SVNAS 8e ISM Chapter 10

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SVNAS 8th Edition Annotated Solutions
Chapter 10
Siig T , P   Siig T , pi   Siig T , P   R ln yi
N
N
N
i 1
i 1
i 1
ig
Smix
 S ig   yi Siig   R yi ln yi  R yi ln
1
yi


Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10









Siig T , P   Siig T , pi   Siig T , P   R ln yi
N
N
N
i 1
i 1
i 1
ig
Smix
 S ig   yi Siig   R yi ln yi  R yi ln
Updated 4/5/2017
1
yi
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SVNAS 8th Edition Annotated Solutions
Chapter 10

Wideal  H  T S







N
ig
Smix
 R yi ln
i 1
1
yi




Updated 4/5/2017

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SVNAS 8th Edition Annotated Solutions
T1 (K)
448.15
T2 (K)
308.15
T2
ICPH 
R
Cp
A
1.702
dT  A  T2  T1  
T1
T1 (K)
448.15
T2 (K)
308.15
T2
ICPS 
A
1.702
Chapter 10
C (1/K2)
D (K2)
B (1/K)
9.08E-03 -2.16E-06 0.00E+00



T2 (K)
308.15
T2
ICPH 
C
2
2
2
1
A
1.131
B
2
1
2
2

 T12 
T1
T1 (K)
448.15
T2 (K)
308.15
T2
ICPS 

T1
A
1.131
T
dT  A ln  2
RT
 T1
Cp

ICPH (K)
-1.06E+03

H (J/mol)
-8843


C 2
D 2
2
T2  T1 2
  B T2  T1   T2  T1 
2
2


S (J/(mol K))
-23.45
ICPS
-2.821





Updated 4/5/2017

 1 1
C 3
T2  T13  D   
3
 T2 T1 
C (1/K2)
D (K2)
B (1/K)
1.92E-02 -5.56E-06 0.00E+00

S (J/(mol K))
-14.92
ICPS
-1.794
D 2
T2  T1 2
2
C (1/K2)
D (K2)
B (1/K)
1.92E-02 -5.56E-06 0.00E+00
 R dT  AT  T   2 T
Cp

 T12 
1
T1 (K)
448.15

C (1/K2)
D (K2)
B (1/K)
9.08E-03 -2.16E-06 0.00E+00
 T2 
T1
H (J/mol)
-5614
 1 1
B 2
C 3
T2  T12 
T2  T13  D   
2
3
 T2 T1 
 RT dT  A ln  T   B T  T   2 T
Cp
ICPH (K)
-6.75E+02

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SVNAS 8th Edition Annotated Solutions
Chapter 10


Wideal  H  T S
t  work required  
Wideal
Ws
 0.05
Ws  20Wideal  20  H  T S 
N
N
N
i 1
i 1
i 1
ig
Smix
 S ig   yi Siig   R yi ln yi  R yi ln
1
yi

Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
Ϻ
Ϻ
Ϻ
Ϻ
Ϻ
Ϻ
Ϻ
Ϻ
Ϻ
Ϻ
Ϻ
Ϻ
Ϻ
Updated 4/5/2017
Ϻ
Ϻ
Ϻ
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SVNAS 8th Edition Annotated Solutions
Chapter 10
M1  M  x2
dM
dM
 M  1  x1 
dx1
dx1
M 2  M  x1
dM
dx1
V
1


1
a0  a1 x1  a2 x12
dV
a1  2a2 x1

2
dx1
 a0  a1 x1  a2 x12 
V1  V  1  x1 
V1 
1  x1  a1  2a2 x1 
dV
1


2
dx1 a0  a1 x1  a2 x1  a  a x  a x 2 2
0
1 1
2 1
a0  a1 x1  a2 x12  a1  2a2 x1  a1 x1  2a2 x12
a
0
V1 
2
a0  a1  2a1 x1  2a2 x1  3a2 x12
a
0
V2  V  x1
V2 
 a1 x1  a2 x12 
 a1 x1  a2 x12 
2
x1  a1  2a2 x1 
dV
1


