H2, but for the time being wedetermine the mixture parameters on the basis of 1 mole% CO2.
Iteration 1: y1 = 0.01
This gives:
a = (0.012 × 3.640× 106 + 2× 0.01× 0.99×√
3.640× 106 × 0.2476× 106
+ 0.992 × 0.2476× 106) atm cm6 mol−2
= 2.618 3× 105 atm cm6 mol−2
b = (0.01× 42.67+ 0.99× 26.61) cm3 mol−1
= 26.77 cm3 mol−1
At 60 bar, the molar volume in Eq. 14 (note that we have to solvea cubic equation inv) isv = 250.69 cm3 mol−1 andz = 1.048. Inserted into Eq. 15:
lnϕ1 = ln
(
250.69250.69− 26.77
)
+42.67
250.69− 26.77
−2√
3.640× 106(
0.01×√
3.640× 106 + 0.99×√
0.2476× 106)
250.69× 82.0567× 173− ln 1.048
= −0.292
∴ ϕ1 = exp(−0.292)= 0.7468
14
Inserting the last result into Eq. 13 gives the change in chemical potential of CO2 in the gasphase:
∆µ1(g) = 83.1439 cm3 bar mol-1 K-1 × 173 K× ln
(
0.7468× y1 × 60 bar0.1392 bar
)
= 14428 cm3 bar mol-1 × ln (321.9y1)
The change in chemical potential for CO2 in the solid phase is
∆µ1(s) =
c∫
⋆
vsdP = vs
60 bar∫
0.1392 bar
dP = 27.6 cm3 mol−1 × (60− 0.1392) bar
= 1652.16 cm3 bar mol-1
Here, we have assumed that the solid is incompressible and that H2 does not dissolves inthe solid. At equilibrium, in the compressed state, we have∆µ1(g) = ∆µ1(s) which gives
14428 cm3 bar mol-1 × ln (321.9y1) = 1652.16 cm3 bar mol-1
y1 =1
321.9exp
(
1652.1614428
)
= 3.4835× 10−3
At this point, we conclude that Iteration 1 is slightly off, and try another calculation:
Iteration 2: y1 = 3.4835× 10−3
This gives
a = (0.00352 × 3.640× 106 + 2× 0.0035× 0.9965×√
3.640× 106 × 0.2476× 106
+ 0.99652 × 0.2476× 106) atm cm6 mol−2
= 2.5254× 105 atm cm6 mol−2
b = (0.0035× 42.67+ 0.9965× 26.61) cm3 mol−1
= 26.67 cm3 mol−1
andv = 251.17 cm3 mol−1, andz = 1.048. Inserting these numbers into Eq. 15 gives:
lnϕ1 = ln
(
251.17251.17− 26.77
)
+42.67
251.17− 26.77
−2√
3.640× 106(
0.0035×√
3.640× 106 + 0.9965×√
0.2476× 106)
251.17× 82.0567× 173− ln 1.048
= −0.282
∴ ϕ1 = exp(−0.282)= 0.7543
A small change inϕ1 from 0.7468 to 0.7543 is observed. The improved value of the changein chemical potential for the gas phase is calculated (once more) from Eq. 13):
∆µ1(g) = 83.1439 cm3 bar mol-1 K-1 × 173 K× ln
(
0.7543× y1 × 60 bar0.1392 bar
)
= 14428 cm3 bar mol-1 × ln (325.1y1)
15
The solid phase chemical potential is the same as above, and we find
y1 =1
325.1exp
(
1652.1614428
)
= 3.4492× 10−3
We consider Iteration 2 to be good, and accept this as the finalresult. To conclude thecalculation, we need a material balance, using one mole material in total as basis:
nCO2(g) + nCO2(s) + nH2(g) = 1 mol (16)
nH2(g) = 0.99 mol (17)
The mole fraction of CO2 in the gas phase is
yCO2 =nCO2(g)
nCO2(g) + nH2(g)= 0.00345 (18)
Solution of Eqs. 16–18 gives the amount of precipitated CO2:
nCO2(s) = 0.0066 mol
Problem 5.3
The situation is shown in Figure 4. The question is whether some of the CO2 will condenseinto liquid or solid, or not. This depends on the pressure andtemperature in the gas duringthe isenthalpic expansion. In the following, we will denoteCH4 by 1 and CO2 by 2. The
30 mol% CO2 (g)70 mol% CH4 (g)70 bar, 313 K
CO2 (s) or CO2 (l)?CH4 (g)1 bar, T=?
Figure 4: The given process.
critical point of CO2 is atPc = 73.8 bar andTc = 304.2 K. The triple point is atPT = 5.18bar andTT = 216.8 K. If the gas cools to a temperature in the range between the criticalpoint and the triple point, some CO2 may possibly condense to a liquid, if it cools toT < 216.8 K, it may condense to a solid. Because the equation of state is given as avirial expansion in molar density, it is most convenient to express the enthalpy of the gas as
H =
∞∫
V
[
P − T
(
∂P∂T
)
V,nT
]
dV + PV +∑
i
niu0i
whereni is the number of moles of componenti andu0i is the molar internal energy of pure
i in the ideal gas state at the actual temperature. The strategy is to find howH depends
16
on V andT , then constrainH to be constant, and finally to use this constraint to find theisenthalpic equation of state for the gas. This equation will tell us how much the gas coolson expanding. The given data for the mixture are of the form
zmix =PvRT= 1+
Bmix
v(19)
whereBmix =
∑
i
∑
j
yiy jBi j (20)
andyi is the mole fraction of componenti. EachBi j is of the form
Bi j = ai j +bi j
T+
ci j
T 2,
which meansBmix will be on the same form,
Bmix = a +bT+
cT 2
where
a =∑
i
∑
j
yiy jai j = (0.72 × 42.5+ 2× 0.7× 0.3× 41.4+ 0.32 × 40.4) cm3 mol-1
= 41.85 cm3 mol-1
b =∑
i
∑
j
yiy jbi j = −(0.72 × 16.75+ 2× 0.7× 0.3× 19.50+ 0.32 × 25.39)× 103 cm3 K mol-1
= −18.683× 103 cm3 K mol-1
c =∑
i
∑
j
yiy jci j = −(0.72 × 25.05+ 2× 0.7× 0.3× 37.3+ 0.32 × 68.7)× 105 cm3 K2 mol-1
= −34.124× 105 cm3 K2 mol-1
We then find for one mole of gas
P − T
(
∂P∂T
)
V,nT
=Rv2
(
b +2cT
)
and for the molar enthalpy:
h =Rv
(
aT + 2b +3cT
)
+∑
i
yih0i (21)
We note that the ideal-gas contribution to enthalpy is (constantcP assumption):∑
i
yih0i = c0
PT
where
c0P =
∑
i
yic0P,i = (0.7× 35.8+ 0.3× 37.2) J mol-1 K−1 = 36.22 J mol-1 K−1
17
The energy constraint tells thath is constant and solving Eq. 21 forv gives:
v(T ) =R
(
aT + 2b + 3cT
)
h − c0PT
(22)
At the upstream conditions,
Bmix =
(
41.85−18.683× 103
313−
34.124× 105
3132
)
cm3 mol-1 = −52.672 cm3 mol-1
and from Eq. 19:
70 bar× v
83.1439 bar cm3 mol-1 K−1 × 313 K= 1−
52.672 cm3 mol-1
v
which givesv1 = 308.24 cm3 mol-1. The constant enthalpy in Eq. 21 can now be deter-mined to
h1 =8.314 J mol-1 K-1
(
41.85× 313− 2× 18.683× 103 − 3×34.124×105
313
)
K cm3 mol-1
308.24 cm3 mol-1
+ 36.22 J mol-1 K−1 × 313 K
= 9800 J mol-1
The question is whether the temperature-pressure relationship will be such that the gas iscooled to a temperature below the condensation point for CO2, or in other words, will thefugacity CO2 in the gas be higher than the equilibrium value in contact with pure CO2? Thefugacity of CO2 in the expanding gas is given by
f2 = ϕ2y2P,
which should be compared with the fugacity for saturated gasin equilibrium with liquid orsolid:
f sat2 = Psat
2 exp
(
vl∆PRT
)
(3.5)
The saturation pressure of pure CO2 is
ln Psat(bar)= 10.807−1980.24
T(23)
andvl is the molar volume of pure CO2 (l). The fugacity coefficient is given by
lnϕ2 =2v
∑
j
y jB2 j − ln zmix
Most likely the Poynting correction is quite small, and the fugacity coefficient is probablyless than 1. This implies that CO2 is more volatile in the compressed gas than in the pureform (whereϕ2 ≈ 1), and if we find that if the gas is unsaturated in pure form, noCO2 willcondense. Under this assumption, we solve Eqs. 19, 20, 22, 23and calculate the data shownin Table 1: The outlet pressure of 1 bar is reached before the temperature drops below the
18
Table 1: Saturation pressure(s) of CO2.T /K v/cm3 mol−1 B/cm3 mol−1 P/bar P2/bar Psat
2 /bar271 35182 -73.6 0.639 0.192 33.1... ... ... ... ... ...304 400.6 -56.5 54.2 16.26 73.2304.2 398 -56.4 54.5 16.36 73.8305 387.9 -56.1 55.9 16.78 74.8... ... ... ... ... ...313 308.2 -52.7 70 21 (N/A)
T, K
270 280 290 300
P, b
ar
0
20
40
60
80
100
P2sat
P2
Figure 5: Saturation pressure and partial pressure of CO2.
