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November 14, 2016 | Author: Eduardo Paulini Villanueva | Category: N/A
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Advanced Thermodynamics KP8108 — Solutions to Selected Problems (a companion to J. M. Prausnitz, R. N. Lichtenthaler and E. G. de Azevedo Molecular Thermodynamics, 2nd ed.) Prof. Bjørn Hafskjold, Dep. of Chemistry (NTNU) Assoc. Prof. Tore Haug-Warberg, Dep. of Chem. Eng. (NTNU) May 16, 2007
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Contents Problem 2.2 . . . . . . . . . . . . . . . . . . Problem 2.5 . . . . . . . . . . . . . . . . . . Problem 2.6 . . . . . . . . . . . . . . . . . . Problem 2.7 . . . . . . . . . . . . . . . . . . Detailed derivation of Eqs. (3-53) and (3-54) . Problem 3.2 . . . . . . . . . . . . . . . . . . Problem 3.3 . . . . . . . . . . . . . . . . . . Problem 3.7 . . . . . . . . . . . . . . . . . . Problem 3.9 . . . . . . . . . . . . . . . . . . Problem 5.2 . . . . . . . . . . . . . . . . . . Iteration 1: y1 = 0.01 . . . . . . . . . . Iteration 2: y1 = 3.4835 × 10−3 . . . . . Problem 5.3 . . . . . . . . . . . . . . . . . . Problem 5.16 . . . . . . . . . . . . . . . . . Problem 5.17 . . . . . . . . . . . . . . . . . Iteration 1: y3 = 0 . . . . . . . . . . . . Iteration 2: y3 = 1.04 × 10−4 . . . . . . Problem 6.2 . . . . . . . . . . . . . . . . . . Problem 6.16 . . . . . . . . . . . . . . . . . Problem 7.1 . . . . . . . . . . . . . . . . . . Problem 7.3 . . . . . . . . . . . . . . . . . . Problem 10.3 . . . . . . . . . . . . . . . . . Method 1 . . . . . . . . . . . . . . . . Method 2 . . . . . . . . . . . . . . . . Problem 10.6 . . . . . . . . . . . . . . . . . Problem 10.9 . . . . . . . . . . . . . . . . . Question a . . . . . . . . . . . . . . . . Question b . . . . . . . . . . . . . . . . Question c (method I) . . . . . . . . . . Question c (method II) . . . . . . . . . Problem 11.1 . . . . . . . . . . . . . . . . . Problem 11.6 . . . . . . . . . . . . . . . . . Problem 11.9 . . . . . . . . . . . . . . . . . Problem 12.3 . . . . . . . . . . . . . . . . . Problem 12.4 . . . . . . . . . . . . . . . . . Problem 12.7 . . . . . . . . . . . . . . . . .
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1 2 3 4 5 7 8 10 11 13 14 15 16 20 24 25 26 27 30 35 38 41 41 44 44 45 45 47 50 51 53 54 55 57 58 60
iv
Chapter 2 Problem 2.2 For the given equation of state, P Vn − b = RT , and by using the Maxwell relation ∂2 A/∂T ∂V = ∂2 A/∂V∂T , we find: ! ! P R ∂P ∂S = . = = V ∂V T ∂T V −b T n
For an isothermal change: ! Z Z ∂S dV = ∆S = ∂V T
V n
! ! R P1 V2 − nb = nR ln . dV = nR ln V1 − nb P2 −b
Alternatively, from ∂2G/∂T ∂P = ∂2G/∂P∂T , we find: ! ! nR ∂S ∂V =− =− . ∂P T ∂T P P
This formula integrates to the same expression for ∆S as above. For U we know the differential dU = T dS − pdV and hence (∂U/∂V)T = T (∂S /∂V)T − P. Using the first of the Maxwell rules from above we get ! ! ∂P ∂U =T − P = 0. ∂V T ∂T V
Consequently, ∆U = 0. Alternatively, we can write (∂U/∂P)T = T (∂S /∂P)T − P(∂V/∂P)T which gives: ! ! ! ∂U ∂S ∂V =T −P ∂P T ∂P T ∂P T ! ! ∂V ∂V =T −P ∂T P ∂P T RT RT =− + P P = 0. Again we find that ∆U = 0. For the next derivative we shall use H = U + PV and ! ! ! ∂H ∂U ∂V = +V +P ∂P T ∂P T ∂P T nRT nRT + nb − P 2 =0+ P P = nb. 1
For an isothermal change: ∆H =
Z
∂H ∂P
!
dP =
T
Z
nbdP = nb (P2 − P1 ) .
The changes in Gibbs energy (G) and Helmholtz energy (A) are found from the definitions of G and A, and the results above: ! P1 , ∆G = ∆H − T ∆S = nb (P2 − P1 ) − nRT ln P2 ! P1 ∆A = ∆U − T ∆S = −nRT ln . P2
Problem 2.5 Because entropy is constant it would be natural to express the result as an explicit function of S . However, this is not feasible because the equation of state is expressed in terms of T and V. We shall therefore use implicit differentiation and first write entropy as a function of volume and temperature, differentiate, use some of the Maxwell relations, and then put the differential of S (V, T ) equal to zero: ! ! ! ∂S ∂P CV ∂S dV + dT = dV + dS = dT = 0 ∂V T ∂T V ∂T V T Thermodynamically, the exhaust temperature does not depend on the gas flow rate through the 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: ! R a ∂P RT = − 2 ⇒ P= v−b v ∂T V v − b Inserted into the expression for dS = 0 this gives the following differential equation between v and T at constant entropy: cV dT dv =− v−b R T This is the adiabatic equation of state for a van der Waal gas. Integration at constant cv gives ! ! v2 − b cV T2 ln = − ln v1 − b R T1
where T 2 and v2 are unknown. It is given that the final pressure is atmospheric, and by using the equation of state, we could express v2 by P2 . This is, however, a little cumbersome, but since the exhaust pressure is low, we try first by approximating the exhaust gas is an ideal gas. This gives: ! ! − b RT 2 T2 R vig R 2 ≈ − ln ln = − ln T1 cV v1 − b cV P2 (v1 − b) 2
ig where we have approximated vig 2 − b ≈ v2 = RT 2 /P2 . Rearrangement gives ! ! ! RT 1 T2 R T1 R ln = ln − ln T1 cV T2 cV P2 (v1 − b) ! R RT 1 ln =− P2 (v1 − b) cV 1 + cRV ! R RT 1 = − ln cP P2 (v1 − b)
where we have made use of cigP = cig V + R. With the given numbers, we find ! ! 82.06cm3 atm mol-1 K-1 (350+273) K T2 8.314 J K-1 mol-1 ln ln =− (350 + 273) K 33.5J K-1 mol-1 1 atm (600-45) cm3 mol-1 = −1.123 ∴ T 2 = 623 exp(−1.123) = 203 K Checking the ideal-gas assumption: ig
RT 2 82.06 cm3 atm mol-1 K-1 203 K = = 16700 cm3 mol-1 P2 1 atm = 16400 cm3 mol-1
V2 = V2vdw
Note: The van der Waal result was found by an iterative solution of the cubic equation in V at 203 K and 1 atm. If we want to improve the ideal-gas assumption, we may insert V2vdw = 16400 cm3 mol-1 into the equation above: ! ! − b 8.314 J K-1 mol-1 (16400 − 45) cm3 mol-1 T2 R vvdw 2 = − ln ln = − ln T1 cV v1 − b 25.2 J K-1 mol-1 (600-45) cm3 mol-1 = −1.117 ∴ T 2 = 623 exp(−1.117) = 204 K
Problem 2.6 The virial expansion of the compression factor is Pv B C = 1+ + 2 + ··· RT v v The compression factor for the van der Waal gas is a 1 v RT a 1 − zvdw = − 2 = b RT v − b v 1 − v RT v z=
We recognize the first term on the right-hand side as the sum of a geometric series: !2 b 1 b +··· =1+ + v v 1 − bv 3
Ordering the van der Waal compression fcator in increasing powers of z
vdw
a 1 b =1+ b− + RT v v
and we conclude that B=b−
!2
+ ···
a . RT
Problem 2.7 We start by recovering one result from Problem 2.2: ! ! ! ∂V ∂V ∂U = −T −P . ∂P T ∂T P ∂P T From the given equation of state, the volume is Pv BP =1+ RT RT BP RT b RT 1+ = +a− 2 ∴v= P RT P T z=
From this we find: ! ∂V R 2b = + 3 ∂T P P T ! RT ∂V =− 2 ∂P T P ! ! RT ∂U R 2b 2b = −T + 3 −P − 2 =− 2 ∂P T P T P T Integration at the constant temperature τ then gives: ∆U(τ) =
Zπ 0
∂U ∂P
!
4
T =τ
dP = −
2b π τ2
1 v
gives
Chapter 3 Detailed derivation of Eqs. (3-53) and (3-54) Start with this definition of the chemical potential: ! ∂A µi = ∂ni V Applied to Eq. (3-50) the definition leads to ! !# " Z∞ RT P◦ V ∂ X ∂P µi = − + u◦i − T s◦i n j ln dV − RT ∂ni T,V,n j V ∂ni j n j RT
(1)
V
We note that j = 1 is the only term that survives the differentiation of the sum. This term is " !# ! ∂ P◦ V P◦ V ni ln = ln −1 (2) ∂ni ni RT ni RT
Furthermore, the standard state of component i is pure ideal gas at temperature T and pressure P◦ = 1 bar. The chemical potential of i in this state is µ◦i = g◦i = u◦i + RT − T s◦i Eqs. 2 and 3 into 1 gives ! ! # " Z∞ RT P◦ V ∂P µi = − − 1 + µ◦i − RT dV − RT ln ∂ni T,V,n j V ni RT
(3)
V
! ! Z∞ RT P◦ V ∂P ◦ − µi − µi = dV − RT ln ∂ni T,V,n j V ni RT
(4)
V
The relation between the chemical potential and fugacity is ! fi ◦ µi − µi = RT ln ◦ fi
(5)
The fugacity of the standard state is P◦ = 1 bar. The fugacity is further expressed by the fugacity coefficient as fi = ϕi Pi = ϕi yi P fi ϕi = yi P 5
(6)
Combination of Eqs. 4, 5, and 6 yields ! Z∞ ! ! ∂P ϕi yi P RT P V ◦ dV − RT ln RT ln = − P◦ ∂ni T,V,n j V ni RT V
! ! ! Z∞ P◦ V RT yi P ∂P − − RT ln RT ln ϕi = dV − RT ln ∂ni T,V,n j V ni RT P◦ V
! Z∞ PV RT ∂P dV − RT ln − = ∂n V nRT i T,V,n j V
! Z∞ RT ∂P − = dV − RT ln z ∂ni T,V,n j V
(3-53)
V
For a pure component, the chemical potential is ! G fi ◦ µi = = µi + RT ln ◦ ni fi ! ! P fi ◦ + RT ln ◦ = µi + RT ln P fi ! ! fi P RT ln = µi − µ◦i − RT ln ◦ P pure i fi ! P G G◦ − RT ln = − ni ni P◦
(7)
The Gibbs energy is taken from Eq. (3-51) in the book:
G = ni
Z∞ V
! ! P◦ V PV P RT dV − RT ln + − + u◦i − T s◦i ni V ni RT ni
(8)
Note that G◦ = h◦i − T s◦i = u◦i + RT − T s◦i ni
(9)
PV = RT z ni
(10)
and
6
Eqs. 8, 9, and 10, inserted into 7, combine to: ! ! fi G G◦ P RT ln = − − RT ln P pure i ni ni P◦ ! ! Z∞ P RT P◦ V = − dV − RT ln + ni V ni RT V ! P ◦ ◦ − ui + RT − T si − RT ln P◦ ! ! Z∞ PV P RT − dV − RT ln + = ni V ni RT
PV + u◦i − T s◦i ni
PV − RT ni
V
=
Z∞ V
! P RT dV − RT ln z + RT z − RT − ni V
=
Z∞
! P RT dV − RT ln z + RT (z − 1) − ni V
V
(3-54)
Note: There is probably an easier way to understand this equation. It is illogical to bring in expressions for both A and G and pretend they are independent functions. The elegant solution would be to bring in the homogeneous properties of P = P(T, V, N) and develop Eq. 3-54 directly from Eq. 3-53. Suggestions are welcome.