2
dx1 a0  a1 x1  a2 x1  a  a x  a x 2 2
0
1 1
2 1
a0  2a1 x2  3a2 x12
a
0
Updated 4/5/2017
 a1 x1  a2 x12 
2
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SVNAS 8th Edition Annotated Solutions
Chapter 10
   nM  
Mi  

 ni  T , P ,n ji
   nM  
M1  

 n1  T , P ,n2 ,n3
M2
M3
M1
M2
M3
x1M1  x2 M 2  x3M 3
lim x1 0 M1
lim x2 0 M 2
lim x3 0 M 3
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Updated 4/5/2017
Chapter 10
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SVNAS 8th Edition Annotated Solutions
Chapter 10
and
   nV  
Vi  

 ni  T , P ,n j  i
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
V1 
V1 
V1 
V1 
V1 
V2 
V2 
V2 
V2 
V2 
x1V1  x2V2
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
N
 x dV
i 1
i
i
Chapter 10
0
x1dV1  x2 dV2  0
V1 
V1 
V1 
V2 
V2 
V2 
dV1 
dV2 
x1dV1  x2 dV2  0
x1dV1  x2 dV2
x1dV1  x2 dV2
dV1 
dV2 
dV1
dx1
dV2
dx1
 2  40 1  42 1  0
x1 1
 2  0   42  0   0
x1  0
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
Total or Partial Molar Volume (cm 3/mol)
SVA problem 11.13(e)
140
V1,  128
130
V1  120
V1
120
110
100
V
V2,  85
90
80
V2  70
V2
70
60
0
0.2
0.4
0.6
0.8
1
x1
   nH  
Hi  

 ni  T , P ,n ji


n1 
n2 
nH  n1  a1  b1
  n2  a2  b2

n1  n2 
n1  n2 


   nH  

n1 
n1n2
n22
H1  

a

b

b

b

 1 1
 1
2
2
2
n1  n2 
 n1  n2 
 n1  n2 
 n1  T , P ,n2 
H1   a1  b1 x1   b1 x1 x2  b2 x22
H 2   a2  b2 x2   b2 x1 x2  b1 x12
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
x1 H1  x2 H 2  x1  a1  b1 x1   b1 x12 x2  b2 x1 x22  x2  a2  b2 x2   b2 x1 x22  b1 x12 x2
x1 H1  x2 H 2  x1  a1  b1 x1   x2  a2  b2 x2 
H1  H  x2
dH
dx1
H 2  H  x1
dH
dx1
Ϻ
Ϻ
Ϻ
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
P
ln i 

Chapter 10
Zi  1
dP
P
0


Z
0.9850
0.9700
0.9420
0.9130
0.8850
0.8690
0.7650
0.7620
0.8240
0.9100
(Z-1)/P
(1/bar)
-0.001500
-0.001500
-0.001450
-0.001450
-0.001438
-0.001310
-0.001175
-0.000793
-0.000440
-0.000180
ln(Phi)
-0.0150
-0.0300
-0.0595
-0.0885
-0.1174
-0.1449
-0.2691
-0.3675
-0.4292
-0.4602
Phi
0.9851
0.9704
0.9422
0.9153
0.8893
0.8652
0.7641
0.6925
0.6510
0.6312
f
(bar)
9.85
19.41
37.69
54.92
71.14
86.52
152.81
207.74
260.42
315.58
0.000000
-0.000200
0
100
200
300
400
500
600
-0.000400
(Z - 1)/P (1/bar)
P
(bar)
10
20
40
60
80
100
200
300
400
500
-0.000600
-0.000800
-0.001000
-0.001200
-0.001400
-0.001600
Pressure (bar)

Updated 4/5/2017

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SVNAS 8th Edition Annotated Solutions
T (K)
P (bar)
423.15
Chapter 10
Tc (K) Pc (bar)
10
304.2
73.83

0.224
Tr
Pr
B0
B1

1.3910
0.1354
-0.1659
0.0960
0.9860
B0  0.083 
B1  0.139 
0.422



Tr1.6
  exp  B 0   B 1
0.172
 TP
r
r



Tr4.2
Pale blue boxes are input fields, pink box is the final output.
T (K)
P (bar)
423.15
200
Tc (K) Pc (bar)
304.2
73.83