triple point temperature, and any condensed CO2 will be in liquid form. We note that fromthe table∆P = P − Psat
2 is negative, which means the Poynting correction is less than unity.The effect of the inert gas is therefore to reduce the saturation fugacity as compared withthe saturation pressure. A plot ofP2 andPsat
2 is given in Figure 5. The figure shows thatthe saturation pressure is higher than the actual partial pressure of CO2, and consequentlythat CO2(l) will not condense. To improve the assumption of the idealgas phase we needto estimate the molar volume of liquid CO2. Data from Reidet al. indicate thatρl(T = 293K) = 0.777 g cm3. The molecular weight of CO2 is M = 44.01 g mol−1. This gives
vl =Mρ=
44.01 g mol-1
0.777 g cm3= 56.6 cm3 mol-1
Assuming that the liquid density varies only a little with temperature, the saturation fugac-ity is easily calculated:
f sat2 = Psat
2 exp
(
56.6 cm3 mol-1∆P
83.1439 cm3 bar mol-1 K-1 T
)
19
The fugacity coefficient in the gas phase is
lnϕ2 =2v
∑
j
y jB2 j − ln zmix =2v
(
a2 +b2
T+
c2
T 2
)
− ln(
1+Bmix
v
)
(24)
where
a2 =∑
j
y ja2 j = (0.7× 41.4+ 0.3× 40.4) cm3 mol-1
= 41.1 cm3 mol-1
b2 =∑
j
y jb2 j = −(0.7× 19.50+ 0.3× 25.39)× 103 cm3 K mol-1
= −21.267× 103 cm3 K mol-1
c2 =∑
j
y jc2 j = −(0.7× 37.3+ 0.3× 68.7)× 105 cm3 K2 mol-1
= −46.72× 105 cm3 K2 mol-1
The results for the fugacity may now be calculated from Eqs. 19, 20, 22–24, and some ofthe results are given in Table 2: A plot off2 and f sat
2 is shown in Figure 6. The figure
Table 2: Fugacities of CO2.T /K ϕ2 f2/bar f sat
2 /bar271 1.365 0.262 30.52... ... ... ...304 0.826 13.43 70.14304.2 0.824 13.48 70.45305 0.816 13.69 71.69... ... ... ...313 0.741 15.57 (N/A)
confirms that no CO2(l) will be formed.
Problem 5.16
This problem is similar to problem 5.2. We consider a processas illustrated in Figure 7.The ethylene is inert in the sense that it does not dissolve inthe solid phase. Like in Problem5.2, we compute the change in chemical potential for naphtalene in the gas phase as
∆µ1(g) = RT
c∫
⋆
d ln f1 = RT[
ln f1]c⋆
The fugacity of pure napthalene vapour can be assumed to be equal to the vapour pres-sure, 2.80×10−4 bar, because the vapour pressure is low. The fugacity of napthalene in thecompressed state is determined as
f1 = ϕ1y1P (25)
20
T, K
270 280 290 300
f, ba
r
0
20
40
60
80
100
f2sat
f2
Figure 6: Saturation fugacity and fugacity of CO2.
Naphtalene (g) (1) Naphtalene (g)+ Ethylene (g) (2)
P∗1 = 2.80× 10−4 bar ∆µ1 (g) P1 + P2 = 30 bar−→ P1 = ?
T = 308 K T = 308 K
Naphtalene (s) ∆µ1 (s) Naphtalene (s)T = 308 K −→ T = 308 K
Figure 7: Naphtalene dissolving in ethylene.
with the usual notation. The change in chemical potential for naphtalene in the solid phaseis
∆µ1(s) =
c∫
⋆
vsdP = vs
30 bar∫
2.80×10−4 bar
dP
The molar volume of the solid is given by the density and the molecular weight:
vs =Mρ=
128.2 g mol-1
1.145 g cm−3= 112.0 cm3 mol-1
This gives
∆µ1(s) = 112.0 cm3 mol-1(30− 2.80× 10−4) bar= 3360 cm3 bar mol-1
21
At equilibrium in the compressed state, we must have
∆µ1(g) = ∆µ1(s) (26)
We now consider the two cases, (a) The vapour is an ideal gas, and (b) the gas obeys thevirial expansion truncated after the second term.a) With the ideal gas law
∆µ1(g) = RT ln( P1
2.80× 10−4 bar
)
= 3360 cm3 bar mol-1
P1 = 2.80× 10−4 bar× exp
(
3360 cm3 bar mol-1
83.1439 cm3 bar mol-1 K-1308 K
)
= 3.19× 10−4 bar
y1 =P1
P=
3.19× 10−4 bar30 bar
= 1.06× 10−5
b) With the virial expansion truncated after the second term. The virial expansion reads
z =PV
nRT= 1+ B
nV∴ P =
RTV
(
n +BV
n2)
(27)
The second virial coefficient is given by the van der Waals result,
B = b −a
RT(28)
where
a =∑
i
∑
j
yiy j√
aia j
b =∑
i
yibi
It is convenient to write the expression forb as
b =∑
i
∑
j
yiy jbi j
where
bi j =12
(
bi + b j
)
,
because we can then use Eq. (5-29) in Prausnitz et al.:
lnϕ1 =2v
(y1B11+ y2B12) − ln z (29)
with
Bi j = bi j −√
aia j
RT(30)
andB =
∑
i
∑
j
yiy jBi j (31)
22
We will need the numerical values fora andb, which we find from the critical values:
a =27(RTc)2
64Pc
b =RTc
8Pc
The critical values (See e.g. Reid et al.) are listed in Table3, which also gives the computedvalues fora andb. The values forai j, bi j, andBi j are given in Table 4. We now make a
Table 3: Critical data for napthalene and ethylene.Tc/K Pc/atm a/cm6 atm mol−2 b/cm3 mol−1
Naphtalene (1) 748.4 40.0 3.98×107 191.91Ethylene (2) 282.4 49.7 4.56×106 58.28
Table 4: Interaction parameters for napthalene and ethylene.Combination ai j/cm6 atm mol−2 bi j/cm3 mol−1 Bi j/cm3 mol−1 b/cm3 mol−1
11 3.98×107 191.91 -1382.9 191.9112 1.35×107 125.10 -409.06 58.2822 4.56×106 58.28 -122.15
guess for the mole fraction of naphtalene to obtain a first estimate of the viral coefficientB. We can for example assume thatP1 is little affected by the presence of ethylene whichgives the starting value
y1 =P∗1P=
2.80 · 10−4bar30bar
= 9.3 · 10−6
(Other qualified guesses include the answer from a) ory1 = 0.) The equations 25–31 givean iterative procedure to findy1:
B = y21B11+ 2y1 (1− y1) B12+ (1− y1)
2 B22
PRT
v2 − v − B = 0
z = 1+Bv
lnφ1 =2v
(y1B11+ (1− y1) B12) − ln z
RT ln
(
φ1Py1
P∗1
)
= ∆µ
B, v, z andφ1 are estimated from the previousy1 and a better estimate is obtained in thelast equation. WithBi j from Table 4,∆µ=3360cm3bar/mol, T = 308K, P = 30bar andP∗1 = 2.8·10−4bar the two first iterations give the values shown in Table 5. We can thereforeconclude thaty1 = 2.80 · 10−5.
23
Table 5: Iteration sequence.iteration 1 iteration 2
B -122.155 -122.166v 705.893 705.876z 0.826949 0.826930φ1 0.379462 0.379441y1 2.80447·10−5 2.80462·10−5
Problem 5.17
This problem is also similar to problem 5.2. Instead of considering the cooling, we considercompressing water vapour by adding air at 263 K. This is illustrated in Figure 8. Like in
H2O (g) (3) N2 (1) + O2 (2) (g)+ H2O (g)
P∗1 = 1.95 torr ∆µ3 (g) P = 30 bar−→ y3 = ?
T = 263 K T = 263 K
H2O (s) ∆µ3 (s) H2O (s)ρs = 0.92 g cm−3 −→
Figure 8: Available data for the state changes.
Problem 5.16, we compute the change in chemical potential for water vapour in the gasphase as
∆µ3(g) = RT
c∫
⋆
d ln f3 = RT[
ln f3]c⋆ = RT ln
f3P∗3
The fugacity of pure water vapour is again assumed to be equalto the vapour pressure, 1.95torr = 1.95
760 atm= 2.57×10−3 atm= 2.60×10−3 bar, because the vapour pressure is low. Thefugacity of water vapour in the compressed state is
f3 = ϕ3y3P.
The change in chemical potential for ice is
∆µ3(s) =
c∫
⋆
vsdP = vs
30 bar∫
2.60×10−3 bar
dP
24
The molar volume of ice is the molecular weight divided by thedensity:
vs =Mρ=
18.0 g mol-1
0.92 g cm−3= 19.57 cm3 mol-1
This gives
∆µ1(s) = 19.57 cm3 mol-1(30− 2.60× 10−3) bar= 587.1 cm3 bar mol-1
At equilibrium in the compressed state, we must have
∆µ1(g) = ∆µ1(s)
We have to determine the second virial coefficient, the compression factor, and the fugacitycoefficient:
Bmix =∑
i
∑
j
yiy jBi j
zmix =PvRT= 1+
Bmix
v
lnϕ3 =2v
∑
j
y jB3 j − ln zmix
A material balance for N2 (g), O2 (g), and H2O (g) gives
y1 = 0.8(1− y3), y2 = 0.2(1− y3)
Again, we have to solve the problem by iteration, which is considered below.