Problem 3.2 The gas mixture is illustrated in Figure 1. If we consider the gas to be a pseudo pure A, B xA = 0.25, xB = 0.75 P = 50 bar T = 373 K ϕA = 0.65, ϕB = 0.90 Figure 1: Properties of the gas mixture. substance, the relation between the chemical potential and the fugacity would be ! fM ⊖ µM = µM + RT ln ⊖ fM where µ⊖M and f M⊖ are the chemical potential and the fugacity, respectively, in the standard state. This would correspond to a molar Gibbs energy for the mixture, G M = µM . On the 7
other hand, the molar Gibbs energy for the mixture is given by X Gm = µi xi i
where the chemical potential of each component is ! ! ϕ i Pi fi ⊖ ⊖ µi = µi + RT ln ⊖ = µi + RT ln fi fi⊖ Here, ϕi is the fugacity coefficient of component i. Equating G M and Gm gives ! X ! X fM ϕ i Pi ⊖ ⊖ µM + RT ln ⊖ = xi µi + RT xi ln fM fi⊖ i i
(11)
which we can use to express f M in terms of the component properties. We should, however, first make a remark on the relation between the standard states. The µ⊖M is not the same as µ⊖i even if they refer to the same pressure, because µ⊖M contains the properties of an ideal mixture: X X xi ln xi (12) xi µ⊖i + RT µ⊖M = i
i
The standard fugacities, which are usually taken to be ideal gas at 1 bar pressure, are the same: f M⊖ = fi⊖ . Equations 11 and 12 give: ! X ! X X X fM ϕ i Pi ⊖ ⊖ xi ln xi + RT ln ⊖ = xi µi + RT xi µi + RT xi ln f fi⊖ M i i i i ! X ϕP X fM i i = − ∴ ln xi ln xi ln xi 1 bar 1 bar i i With the given data, we find: ! fM = 0.25 × ln (0.65 × 0.25 × 50) + 0.75 × ln (0.90 × 0.75 × 50) ln 1 bar −0.25 × ln 0.25 − 0.75 × ln 0.75 = 3.725 3 f M = 1 × exp(3.7253) = 41.5 bar
Problem 3.3 The two situations are shown schematically in Figure 2. Equilibrium between gas and liquid at 1 bar may be expressed as gas
liq
µC2 H6 (1bar) = µC2 H6 (xC2 H6 = 0.33 × 10−4 ) The same equilibrium at 35 bar may be expressed as gas
liq
µC2 H6 (35bar) = µC2 H6 (xC2 H6 = x) 8
gas H2 O, C2 H6 P = 1 bar PH2 O = 0.0316 bar PC2 H6 ≈ 1 bar T = 298 K
gas H2 O, C2 H6 P = 35 bar PH2 O = 0.0316 bar PC2 H6 ≈ 35 bar T = 298 K
liquid H2 O, C2 H6 xH2 O ≈ 1.0 xC2 H6 = 0.33 × 10−4
liquid H2 O, C2 H6 xH2 O ≈ 1.0 xC2 H6 =?
Figure 2: The two equilibrium situations.
where xC2 H6 is the unknown denoted by x. In the gas phase, the effect of a change in the chemical potential at constant temperature is dµi = v¯ i dP where v¯ i is the partial molar volume of component i. Since there is so little water in the gas phase, we assume it to be pure C2 H6 , in which case we can neglect the effect of a change in composition, and vi is the molar volume of C2 H6 , i.e. v¯ i = vC2 H6 . The molar volume is given by the equation of state, v=
RT 1 − aP − bP2 P
where a = 7.63×10−3 bar−1 and b = 7.22×10−5 bar−2 . Integration of the chemical potential at constant T gives 35 Z bar
!
35 Z bar
! 1 = − dP = RT − a − bP dP P T 1 bar !1 bar # " 1 35 bar 2 2 2 − a 34 bar − b 35 − 1 bar = RT ln 1 bar 2
µgas C2 H6 (35bar)
µgas C2 H6 (1bar)
∂µC2 H6 dP
In the liquid phase, the change in chemical potential is the sum of two contributions; one due to the pressure change, and one due to the composition change. The effect of the change in pressure is dµC2 H6 = v¯ C2 H6 dP, where v¯ C2 H6 is the partial molar volume of ethane in the liquid. Since the liquid is practically incompressible, it may be assumed to be constant over the pressure change in question here, and in any case the contribution is negligibly small for the liquid. The effect of the change in composition is dµC2 H6 = RT d ln aC2 H6 , where aC2 H6 = fC2 H6 / fC⊖2 H6 is the activity of ethane. We have to assume that the activity 9
coefficient of ethane, defined by γC2 H6 = aC2 H6 /xC2 H6 , is constant over the concentration range considered here, which leads to liq
liq
=
µC2 H6 (x) − µC2 H6 (x = 0.33 × 10−4 ) xC2 H6 =x
35 Z bar
Z
vC2 H6 dP + RT
1 bar
d ln γC2 H6 + RT
xC2 H6 =x
Z
d ln xC2 H6
xC2 H6 =0.33×10−4
xC2 H6 =0.33×10−4 xC2 H6 =x
≈ RT
Z
d ln xC2 H6 = RT ln
xC2 H6 =0.33×10−4
x 0.33 × 10−4
Equating the changes in the chemical potential of the gas and liquid phases give " # 1 x −3 −5 2 2 = RT ln(35) − 7.63 × 10 × 34 − × 7.22 × 10 35 − 1 RT ln 0.33 × 10−4 2 x 1 −3 −5 2 2 ln = ln(35) − 7.63 × 10 × 34 − × 7.22 × 10 35 − 1 = 3.2517 0.33 × 10−4 2 ∴ 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 of enthalpy in this problem. Since the equation of state is given explicitly in the volume as function of P and T , we find the enthalpy from the following relation: ! # ZP " X ∂V dP + ni h0i H= V −T ∂T P,nT i 0
where h0i is the molar enthalpy of pure i at the actual temperature. The given volumetric data for the mixture leads to " ! # b ∂V V −T =a+ ∂T P,nT T where a = 5 × 10−5 m3 mol−1 and b = −0.2 m3 K mol−1 . Integration leads to the molar enthalpy ! X b h= a+ P+ xi h0i T i where xi is the mole fraction of component i. The condition hin = hout gives ! ! X X b b 0 Pin + xi hi,in = a + Pout + xi h0i,out a+ T in T out i i ! ! X b b a+ Pin = a + Pout + xi h0i,out − h0i,in T in T out i 10
We have h0i,out − h0i,in = c0P,i (T out − T in ) and c0P =
P i
gives Pin =
a+
b T out
xi c0P,i = 33.5 J mol−1 K−1 . Solving for Pin
Pout + c0P (T out − T in ) a + Tbin
5 · 10−5 m3 mol−1 − 0.2m3 Kmol−1 /200K 105 Pa + 33.5Jmol−1 K−1 (200K − 300K) = 5 · 10−5 m3 mol−1 − 0.2m3 Kmol−1 /300K
= 55.9 · 105 Pa = 55.9 bar
Problem 3.9 The internal energy of mixing is (using one mole as basis): ∆mix u = u M − x1 u1 − x2 u2 Since the van der Waal equation is explicit in P: P=
RT a − 2, v−b v
it is convenient to find the internal energy from u=
Z∞ "
∂P P−T ∂T
v
where
∂P P−T ∂T
!
=
V,nT
!
V,nT
#
dv +
X
xi u0i
i
RT a RT a − 2− =− 2 v−b v v−b v
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 the mixture and the two pure components. The molar internal energy for the mixture and each of the pure components are: u M = −a M
Z∞
vM
u1 u2
aM X 0 dv X 0 + + x u = − xi ui i i v2 vM i i
a1 = − + u01 v1 a2 = − + u02 v2
The change in internal energy for the mixture is ! ! aM X 0 a2 a M x1 a1 x2 a2 a1 0 0 ∆mix u = − + − u1 + x2 − u2 = − + + xi ui + x1 vM v1 v2 vM v1 v2 i 11
With the given data, we find XX √ aM = xi x j ai a j (1 − ki j ) i
j
h i √ = 0.5 × 106 1.04 + 2 × 1.04 × 4.17(1.0 − 0.1) + 4.17 bar cm6 mol−2 2
vM
= X 2.24 × 106 bar cm6 mol−2 = xi vi = 0.5(32.2 + 48.5) cm3 mol−1 = 40.35 cm3 mol−1 i
∆mix u =
! bar cm6 mol−2 2.24 0.5 × 1.04 0.5 × 4.17 106 − + + = 363 J mol−1 40.35 32.2 48.5 cm3 mol−1
12
Chapter 5 Problem 5.2 The question here is whether the fugacity of CO2 in the compressed gas is higher than the fugacity at saturation conditions. The actual temperature is lower than the triple-point temperature, 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 of pure CO2 (component 1) to its new value after injection of H2 (component 2) until the total pressure is 60 bar. We therefore consider a charging process as illustrated in Figure 3. The H2 is inert in this case, in the sense that it does not dissolve in the solid phase. The change CO2 (g) P∗CO2 = 0.1392 bar T = 173 K
∆µ (g) −→
CO2 (s) T = 173 K
∆µ (s) −→
CO2 (g) + H2 (g) PCO2 + PH2 = 60 bar T = 173 K
CO2 (s) T = 173 K
Figure 3: Available data for the state changes.
