Question 2. A mixture of 1-butanol (1) + water (2) forms an azeotrope where x₁ = 0.807 and T = 335.15 K. Assuming the following relations apply for the activity coefficients: In y₂ = A(x₁)² In Y₁ = A(x₂)² Given: Psat = 8.703 kPa and Pat = 21.783 kPa (a) Derive an expression for G/RT as a function of A and x₁ (b) Determine the numerical value of the constant A (c) Using modified Raoult's law, determine the pressure at x = 0.807 and T = 335.15 K.
Question 2. A mixture of 1-butanol (1) + water (2) forms an azeotrope where x₁ = 0.807 and T = 335.15 K. Assuming the following relations apply for the activity coefficients: In y₂ = A(x₁)² In Y₁ = A(x₂)² Given: Psat = 8.703 kPa and Pat = 21.783 kPa (a) Derive an expression for G/RT as a function of A and x₁ (b) Determine the numerical value of the constant A (c) Using modified Raoult's law, determine the pressure at x = 0.807 and T = 335.15 K.
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
Related questions
Question
I need solve this question of thermo
![αij
In K(T₂)- In K(T₁) = -
[],
GE
Jp.x
Ki Yi/xi
=
K₁y₁/x
RT
GE
RT
Π(8P)
GR = RT In fi
Po
Pr
In (Bº+wB¹)
Tr
K
A,Gº=A,H298 + R
"
AMig = Mig_
"mix
0.422
T,1.6
Hi G + RT In yixi
HT(T) + RT In (f)
B® = 0.083
F=2-+N-r-s
Y₁P = x₁Y₁Psat
=
dT+
YiPsat (T)
Ypsat (T)
Δ,Η
71
-(2-7)
G
RT
R
= Ax₁x₂
d² (GE/RT)
dx²
formula sheet
-Σy. Mg
Y₁
= (A21x1 + A12X₂)X1X2
M₁ = M+x₂
A₁2A21
A12X1 + A21x2,
ƏM
Lap Tx
SE = AmixS + R
(dGtot)T,P ≤ 0
1
X1X2
R= 8.314-
298
dp-x₁dM₁ = 0
dM
² dx₁
X1X2
mol. K
x₁ In xi
4-C
R
(const T, P)
-RT In K = A,Gº
A,GO A,HO TA,SO
R
R
R
-dT-TA-S298 - RT
=
K=
In Y₁ = A(x₂)²
Given: Psat = 8.703 kPa and Psat = 21.783 kPa
fi
ΦΙΞ P
fi (P): = psat psat exp
A,Cp dT
RT
(1+x²
298
y₁P = x₁psat
B¹ = 0.139-
[V(P-pat)
RT
H = G+ RT In y¡ = [¡(T) + RT ln(y;P)
fi
GE=G₁ - Gid = RT In- = RT In Yi
xifi
= M (T.P.)
M (T,P) =
M =
d²(AG/RT)
dx²
d In Y₁
dx₁
Iny₁ = Ax²
In y₂ = Ax
In y₁=x[A12 + 2(A21-A12)x1]
In y2 = x²421 +2(A12 - A21)x2]
In y₁= A12 (1+
A₁2x1
A21x2
>
0.172
T.4.2
M₂M-X1
GE=AmixG-RT
x₁ M₁
dM
dx₁
x₁ In x₁
>0
1
X1
1 bar = 105 Pa
1m³= 106 cm³
(const T,P)
(const T, P)
Question 2.
A mixture of 1-butanol (1) + water (2) forms an azeotrope where x₁¹² = 0.807
and T = 335.15 K. Assuming the following relations apply for the activity coefficients:
In y₂ = A(x₁)²
(a) Derive an expression for GE/RT as a function of A and x₁
(b) Determine the numerical value of the constant A
(c) Using modified Raoult's law, determine the pressure at x = 0.807 and T = 335.15 K.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F207d9547-25b3-4a81-9737-9c60140c78ef%2Fdb96518c-4d70-4f21-a0bf-e211ec2d8368%2Fjyt5svl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:αij
In K(T₂)- In K(T₁) = -
[],
GE
Jp.x
Ki Yi/xi
=
K₁y₁/x
RT
GE
RT
Π(8P)
GR = RT In fi
Po
Pr
In (Bº+wB¹)
Tr
K
A,Gº=A,H298 + R
"
AMig = Mig_
"mix
0.422
T,1.6
Hi G + RT In yixi
HT(T) + RT In (f)
B® = 0.083
F=2-+N-r-s
Y₁P = x₁Y₁Psat
=
dT+
YiPsat (T)
Ypsat (T)
Δ,Η
71
-(2-7)
G
RT
R
= Ax₁x₂
d² (GE/RT)
dx²
formula sheet
-Σy. Mg
Y₁
= (A21x1 + A12X₂)X1X2
M₁ = M+x₂
A₁2A21
A12X1 + A21x2,
ƏM
Lap Tx
SE = AmixS + R
(dGtot)T,P ≤ 0
1
X1X2
R= 8.314-
298
dp-x₁dM₁ = 0
dM
² dx₁
X1X2
mol. K
x₁ In xi
4-C
R
(const T, P)
-RT In K = A,Gº
A,GO A,HO TA,SO
R
R
R
-dT-TA-S298 - RT
=
K=
In Y₁ = A(x₂)²
Given: Psat = 8.703 kPa and Psat = 21.783 kPa
fi
ΦΙΞ P
fi (P): = psat psat exp
A,Cp dT
RT
(1+x²
298
y₁P = x₁psat
B¹ = 0.139-
[V(P-pat)
RT
H = G+ RT In y¡ = [¡(T) + RT ln(y;P)
fi
GE=G₁ - Gid = RT In- = RT In Yi
xifi
= M (T.P.)
M (T,P) =
M =
d²(AG/RT)
dx²
d In Y₁
dx₁
Iny₁ = Ax²
In y₂ = Ax
In y₁=x[A12 + 2(A21-A12)x1]
In y2 = x²421 +2(A12 - A21)x2]
In y₁= A12 (1+
A₁2x1
A21x2
>
0.172
T.4.2
M₂M-X1
GE=AmixG-RT
x₁ M₁
dM
dx₁
x₁ In x₁
>0
1
X1
1 bar = 105 Pa
1m³= 106 cm³
(const T,P)
(const T, P)
Question 2.
A mixture of 1-butanol (1) + water (2) forms an azeotrope where x₁¹² = 0.807
and T = 335.15 K. Assuming the following relations apply for the activity coefficients:
In y₂ = A(x₁)²
(a) Derive an expression for GE/RT as a function of A and x₁
(b) Determine the numerical value of the constant A
(c) Using modified Raoult's law, determine the pressure at x = 0.807 and T = 335.15 K.
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