Iteration 1: y3 = 0
This is for the saturated vapour. With the numerical values given in the problem, we find
Bmix =[
0.8(1− y3)]2 B11 + 2× 0.8(1− y3) × 0.2(1− y3)B12+ 2× 0.8(1− y3)y3B13
+[
0.2(1− y3)]2 B22 + 2× 0.2(1− y3)y3B23+ y3y3B33
= −(
0.82 × 25+ 2× 0.8× 0.2× 15+ 0.22 × 12)
cm3 mol-1 = −21.3 cm3 mol-1
30 barv
83.1451 cm3 bar mol-1 K-1 × 263 K= 1−
21.3 cm3 mol-1
v
∴ v = 706.9 cm3 mol-1
zmix = 1−21.3 cm3 mol-1
706.9 cm3 mol-1= 0.9699
lnϕ3 =2
706.9 cm3 mol-1(−0.8× 63− 0.2× 72) cm3 mol-1 − ln (0.9699) = −0.1528
∴ ϕ3 = 0.8583
The maximum permissible moisture content is therefore
y3 =2.60× 10−3
0.8583× 30× exp
(
587.183.1451× 263
)
= 1.04× 10−4
If we are in doubt concerning the accuracy of this result, we can make another iteration asshown below.
25
Iteration 2: y3 = 1.04× 10−4
This gives
Bmix =[
0.8(1− y3)]2 B11+ 2× 0.8(1− y3) × 0.2(1− y3)B12+ 2× 0.8(1− y3)y3B13
+[
0.2(1− y3)]2 B22+ 2× 0.2(1− y3)y3B23+ y3y3B33
=[
0.8(1− 1.04× 10−4)]2
B11+ 2× 0.8(1− 1.04× 10−4) × 0.2(1− 1.04× 10−4)B12
+ 2× 0.8(1− 1.04× 10−4) × 1.04× 10−4B13+[
0.2(1− 1.04× 10−4)]2
B22
+ 2× 0.2(1− 1.04× 10−4)1.04× 10−4B23+ 1.04× 10−4 × 1.04× 10−4B33
= 0.6399B11+ 0.3199B12+ 1.6638× 10−4B13+ 0.0400B22+ 4.1596× 10−5B23
+ 1.0816× 10−8B33
= −21.289 cm3 mol-1
30 barv
83.1439 cm3 bar mol-1 K-1 × 263 K= 1−
21.289 cm3 mol-1
v∴ v = 706.9 cm3 mol-1
zmix = 1−21.289 cm3 mol-1
706.9 cm3 mol-1= 0.9699
lnϕ3 =2
706.9 cm3 mol-1(−0.8× 63− 0.2× 72) cm3 mol-1 − ln (0.9699) = −0.1528
∴ ϕ3 = 0.8583
The maximum permissible moisture content is therefore
y3 =2.60× 10−3
0.8583× 30× exp
(
587.183.1451× 263
)
= 1.04× 10−4
26
Chapter 6
Problem 6.2
An azeotrope is a liquid mixture that does not change composition when it evaporates. Ifwe plot a liquid-vapour phase diagram for a mixture that forms an azeotrope, it will looklike shown in Fig. 6-12 in the textbook (for aP − x -diagram) or in Fig. 6-22 (for aT − x-diagram). The point is that in the minimum or maximum of the curves in either of thediagrams, the bubble-point and dew-point lines go through the same point. In order to findthe bubble-point and dew-point lines, we have to consider the phase equilibrium shown inFigure 9. In this figure, we have arbitrarily chosen to illustrate the change in chemicalpotential of component 1 from pure state to the mixed state. The change in chemical
Pure 1 (g) Mixture 1+2 (g)Ps
1 ∆µ1 (g) P1 + P2 = PT −→ T
Pure 1 (l) ∆µ1 (l) Mixture 1+2 (l)T −→ T
Figure 9: Schematic illustration of a mixing process.
potential for component 1 in the gas phase is
∆µ1(g) = RT
mixture∫
pure 1
d ln f1 = RT ln
(
P1
Ps1
)
= RT ln
(
y1PPs
1
)
where we have used the given information that the gas phase may be considered ideal. Inthe liquid phase, the change in chemical potential is
∆µ1(l) =
pure 1 atP∫
pure 1 atPs1
v1dP + RT
mixture∫
pure 1 atP
d ln a1 (32)
27
wherev1 is the molar volume of pure component 1. The first term in Eq. 32is the Poyntingcorrection, which we will neglect here. The change∆µ1(l) is then
∆µ1(l) = RT ln a1 = RT ln (γ1x1)
whereγ1 is the activity coefficient for component 1. By equating the changes in chemicalpotentials, we get
RT ln
(
y1PPs
1
)
= RT ln (γ1x1)
y1PPs
1
= γ1x1
From the given expression for the excess Gibbs energy for themixture, we find (using Eqs.(6-46) and (6-47) in the textbook):
ln γ1 =A
RTx2
2
ln γ2 =A
RTx2
1
For the azeotrope,x1 = y1 which gives the bubble-point at the azeotrope
x1PPs
1
= exp( ART
x22
)
x1
PPs
1
= exp( ART
x22
)
(33)
The same consideration for component 2 gives
PPs
2
= exp( ART
x21
)
(34)
Dividing Eq. 33 by Eq. 34 and using the information thatPs
2Ps
1= 1.649, gives
Ps2
Ps1
= exp[ ART
(
x22 − x2
1
)
]
= 1.649
∴
ART
(
x22 − x2
1
)
= ln 1.649=12
In other words, for an azeotrope to occur, the requirement isthat for any value ofx1 in therange 0≤ x1 ≤ 1, A must satisfy
ART=
1
2(
x22 − x2
1
) =1
2[
(1− x1)2 − x21
] =1
2(1− 2x1)
A plot of the allowed values ofA/RT is shown in Figure 10. This means that fore.g.A
RT = 1, there will be an azeotrope atx1 = 0.25. To illustrate this, we can construct aP − x
28
x1
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
A/R
T
-20
-10
0
10
20
Figure 10: The lines show allowed values ofA/RT .
diagram as follows: The total vapour pressure (the bubble-point pressure) is the sum of thepartial pressures:
P = P1 + P2 = Ps1 exp
( ART
x22
)
x1 + Ps2 exp
( ART
x21
)
x2
If we set ART = 1, we find
PPs
1
= exp(
x22
)
x1 +Ps
2
Ps1
exp(
x21
)
x2
The dew-point line is given by
y1 =P1
P=
Ps1 exp
(
ART x2
2
)
x1
Ps1 exp
(
ART x2
2
)
x1 + Ps2 exp
(
ART x2
1
)
x2
Again setting ART = 1 as an example, we find
y1 =exp
(
x22
)
x1
exp(
x22
)
x1 +Ps
2Ps
1exp
(
x21
)
x2
The bubble-point and dew-point curves are shown in Figure 11for this particular case. Wenow have to consider the question of liquid-liquid miscibility. This is illustrated in Figure12, showing the change in Gibbs energy on mixing as function of composition for fourdifferent values ofA; A
RT = 0, 1, 2, 3. The Gibbs energy of mixing is given by
∆mixG = ∆mixGideal +Gexcess = RT
2∑
i=1
xi ln xi + Ax1x2 (35)
29
x1
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
P/P
1*
0.8
1.0
1.2
1.4
1.6
1.8bubble-point linedew-point line
gas
2-phase
liquid
azeotrope
Figure 11: Bubble- and dew-point lines forART = 1.
Miscibility occurs for all compositions if
∂2∆mixG
∂x21
≥ 0 for all x1
If we use this condition for the model in Eq. 35, we find
∂2
∂x21
[RT (x1 ln x1 + x2 ln x2) + Ax1x2] =RTx1x2
− 2A ≥ 0 ∴
ART≤
12x1x2
The values of 12x1x2
are plotted in Figure 13, together with the conditions onA shown inFigure 10. We see that the minimum value of12x1x2
occurs atx1 = 0.5, when 12x1x2= 2. The
conclusion is that azeotropes may occur for completely miscible liquids ifA ≤ 2RT . Forsuch values ofA, an azeotrope will occur in the ranges
0 ≤ x1 ≤38
and12< x1 ≤ 1
Problem 6.16
The situation is illustrated in Figure 14, showing the partial pressures as given by Henry’sand Raoult’s laws. Henry’s law,
Pi = Hi, jxi
applies in the limit of zero solute concentration,xi → 0, and Raoult’s law,
Pi = Psi xi
applies in the limit of pure solvent,xi → 1. The problem here is to determine the vapourpressure atx1 = 0.5 and the corresponding vapour composition,
y1 =P1
P
30
x1
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
∆ mixG
/RT
-0.8
-0.6
-0.4
-0.2
0.0
0.2
∆mixGideal/RT+3x1x2
∆mixGideal/RT+2x1x2
∆mixGideal/RT+x1x2
∆mixGideal/RT
Figure 12: Change in Gibbs energy of mixing. Complete miscibility occurs if ∂2∆mixG∂x2
1≥ 0.
Between the limits of dilute solution and pure solvent, we have to use a model to estimatethe vapour pressure. Assuming ideal gas phase, the partial pressure is given by
Pi = Psi γixi (36)
We have four given pieces of information,Ps1, Ps
2, H1,2, andH1,2. Becauseγ1 andγ2 are re-lated by the Gibbs-Duhem equation, the model forgE (andγi) may contain two parametersto be determined. We can also understand that at least two parameters are required becausethe properties of the two components are not symmetric, as seen for instance by the ratiosH1,2
Ps1
and H2,1
Ps2
. Thevan Laar model satisfies this, and we set
ln γ1 =A′
(
1+ A′x1B′x2
)2
ln γ2 =B′
(
1+ B′x2A′x1
)2
Using this model with Eq. 36 in the limitxi → 0 gives
ln γ1 = A′ = ln
(
H1,2
Ps1
)
= ln
(
2 bar1.07 bar
)
= 0.6255
ln γ2 = B′ = ln
(
H2,1
Ps2
)
= ln
(
1.60 bar1.33 bar
)
= 0.1848
For an equimolar mixture, we find
ln γ1 =A′
(
1+ A′
B′
)2=
0.6255(
1+ 0.62550.1848
)2= 0.0325∴ γ1 = 1.0330
ln γ2 =B′
(
1+ B′
A′
)2=
0.1848(
1+ 0.18480.6255
)2= 0.1101∴ γ2 = 1.1164
31
x1
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
A/R
T
-20
-10
0
10
20
Figure 13: Allowable values ofA/RT (solid line) and upper limit for complete miscibilityin the liquid phase (dashed line).