in chemical potential of CO2 in the gas phase is ∆µ1 (g) = RT
Zc ⋆
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 be equal to its vapour pressure (the saturation pressure of 0.1392 bar is quite low), but please validate this assumption by numerical calculations. The fugacity of CO2 in the compressed state must be calculated somehow, and we shall be using f1 = ϕ1 y1 P where ϕ1 and y1 are the the fugacity coefficient and the mole fraction of CO2 , respectively, and P is the total pressure (equal to 60 bar). The fugacity coefficient is estimated from van 13
der Waal’s equation: z= The mixing rules are
Pv v a = − RT v − b RT v
a =
XX i
b =
X
(14)
√ yi y j ai a j
j
yi bi
i
The fugacity coefficient of component 1 in the mixture is calculated from Eq. 3-70 in Prausnitz et al.: √ P √ 2 a1 j y j a j v b1 ln ϕ1 = ln + − − ln z (15) v−b v−b RT v 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: CO2 (1) H2 (2)
a/atm cm6 mol−2 3.640×106 0.2476×106
b/cm3 mol−1 42.67 26.61
Alternatively we may use the corresponding state principle and calculate a = 27(RT c )2 /64Pc and b = RT c /8Pc from critical data. Do this on your own and compare the results with the table. Because the mole fraction of CO2 is low, at most 0.01, we make a small error if we choose the van der Waal parameters to be those of pure H2 , but for the time being we determine 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 × + 0.992 × 0.2476 × 106 ) atm cm6 mol−2
√ 3.640 × 106 × 0.2476 × 106
= 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 solve a cubic equation in v) is v = 250.69 cm3 mol−1 and z = 1.048. Inserted into Eq. 15: ! 42.67 250.69 + ln ϕ1 = ln 250.69 − 26.77 250.69 − 26.77 √ √ √ 2 3.640 × 106 0.01 × 3.640 × 106 + 0.99 × 0.2476 × 106 − − ln 1.048 250.69 × 82.0567 × 173 = −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 gas phase: ! 0.7468 × y1 × 60 bar -1 3 -1 ∆µ1 (g) = 83.1439 cm bar mol K × 173 K × ln 0.1392 bar -1 3 = 14428 cm bar mol × ln (321.9y1 ) The change in chemical potential for CO2 in the solid phase is ∆µ1 (s) =
Zc ⋆
vs dP = vs
60 Z bar
dP = 27.6 cm3 mol−1 × (60 − 0.1392) bar
0.1392 bar
= 1652.16 cm3 bar mol-1 Here, we have assumed that the solid is incompressible and that H2 does not dissolves in the 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 ! 1652.16 1 = 3.4835 × 10−3 exp y1 = 321.9 14428 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 = + = b = =
(0.00352 × 3.640 × 106 + 2 × 0.0035 × 0.9965 × 0.99652 × 0.2476 × 106 ) atm cm6 mol−2 2.5254 × 105 atm cm6 mol−2 (0.0035 × 42.67 + 0.9965 × 26.61) cm3 mol−1 26.67 cm3 mol−1
√ 3.640 × 106 × 0.2476 × 106
and v = 251.17 cm3 mol−1, and z = 1.048. Inserting these numbers into Eq. 15 gives: ! 42.67 251.17 + ln ϕ1 = ln 251.17 − 26.77 251.17 − 26.77 √ √ √ 2 3.640 × 106 0.0035 × 3.640 × 106 + 0.9965 × 0.2476 × 106 − − ln 1.048 251.17 × 82.0567 × 173 = −0.282 ∴ ϕ1 = exp(−0.282) = 0.7543 A small change in ϕ1 from 0.7468 to 0.7543 is observed. The improved value of the change in chemical potential for the gas phase is calculated (once more) from Eq. 13): ! 0.7543 × y1 × 60 bar -1 3 -1 ∆µ1 (g) = 83.1439 cm bar mol K × 173 K × ln 0.1392 bar -1 3 = 14428 cm bar mol × ln (325.1y1 ) 15
The solid phase chemical potential is the same as above, and we find ! 1652.16 1 = 3.4492 × 10−3 exp y1 = 325.1 14428 We consider Iteration 2 to be good, and accept this as the final result. To conclude the calculation, we need a material balance, using one mole material in total as basis: nCO2 (g) + nCO2 (s) + nH2 (g) = 1 mol nH2 (g) = 0.99 mol
(16) (17)
The mole fraction of CO2 in the gas phase is yCO2 =
nCO2 (g) = 0.00345 nCO2 (g) + nH2 (g)
(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 condense into liquid or solid, or not. This depends on the pressure and temperature in the gas during the isenthalpic expansion. In the following, we will denote CH4 by 1 and CO2 by 2. The
CO2 (s) or CO2 (l)? CH4 (g) 1 bar, T=?
30 mol% CO2 (g) 70 mol% CH4 (g) 70 bar, 313 K
Figure 4: The given process.
critical point of CO2 is at Pc = 73.8 bar and T c = 304.2 K. The triple point is at PT = 5.18 bar and T T = 216.8 K. If the gas cools to a temperature in the range between the critical point and the triple point, some CO2 may possibly condense to a liquid, if it cools to T < 216.8 K, it may condense to a solid. Because the equation of state is given as a virial expansion in molar density, it is most convenient to express the enthalpy of the gas as H=
Z∞ " V
∂P P−T ∂T
!
V,nT
#
dV + PV +
X
ni u0i
i
where ni is the number of moles of component i and u0i is the molar internal energy of pure i in the ideal gas state at the actual temperature. The strategy is to find how H depends 16
on V and T , then constrain H to be constant, and finally to use this constraint to find the isenthalpic equation of state for the gas. This equation will tell us how much the gas cools on expanding. The given data for the mixture are of the form zmix =
Pv Bmix =1+ RT v
where Bmix =
XX i
(19)
yi y j Bi j
(20)
j
and yi is the mole fraction of component i. Each Bi j is of the form bi j ci j + 2, T T
Bi j = a i j + which means Bmix will be on the same form, Bmix = a +
b c + 2 T T
where a =
XX i
j
yi y j ai j = (0.72 × 42.5 + 2 × 0.7 × 0.3 × 41.4 + 0.32 × 40.4) cm3 mol-1
-1 3 = X 41.85 Xcm mol b = yi y j bi j = −(0.72 × 16.75 + 2 × 0.7 × 0.3 × 19.50 + 0.32 × 25.39) × 103 cm3 K mol-1 i
j
-1 3 3 = X −18.683 X × 10 cm K mol c = yi y j ci j = −(0.72 × 25.05 + 2 × 0.7 × 0.3 × 37.3 + 0.32 × 68.7) × 105 cm3 K2 mol-1 i
j
= −34.124 × 105 cm3 K2 mol-1
We then find for one mole of gas ∂P P−T ∂T
!
V,nT
R 2c = 2 b+ v T
!
and for the molar enthalpy: ! X 3c R aT + 2b + + h= yi h0i v T i We note that the ideal-gas contribution to enthalpy is (constant cP assumption): X yi h0i = c0P T i
where c0P =
X i
yi c0P,i = (0.7 × 35.8 + 0.3 × 37.2) J mol-1 K−1 = 36.22 J mol-1 K−1 17
(21)
The energy constraint tells that h is constant and solving Eq. 21 for v gives: R aT + 2b + 3c T v(T ) = 0 h − cP T
(22)
At the upstream conditions, Bmix
! 18.683 × 103 34.124 × 105 cm3 mol-1 = −52.672 cm3 mol-1 − = 41.85 − 313 3132
and from Eq. 19: 70 bar × v 52.672 cm3 mol-1 = 1 − v 83.1439 bar cm3 mol-1 K−1 × 313 K which gives v1 = 308.24 cm3 mol-1 . The constant enthalpy in Eq. 21 can now be determined to 5 8.314 J mol-1 K-1 41.85 × 313 − 2 × 18.683 × 103 − 3×34.124×10 K cm3 mol-1 313 h1 = 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 is cooled to a temperature below the condensation point for CO2 , or in other words, will the fugacity CO2 in the gas be higher than the equilibrium value in contact with pure CO2 ? The fugacity of CO2 in the expanding gas is given by f2 = ϕ2 y2 P, which should be compared with the fugacity for saturated gas in equilibrium with liquid or solid: ! vl ∆P sat sat (3.5) f2 = P2 exp RT The saturation pressure of pure CO2 is ln Psat (bar) = 10.807 −
1980.24 T
(23)
and vl is the molar volume of pure CO2 (l). The fugacity coefficient is given by ln ϕ2 =
2X y j B2 j − ln zmix v j
Most likely the Poynting correction is quite small, and the fugacity coefficient is probably less than 1. This implies that CO2 is more volatile in the compressed gas than in the pure form (where ϕ2 ≈ 1), and if we find that if the gas is unsaturated in pure form, no CO2 will condense. Under this assumption, we solve Eqs. 19, 20, 22, 23 and calculate the data shown in Table 1: The outlet pressure of 1 bar is reached before the temperature drops below the 18
T /K 271 ... 304 304.2 305 ... 313
Table 1: Saturation pressure(s) of CO2 . v/cm3 mol−1 B/cm3 mol−1 P/bar P2 /bar 35182 -73.6 0.639 0.192 ... ... ... ... 400.6 -56.5 54.2 16.26 398 -56.4 54.5 16.36 387.9 -56.1 55.9 16.78 ... ... ... ... 308.2 -52.7 70 21
Psat 2 /bar 33.1 ... 73.2 73.8 74.8 ... (N/A)
100
P, bar
80
P2sat
60 40
P2
20 0 270
280
290
300
T, K
Figure 5: Saturation pressure and partial pressure of CO2 .
triple point temperature, and any condensed CO2 will be in liquid form. We note that from the 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 with the saturation pressure. A plot of P2 and Psat 2 is given in Figure 5. The figure shows that the saturation pressure is higher than the actual partial pressure of CO2 , and consequently that CO2 (l) will not condense. To improve the assumption of the ideal gas phase we need to estimate the molar volume of liquid CO2 . Data from Reid et al. indicate that ρl (T = 293 K) = 0.777 g cm3 . The molecular weight of CO2 is M = 44.01 g mol−1 . This gives vl =
M 44.01 g mol-1 = = 56.6 cm3 mol-1 ρ 0.777 g cm3
Assuming that the liquid density varies only a little with temperature, the saturation fugacity is easily calculated: ! 56.6 cm3 mol-1 ∆P sat sat f2 = P2 exp 83.1439 cm3 bar mol-1 K-1 T 19
The fugacity coefficient in the gas phase is ! 2X b2 c2 Bmix 2 ln ϕ2 = a2 + + 2 − ln 1 + y j B2 j − ln zmix = v j v T T v
(24)
where a2 =
X j
b2
y j a2 j = (0.7 × 41.4 + 0.3 × 40.4) cm3 mol-1
= X 41.1 cm3 mol-1 = y j b2 j = −(0.7 × 19.50 + 0.3 × 25.39) × 103 cm3 K mol-1 j
c2
= X −21.267 × 103 cm3 K mol-1 = y j c2 j = −(0.7 × 37.3 + 0.3 × 68.7) × 105 cm3 K2 mol-1 j
= −46.72 × 105 cm3 K2 mol-1
The results for the fugacity may now be calculated from Eqs. 19, 20, 22–24, and some of the results are given in Table 2: A plot of f2 and f2sat is shown in Figure 6. The figure Table 2: Fugacities of CO2 . T /K ϕ2 f2 /bar f2sat /bar 271 1.365 0.262 30.52 ... ... ... ... 304 0.826 13.43 70.14 304.2 0.824 13.48 70.45 305 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 process as illustrated in Figure 7. The ethylene is inert in the sense that it does not dissolve in the solid phase. Like in Problem 5.2, we compute the change in chemical potential for naphtalene in the gas phase as Zc ∆µ1 (g) = RT d ln f1 = RT ln f1 c⋆ ⋆
The fugacity of pure napthalene vapour can be assumed to be equal to the vapour pressure, 2.80×10−4 bar, because the vapour pressure is low. The fugacity of napthalene in the compressed state is determined as f1 = ϕ1 y1 P (25) 20
100
f, bar
80
f2sat
60 40
f2
20 0 270
280
290
300
T, K
Figure 6: Saturation fugacity and fugacity of CO2 .