This gives
P1 = 1.07 bar × 1.0330× 0.5 = 0.5527 bar
P2 = 1.33 bar × 1.1164× 0.5 = 0.7424 bar
P = P1 + P2 = 1.2951 bar
The vapour (dew-point) composition is
y1 =0.5527 bar1.2951 bar
= 0.4268
Data generated with this model are shown in Figure 15. Alternatively, we can use thethree-suffix Margules equation,
ln γ1 =1
RT
[
(A + 3B) x22 − 4Bx3
2
]
ln γ2 =1
RT
[
(A − 3B) x21 + 4Bx3
1
]
Using this model with Eq. 36 in the limitxi → 0 gives
ln γ1 =A − B
RT= ln
(
H1,2
Ps1
)
= ln
(
2 bar1.07 bar
)
= 0.6255
ln γ2 =A + B
RT= ln
(
H2,1
Ps2
)
= ln
(
1.60 bar1.33 bar
)
= 0.1848
ART
=0.6255+ 0.1848
2= 0.4052
BRT
=0.1848− 0.6255
2= −0.2204
32
x1
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
P, b
ar
0.0
0.5
1.0
1.5
2.0
P1
P2Henry's
law
Raoult's law
2.00
1.07
1.60
1.33
Henry's lawRaoult's law
Figure 14: The given information.
x1
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
P, b
ar
0.0
0.5
1.0
1.5
2.0
P1 P2
PBubble-point line
Dew-point line
2.00
1.07
1.60
1.33
Figure 15: Partial pressures and total vapour pressure computed from the van Laar model.
For an equimolar mixture, we find
ln γ1 = (0.4052− 3× 0.2204) × 0.52 + 4× 0.2204× 0.53 = 0.0462∴ γ1 = 1.0473
ln γ2 = (0.4052+ 3× 0.2204) × 0.52 − 4× 0.2204× 0.53 = 0.1564∴ γ1 = 1.1693
This gives
P1 = 1.07 bar × 1.0473× 0.5 = 0.5603 bar
P2 = 1.33 bar × 1.1693× 0.5 = 0.7776 bar
P = P1 + P2 = 1.3379 bar
Tha vapour (dew-point) composition is
y1 =0.5603 bar1.3379 bar
= 0.4188
Data generated with this model are shown in Figure 16.
33
x1
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
P, b
ar
0.0
0.5
1.0
1.5
2.0
P1 P2
P Bubble-point line
Dew-point line
2.00
1.07
1.60
1.33
Figure 16: Partial pressures and total vapour pressure computed from the three-suffix Mar-gules equation.
34
Chapter 7
Problem 7.1
The situation is shown in theP − x diagram, Figure 1 for the mixture A-CS2. From the
xA
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
P, k
Pa
0
2
4
6
8
10
12
14
PA
PCS2
P
Figure 17: Plot of the partial pressures and total pressure for the mixture A-CS2. The curvesare based on the van Laar model.
given information, we can determine the activity coefficient atx1 = 0.5 andT = 283 K:
PA = PsAγAxA
∴ γA =PA
PsAxA=
8 kPa13.3 kPa× 0.5
= 1.2030
ln γA = 0.1848
The given solubility parameters suggest that we use regularsolution theory. The relation tothe van Laar model,
ln γ1 =A′
(
1+ A′x1B′x2
)2
ln γ2 =B′
(
1+ B′x2A′x1
)2
35
is
A′ =v1
RT(δ1 − δ2)2
B′ =v2
RT(δ1 − δ2)2
wherevi andδi are the molar volume and the solubility parameter, respectively, of compo-nenti. With the given information, we find that
vA =VnA=
V MA
m=
MA
ρA=
160 g mol-1
0.8 g cm-3= 200 cm3 mol-1
A′
B′=
vA
vCS2
=200 cm3 mol-1
61 cm3 mol-1
A′ = ln γA
(
1+A′
B′
)2
= 0.1848×(
1+20061
)2
= 3.3832
Assuming thatv1 (δ1 − δ2)2 is independent of the temperature, we find
vA(
δA − δCS2
)2 @(T = 298 K)= vA(
δA − δCS2
)2 @(T = 283 K)= A′RT
δA = δCS2 ±
√
A′RTvA= 20.5 (J cm-3)
12 ±
√
3.3832× 8.3145 J mol-1 K-1 × 283 K
200 cm3 mol-1
= 20.5 (J cm-3)12 ± 6.3 (J cm-3)
12
We will return to the question of which sign to choose, but forthe time being assume thatthe correct sign is -. This gives the complete set of properties shown in Table 6. For the
Table 6: Solubility parameters.A CS2 Toluene
Solubility parameter at 298 K, (J cm-3)12 14.2 20.5 18.2
Liquid molar volume at 298 K, cm3 mol-1 200 61 107Saturation pressure at 283 K, kPa 13.3 1.73 25.5
mixture A-T (Toluene), we now find
A′ =vA
RT(δA − δT )2
=200 cm3 mol-1
8.3145 J mol-1 K-1 × 283 K(14.2− 18.2)2 J cm-3 = 1.36
B′ =vT
RT(δA − δT )2
=107 cm3 mol-1
8.3145 J mol-1 K-1 × 283 K(14.2− 18.2)2 J cm-3 = 0.7276
36
At xA = xT = 0.5:
ln γA =A′
(
1+ A′xA
B′xT
)2=
1.36(
1+ 1.36×0.50.7276×0.5
)2= 0.1652
∴ γA = 1.1796
ln γT =B′
(
1+ B′xTA′xA
)2=
0.7276(
1+ 0.7276×0.51.36×0.5
)2= 0.3088
∴ γT = 1.3618
PA = PsAγAxA = 13.3 kPa× 1.1796× 0.5 = 7.84 kPa
PT = PsTγT xT = 25.5 kPa× 1.3618× 0.5 = 17.36 kPa
P = PA + PT = 7.84 kPa+ 17.36 kPa= 25.20 kPa
yA =PA
P=
7.84 kPa25.20 kPa
= 0.31
The full phase diagram is shown in Figure 2. An alternative representation of the data is
xA
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
P, k
Pa
0
5
10
15
20
25
30
PA
PT
PBubble-point line
Dew-point line
Figure 18:P − x diagram for the mixture A-Toluene at 283 K.
shown in Figure 3. The alternative positive sign for√
A′RTvA
would give
δA = 20.5 (J cm-3)12 + 6.3 (J cm-3)
12 = 26.8 (J cm-3)
12
This would correspond to a change in molar internal energy onvaporization (Eq. (7-33) inthe textbook) of
∆vapuA = δ2AvA = 26.82 J cm-3 × 200 cm3 mol-1 = 143.6 kJ mol-1
The density of 0.8 g cm−3 indicates that A is a hydrocarbon. Typical values for the enthalpyof vaporization of a hydrocarbon is 30-50 kJ mol−1. The value corresponding toδA = 26.8(J cm-3)
12 is therefore not acceptable.
37
xA
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
y A
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Figure 19: Vapour-liquid equilibria for the mixture A-Toluene at 283 K.
Problem 7.3
We denote n-Hexane by 1 and benzene by 2. Using regular solution theory and the givendata
A′ =v1
RT(δ1 − δ2)2
=132 cm3 mol-1
8.3145 J mol-1 K-1 × 323 K(14.9− 18.8)2 J cm-3 = 0.7476
A′
B′=
v1
v2=
132 cm3 mol-1
89 cm3 mol -1= 1.4831
B′
A′=
(
A′
B′
)−1
= 0.6742
B′ =B′
A′A′ = 0.6742× 0.7476= 0.5041
we find:
ln γ1 =A′
(
1+ A′x1B′x2
)2=
0.7476(
1+ 1.48310.30.7
)2= 0.2795
∴ γ1 = 1.3224
ln γ2 =B′
(
1+ B′x2A′x1
)2=
0.5041(
1+ 0.67420.70.3
)2= 0.0761
∴ γ2 = 1.0791
38
At the given composition, we then find
P1 = Ps1γ1x1 = 0.533 bar× 1.3224× 0.3 = 0.2115 bar
P2 = Ps2γ2x2 = 0.380 bar× 1.0791× 0.7 = 0.2870 bar
P = P1 + P2 = 0.2115 bar+ 0.2870 bar= 0.4985 bar
and finally
K1 =y1
x1=
P1
Px1=
0.2115 bar0.4985 bar× 0.3
= 1.414
K2 =y2
x2=
P2
Px2=
0.2870 bar0.4985 bar× 0.7
= 0.822
39
40
Chapter 10
Problem 10.3
This problem can be solved in several ways.
Method 1
Inspired by the Hildebrand-Scott plots (Figure 10-11 in thetextbook), we may estimate thesolubility from a correlation of the type
ln xH2 = Aδi + B (37)
where xH2 is the mole fraction (solubility) of H2, δi is the solubility parameter for thesolvent, andA andB are (temperature dependent) constants. To determineA andB , we usedata for CH4 and CO. Air is a mixture of 78% N2, 21% O2, and 1% Ar. In terms of solubilityproperties, Ar is like O2 (see Table 7-1 in the textbook), and we take the composition to be78% N2 and 22% O2. For the mixed solvent (air), we replaceδ1 in Eq. 37 by
δ̄ =∑
i
Φiδi
whereΦi is the volume fraction of componenti . The sum includes all components, but weassume the solubility of H2 to be so low that its contribution can be neglected. The volumefraction is defined as
Φi =xivi
∑
j x jv j
We will need data for the solubility parameter, defined by Eq.(7-33) in the textbook:
δi =
(
∆vapu
v
)12
i
where∆vapu is the change in internal energy on ”complete” vaporization, i.e. to an ideal gasstate. Data are given for∆vaph, which is related to∆vapu by
∆vapu = ∆vaph − ∆vap(Pv) ≈ ∆vaph − RT
Data for the various properties are given in Table 7, all dataat 90 K. The values forA andB in Eq. 37 are determined from Table 8. which givesA = −0.435 (J cm−3)−
12 , B = −0.982.
(If we instead use the data from Table 7-1 in the textbook, we find A = −0.461 (J cm−3)−12 ,
41
Table 7: Physical properties of gases.From given data From Table 7-1
Comp. ∆vaph
J mol−1
∆vapu
J mol−1vi
cm3mol−1δi
(J cm−3)0.5vi
cm3mol−1δi
(J cm−3)0.5
CH4 8750 8000 35.6 15.0 35.3 15.1CO 5530 4780 37.0 11.4 37.1 11.7N2 5110 4360 37.5 10.8 38.1 10.8O2 6550 5800 27.9 14.4 28.0 14.7
Table 8: Solubility parameters.Component xH2 ln xH2 δi/(J cm−3)0.5
CH4 5.49×10−4 -7.507 15.0CO 2.63×10−3 -5.941 11.4
B = −0.598.) We shall, however, see below that, due to a numerical coincidence, we shallnot need these numbers. The average solubility parameter for liquid air at 90 K is calculatedfrom the data in Table 9. which gives̄δ = 11.4 (J cm−3)
12 . (If we instead use the data from
Table 7-1, we find̄δ = 11.5 (J cm−3)12 .) We note that this is the same solubility parameter
as for CO, and according to this method, we would expect the solubility in air to be thesame as in CO,
xH2 = 2.63× 10−3
Using data from Table 7-1 gives
xH2 = 2.7× 10−3
We might be concerned about the vapour pressure and how well the vapour is representedby an ideal gas. The normal boiling points for N2 and O2 are listed in Table 10. At 90 K, wemust therefore expect the vapour pressure to be above 1 atm (1.013 bar), and the solubilitywill be smaller than the value found above. The saturation pressure for each component at90 K can be found from the Clausius-Clapeyron equation:
Psi (T ) = Ps
i (Tb) exp
[
−∆vaph
R
(
1T−
1Tb
)]
Using this, we find the values listed in Table 4. To compute thevapour pressure of liquidair at 90 K, we first determine the activity coefficients from the van Laar model (we neglect
Table 9: Solubility parameters for liquid airComponent xi vi/cm3 mol−1 Φi δi/(J cm−3)0.5
N2 0.78 37.5 0.827 10.8O2 0.22 27.9 0.173 14.4
42
Table 10: Normal boiling point.Component Tb/K Ps
i /barCH4 111.5 1.106CO 83 1.89N2 77 3.209O2 90 1.013
H2 in the liquid phase becausexH2 is low):
ln γN2 =A′
(
1+A′xN2B′xO2
)2
ln γO2 =B′
(
1+B′xO2A′xN2
)2
where
A′ =vN2
RT
(
δN2 − δO2
)2
=37.5 cm3 mol-1
8.3145 J mol-1 K-1 × 90 K(10.8− 14.4)2 J cm-3 = 0.6495
B′ =vO2
RT
(
δN2 − δO2
)2
=27.9 cm3 mol-1
8.3145 J mol-1 K-1 × 90 K(10.8− 14.4)2 J cm-3 = 0.4832
which gives
ln γN2 =0.6495
(
1+ 0.6495×0.780.4832×0.22
)2= 0.0195
∴ γN2 = 1.0197
ln γO2 =0.4832
(
1+ 0.4832×0.220.6495×0.78
)2= 0.330
∴ γO2 = 1.391
and finally
PN2 = PsN2γN2 xN2 = 3.209 bar× 1.0197× 0.78= 2.55 bar
PO2 = PsO2γO2 xO2 = 1.013 bar× 1.391× 0.22= 0.31 bar
P = PN2 + PO2 + PH2 = 2.55 bar+ 0.31 bar+ 1.00 bar= 3.86 bar
We conclude that the pressure is relatively low, and that thegas phase is ideal.
43
Method 2
Using the two-suffix Margules expansion for the excess Gibbs energy, see Chapter 10-6 inthe textbook leading to Eq. (10-54):
ln HH2,air = xN2 ln HH2,N2 + xO2 ln HH2,O2 − aN2,O2 xN2 xO2
whereaN2,O2 is estimated by (see footnote on p. 615 in the textbook)
aN2,O2 ≈(
δN2 − δO2
)2 (
vN2 + vO2
)
2RT
=(10.8− 14.4)2 J cm-3 (37.5+ 27.9) cm3 mol-1
2× 8.3145 J mol-1 K-1 × 90 K= 0.566
Data for Henry’s constantHH2,N2 are found in Table 10-4 in the textbook:
T /K HH2,N2/bar79 45695 345
Interpolated to 90 K;HH2,N2 = 380 bar; lnHH2,N2 = 5.940. We have no data forHH2,O2, andestimate it from the correlation found under Method 1:
ln xH2 = ln
(
PH2
HH2,O2
)
= AδO2 + B
ln HH2,O2 = ln PH2 − AδO2 − B = 0+ 0.435× 14.4+ 0.982= 7.246
This gives:
ln HH2,air = 0.78× 5.940+ 0.22× 7.246− 0.566× 0.22× 0.78= 6.130
HH2,air = 460 bar
xH2 =1 bar
460 bar= 2.2× 10−3
Problem 10.6
We are supposed to use the Orentlicher correlation,
lnf2x2= ln H
(Ps1)
2,1 +A
RT
(
x21 − 1
)
+v̄∞2
(
P − Ps1
)
RT(6.1)
with the following data for H2 in N2 around 77 K taken from Table 10-4 in the textbook:
T /K H(Ps
1)2,1 /bar v̄∞2 /cm3 mol−1 A/J mol−1
68 547 30.4 70479 456 31.5 704
44
The normal boiling point of N2 is 77 K, which means thatPs1 = 1 atm= 1.013 bar. By
linear interpolation, we find forT = 77 K: H(Ps
1)2,1 = 473 bar and ¯v∞2 = 31.3 cm3 mol−1. We
then find from Eq. (10.1):
lnf2x2= ln 473+
704 J mol-1
8.3145 J mol-1 K-1 77 K
[
(1− x2)2 − 1
]
ref Eq.6.2 (38)
+31.3 cm3 mol−1 (100− 1.013) bar
83.1451 bar cm3 mol-1 K-1 77 K= 6.6430+ 1.0996
(
x22 − 2x2
)
ln x2 = ln f2 − 6.6430− 1.0996(
x22 − 2x2
)
The fugacity of hydrogen in the gas phase is
f2 = ϕ2P2 = 0.88× (100− 1.013) bar= 87.1 bar
∴ ln f2 = 4.4672
We now set up an iterative solution method, starting withx2 = 0:
Guessedx2 ln x2 from Eq. (6.2) New value ofx2
0 -2.1758 0.1140.114 -1.9404 0.1440.144 -1.8826 0.1520.152 -1.8666 0.1550.155 -1.8620 0.155
Conclusion: The solubility isx2 = 0.16.
Problem 10.9
Question a
We consider the expressions for the fugacities of nitrogen (component 2) inn-butane (com-ponent 1) in gas and liquid:
f g2 = H2,1x2
f l2 = ϕ
l2x2P
in the limit of x2→ 0. At equilibrium, the two fugacities are equal,i.e.