Naphtalene (g) (1) P∗1 = 2.80 × 10−4 bar T = 308 K
Naphtalene (s) T = 308 K
∆µ1 (g) −→
∆µ1 (s) −→
Naphtalene (g) + Ethylene (g) (2) P1 + P2 = 30 bar P1 = ? T = 308 K
Naphtalene (s) T = 308 K
Figure 7: Naphtalene dissolving in ethylene.
with the usual notation. The change in chemical potential for naphtalene in the solid phase is 30 Zc Z bar ∆µ1 (s) = vs dP = vs dP ⋆
2.80×10−4 bar
The molar volume of the solid is given by the density and the molecular weight: M 128.2 g mol-1 = = 112.0 cm3 mol-1 vs = ρ 1.145 g cm−3 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 the virial expansion truncated after the second term. a) With the ideal gas law P1 = 3360 cm3 bar mol-1 ∆µ1 (g) = RT ln −4 2.80 × 10 bar ! 3360 cm3 bar mol-1 −4 P1 = 2.80 × 10 bar × exp = 3.19 × 10−4 bar -1 -1 3 83.1439 cm bar mol K 308 K −4 P1 3.19 × 10 bar = = 1.06 × 10−5 y1 = P 30 bar b) With the virial expansion truncated after the second term. The virial expansion reads PV B 2 n RT z= n+ n (27) =1+B ∴ P= nRT V V V The second virial coefficient is given by the van der Waals result, B=b−
a RT
(28)
where a =
XX i
b =
X
√ yi y j ai a j
j
yi bi
i
It is convenient to write the expression for b as XX b= yi y j bi j i
j
where
1 bi + b j , 2 because we can then use Eq. (5-29) in Prausnitz et al.: bi j =
ln ϕ1 =
2 (y1 B11 + y2 B12 ) − ln z v
with Bi j = b i j − and B=
XX i
j
22
√ ai a j RT yi y j Bi j
(29)
(30) (31)
We will need the numerical values for a and b, which we find from the critical values: 27(RT c )2 64Pc RT c b = 8Pc The critical values (See e.g. Reid et al.) are listed in Table 3, which also gives the computed values for a and b. The values for ai j , bi j , and Bi j are given in Table 4. We now make a a =
Table 3: Critical data for napthalene and ethylene. T c /K Pc /atm a/cm6 atm mol−2 b/cm3 mol−1 Naphtalene (1) 748.4 40.0 3.98×107 191.91 6 Ethylene (2) 282.4 49.7 4.56×10 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 -1382.9 191.91 12 1.35×107 125.10 -409.06 58.28 6 22 4.56×10 58.28 -122.15
guess for the mole fraction of naphtalene to obtain a first estimate of the viral coefficient B. We can for example assume that P1 is little affected by the presence of ethylene which gives the starting value P∗1 2.80 · 10−4 bar y1 = = = 9.3 · 10−6 P 30bar (Other qualified guesses include the answer from a) or y1 = 0.) The equations 25–31 give an iterative procedure to find y1 : B = y21 B11 + 2y1 (1 − y1 ) B12 + (1 − y1 )2 B22
P 2 v −v−B = 0 RT
z = 1+
B v
2 (y1 B11 + (1 − y1 ) B12 ) − ln z ln φ1 = v ! φ1 Py1 = ∆µ RT ln P∗1 B, v, z and φ1 are estimated from the previous y1 and a better estimate is obtained in the last equation. With Bi j from Table 4, ∆µ=3360cm3 bar/mol, T = 308K, P = 30bar and P∗1 = 2.8 · 10−4 bar the two first iterations give the values shown in Table 5. We can therefore conclude that y1 = 2.80 · 10−5 . 23
B v z φ1 y1
Table 5: Iteration sequence. iteration 1 iteration 2 -122.155 -122.166 705.893 705.876 0.826949 0.826930 0.379462 0.379441 −5 2.80447·10 2.80462·10−5
Problem 5.17 This problem is also similar to problem 5.2. Instead of considering the cooling, we consider compressing water vapour by adding air at 263 K. This is illustrated in Figure 8. Like in
H2 O (g) (3) P∗1 = 1.95 torr
N2 (1) + O2 (2) (g) + H2 O (g) P = 30 bar y3 = ? T = 263 K
∆µ3 (g) −→
T = 263 K
H2 O (s) ρ s = 0.92 g cm−3
∆µ3 (s) −→
H2 O (s)
Figure 8: Available data for the state changes.
Problem 5.16, we compute the change in chemical potential for water vapour in the gas phase as Zc f3 d ln f3 = RT ln f3 c⋆ = RT ln ∗ ∆µ3 (g) = RT P3 ⋆
The fugacity of pure water vapour is again assumed to be equal to the vapour pressure, 1.95 atm = 2.57×10−3 atm = 2.60×10−3 bar, because the vapour pressure is low. The torr = 1.95 760 fugacity of water vapour in the compressed state is f3 = ϕ3 y3 P. The change in chemical potential for ice is
∆µ3 (s) =
Zc
vs dP = vs
⋆
30 Z bar
2.60×10−3 bar
24
dP
The molar volume of ice is the molecular weight divided by the density: vs =
M 18.0 g mol-1 = = 19.57 cm3 mol-1 −3 ρ 0.92 g cm
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 fugacity coefficient: XX Bmix = yi y j Bi j i
j
Bmix Pv =1+ RT v 2X ln ϕ3 = y j B3 j − ln zmix v j zmix =
A material balance for N2 (g), O2 (g), and H2 O (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 )y3 B13 + 0.2(1 − y3 ) 2 B22 + 2 × 0.2(1 − y3 )y3 B23 + y3 y3 B33 = − 0.82 × 25 + 2 × 0.8 × 0.2 × 15 + 0.22 × 12 cm3 mol-1 = −21.3 cm3 mol-1 21.3 cm3 mol-1 30 bar v = 1 − v 83.1451 cm3 bar mol-1 K-1 × 263 K ∴ v = 706.9 cm3 mol-1
zmix = 1 −
21.3 cm3 mol-1 = 0.9699 706.9 cm3 mol-1
2 (−0.8 × 63 − 0.2 × 72) cm3 mol-1 − ln (0.9699) = −0.1528 -1 3 706.9 cm mol = 0.8583
ln ϕ3 = ∴ ϕ3
The maximum permissible moisture content is therefore ! 587.1 2.60 × 10−3 = 1.04 × 10−4 × exp y3 = 0.8583 × 30 83.1451 × 263
If we are in doubt concerning the accuracy of this result, we can make another iteration as shown 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 )y3 B13 + 0.2(1 − y3 ) 2 B22 + 2 × 0.2(1 − y3 )y3 B23 + y3 y3 B33 h i2 = 0.8(1 − 1.04 × 10−4 ) B11 + 2 × 0.8(1 − 1.04 × 10−4 ) × 0.2(1 − 1.04 × 10−4 )B12 h i2 + 2 × 0.8(1 − 1.04 × 10−4 ) × 1.04 × 10−4 B13 + 0.2(1 − 1.04 × 10−4 ) B22 + = + =
2 × 0.2(1 − 1.04 × 10−4 )1.04 × 10−4 B23 + 1.04 × 10−4 × 1.04 × 10−4 B33 0.6399B11 + 0.3199B12 + 1.6638 × 10−4 B13 + 0.0400B22 + 4.1596 × 10−5 B23 1.0816 × 10−8 B33 −21.289 cm3 mol-1
30 bar v 21.289 cm3 mol-1 = 1 − ∴ v = 706.9 cm3 mol-1 v 83.1439 cm3 bar mol-1 K-1 × 263 K zmix = 1 −
21.289 cm3 mol-1 = 0.9699 706.9 cm3 mol-1
2 (−0.8 × 63 − 0.2 × 72) cm3 mol-1 − ln (0.9699) = −0.1528 -1 3 706.9 cm mol = 0.8583
ln ϕ3 = ∴ ϕ3
The maximum permissible moisture content is therefore ! 587.1 2.60 × 10−3 = 1.04 × 10−4 × exp y3 = 0.8583 × 30 83.1451 × 263
26
Chapter 6 Problem 6.2 An azeotrope is a liquid mixture that does not change composition when it evaporates. If we plot a liquid-vapour phase diagram for a mixture that forms an azeotrope, it will look like shown in Fig. 6-12 in the textbook (for a P − x -diagram) or in Fig. 6-22 (for a T − x -diagram). The point is that in the minimum or maximum of the curves in either of the diagrams, the bubble-point and dew-point lines go through the same point. In order to find the bubble-point and dew-point lines, we have to consider the phase equilibrium shown in Figure 9. In this figure, we have arbitrarily chosen to illustrate the change in chemical potential of component 1 from pure state to the mixed state. The change in chemical
Pure 1 (g) P1s T
Mixture 1+2 (g) ∆µ1 (g) P1 + P2 = P −→ T
Pure 1 (l) T
∆µ1 (l) −→
Mixture 1+2 (l) T
Figure 9: Schematic illustration of a mixing process.
potential for component 1 in the gas phase is
∆µ1 (g) = RT
mixture Z
pure 1
! ! y1 P P1 d ln f1 = RT ln s = RT ln P1 P1s
where we have used the given information that the gas phase may be considered ideal. In the liquid phase, the change in chemical potential is
∆µ1 (l) =
pure Z1 at P
v1 dP + RT
pure 1 at P1s
mixture Z
pure 1 at P
27
d ln a1
(32)
where v1 is the molar volume of pure component 1. The first term in Eq. 32 is the Poynting correction, which we will neglect here. The change ∆µ1 (l) is then ∆µ1 (l) = RT ln a1 = RT ln (γ1 x1 ) where γ1 is the activity coefficient for component 1. By equating the changes in chemical potentials, we get ! y1 P = RT ln (γ1 x1 ) RT ln P1s y1 P = γ1 x1 P1s From the given expression for the excess Gibbs energy for the mixture, we find (using Eqs. (6-46) and (6-47) in the textbook): A 2 x RT 2 A 2 x = RT 1
ln γ1 = ln γ2
For the azeotrope, x1 = y1 which gives the bubble-point at the azeotrope x1 P A 2 x x1 = exp P1s RT 2 A P x2 = exp P1s RT 2
(33)
The same consideration for component 2 gives A 2 P x = exp P2s RT 1 Dividing Eq. 33 by Eq. 34 and using the information that
(34) P2s P1s
= 1.649, gives
A P2s 2 2 x − x1 = 1.649 = exp P1s RT 2 A 2 1 ∴ x2 − x21 = ln 1.649 = RT 2
In other words, for an azeotrope to occur, the requirement is that for any value of x1 in the range 0 ≤ x1 ≤ 1, A must satisfy 1 1 1 A = h i= = 2 2 2 RT 2 (1 − 2x1 ) 2 x2 − x1 2 (1 − x1 )2 − x1
A plot of the allowed values of A/RT is shown in Figure 10. This means that for e.g. A = 1, there will be an azeotrope at x1 = 0.25. To illustrate this, we can construct a P − x RT 28
20
A/RT
10 0 -10 -20 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x1
Figure 10: The lines show allowed values of A/RT .
diagram as follows: The total vapour pressure (the bubble-point pressure) is the sum of the partial pressures: P = P1 + P2 = P1s exp If we set
A RT
A 2 A 2 x2 x1 + P2s exp x1 x2 RT RT
= 1, we find P2s P 2 = exp x x + exp x21 x2 1 2 P1s P1s
The dew-point line is given by
Again setting
A RT
A 2 x2 x1 P1s exp RT P1 = y1 = A 2 A 2 P x2 x1 + P2s exp RT x1 x2 P1s exp RT
= 1 as an example, we find
y1 =
exp x22 x1
exp x22 x1 +
P2s P1s
exp x21 x2
The bubble-point and dew-point curves are shown in Figure 11 for this particular case. We now have to consider the question of liquid-liquid miscibility. This is illustrated in Figure 12, showing the change in Gibbs energy on mixing as function of composition for four A = 0, 1, 2, 3. The Gibbs energy of mixing is given by different values of A; RT ∆mixG = ∆mix G
ideal
+G
excess
= RT
2 X i=1
29
xi ln xi + Ax1 x2
(35)
1.8 bu bb lepo int
1.6
2-phase
line
e in
P/P1*
1.4
l nt oi -p w de
azeotrope
liquid
1.2
gas
1.0 0.8 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x1
Figure 11: Bubble- and dew-point lines for
A RT
= 1.