H2,1 = ϕl2P (39)
The fugacity coefficient in the liquid phase must be determined from the Peng-Robinsonequation of state. The Peng-Robinson equation of state reads (Eq. (12-59) in the textbook):
P =RT
v − b−
a(T )v(v + b) + b(v − b)
(40)
45
where the coefficients for each of the components are:
a(T ) = a(Tc)α(T )
a(Tc) = 0.45724(RTc)
2
Pc
α(T ) =[
1+ β(
1−√
T/Tc
)]2(41)
β = 0.37464+ 1.54226ω − 0.26992ω2
b = 0.07780RTc
Pc
The following physical properties are found in Reid et al. (the last five columns are com-puted from Eq. 41:
Component Tc/K Pc/bar ω β
n-butane 425.2 38.0 0.199 0.6709N2 126.2 33.9 0.039 0.4344
Component α(T = 250K) a(Tc)106cm6bar mol−2
a(T=250K)106cm6bar mol−2
bcm3mol−1
n-butane 1.3374 15.039 20.113 72.38N2 0.6773 1.485 1.006 24.08
The mixing rules are:
a =∑
i
∑
j
xix j√
aia j(1− ki j)
b =∑
i
xibi
wherek11 = k22 = 0 andk12 = 0.0867. The expression for the fugacity coefficient is:
lnϕk =bk
b
( PvRT− 1
)
− lnP(v − b)
RT−
a
2√
2bRT
[
2∑
i xiaik
a−
bk
b
]
lnv + (1+
√2)b
v + (1−√
2)b
In the limit of x2→ 0, we get the following relation:
H2,1 = limx2→0
(
f2x2
)
= limx2→0
(ϕ2P)
The following results are found from Eq. 42:
a = a1
b = b1
P = Ps1 = 0.392 bar
v = vs1 = 93.035 cm3 mol-1
lnϕ2 =b2
b1
(
Ps1vs
1
RT− 1
)
− lnPs
1(vs1 − b1)
RT−
a1
2√
2b1RT
[
2a12
a1−
b2
b1
]
lnvs
1 + (1+√
2)b1
vs1 + (1−
√2)b1
46
The value of the fugacity coefficient is:
lnϕl2 =
24.08 cm3 mol-1
72.38 cm3 mol-1
(
0.392 bar× 93.035 cm3 mol-1
83.145 bar cm3 mol-1 K-1 × 250 K− 1
)
− ln0.392 bar× (93.035− 72.38) cm3 mol-1
83.145 bar cm3 mol-1 K-1 × 250 K
−20.113× 106cm6 bar mol-2
2×√
2× 72.38 cm3 mol-1 × 83.145 bar cm3 mol-1 K-1 × 250 K
×
2√
1.006× 20.113× 106 × (1− 0.0867) cm6 bar mol-2
20.113× 106cm6 bar mol-2−
24.08 cm3 mol-1
72.38 cm3 mol-1
× ln
[
93.035+ (1+√
2)72.38]
cm3 mol-1[
93.035+ (1−√
2)72.38]
cm3 mol-1
= 7.0002
∴ ϕl2 = exp(7.0002)= 1097
From Eq. 39 we then findH2,1 = 1097× 0.392 bar= 430 bar
Question b
The partial molar volume is defined as
v2 =
(
∂V∂n2
)
P,T,n1
Since the equation of state is not explicit inv, It is more convenient to consider how thepressure varies when we varyT , P, n1, andn2:
dP =
(
∂P∂T
)
V,n1,n2
dT +
(
∂P∂V
)
T,n1,n2
dV +
(
∂P∂n1
)
V,T,n2
dn1 +
(
∂P∂n2
)
V,T,n1
dn2
At constantP, T , andn1, we have
dP =
(
∂P∂V
)
T,n1,n2
dV +
(
∂P∂n2
)
V,T,n1
dn2 = 0
so under these conditions,
v2 = −(∂P/∂n2)V,T,n1
(∂P/∂V)T,n1,n2
47
We can now evluate the differentials ofP in the limit x2 → 0. From Eq. 40:
P =nRT
V − nb−
n2a(T )V2 + 2Vnb − n2b2
∂P∂n2
=(V − nb) n′RT − nRT (−nb′ − n′b)
(V − nb)2
−
(
V2 + 2Vnb − n2b2)
(2nn′a + n2a′) − n2a[2V(n′b + nb′) − (2nn′b2 + 2n2bb′)](
V2 + 2Vnb − n2b2)2
=RT (Vn′ + n2b′)
(V − nb)2
−V2n(2n′a + na′) + 2Vn2(bn′a + nba′ − nb′a) + bn4(−ba′ + 2b′a)
(
V2 + 2Vnb − n2b2)2
=RT (vn′ + nb′)
n (v − b)2−
v2(2n′a + na′) + 2v(bn′a + nba′ − nb′a) + b(−bna′ + 2nb′a)
n(
v2 + 2vb − b2)2
∂P∂V
= −nRT
(V − nb)2−
(
V2 + 2Vnb − n2b2)
× 0− n2a(2V + 2nb)(
V2 + 2Vnb − n2b2)2
= −RT
n (v − b)2+
2a(v + b)
n(
v2 + 2vb − b2)2
Here,
n′ =∂n∂n2= 1
a′ =∂a∂n2=∂
∂n2
[
n21a1 + 2n1n2
√a1a2(1− k12) + n2
2a2
(n1 + n2)2
]
=(n1 + n2)
2(
2n1√
a1a2(1− k12) + 2n2a2
)
−(
n21a1 + 2n1n2
√a1a2(1− k12) + n2
2a2
)
2(n1 + n2)
(n1 + n2)4
=1n
[(
2x1√
a1a2(1− k12) + 2x2a2
)
− 2(
x21a1 + 2x1x2
√a1a2(1− k12) + x2
2a2
)]
na′ = 2[
x1√
a1a2(1− k12)(1− 2x2) − x21a1 + x2a2(1− x2)
]
= 2x1
[
−x1a1 +√
a1a2(1− k12)(x1 − x2) + x2a2
]
b′ =∂b∂n2=∂
∂n2
(
n1b1 + n2b2
n1 + n2
)
=(n1 + n2) b2 − (n1b1 + n2b2)
(n1 + n2)2
=1n
(b2 − x1b1 − x2b2)
nb′ = x1 (b2 − b1)
In the limit x2→ 0, the expressions reduce to
na′ = 2[
−a1 +√
a1a2(1− k12)]
nb′ = b2 − b1
48
∂P∂n2
=RT (v1 + b2 − b1)
n (v1 − b1)2−
v21
[
2a1 + 2(
−a1 +√
a1a2(1− k12))]
n(
v21 + 2v1b1 − b2
1
)2
−2v1
[
b1a1 + 2b1
(
−a1 +√
a1a2(1− k12))
− a1(b2 − b1)]
n(
v21 + 2v1b1 − b2
1
)2
−b1
[
−b12(
−a1 +√
a1a2(1− k12))
+ 2a1 (b2 − b1)]
n(
v21 + 2v1b1 − b2
1
)2
=RT (v1 + b2 − b1)
n (v1 − b1)2
− 2v2
1
√a1a2(1− k12) + v1
[
2b1√
a1a2(1− k12) − a1b2
]
+ b1
[
−b1√
a1a2(1− k12) + a1b2
]
n(
v21 + 2v1b1 − b2
1
)2
=RT (v1 + b2 − b1)
n (v1 − b1)2− 2
√a1a2(1− k12)
n(
v21 + 2v1b1 − b2
1
) + 2a1b2 (v1 − b1)
n(
v21 + 2v1b1 − b2
1
)2
∂P∂V
= −RT
n (v1 − b1)2+
2a1(v1 + b1)
n(
v21 + 2v1b1 − b2
1
)2
Evaluated with the given data:
∂P∂n2
=83.145 bar cm3 mol-1 K-1 × 250 K× (93.035+ 24.08− 72.38) cm3 mol-1
n (93.035− 72.38)2 cm6 mol-2
−2×√
20.113× 1.006× 106 × (1− 0.0867) cm6 bar mol-2
n(
93.0352 + 2× 93.035× 72.38− 72.382)
cm6 mol-2
+2× 20.113× 106 cm6 bar mol-2 × 24.08 cm3 mol-1 × (93.035− 72.38) cm3 mol-1
n(
93.0352 + 2× 93.035× 72.38− 72.382)2 cm12 mol-4
=1763 bar
n∂P∂V
= −83.145 bar cm3 mol-1 K-1 × 250 K
n (93.035− 72.38)2 cm6 mol-2
+2× 20.113× 106 cm6 bar mol-2 × (93.035+ 72.38) cm3 mol-1
n(
93.0352 + 2× 93.035× 72.38− 72.382)2 cm12 mol-4
= −25.38 bar cm-3 mol
n
The partial molar volume is then
v̄∞2 =1763 bar/n
25.38 bar cm-3 mol/n= 69.5 cm3 mol-1
49
Question c (method I)
The parameter in the two-suffix Margules equation is defined by
ln γ2 =A
RTx2
1
The activity coefficient γ2 refers to pure component 2 as reference state (γ2 = 1whenx2 = 1). In this problem, the properties are given atx2 = 0, and we must change the refer-ence state for component 2. Alternatively, we might think that we could instead considercomponent 1 and use
ln γ1 =A
RTx2
2
∴
ART
=ln γ1
x22
= limx2→0
ln γ1
x22
This wil, however, also raise a problem because lnγ1 → 0 in this limit. The change ofreference state is made with use of the Gibbs-Duhem equationat constantP andT :
x1d ln γ1 + x2d ln γ2 = 0
d ln γ2 = −x1
x2d ln γ1 = −
ART
x1
x22x2dx2 = −
ART
2(1− x2)dx2
We may note that if we use pure component 2 as reference state,we find
ln γ2 =
x2∫
x2=1
d ln γ2 = −A
RT2
x2∫
x2=1
(1− x2)dx2 = −A
RT2
[
(x2 − 1)−12
(x22 − 1)
]
= −A
RTx2
1
We now use pure componnet 1 as reference state:
ln γ∗2 =
x2∫
x2=0
d ln γ2 = −A
RT2
x2∫
x2=0
(1− x2)dx2 = −A
RT2
(
x2 −12
x22
)
= −A
RT(2x2 − x2
2)
In this limit we may therefore write
f2 = f2(x2 = 0)a2 = H2,1γ∗2x2
∴
f2x2= H2,1γ
∗2
lnf2x2= ln H2,1 + ln γ∗2 = ln H2,1 −
2ART
x2 +A
RTx2
2 (42)
If we consider lnf2x2
as function ofx2 and expand aroundx2 = 0 at constantP andT , we get
lnf2x2=
(
lnH2,1x2
x2
)
x2=0
+
∂ ln f2x2
∂x2
P,T,x2=0
x2 + ...
= ln H2,1 +
(
∂ ln H2,1
∂x2
)
P,T,x2=0
x2 + ... ref Eq.9.5 (43)
50
Comparing Eq. 42 with and using Eq. 39, we find
−2ART=
(
∂ ln H2,1
∂x2
)
P,T,x2=0
=
[
∂
∂x2(lnϕ2 + ln P)
]
P,T,x2=0
=
(
∂ lnϕ2
∂x2
)
x2=0
Note that the differentiation with respect tox2 reprsents the effect of variation in composi-tion only, and thatP andTare held constant (P = Ps
1). We must consider bothx1 andx2 asvariables. IfP andT are constants while the composition varies, the volume has to vary.The molar volume of the mixture may be expressed as
v = v̄∞1 x1 + v̄∞2 x2 = v̄∞1 + (v̄∞2 − v̄∞1 )x2
Question c (method II)
An alternative expression for the fugacity of N2 is
f2 = γ∗2x2 f 0
2
Since the solubility of N2 in n-butane is small, pure solvent is chosen as reference state(indicated with the asterisk),γ∗2 = 1. The parameter in the two-suffix Margules equation isdefined by
ln γ∗2 =A
RT(x2
1 − 1) = −A
RT(1+ x1)x2
Combining Eqs. 39 and gives
f2x2= γ∗2 f 0
2 = ϕ2P
lnf2x2= ln γ∗2 + ln f 0
2 + lnϕ2 + ln P
Expanding lnf2x2
in powers ofx2 and equating the coefficients of the linear terms, gives
−2ART=
[
∂ lnϕ2
∂x2
]
P=Ps1,x2=0
The expression for lnϕ2 is
lnϕ2 =bk
b
( PvRT− 1
)
−lnP(v − b)
RT−
a
2√
2bRT
2∑
i xiaik
a−
bk
bln
v + (1+√
2)bv
+ (1−√
2)b
This depends onx2 in a complex way. It is assumed in Eq. thatP andT are constant (thetemperature is held constant and the pressure variation gives rise to the term
v∞2RT (P − Ps
1)).Under these conditions,v has to vary whenx2 varies. Using the chain rule, we get
d lnϕ2
dx2=∂ lnϕ2
∂x2+∂ lnϕ2
∂x1
dx1
dx2+∂ lnϕ2
∂vdvdx2
It is understood here that all differentials are at constantP andT so that
f2 = γ∗2x2H2,1 = ϕ2x2P
51
or
γ∗2 = ϕ2P
H2,1
ln γ∗2 = ln(ϕ2P) − lnH2,1
For small values ofx2 we may expand
f2 = f2(x2 = 0)a2 = H(Ps
1)2,1 γ
∗2x2
∴
f2x2= H
(Ps1)
2,1 γ∗2
lnf2x2= ln H
(Ps1)
2,1 + ln γ∗2 = ln H(Ps
1)2,1 −
2ART
x2 +A
RTx2
2 (9.4)
If we consider lnf2x2
as function ofx2 and expand aroundx2 = 0 at constantP andT , we get
lnf2x2=
lnH
(Ps1)
2,1 x2
x2
P,T,x2=0
+
∂ ln f2x2
∂x2
P,T,x2=0
x2 + ...
= ln H(Ps
1)2,1 +
∂ ln H(Ps
1)2,1
∂x2
P,T,x2=0
x2 + ... (9.5)
Comparing Eq. 42 with and using Eq. 39, we find
−2ART=
∂ ln H(Ps
1)2,1
∂x2
P,T,x2=0
=
[
∂
∂x2(ln ϕ2 + ln P)
]
P,T,x2=0
=
(
∂ lnϕ2
∂x2
)
x2=0
Note that the differentiation with respect tox2 represents the effect of variation in compo-sition only, and thatP andTare held constant (P = Ps
1). We must consider bothx1 andx2
as variables. IfP andT are constants while the composition varies, the volume has to vary.The molar volume of the mixture may be expressed as
v = v̄∞1 x1 + v̄∞2 x2 = v̄∞1 + (v̄∞2 − v̄∞1 )x2
52
Chapter 11
Problem 11.1
We consider a solvent (component 1) with a solute (component2). The change in chemicalpotential of component 1 as a result of changing the composition and temperature of theliquid mixture is
dµ1(l) = −s1(l)dT + RTd ln x1
Suppose the liquid mixture is in equilibrium with a solid, which is assumed to be puresolvent. The corresponding change in chemical potential for the solid is
dµ1(s) = −s1(s)dT
If equilibrium is established after the change, the two changes in chemical potential mustbe equal:
−s1(l)dT + RTd ln x1 = −s1(s)dT
or
RTd ln x1 = [s1(l) − s1(s)] dT = ∆fuss1dT =∆fush1
TdT (44)
Equation 44 is the basic equation for freezing-point depression. Integration from puresolvent to an actual mixture composition ofx1 gives
[ln x1]x11 = −
∆fush1
R
[
1T
]T
T ∗
ln x1 = −∆fush1
R
(
1T−
1T ∗
)
whereT ∗ is the melting point of pure solvent. This equation can be solved forT :
1T=
1T ∗−
R∆fush1
ln x1
Since the mixture is rich in benzene, and the melting points of the two components are notvery different, we may assume that benzene freezes out first, and consider it as the solvent(component 1):
1T=
1278.7 K
−8.3145 J mol-1 K-1
9843 J mol-1ln 0.95= 0.003631
∴ T = 275.4 K
The phase diagram for the system is shown in Figure 20.
53
x1
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
T, K
260
280
300
320
340
360
Figure 20: Liquidus curve for the naphtalene-benzene system. Benzene is component 1.
Problem 11.6
Also in this case we consider a solvent (component 1) and a solute (component 2). Since thenecessary information for computing the activity coefficients is available, we may assumea nonideal mixture in this case. The expression forT is, by analogy to problem 11.1:
1T=
1T ∗−
R∆fush1
ln a1 =1
T ∗−
R∆fush1
(ln γ1 + ln x1)
In this case, it is not clear which of the components will freeze out first at the given compo-sition. Even if there is much less benzene thann-heptane in the mixture, the freezing pointof benzene is so much higer than that ofn-heptane that it might still freeze out first. To besure, we can construct the phase diagram like in Problem 11.1, except that here, we use thevan Laar model for the activity coefficients:
ln γ1 =A′
(
1+ A′x1B′x2
)2
ln γ2 =B′
(
1+ B′x2A′x1
)2
A′ =v1
RT(δ1 − δ2)2
B′ = A′v2
v1
A′
B′=
v1
v2
The phase diagram for this system is shown in Figure 21. We seefrom the figure thatx1 = 0.1 is at the benzene-rich side of the eutectic point. We therefore consider benzene as
54
x1
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
T, K
160
180
200
220
240
260
280
300
Figure 21: Liquidus curve for then-heptane-benzene system. Benzene is component 1.
the solvent (component 1) and find:
A′ =89 cm3 mol-1
8.3145 J mol-1 K-1 T(18.8− 15.1)2 J cm-3 =
146.54 KT
B′ = A′148 cm3 mol-1
89 cm3 mol-1=
243.69 KT
A′
B′=
89 cm3 mol-1
148 cm3 mol-1= 0.6014
ln γ1 =A′
(
1+ A′x1B′x2
)2=
146.54 K
T(
1+ 0.60140.10.9
)2=
128.76 KT
ln γ2 =B′
(
1+ B′x2A′x1
)2=
243.69 K
T(
1+ 1.66290.90.1
)2=
0.9559 KT
This gives the following expression for the freezing temperature:
1T=
1278.7 K
−8.3145 J mol-1 K-1
9843 J mol-1
(
128.76 KT
+ ln 0.1
)
(
1+8.3145× 128.76
9843
)
1T=
1278.7 K
−8.31459843 K
ln 0.1
∴ T =1+ 8.3145×128.76
98431
278.7 K −8.31459843 K ln 0.1
= 200.39 K
Problem 11.9
The condition for equilibrium isf2(g) = f2(l)
that is, the fugacity of the CO2 (component 2) is the same in gas and liquid phases. At194.3 K andx2 = 0.05, the solution is undersaturated with respect to CO2. If the mole
55
fraction x2 were equal to 0.25, the fugacity of the liquid would have beenequal to that ofthe solid,i.e.