Miscibility occurs for all compositions if ∂2 ∆mixG ≥ 0 for all x1 ∂x21 If we use this condition for the model in Eq. 35, we find RT ∂2 − 2A ≥ 0 [RT (x1 ln x1 + x2 ln x2 ) + Ax1 x2 ] = 2 x1 x2 ∂x1
∴
A 1 ≤ RT 2x1 x2
The values of 2x11 x2 are plotted in Figure 13, together with the conditions on A shown in Figure 10. We see that the minimum value of 2x11 x2 occurs at x1 = 0.5, when 2x11 x2 = 2. The conclusion is that azeotropes may occur for completely miscible liquids ifA ≤ 2RT . For such values of A, an azeotrope will occur in the ranges 0 ≤ x1 ≤
3 8
and
1 < x1 ≤ 1 2
Problem 6.16 The situation is illustrated in Figure 14, showing the partial pressures as given by Henry’s and Raoult’s laws. Henry’s law, Pi = Hi, j xi applies in the limit of zero solute concentration, xi → 0, and Raoult’s law, Pi = Pis xi applies in the limit of pure solvent, xi → 1. The problem here is to determine the vapour pressure at x1 = 0.5 and the corresponding vapour composition, y1 = 30
P1 P
0.2
∆mixG/RT
0.0
∆mixGideal/RT+3x1x2
-0.2
∆mixGideal/RT+2x1x2
-0.4 ∆mixG
ideal
/RT+x1x2
-0.6
∆mixG
ideal
/RT
-0.8 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x1
Figure 12: Change in Gibbs energy of mixing. Complete miscibility occurs if
∂2 ∆mix G ∂x21
≥ 0.
Between the limits of dilute solution and pure solvent, we have to use a model to estimate the vapour pressure. Assuming ideal gas phase, the partial pressure is given by Pi = Pis γi xi
(36)
We have four given pieces of information, P1s , P2s , H1,2 , and H1,2 . Because γ1 and γ2 are related by the Gibbs-Duhem equation, the model for gE (and γi ) may contain two parameters to be determined. We can also understand that at least two parameters are required because the properties of the two components are not symmetric, as seen for instance by the ratios H1,2 H and P2,1s . The van Laar model satisfies this, and we set Ps 1
2
A′ ln γ1 = 2 ′ 1 + BA′ xx21 B′ ln γ2 = 2 ′ 1 + AB′ xx12
Using this model with Eq. 36 in the limit xi → 0 gives ! ! H1,2 2 bar ′ ln γ1 = A = ln = 0.6255 = ln P1s 1.07 bar ! ! H2,1 1.60 bar ′ = 0.1848 = ln ln γ2 = B = ln P2s 1.33 bar
For an equimolar mixture, we find
A′ 0.6255 ln γ1 = 2 = = 0.0325 ∴ γ1 = 1.0330 ′ 0.6255 2 1 + BA′ 1 + 0.1848 B′ 0.1848 ln γ2 = 2 = = 0.1101 ∴ γ2 = 1.1164 ′ 0.1848 2 1 + AB′ 1 + 0.6255 31
20
A/RT
10 0 -10 -20 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x1
Figure 13: Allowable values of A/RT (solid line) and upper limit for complete miscibility in 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 bar = 0.4268 1.2951 bar
Data generated with this model are shown in Figure 15. Alternatively, we can use the three-suffix Margules equation, 1 RT 1 = RT
ln γ1 = ln γ2
i h (A + 3B) x22 − 4Bx32 i h (A − 3B) x21 + 4Bx31
Using this model with Eq. 36 in the limit xi → 0 gives ! ! H1,2 2 bar A−B = 0.6255 = ln = ln ln γ1 = RT P1s 1.07 bar ! ! H2,1 1.60 bar A+B = 0.1848 = ln = ln ln γ2 = RT P2s 1.33 bar A RT B RT
0.6255 + 0.1848 = 0.4052 2 0.1848 − 0.6255 = = −0.2204 2 =
32
2.0
2.00
1.60
1.5
P, bar
1.33
Ra oul
1.0
's nry He
He nry 's l P2 aw
law
1.07
t's law u Rao
P1
0.5
a lt's l
w
0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x1
Figure 14: The given information. 2.0
2.00
1.60
1.5
P
Bubble-poi
P, bar
1.33
nt line
Dew-poin t line
1.0
P1
0.5
1.07
P2
0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x1
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 0.5603 bar = 0.4188 1.3379 bar Data generated with this model are shown in Figure 16. y1 =
33
2.0
2.00
1.60
1.5
P
Bubble-po int lin
P, bar
1.33
Dew-poin
1.0
0.5
e
P1
t line
1.07
P2
0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x1
Figure 16: Partial pressures and total vapour pressure computed from the three-suffix Margules equation.
34
Chapter 7 Problem 7.1 The situation is shown in the P − x diagram, Figure 1 for the mixture A-CS2 . From the 14 12
P, kPa
10
P
PA
8 6 4
PCS
2
2
0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
xA
Figure 17: Plot of the partial pressures and total pressure for the mixture A-CS2 . The curves are based on the van Laar model.
given information, we can determine the activity coefficient at x1 = 0.5 and T = 283 K: PA = PAs γA xA 8 kPa PA = = 1.2030 ∴ γA = s PA xA 13.3 kPa × 0.5 ln γA = 0.1848 The given solubility parameters suggest that we use regular solution theory. The relation to the van Laar model, A′ ln γ1 = 2 ′ 1 + BA′ xx21 B′ ln γ2 = 2 ′ 1 + AB′ xx12 35
is v1 (δ1 − δ2 )2 RT v2 (δ1 − δ2 )2 = RT
A′ = B′
where vi and δi are the molar volume and the solubility parameter, respectively, of component i. With the given information, we find that vA A′ B′ A′
V MA MA 160 g mol-1 V = = = = 200 cm3 mol-1 = nA m ρA 0.8 g cm-3 200 cm3 mol-1 vA = = vCS2 61 cm3 mol-1 !2 !2 200 A′ = 3.3832 = ln γA 1 + ′ = 0.1848 × 1 + B 61
Assuming that v1 (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 ±
r
A′ RT 1 = 20.5 (J cm-3 ) 2 ± vA 1
s
3.3832 × 8.3145 J mol-1 K-1 × 283 K 200 cm3 mol-1
1
= 20.5 (J cm-3 ) 2 ± 6.3 (J cm-3 ) 2
We will return to the question of which sign to choose, but for the time being assume that the correct sign is -. This gives the complete set of properties shown in Table 6. For the Table 6: Solubility parameters. A CS2 Toluene -3 21 Solubility parameter at 298 K, (J cm ) 14.2 20.5 18.2 -1 3 Liquid molar volume at 298 K, cm mol 200 61 107 Saturation pressure at 283 K, kPa 13.3 1.73 25.5
mixture A-T (Toluene), we now find vA (δA − δT )2 RT 200 cm3 mol-1 (14.2 − 18.2)2 J cm-3 = 1.36 = -1 -1 8.3145 J mol K × 283 K vT (δA − δT )2 = RT 107 cm3 mol-1 (14.2 − 18.2)2 J cm-3 = 0.7276 = 8.3145 J mol-1 K-1 × 283 K
A′ =
B′
36
At xA = xT = 0.5: ln γA = ∴ γA
A′
2 = 1 + B′ xTA 1+ = 1.1796 A′ x
1.36 1.36×0.5 2 0.7276×0.5
= 0.1652
B′ 0.7276 ln γT = = = 0.3088 2 B′ xT 0.7276×0.5 2 1 + A′ xA 1 + 1.36×0.5 ∴ γT = 1.3618 PA = PAs γA xA = 13.3 kPa × 1.1796 × 0.5 = 7.84 kPa PT = PTs γT xT = 25.5 kPa × 1.3618 × 0.5 = 17.36 kPa P = PA + PT = 7.84 kPa + 17.36 kPa = 25.20 kPa 7.84 kPa PA = = 0.31 yA = P 25.20 kPa The full phase diagram is shown in Figure 2. An alternative representation of the data is 30
P
B ub
25
P, kPa
20
PT
bl e-
D ew - po
15
poi n
int li
t lin
e
ne
10 5
PA
0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
xA
Figure 18: P − x diagram for the mixture A-Toluene at 283 K. shown in Figure 3. The alternative positive sign for
q 1
1
A′ RT vA
would give 1
δA = 20.5 (J cm-3 ) 2 + 6.3 (J cm-3 ) 2 = 26.8 (J cm-3 ) 2
This would correspond to a change in molar internal energy on vaporization (Eq. (7-33) in the textbook) of ∆vap uA = δ2A vA = 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 enthalpy of vaporization of a hydrocarbon is 30-50 kJ mol−1 . The value corresponding to δA = 26.8 1 (J cm-3 ) 2 is therefore not acceptable. 37
1.0 0.9 0.8 0.7
yA
0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
xA
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 given data A′ = = A′ = B′ B′ = A′ B′ =
v1 (δ1 − δ2 )2 RT 132 cm3 mol-1 (14.9 − 18.8)2 J cm-3 = 0.7476 -1 -1 8.3145 J mol K × 323 K v1 132 cm3 mol-1 = = 1.4831 v2 89 cm3 mol -1 !−1 A′ = 0.6742 B′ B′ ′ A = 0.6742 × 0.7476 = 0.5041 A′
we find: 0.7476 A′ = ln γ1 = 2 = 0.2795 2 ′ 1 + BA′ xx21 1 + 1.4831 0.3 0.7 ∴ γ1 = 1.3224 B′ 0.5041 ln γ2 = = = 0.0761 2 B′ x2 0.7 2 1 + A′ x1 1 + 0.6742 0.3 ∴ γ2 = 1.0791 38
At the given composition, we then find P1 = P1s γ1 x1 = 0.533 bar × 1.3224 × 0.3 = 0.2115 bar P2 = P2s γ2 x2 = 0.380 bar × 1.0791 × 0.7 = 0.2870 bar P = P1 + P2 = 0.2115 bar + 0.2870 bar = 0.4985 bar and finally P1 0.2115 bar y1 = = = 1.414 x1 Px1 0.4985 bar × 0.3 P2 0.2870 bar y2 = = = 0.822 = x2 Px2 0.4985 bar × 0.7
K1 = K2
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 the textbook), we may estimate the solubility 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 the solvent, and A and B are (temperature dependent) constants. To determine A and B , we use data for CH4 and CO. Air is a mixture of 78% N2 , 21% O2 , and 1% Ar. In terms of solubility properties, Ar is like O2 (see Table 7-1 in the textbook), and we take the composition to be 78% N2 and 22% O2 . For the mixed solvent (air), we replace δ1 in Eq. 37 by X δ¯ = Φi δi i
where Φi is the volume fraction of component i . The sum includes all components, but we assume the solubility of H2 to be so low that its contribution can be neglected. The volume fraction is defined as xi vi Φi = P j x jv j We will need data for the solubility parameter, defined by Eq. (7-33) in the textbook: ∆vap u δi = v
! 21 i
where ∆vap u is the change in internal energy on ”complete” vaporization, i.e. to an ideal gas state. Data are given for ∆vap h, which is related to ∆vap u by ∆vap u = ∆vap h − ∆vap (Pv) ≈ ∆vap h − RT Data for the various properties are given in Table 7, all data at 90 K. The values for A and 1 B in Eq. 37 are determined from Table 8. which gives A = −0.435 (J cm−3 )− 2 , B = −0.982. 1 (If we instead use the data from Table 7-1 in the textbook, we find A = −0.461 (J cm−3 )− 2 , 41
Table 7: Physical properties of gases. From given data From Table 7-1 Comp. CH4 CO N2 O2
∆vap h J mol−1
∆vap u J mol−1
vi cm3 mol−1
δi (J cm−3 )0.5
vi cm3 mol−1
δi (J cm−3 )0.5
8750 5530 5110 6550
8000 4780 4360 5800
35.6 37.0 37.5 27.9
15.0 11.4 10.8 14.4
35.3 37.1 38.1 28.0
15.1 11.7 10.8 14.7
Table 8: Solubility parameters. Component xH2 ln xH2 δi /(J cm−3 )0.5 CH4 5.49×10−4 -7.507 15.0 −3 CO 2.63×10 -5.941 11.4
B = −0.598.) We shall, however, see below that, due to a numerical coincidence, we shall not need these numbers. The average solubility parameter for liquid air at 90 K is calculated 1 from the data in Table 9. which gives δ¯ = 11.4 (J cm−3 ) 2 . (If we instead use the data from 1 Table 7-1, we find δ¯ = 11.5 (J cm−3 ) 2 .) 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 the same 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 represented by an ideal gas. The normal boiling points for N2 and O2 are listed in Table 10. At 90 K, we must therefore expect the vapour pressure to be above 1 atm (1.013 bar), and the solubility will be smaller than the value found above. The saturation pressure for each component at 90 K can be found from the Clausius-Clapeyron equation: !# " ∆vap h 1 1 s s − Pi (T ) = Pi (T b ) exp − R T Tb Using this, we find the values listed in Table 4. To compute the vapour pressure of liquid air at 90 K, we first determine the activity coefficients from the van Laar model (we neglect
Table 9: Solubility parameters for liquid air Component xi vi /cm3 mol−1 Φi δi /(J cm−3 )0.5 N2 0.78 37.5 0.827 10.8 O2 0.22 27.9 0.173 14.4
42
Table 10: Normal boiling point. Component T b /K Pis /bar CH4 111.5 1.106 CO 83 1.89 N2 77 3.209 O2 90 1.013
H2 in the liquid phase because xH2 is low): ln γN2 = ln γO2 =
A′ 1+
A′ xN2 B′ xO2
B′ 1+
B′ xO2 A′ xN2
2 2
where vN2 δN2 − δO2 2 RT 37.5 cm3 mol-1 (10.8 − 14.4)2 J cm-3 = 0.6495 = -1 -1 8.3145 J mol K × 90 K vO2 = δN2 − δO2 2 RT 27.9 cm3 mol-1 (10.8 − 14.4)2 J cm-3 = 0.4832 = -1 -1 8.3145 J mol K × 90 K
A′ =
B′
which gives ln γN2 = ∴ γ N2
ln γO2 ∴ γ O2
0.6495
2 = 0.0195 1 + 0.6495×0.78 0.4832×0.22 = 1.0197 0.4832 = = 0.330 0.4832×0.22 2 1 + 0.6495×0.78 = 1.391
and finally PN2 = PNs 2 γN2 xN2 = 3.209 bar × 1.0197 × 0.78 = 2.55 bar PO2 = POs 2 γO2 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 the gas phase is ideal. 43
Method 2 Using the two-suffix Margules expansion for the excess Gibbs energy, see Chapter 10-6 in the textbook leading to Eq. (10-54): ln HH2 ,air = xN2 ln HH2 ,N2 + xO2 ln HH2 ,O2 − aN2 ,O2 xN2 xO2 where aN2 ,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 = = 0.566 2 × 8.3145 J mol-1 K-1 × 90 K
Data for Henry’s constant HH2 ,N2 are found in Table 10-4 in the textbook: T /K 79 95
HH2 ,N2 /bar 456 345
Interpolated to 90 K; HH2 ,N2 = 380 bar; ln HH2 ,N2 = 5.940. We have no data for HH2 ,O2 , and estimate it from the correlation found under Method 1: ! PH 2 = AδO2 + B ln xH2 = ln HH2 ,O2 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 1 bar xH2 = = 2.2 × 10−3 460 bar
Problem 10.6 We are supposed to use the Orentlicher correlation, ln
f2 = ln H2,1 x2
(P1s )
s ∞ v ¯ P − P A 2 2 1 + x1 − 1 + RT RT
(6.1)
with the following data for H2 in N2 around 77 K taken from Table 10-4 in the textbook: T /K 68 79
(P s )
H2,11 /bar 547 456
3 −1 v¯ ∞ 2 /cm mol 30.4 31.5
44
A/J mol−1 704 704
The normal boiling point of N2 is 77 K, which means that P1s = 1 atm = 1.013 bar. By (P s ) 3 −1 linear interpolation, we find for T = 77 K: H2,11 = 473 bar and v¯ ∞ 2 = 31.3 cm mol . We then find from Eq. (10.1): h i 704 J mol-1 f2 2 = ln 473 + (1 − x ) − 1 2 x2 8.3145 J mol-1 K-1 77 K 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
ln
ref Eq.6.2
(38)
The fugacity of hydrogen in the gas phase is
f2 = ϕ2 P2 = 0.88 × (100 − 1.013) bar = 87.1 bar ∴ ln f2 = 4.4672 We now set up an iterative solution method, starting with x2 = 0: Guessed x2 0 0.114 0.144 0.152 0.155
ln x2 from Eq. (6.2) New value of x2 -2.1758 0.114 -1.9404 0.144 -1.8826 0.152 -1.8666 0.155 -1.8620 0.155
Conclusion: The solubility is x2 = 0.16.
Problem 10.9 Question a We consider the expressions for the fugacities of nitrogen (component 2) in n-butane (component 1) in gas and liquid: f2g = H2,1 x2 f2l = ϕl2 x2 P in the limit of x2 → 0. At equilibrium, the two fugacities are equal, i.e. H2,1 = ϕl2 P
(39)
The fugacity coefficient in the liquid phase must be determined from the Peng-Robinson equation of state. The Peng-Robinson equation of state reads (Eq. (12-59) in the textbook): P=
RT a(T ) − v − b v(v + b) + b(v − b) 45
(40)
where the coefficients for each of the components are: a(T ) = a(T c )α(T ) (RT c )2 a(T c ) = 0.45724 Pc i2 h p α(T ) = 1 + β 1 − T/T c
(41)
β = 0.37464 + 1.54226ω − 0.26992ω2 RT c b = 0.07780 Pc
The following physical properties are found in Reid et al. (the last five columns are computed from Eq. 41: Component T c /K Pc /bar n-butane 425.2 38.0 N2 126.2 33.9 Component α(T = 250K) n-butane 1.3374 N2 0.6773
ω β 0.199 0.6709 0.039 0.4344
a(T c ) 106 cm6 bar mol−2
15.039 1.485
a(T =250K) 106 cm6 bar mol−2
20.113 1.006
b cm3 mol−1
72.38 24.08
The mixing rules are: a =
XX i
b =
X
j
√ xi x j ai a j (1 − ki j )
xi bi
i
where k11 = k22 = 0 and k12 = 0.0867. The expression for the fugacity coefficient is: √ " P # v + (1 + 2)b 2 i xi aik bk P(v − b) a bk Pv ln − 1 − ln − √ − ln ϕk = √ b RT RT a b 2 2bRT v + (1 − 2)b In the limit of x2 → 0, we get the following relation: ! f2 H2,1 = lim = lim (ϕ2 P) x2 →0 x2 x2 →0 The following results are found from Eq. 42: a b P v ln ϕ2
= = = =
a1 b1 P1s = 0.392 bar v1s = 93.035 cm3 mol-1 √ # ! " v1s + (1 + 2)b1 P1s (v1s − b1 ) b2 P1s v1s a1 2a12 b2 = ln − 1 − ln − √ − √ b1 RT RT b1 2 2b1 RT a1 v1s + (1 − 2)b1 46
The value of the fugacity coefficient is:
ln ϕl2
= − − × ×
∴ ϕl2
= =
! 24.08 cm3 mol-1 0.392 bar × 93.035 cm3 mol-1 −1 72.38 cm3 mol-1 83.145 bar cm3 mol-1 K-1 × 250 K 0.392 bar × (93.035 − 72.38) cm3 mol-1 ln 83.145 bar cm3 mol-1 K-1 × 250 K 20.113 × 106 cm6 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 24.08 cm3 mol-1 − 20.113 × 106 cm6 bar mol-2 72.38 cm3 mol-1 h i √ 93.035 + (1 + 2)72.38 cm3 mol-1 i ln h √ 93.035 + (1 − 2)72.38 cm3 mol-1 7.0002 exp(7.0002) = 1097
From Eq. 39 we then find H2,1 = 1097 × 0.392 bar = 430 bar
Question b The partial molar volume is defined as ∂V v2 = ∂n2
!
P,T,n1
Since the equation of state is not explicit in v, It is more convenient to consider how the pressure varies when we vary T , P, n1 , and n2 : ∂P dP = ∂T
!
∂P dT + ∂V V,n1 ,n2
!
∂P dV + ∂n1 T,n1 ,n2
!
∂P dn1 + ∂n2 V,T,n2
At constant P, T , and n1 , we have ∂P dP = ∂V
!
∂P dV + ∂n2 T,n1 ,n2
!
V,T,n1
so under these conditions, v2 = −
(∂P/∂n2 )V,T,n1 (∂P/∂V)T,n1 ,n2 47
dn2 = 0
!