f2(g) = f2(l) = f2(s) (45)
and the saturation pressure would in that case have beenPs2 = 0.99 bar. In the actual case
with x2 = 0.05, the fugacity must be lower than this, the vapour pressuremust be lowerthan 0.99 bar. We may therefore safely assume that the gas is ideal, and consequently
P2 = f2(g) = f2(l) (46)
The fugacity in the liquid phase is the activity times the fugacity of a standard state,
f2(l) = a2 f L2 = γ2x2 f L
2 (47)
where f L2 is the fugacity of (the hypothetical state) pure liquid CO2 at T = 194.3 K. This
implies that the standard state corresponding to the activity coefficient γ2 is pure liquidCO2. In order to use the information provided by Eq. 45, it is convenient, however, to usepure solid CO2 at T = 194.3 K as the standard state. This gives
f2(l) = γ2x2 f L2 = γ2x2 f S
2
f L2
f S2
= f2(s) = f S2 = Ps
2
where f S2 is the fugacity of (pure) solid CO2 atT = 194.3 K. Combining Eqs. 45–47 gives
P2 = γ2x2 f S2
f L2
f S2
for x2 = 0.05
Ps2 = γ2x2Ps
2f L2
f S2
for x2 = 0.25
We first determine the activity coefficient from the data atx2 = 0.25:
γ2(x2 = 0.25)=1x2
f S2
f L2
=0.560.25
= 2.24
The activity coefficient atx2 = 0.05 must be estimated from this value, which means thatwe can use a model with at most one parameter. One such model isthe two-suffix Margulesequation:
ln γ2 =A
RTx2
1
∴
ART=
ln γ2
x21
=ln 2.240.752
= 1.4337
Hence,
ln γ2(x2 = 0.05)=A
RT0.952 = 1.4337× 0.952 = 1.2939
∴ γ2(x2 = 0.05) = 3.6471
The partial pressure of CO2 at x2 = 0.05 is therefore
P2 = 3.6471× 0.05× 0.99 bar×1
0.56= 0.322 bar
56
Chapter 12
Problem 12.3
The thermodynamic condition for equilibrium is
µ′A = µ′′A (48)
whereµA is the chemical potential of the alcohol. Prime and double prime refer to thehexane phase and DMSO phase, respectively. The chemical potential in each phase is
µA = µoA + RT ln xA + RT ln γA (49)
whereµoA is the chemical potential in a reference state,xA is the mole fraction of the alcohol,
andγA is the activity coefficient of the alcohol. We note that the excess Gibbs energy is ofthe form
gE = AxAxB
whereA is a constant andxB is the mole fraction of the solvent (hexane or DMSO). Thisleads to the following form for the activity coefficient:
RT ln γA = Ax2B (50)
Combining Eqs. 48–50 gives
RT ln x′A + A′(
x′B)2= RT ln x′′A + A′′
(
x′′B)2
RT ln
(
x′Ax′′A
)
= A′′(
1− x′′A)2 − A′
(
1− x′A)2 (51)
ln K = ln
limx′A→0x′′A→0
(
x′Ax′′A
)
=A′′ − A′
RT
The problem is now to find the constantsA′′ andA′ at 30◦C and 100 bar. For this, we usethe fact that Gibbs energy depends on temperature and pressure as
dG = VdP − S dT
For the excess properties, we then find
dgE = vEdP − sEdT (52)
57
This can be used to compute the difference in excess Gibbs energy when the conditions arechanged from 0◦C and 1 bar to 30◦C and 100 bar. The last term in Eq. 52 is, however,not convenient because we do not have data forsE available. Instead we use the Gibbs-Helmholtz equation for the temperature variation:
d(gE/T ) = hEd(1/T )
The effect of the pressure change is (assuming constant excess volumes over the pressurerange in question):
∆gE = vE∆P
The effect of the temperature change is (assuming constant excess enthalpies over the tem-perature range in question):
∆
(
gE
T
)
= hE∆
(
1T
)
Inserting the given numbers leads to for the hexane phase andthe DMSO phase at the actualpressure:
gE = 2400x′Ax′H J mol-1 + 16x′Ax′H cm3 mol-1 × 990 N cm-2 = 2558.4x′Ax′H J mol-1
gE = 320x′′A x′′D J mol-1 − 10x′′A x′′D cm3 mol-1 × 990 N cm-2 = 221.0x′′A x′′D J mol-1
Similarly at the actual temperature:
gE = 2558.4x′Ax′H J mol-1 ×303 K273 K
+ 4800x′Ax′H J mol-1 ×(
1−303 K273 K
)
= 2312.1x′Ax′H J mol-1
∴ A′ = 2312.1 J mol-1
gE = 221.0x′′A x′′D J mol-1 ×303 K273 K
+ 600x′′A x′′D J mol-1 ×(
1−303 K273 K
)
= 179.4x′′A x′′D J mol-1
∴ A′′ = 179.4 J mol-1
Used in Eq. 51, we then find:
ln K =(179.4− 2312.1) J mol-1
8.3145 J mol-1 K-1 × 303 K= −0.8465
∴ K = 0.429
Problem 12.4
The phase diagram (P − T -diagram) is shown in Figure 22. The problem here is to deter-mine the temperature at which the solid-liquid equilibriumpressure is 200 bar (indicatedwith the dashed line in Figure 22. If we consider equilibriumbetween solid and liquid at acertainP andT , given by equality in chemical potentails,
µl(P, T ) = µs(P, T )
58
T, K
265 270 275 280 285 290 295 300
P, b
ar0.004
0.006
0.008
50.000
100.000
150.000
200.000
Figure 22:P − T diagram for benzene.
the equilibrium atP + dP andT + dT will be given by
µl(P + dP, T + dT ) = µs(P + dP, T + dT )
The corresponding changes in the chemical potentials must be the same in the two phases,
dµl = µl(P + dP, T + dT ) − µl(P, T ) = dµs
The change in chemical potential is given by
dµl = vldP − sldT
and similarly for the solid phase. This gives the following differential equation for theliquidus curve:
(
vl − vs)
dP =(
sl − ss)
dT (53)
The molar volumes at 278.7 K may be found from the given densities:
vl =Mρl=
78.1 g mol-1
0.891 g cm-3= 87.65 cm3 mol-1
vs =Mρs=
78.1 g mol-1
1.010 g cm-3= 77.33 cm3 mol-1
vl − vs = (87.65− 77.33) cm3 mol-1 = 10.32 cm3 mol-1 (54)
At equilibrium, the change in entropy on fusion is equal to the change in enthalpy on fusiondivided by the melting temperature. The enthalpy on fusion is not given, but the enthalpieson vaporization and sublimation may be deducted from the vapour pressure curves by com-paring the given vapour-pressure equations with the Clausius-Clapeyron equation:
ln P = ln Po −∆vaph
R
(
1T−
1T o
)
59
This version of the Clausius-Clapeyron equation assumes that ∆vaph is constant over thetemperature range in question, which is consistent with thegiven vapour-pressure equa-tions. We find
∆vaph
R= 1785 K
∆subhR
= 2310 K
Enthalpy is a state function, and at the triple point we must have
∆fush = ∆subh−∆vaph = (2310−1785) K×83.145 bar cm3 mol-1 K-1 = 43651 bar cm3 mol-1
(55)Consistently with the assumption that∆subh and∆vaph are constant over the temperaturerange, we assume that also∆fush is constant. This gives for the entropy of fusion at atemperatureT :
∆fuss =∆fush
T=
43651 bar cm3 mol-1
T(56)
The resukts 54 and 56 inserted into Eq. 53 gives:
10.32 cm3 mol-1dP = 43651 bar cm3 mol-1dTT
∴ P = Po +43651 bar cm3 mol-1
10.32 cm3 mol-1ln
TT o= Po + 4230 bar× ln
TT o
The fixed pointsPo andT o are given by the normal melting point:Po = 1.013 bar andT o = 278.7 K, leading to
P = 1.013 bar+ 4230 bar× lnT
278.7 K
At 200 bar, we then find:
200 bar = 1.013 bar+ 4230 bar× lnT
278.7 K
∴ T = 278.7 Kexp
(
(200− 1.013) bar4230 bar
)
= 292.1 K
Problem 12.7
The stability of the mixture with respect to phase separation is expressed by∆mixg (seeFigures 12-14 and 12-15 in the textbook). The Gibbs energy onmixing for a binary mixtureis
∆mixg = RT (x1 ln x1 + x2 ln x2) + gE(P, T ) (57)
At 1 atm and 273 K, the excess Gibbs energy of the mixture is given as
gE(1 atm, 273 K)= RT × 1.877× x1x2
60
At a pressureP and 273 K, the excess Gibbs energy is
gE(P, 273 K) = gE(1 atm, 273 K)+
P∫
1 atm
vEdP
= RT × 1.877× x1x2 +
P∫
1 atm
x1x2(4.026− 0.233 lnP)dP
= 82.0578 atm cm3 mol-1 K-1 × 1.877× 273 K× x1x2
+[
4.026 cm3 mol-1(P − 1 atm)− 0.233 cm3 mol-1(P ln P − P + 1 atm)]
x1x2
= (42048− 4.026− 0.233) atm cm3 mol-1 × x1x2
+ (4.026+ 0.233− 0.233 lnP) cm3 mol-1 × P × x1x2
= (42044 atm+ 4.259P − 0.233P ln P) × x1x2 cm3 mol-1 (58)
Eq. 58 in 57 gives at 273 K:
∆mixg = 22402 atm cm3 mol-1 (x1 ln x1 + x2 ln x2)
+ (42044 atm+ 4.259P − 0.233P ln P) × x1x2 cm3 mol-1
This expression is plotted for two values ofP, P = 10 atm andP = 2000 atm, in Figure23. We observe that the function is symmetric aroundx1 = x2 = 0.5. To find the minimum
x1
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
∆ mixg,
J m
ol-1
-1800
-1600
-1400
-1200
-1000
-800
-600
-400
-200
0
∆mixgid
∆mixg, P=10 atm
∆mixg, P=2000 atm
Figure 23: Gibbs energy of (regular) mixing.
pressure at which the mixture separates, we have to find the solution to
∂2∆mixg
∂x21
= 0 at x1 = 0.5
With gE(P, 273 K) of the form
gE(P, 273 K)= A(P)x1x2
whereA(P) = (42044 atm+ 4.259P − 0.233P ln P) cm3 mol-1
61
we first find∂∆mixg∂x1
= RT [ln x1 − ln(1− x1)] + A(1− 2x1)
and then∂2∆mixg
∂x21
=RT
x1(1− x1)− 2A
Setting this expression equal to 0 atx1 = 0.5 gives
A =RT
2× 0.52=
82.0578 atm cm3 mol-1 K-1 × 273 K2× 0.52
= 44804 atm cm3 mol-1
This corresponds to a pressure given by the equation
(42044 atm+ 4.259P − 0.233P ln P) cm3 mol-1 = 44804 atm cm3 mol-1
which has the solutionP = 1045 atm=1059 bar. AtP = 1500 atm, the Gibbs energy ofmixing will have two minima (located symmetrically aroundx1 = 0.5), given by
∂∆mixg∂x1
= RT [ln x1 − ln(1− x1)] + A(P = 1500 atm)(1− 2x1) = 0
This gives the following solution forx1:
A(P) = (42044 atm+ 4.259× 1500 atm− 0.233× 1500atm ln 1500)cm3mol-1
= 45877 atm cm3 mol-1
which makes
82.0578 atm cm3 mol-1 K-1 × 273 K [ln x1 − ln(1− x1)]
+45877 atm cm3 mol-1(1− 2x1) = 0
and finallyx′1 = 0.37, x′′1 = 0.63
62
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