V,T,n1
dn2
We can now evluate the differentials of P in the limit x2 → 0. From Eq. 40: P = ∂P = ∂n2 − = − = ∂P = ∂V = Here,
nRT n2 a(T ) − 2 V − nb V + 2Vnb − n2 b2 (V − nb) n′ RT − nRT (−nb′ − n′ b) (V − nb)2 V 2 + 2Vnb − n2 b2 (2nn′ a + n2 a′ ) − n2 a[2V(n′ b + nb′ ) − (2nn′ b2 + 2n2 bb′)] V 2 + 2Vnb − n2 b2 2 RT (Vn′ + n2 b′ ) (V − nb)2 V 2 n(2n′ a + na′ ) + 2Vn2(bn′ a + nba′ − nb′ a) + bn4 (−ba′ + 2b′ a) V 2 + 2Vnb − n2 b2 2 RT (vn′ + nb′ ) v2 (2n′ a + na′) + 2v(bn′ a + nba′ − nb′ a) + b(−bna′ + 2nb′ a) − n (v − b)2 n v2 + 2vb − b2 2 V 2 + 2Vnb − n2 b2 × 0 − n2 a(2V + 2nb) nRT − − (V − nb)2 V 2 + 2Vnb − n2 b2 2 2a(v + b) RT + − 2 2 n (v − b) n v + 2vb − b2 2
∂n =1 ∂n2 " √ # ∂a ∂ n21 a1 + 2n1 n2 a1 a2 (1 − k12 ) + n22 a2 = = ∂n2 ∂n2 (n1 + n2 )2 √ √ (n1 + n2 )2 2n1 a1 a2 (1 − k12 ) + 2n2 a2 − n21 a1 + 2n1 n2 a1 a2 (1 − k12 ) + n22 a2 2 (n1 + n2 ) = (n1 + n2 )4 i √ √ 1 h 2x1 a1 a2 (1 − k12 ) + 2x2 a2 − 2 x21 a1 + 2x1 x2 a1 a2 (1 − k12 ) + x22 a2 = nh i √ = 2 x1 a1 a2 (1 − k12 )(1 − 2x2 ) − x21 a1 + x2 a2 (1 − x2 ) i h √ = 2x1 −x1 a1 + a1 a2 (1 − k12 )(x1 − x2 ) + x2 a2 ! ∂ n1 b1 + n2 b2 ∂b = = ∂n2 ∂n2 n1 + n2 (n1 + n2 ) b2 − (n1 b1 + n2 b2 ) = (n1 + n2 )2 1 (b2 − x1 b1 − x2 b2 ) = n = x1 (b2 − b1 )
n′ = a′
na′
b′
nb′
In the limit x2 → 0, the expressions reduce to i h √ na′ = 2 −a1 + a1 a2 (1 − k12 ) nb′ = b2 − b1 48
i h √ 2 RT (v1 + b2 − b1 ) v1 2a1 + 2 −a1 + a1 a2 (1 − k12 ) ∂P = − 2 ∂n2 n (v1 − b1 )2 n v21 + 2v1 b1 − b21 i h √ 2v1 b1 a1 + 2b1 −a1 + a1 a2 (1 − k12 ) − a1 (b2 − b1 ) − 2 n v21 + 2v1 b1 − b21 i h √ b1 −b1 2 −a1 + a1 a2 (1 − k12 ) + 2a1 (b2 − b1 ) − 2 n v21 + 2v1 b1 − b21 RT (v1 + b2 − b1 ) = n (v1 − b1 )2 h i h i √ √ √ v21 a1 a2 (1 − k12 ) + v1 2b1 a1 a2 (1 − k12 ) − a1 b2 + b1 −b1 a1 a2 (1 − k12 ) + a1 b2 − 2 2 n v21 + 2v1 b1 − b21 √ a1 a2 (1 − k12 ) RT (v1 + b2 − b1 ) a1 b2 (v1 − b1 ) − 2 = +2 2 2 2 2 n (v1 − b1 ) n v1 + 2v1 b1 − b1 n v21 + 2v1 b1 − b21 ∂P RT 2a1 (v1 + b1 ) = − + ∂V n (v1 − b1 )2 n v2 + 2v1 b1 − b2 2 1
1
Evaluated with the given data: ∂P 83.145 bar cm3 mol-1 K-1 × 250 K × (93.035 + 24.08 − 72.38) cm3 mol-1 = ∂n2 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 83.145 bar cm3 mol-1 K-1 × 250 K = − ∂V 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
v¯ ∞ 2 =
1763 bar/n = 69.5 cm3 mol-1 25.38 bar cm-3 mol/n 49
Question c (method I) The parameter in the two-suffix Margules equation is defined by ln γ2 =
A 2 x RT 1
The activity coefficient γ2 refers to pure component 2 as reference state (γ2 = 1when x2 = 1). In this problem, the properties are given at x2 = 0, and we must change the reference state for component 2. Alternatively, we might think that we could instead consider component 1 and use A 2 x RT 2 ln γ1 ln γ1 = = lim 2 2 x2 →0 x x2 2
ln γ1 = ∴
A RT
This wil, however, also raise a problem because ln γ1 → 0 in this limit. The change of reference state is made with use of the Gibbs-Duhem equation at constant P and T : x1 d ln γ1 + x2 d ln γ2 = 0 x1 A x1 A d ln γ2 = − d ln γ1 = − 2x2 dx2 = − 2(1 − x2 )dx2 x2 RT x2 RT We may note that if we use pure component 2 as reference state, we find ln γ2 =
Zx2
A d ln γ2 = − 2 RT
Zx2
x2 =1
x2 =1
" # 1 2 A 2 A x (1 − x2 )dx2 = − 2 (x2 − 1) − (x2 − 1) = − RT 2 RT 1
We now use pure componnet 1 as reference state: ln γ2∗
=
Zx2
A d ln γ2 = − 2 RT
x2 =0
Zx2
x2 =0
! 1 2 A A (1 − x2 )dx2 = − 2 x2 − x2 = − (2x2 − x22 ) RT 2 RT
In this limit we may therefore write f2 = f2 (x2 = 0)a2 = H2,1 γ2∗ x2 f2 ∴ = H2,1 γ2∗ x2 2A A 2 f2 = ln H2,1 + ln γ2∗ = ln H2,1 − x2 + x ln x2 RT RT 2
(42)
If we consider ln xf22 as function of x2 and expand around x2 = 0 at constant P and T , we get ∂ ln xf2 2 x2 + ... + ∂x 2 x2 =0 P,T,x2 =0 ! ∂ ln H2,1 = ln H2,1 + x2 + ... ref Eq.9.5 ∂x2 P,T,x2 =0
f2 = ln x2
H2,1 x2 ln x2
!
50
(43)
Comparing Eq. 42 with and using Eq. 39, we find ! " ! # ∂ ln H2,1 2A ∂ ∂ ln ϕ2 − = = (ln ϕ2 + ln P) = RT ∂x2 P,T,x2 =0 ∂x2 ∂x2 x2 =0 P,T,x2 =0 Note that the differentiation with respect to x2 reprsents the effect of variation in composition only, and that P and T are held constant (P = P1s ). We must consider both x1 and x2 as variables. If P and T are constants while the composition varies, the volume has to vary. The molar volume of the mixture may be expressed as v = v¯ ∞ ¯∞ ¯∞ v∞ ¯∞ 1 x1 + v 2 x2 = v 1 + (¯ 2 −v 1 )x2
Question c (method II) An alternative expression for the fugacity of N2 is f2 = γ2∗ x2 f20 Since the solubility of N2 in n-butane is small, pure solvent is chosen as reference state (indicated with the asterisk), γ2∗ = 1. The parameter in the two-suffix Margules equation is defined by A A 2 (x1 − 1) = − (1 + x1 )x2 ln γ2∗ = RT RT Combining Eqs. 39 and gives f2 = γ2∗ f20 = ϕ2 P x2 f2 ln = ln γ2∗ + ln f20 + ln ϕ2 + ln P x2 Expanding ln xf22 in powers of x2 and equating the coefficients of the linear terms, gives # " 2A ∂ ln ϕ2 − = RT ∂x2 P=Ps ,x2 =0 1
The expression for ln ϕ2 is bk Pv P(v − b) a ln ϕ2 = − 1 −ln − √ b RT RT 2 2bRT
√ P √ 2 i xi aik bk v + (1 + 2)b − ln + (1 − 2)b a b v
This depends on x2 in a complex way. It is assumed in Eq. that P and T are constant (the v∞ temperature is held constant and the pressure variation gives rise to the term RT2 (P − P1s )). Under these conditions, v has to vary when x2 varies. Using the chain rule, we get d ln ϕ2 ∂ ln ϕ2 ∂ ln ϕ2 dx1 ∂ ln ϕ2 dv = + + dx2 ∂x2 ∂x1 dx2 ∂v dx2 It is understood here that all differentials are at constant P and T so that f2 = γ2∗ x2 H2,1 = ϕ2 x2 P 51
or P H2,1 = ln(ϕ2 P) − lnH2,1
γ2∗ = ϕ2 ln γ2∗ For small values of x2 we may expand
(P s )
f2 = f2 (x2 = 0)a2 = H2,11 γ2∗ x2 f2 (P s ) ∴ = H2,11 γ2∗ x2 f2 2A A 2 (P s ) (P s ) ln = ln H2,11 + ln γ2∗ = ln H2,11 − x2 + x x2 RT RT 2
(9.4)
If we consider ln xf22 as function of x2 and expand around x2 = 0 at constant P and T , we get (P1s ) ∂ ln xf2 x H f2 2,1 2 2 = ln ln x2 + ... + x2 x2 ∂x2 P,T,x2 =0 P,T,x2 =0 (P1s ) ∂ ln H (P s ) 2,1 x2 + ... = ln H2,11 + ∂x2
(9.5)
P,T,x2 =0
Comparing Eq. 42 with and using Eq. 39, we find (P1s ) ! # " 2A ∂ ln H2,1 ∂ ln ϕ2 ∂ − = (ln ϕ2 + ln P) = = RT ∂x2 ∂x2 ∂x2 x2 =0 P,T,x2 =0 P,T,x2 =0
Note that the differentiation with respect to x2 represents the effect of variation in composition only, and that P and T are held constant (P = P1s ). We must consider both x1 and x2 as variables. If P and T are constants while the composition varies, the volume has to vary. The molar volume of the mixture may be expressed as v = v¯ ∞ ¯∞ ¯∞ v∞ ¯∞ 1 x1 + v 2 x2 = v 1 + (¯ 2 −v 1 )x2
52
Chapter 11 Problem 11.1 We consider a solvent (component 1) with a solute (component 2). The change in chemical potential of component 1 as a result of changing the composition and temperature of the liquid mixture is dµ1 (l) = −s1 (l)dT + RT d ln x1 Suppose the liquid mixture is in equilibrium with a solid, which is assumed to be pure solvent. 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 must be equal: −s1 (l)dT + RT d ln x1 = −s1 (s)dT or
∆fus h1 dT (44) T Equation 44 is the basic equation for freezing-point depression. Integration from pure solvent to an actual mixture composition of x1 gives " #T ∆fus h1 1 x1 [ln x1 ]1 = − R T T∗ ! 1 ∆fus h1 1 − ln x1 = − R T T∗ RT d ln x1 = [s1 (l) − s1 (s)] dT = ∆fus s1 dT =
where T ∗ is the melting point of pure solvent. This equation can be solved for T : 1 1 R = ∗− ln x1 T T ∆fus h1 Since the mixture is rich in benzene, and the melting points of the two components are not very different, we may assume that benzene freezes out first, and consider it as the solvent (component 1): 1 8.3145 J mol-1 K-1 1 ln 0.95 = 0.003631 = − T 278.7 K 9843 J mol-1 ∴ T = 275.4 K The phase diagram for the system is shown in Figure 20. 53
360 340
T, K
320 300 280 260 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x1
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 the necessary information for computing the activity coefficients is available, we may assume a nonideal mixture in this case. The expression for T is, by analogy to problem 11.1: 1 1 R 1 R = ∗− ln a1 = ∗ − (ln γ1 + ln x1 ) T T ∆fus h1 T ∆fus h1 In this case, it is not clear which of the components will freeze out first at the given composition. Even if there is much less benzene than n-heptane in the mixture, the freezing point of benzene is so much higer than that of n-heptane that it might still freeze out first. To be sure, we can construct the phase diagram like in Problem 11.1, except that here, we use the van Laar model for the activity coefficients: ln γ1 = ln γ2 = A′
B′ A′ B′
A′ ′
1 + AB′ xx12 B′
2
2 ′ 1 + BA′ xx21 v1 (δ1 − δ2 )2 = RT v2 = A′ v1 v1 = v2
The phase diagram for this system is shown in Figure 21. We see from the figure that x1 = 0.1 is at the benzene-rich side of the eutectic point. We therefore consider benzene as 54
300 280
T, K
260 240 220 200 180 160 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x1
Figure 21: Liquidus curve for the n-heptane-benzene system. Benzene is component 1.
the solvent (component 1) and find: 146.54 K 89 cm3 mol-1 (18.8 − 15.1)2 J cm-3 = -1 -1 T 8.3145 J mol K T -1 3 148 cm mol 243.69 K = A′ = -1 3 T 89 cm mol -1 3 89 cm mol = = 0.6014 148 cm3 mol-1 146.54 K 128.76 K A′ = = = 2 2 ′ T T 1 + 0.6014 0.1 1 + A x1
A′ = B′ A′ B′ ln γ1
ln γ2 =
B′ x2 ′
B
1+
B′ x2 A′ x1
0.9
2 =
243.69 K
= 0.9 2 T 1 + 1.6629 0.1
0.9559 K T
This gives the following expression for the freezing temperature: 1 8.3145 J mol-1 K-1 128.76 K 1 = − + ln 0.1 T 278.7 K T 9843 J mol-1 ! 1 8.3145 8.3145 × 128.76 1 = − ln 0.1 1+ 9843 T 278.7 K 9843 K 1 + 8.3145×128.76 9843 = 200.39 K ∴T = 1 8.3145 − 9843 K ln 0.1 278.7 K
!
Problem 11.9 The condition for equilibrium is f2 (g) = f2 (l) that is, the fugacity of the CO2 (component 2) is the same in gas and liquid phases. At 194.3 K and x2 = 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 been equal to that of the solid, i.e. f2 (g) = f2 (l) = f2 (s) (45) and the saturation pressure would in that case have been P2s = 0.99 bar. In the actual case with x2 = 0.05, the fugacity must be lower than this, the vapour pressure must be lower than 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 f2L = γ2 x2 f2L
(47)
where f2L 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 liquid CO2 . In order to use the information provided by Eq. 45, it is convenient, however, to use pure solid CO2 at T = 194.3 K as the standard state. This gives f2 (l) =
γ2 x2 f2L
=
γ2 x2 f2S
f2L f2S
= f2 (s) = f2S = P2s
where f2S is the fugacity of (pure) solid CO2 at T = 194.3 K. Combining Eqs. 45–47 gives f2L f2S L f γ2 x2 P2s f2S 2
P2 = γ2 x2 f2S
for x2 = 0.05
P2s =
for x2 = 0.25
We first determine the activity coefficient from the data at x2 = 0.25: γ2 (x2 = 0.25) =
1 f2S 0.56 = 2.24 = L x2 f2 0.25
The activity coefficient at x2 = 0.05 must be estimated from this value, which means that we can use a model with at most one parameter. One such model is the two-suffix Margules equation: A 2 x RT 1 ln γ2 ln 2.24 A = 2 = = 1.4337 ∴ RT 0.752 x1 ln γ2 =
Hence, ln γ2 (x2 = 0.05) =
A 0.952 = 1.4337 × 0.952 = 1.2939 RT
∴ 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 × 56
1 = 0.322 bar 0.56
Chapter 12 Problem 12.3 The thermodynamic condition for equilibrium is µ′A = µ′′A
(48)
where µA is the chemical potential of the alcohol. Prime and double prime refer to the hexane phase and DMSO phase, respectively. The chemical potential in each phase is µA = µoA + RT ln xA + RT ln γA
(49)
where µoA is the chemical potential in a reference state, xA is the mole fraction of the alcohol, and γA is the activity coefficient of the alcohol. We note that the excess Gibbs energy is of the form gE = AxA xB where A is a constant and xB is the mole fraction of the solvent (hexane or DMSO). This leads 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 ! x′A RT ln ′′ = A′′ 1 − x′′A 2 − A′ 1 − x′A 2 xA ′ ! xA A′′ − A′ = ln K = ln lim x′A →0 x′′A RT ′′
(51)
xA →0
The problem is now to find the constants A′′ and A′ at 30◦ C and 100 bar. For this, we use the fact that Gibbs energy depends on temperature and pressure as dG = VdP − S dT For the excess properties, we then find dgE = vE dP − sE dT 57
(52)
This can be used to compute the difference in excess Gibbs energy when the conditions are changed 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 for sE available. Instead we use the GibbsHelmholtz equation for the temperature variation: d(gE /T ) = hE d(1/T ) The effect of the pressure change is (assuming constant excess volumes over the pressure range in question): ∆gE = vE ∆P The effect of the temperature change is (assuming constant excess enthalpies over the temperature range in question): ! ! gE 1 E ∆ =h ∆ T T Inserting the given numbers leads to for the hexane phase and the DMSO phase at the actual pressure: gE = 2400x′A x′H J mol-1 + 16x′A x′H cm3 mol-1 × 990 N cm-2 = 2558.4x′A x′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: g
E
∴ A′
= = =
gE =
∴A
′′
= =
303 K 303 K J mol × + 4800x′A x′H J mol-1 × 1 − 273 K 273 K -1 ′ ′ 2312.1xA xH J mol 2312.1 J mol-1 ! 303 K 303 K -1 -1 ′′ ′′ ′′ ′′ + 600xA xD J mol × 1 − 221.0xA xD J mol × 273 K 273 K -1 ′′ ′′ 179.4xA xD J mol 179.4 J mol-1
2558.4x′A x′H
-1
!
Used in Eq. 51, we then find: (179.4 − 2312.1) J mol-1 = −0.8465 8.3145 J mol-1 K-1 × 303 K ∴ K = 0.429
ln K =
Problem 12.4 The phase diagram (P − T -diagram) is shown in Figure 22. The problem here is to determine the temperature at which the solid-liquid equilibrium pressure is 200 bar (indicated with the dashed line in Figure 22. If we consider equilibrium between solid and liquid at a certain P and T , given by equality in chemical potentails, µl (P, T ) = µs (P, T ) 58
200.000 150.000
P, bar
100.000 50.000
0.008 0.006 0.004 265
270
275
280
285
290
295
300
T, K
Figure 22: P − T diagram for benzene.
the equilibrium at P + dP and T + 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 = vl dP − sl dT and similarly for the solid phase. This gives the following differential equation for the liquidus curve: (53) vl − vs dP = sl − s s dT
The molar volumes at 278.7 K may be found from the given densities: l
v
vs vl − vs
M 78.1 g mol-1 = = = 87.65 cm3 mol-1 l -3 ρ 0.891 g cm 78.1 g mol-1 M = = 77.33 cm3 mol-1 = ρ s 1.010 g cm-3 = (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 fusion divided by the melting temperature. The enthalpy on fusion is not given, but the enthalpies on vaporization and sublimation may be deducted from the vapour pressure curves by comparing the given vapour-pressure equations with the Clausius-Clapeyron equation: ∆vap h 1 1 ln P = ln P − − o R T T o
59
!
This version of the Clausius-Clapeyron equation assumes that ∆vap h is constant over the temperature range in question, which is consistent with the given vapour-pressure equations. We find ∆vap h = 1785 K R ∆sub h = 2310 K R Enthalpy is a state function, and at the triple point we must have ∆fus h = ∆sub h−∆vap h = (2310−1785) K×83.145 bar cm3 mol-1 K-1 = 43651 bar cm3 mol-1 (55) Consistently with the assumption that ∆sub h and ∆vap h are constant over the temperature range, we assume that also ∆fus h is constant. This gives for the entropy of fusion at a temperature T : ∆fus h 43651 bar cm3 mol-1 = (56) ∆fus s = T T The resukts 54 and 56 inserted into Eq. 53 gives: 10.32 cm3 mol-1 dP = 43651 bar cm3 mol-1 ∴ P = Po +
dT T
T 43651 bar cm3 mol-1 T o ln = P + 4230 bar × ln To To 10.32 cm3 mol-1
The fixed points Po and T o are given by the normal melting point: Po = 1.013 bar and T o = 278.7 K, leading to P = 1.013 bar + 4230 bar × ln
T 278.7 K
At 200 bar, we then find: T 278.7! K (200 − 1.013) bar ∴ T = 278.7 Kexp = 292.1 K 4230 bar
200 bar = 1.013 bar + 4230 bar × ln
Problem 12.7 The stability of the mixture with respect to phase separation is expressed by ∆mix g (see Figures 12-14 and 12-15 in the textbook). The Gibbs energy on mixing for a binary mixture is ∆mix g = 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 × x1 x2 60
At a pressure P and 273 K, the excess Gibbs energy is gE (P, 273 K) = gE (1 atm, 273 K) +
ZP
vE dP
1 atm
= RT × 1.877 × x1 x2 + 3
ZP
x1 x2 (4.026 − 0.233 ln P)dP
1 atm -1 -1
= 82.0578 atm cm mol K × 1.877 × 273 K × x1 x2 h i + 4.026 cm3 mol-1 (P − 1 atm) − 0.233 cm3 mol-1 (P ln P − P + 1 atm) x1 x2
= (42048 − 4.026 − 0.233) atm cm3 mol-1 × x1 x2 + (4.026 + 0.233 − 0.233 ln P) cm3 mol-1 × P × x1 x2 = (42044 atm + 4.259P − 0.233P ln P) × x1 x2 cm3 mol-1
(58)
Eq. 58 in 57 gives at 273 K: ∆mix g = 22402 atm cm3 mol-1 (x1 ln x1 + x2 ln x2 ) + (42044 atm + 4.259P − 0.233P ln P) × x1 x2 cm3 mol-1 This expression is plotted for two values of P, P = 10 atm and P = 2000 atm, in Figure 23. We observe that the function is symmetric around x1 = x2 = 0.5. To find the minimum 0
∆mixg, P=2000 atm
∆mixg, J mol-1
-200 -400 -600
∆mixg, P=10 atm
-800 -1000 -1200 -1400
∆mixg
id
-1600 -1800
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x1
Figure 23: Gibbs energy of (regular) mixing. pressure at which the mixture separates, we have to find the solution to ∂2 ∆mix g = 0 at x1 = 0.5 ∂x21 With gE (P, 273 K) of the form gE (P, 273 K) = A(P)x1 x2 where A(P) = (42044 atm + 4.259P − 0.233P ln P) cm3 mol-1 61
we first find
∂∆mix g = RT [ln x1 − ln(1 − x1 )] + A(1 − 2x1 ) ∂x1
and then
∂2 ∆mix g RT = − 2A 2 x1 (1 − x1 ) ∂x1
Setting this expression equal to 0 at x1 = 0.5 gives 82.0578 atm cm3 mol-1 K-1 × 273 K RT = = 44804 atm cm3 mol-1 A= 2 2 2 × 0.5 2 × 0.5 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 solution P = 1045 atm =1059 bar. At P = 1500 atm, the Gibbs energy of mixing will have two minima (located symmetrically around x1 = 0.5), given by ∂∆mix g = RT [ln x1 − ln(1 − x1 )] + A(P = 1500 atm)(1 − 2x1 ) = 0 ∂x1 This gives the following solution for x1 : A(P) = (42044 atm + 4.259 × 1500 atm − 0.233 × 1500atm ln 1500)cm3 mol-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 finally x′′1 = 0.63
x′1 = 0.37,
62
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