Tr
Pr
0.224 1.391026 2.708926
Table Points
Tr
Tr (1)
Tr (1)
Tr (2)
Tr (2)
Pr(1)
Pr(2)
Pr(1)
Pr(2)
1.30
1.30
1.40
1.40
Pr
2.00
3.00
2.00
3.00
Interpolated for Pr=2
Interpolated for Pr=3
Interpolated to Tr, Pr
Final Value of 
Updated 4/5/2017

0


0.7345
0.6383
0.7925
0.7145
1.1776
1.2853
1.1858
1.2942
0.7873
0.7077
0.7308
0.7699
1.1851
1.2934
1.2619
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SVNAS 8th Edition Annotated Solutions
P
(bar)
10
20
40
60
80
100
200
300
400
500
T (K)
600
Chapter 10
Using appropriate
method
f
Percent
Phi
(bar) difference
0.9860
9.86
0.09%
0.9723 19.45
0.19%
0.9453 37.81
0.33%
0.9191 55.15
0.41%
0.8936 71.49
0.49%
0.8689 86.89
0.43%
0.7699 153.98
0.76%
0.7074 212.22
2.16%
0.6626 265.04
1.78%
0.6422 321.10
1.75%
P (bar)
300
Tc (K)
430.8
Pc (bar)
78.84
Tr
1.30
1.30
1.40
1.40
Pr
3.00
5.00
3.00
5.00
Using 2nd virial
coefficient for all
f
Percent
Phi
(bar) difference
0.9860
9.86
0.09%
0.9723 19.45
0.19%
0.9453 37.81
0.33%
0.9191 55.15
0.41%
0.8936 71.49
0.49%
0.8689 86.89
0.43%
0.7549 150.98
-1.20%
0.6559 196.77
-5.28%
0.5699 227.96
-12.46%
0.4952 247.60
-21.54%

0.245
Tr
1.3928
Pr
3.8052
Table Points
Tr (1)
Tr(1)
Tr(2)
Tr(2)
Pr(1)
Pr(2)
Pr(1)
Pr(2)
0
Z
0.6344
0.7358
0.7202
0.7761
Interpolated Values
Final Values
Z
3
V (cm /mol)
0.7891
131.2
0.7378
R
H /(RT c)
R
H (J/mol)
1
Z
0.2079
0.0875
0.2397
0.1737
0.2092
-2.20103
-7883.38
R 0
(H ) /(RT c)
-2.274
-2.825
-1.857
-2.486
R 1
(H ) /(RT c)
-0.3
-1.066
-0.044
-0.504
-2.1382
-0.2567
R
S /R
R
S (J/(mol K))
R 0
(S ) /R
-1.299
-1.554
-0.99
-1.303
-1.1367
-1.25615
-10.4436
R 1
(S ) /R
-0.481
-1.147
-0.29
-0.73

0.6383
0.5383
0.7145
0.6237

1.2853
1.3868
1.2942
1.4488
0
-0.4876
1
0.6722

f (bar)
0.723998
217.1995
R
-0.32419
0.723111
216.9334
G /(RT)

f (bar)

ln i 
1.3542

GiR H R  TS R

RT
RT


Updated 4/5/2017

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SVNAS 8th Edition Annotated Solutions
Chapter 10


T (K)
P (bar)
553.15
20
Tc (K)

Pc (bar)
417.9
40
0
0.194
1
Tr
Pr
B
B

1.3236
0.5000
-0.1865
0.0860
0.9379

 TP
B 0  0.083 
B1  0.139 
0.422
Tr1.6
0.172


  exp  B 0   B 1
r
r



Tr4.2
Pale blue boxes are input fields, pink boxes are the final output.

Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
T (K)
P (bar)
553.15
100
Chapter 10

Tc (K) Pc (bar)
417.9
40
Tr
0.194 1.323642
Pr
2.5
Table Points
Tr
Tr (1)
Tr (1)
Tr (2)
Tr (2)
Pr(1)
Pr(2)
Pr(1)
Pr(2)
1.30
1.30
1.40
1.40
2.00
3.00
2.00
3.00
Interpolated for Pr=2
Interpolated for Pr=3
Interpolated to Tr, Pr
Final Value of 


Updated 4/5/2017

 



0.7345
0.6383
0.7925
0.7145
1.1776
1.2853
1.1858
1.2942
0.7482
0.6563
0.7023
0.7314
1.1795
1.2874
1.2335
0
Pr

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SVNAS 8th Edition Annotated Solutions
fi  isat Pi sat
Chapter 10

 V sat P  P sat
i
i
exp 

RT

 



T (K)
P (bar)
383.15
5.267
Tc (K)

Pc (bar)
511.8
45.02
0
0.196
1
Tr
Pr
B
B

0.7486
0.1170
-0.5876
-0.4412
0.9000

 TP
B 0  0.083 
B1  0.139 
0.422
Tr1.6
0.172

  exp  B 0   B 1

r
r



Tr4.2
Pale blue boxes are input fields, pink boxes are the final output.
 107.5 cm3 mol-1  275  5.267  bar 

fi  0.9000 *5.267 bar*exp 
 83.145 cm3 bar mol-1 K -1 *383.15 K 


fi  11.78 bar
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
fi
fi sat

fi
isat Pi sat

 V sat P  P sat
i
i
 exp 

RT

 



fi
fi sat
 19.65 cm3 mol-1 150  4.76  bar 
  1.085
 exp 
 83.145 cm3 bar mol-1 K -1 * 423.15 K 


Gi  i T   RT ln fi
G final  Ginitial  i T   RT ln f final  i T   RT ln f initial
 f final 
G final  Ginitial  RT ln 

 finitial 
f final
finitial
 G final  Ginitial 
 G 
 H S 
 exp 
 exp 

  exp 

RT
R 
 RT 
 RT




Updated 4/5/2017

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SVNAS 8th Edition Annotated Solutions
Chapter 10


f final
finitial
f final
finitial

2775.08
31.396 
 H S 

 exp 

 exp 



R 
 RT
 8.3145* 673.15 8.3145 
 exp  0.4958  3.7761  0.0376
Gi  i T   RT ln fi
G final  Ginitial  i T   RT ln f final  i T   RT ln f initial
 f final 
G final  Ginitial  RT ln 

 finitial 
f final
finitial
 G final  Ginitial
 exp 
RT


 G 
 H S 
 exp 

  exp 

R 
 RT 
 RT










f final
finitial
f final
finitial
758.6
6.397 
 H S 

 exp 


  exp 

R 
 RT
 1.986 *1259.7 1.986 
 exp  0.30323  3.2210   0.0541

 V sat P  P sat
i
i
fi  isat Pi sat exp 

RT

Updated 4/5/2017

 


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SVNAS 8th Edition Annotated Solutions
Chapter 10


T (K)
P (bar)
309.2
1.013
Tc (K)

Pc (bar)
469.7
33.7
0
0.252
1
Tr
Pr
B
B

0.6583
0.0301
-0.7408
-0.8568
0.9573

 TP
B 0  0.083 
B1  0.139 
0.422
Tr1.6
0.172


  exp  B 0   B 1
r
r



Tr4.2
Pale blue boxes are input fields, pink boxes are the final output.
 119.4 cm3 mol-1  200  1.01 bar 

fi  0.9573*1.013 bar*exp 
 83.145 cm3 bar mol-1 K -1 *309.2 K 


fi  2.4 bar
fi  isat Pi sat

 V sat P  P sat
i
i
exp 

RT

 



Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
T (K)
Chapter 10
P (bar)
266.3

Tc (K) Pc (bar)
1
417.9
40
0.194
Tr
Pr
B0
B1

0.6372
0.0250
-0.7848
-1.0025
0.9623

 TP 
B 0 0.083
B1 0.139
fi  isat Pi sat
0.422


Tr1.6
  exp  B0   B1
0.172
r
r
Tr4.2

 V sat P  P sat
i
i
exp 

RT

 


 82.0 cm3 mol-1  200  1.01 bar 

fi  0.9623 *1.013 bar*exp 
 83.145 cm3 bar mol-1 K -1 * 266.3 K 


fi  2.04 bar
fi  isat Pi sat

 V sat P  P sat
i
i
exp 

RT

 



Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
T (K)
Chapter 10

P (bar) Tc (K) Pc (bar)
266.9
1.01325
420
40.43
0.191
Tr
Pr
B0
B1

0.6355
0.0251
-0.7887
-1.0158
0.9620

 TP 
B 0 0.083
B1 0.139
0.422
Tr1.6
0.172


  exp  B0   B1
r
r
Tr4.2
Pale blue boxes are input fields, pink boxes are the final output.
fi  isat Pi sat

 V sat P  P sat
i
i
exp 

RT

 


 91.43 cm3 mol-1  200  1.01 bar 

fi  0.9620 *1.013 bar*exp 
 83.145 cm3 bar mol-1 K -1 * 266.9 K 


fi  2.21 bar

Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Psat (bar)
T (K)
473.15
Chapter 10
Tc (K)
22.27
536.4
Tr
B0
B1
0.8821
-0.4328
-0.1523

Pc (bar)
54.72
Vc (cm3/mol)
0.222
Vsat (cm3 /mol)

P 
  exp  B0   B1 r 
T
0.172
r 

B1 0.139
4.2


Tr
Above P sat
fi  isat Pi sat

 V sat P  P sat
i
i
exp 
RT



0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
0.422
Tr1.6
0.293
P (bar)
122.7
Below P sat
B 0 0.083
Zc
239.0
 


f (bar)
1.000
0.995
0.990
0.986
0.981
0.976
0.971
0.967
0.962
0.957
0.953
0.948
0.944
0.939
0.935
0.930
0.926
0.921
0.917
0.912
0.00
0.50
0.99
1.48
1.96
2.44
2.91
3.38
3.85
4.31
4.76
5.22
5.66
6.10
6.54
6.98
7.40
7.83
8.25
8.67
20
1.0
18
0.9
16
0.8
14
0.7
12
0.6
10
0.5
8
0.4
6
0.3
4
0.2
2
0.1
0
Fugacity Coefficient
Fugacity (bar)
Fugacity and Fugacity Coefficient of Chloroform at 200 °C
0.0
0
5
10
15
20
25
30
35
40
Pressure (bar)
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
Spreadsheet for computing fugacity coefficients in a binary mixture (Applied to Problem 11.25)
T (K)
P (bar)
423.15
30
Vc
Mole
3
Fraction
Name
Tc (K)
Pc (bar) (cm /mol)
Zc
Species 1 Ethylene
0.35
282.3
50.4
131
0.281
Species 2 Propylene
0.65
365.6
46.65
188.4
0.289
Cross-Parameters
kij
Vc
Tc (K)
Pc (bar) (cm3/mol)
0 321.26139
48.19
158.0
2nd Virial Coefficients (cm3/mol)
B11
-60
B22
-159
B12
-99
12 (cm /mol)
3
ln ˆ
1
ln ˆ2
ij 
Pcij 
20.83
 ˆ1
0.957
0.875
i   j
2
Zcij RTcij
Tr
1.4989
1.1574
Pr
0.5952
0.6431
B0
-0.1378
-0.2510
B1
0.1076
0.0459
0.1135
Tr
1.3172
Pr
0.6225
B0
-0.1886
B1
0.0849

Tcij  TciTcj 1 kij 
B 0  0.083 
0.422
Zci  Zcj
B1  0.139 
0.172
Zcij 
Vcij
2
Tr1.6
Tr4.2
3
 V 1/3  Vcj1/3 
Vcij   ci


2


-0.04351
-0.13371
ˆ 2
Zc
0.2850
0.087
0.14

Bij 
RTcij
Pcij
B
0
 ij B1 
P
P
 B11  y22 12   RT
 B11  y22  2 B12  B11  B22  
RT
P
P
ln ˆ2 
 B22  y12 12   RT
 B22  y12  2 B12  B11  B22  
RT
ln ˆ1 
fˆethylene  ˆethylene yethylene P  0.957*0.35*30 bar  10.05 bar
fˆpropylene  ˆpropylene y propylene P  0.875*0.65*30 bar  17.06 bar
fˆiid  yi fi  yii P
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
T (K)
P (bar)
423.15
30
Chapter 10
Tc (K)

Pc (bar)
282.3
50.4
0
0.087
1
Tr
Pr
B
B

1.4989
0.5952
-0.1378
0.1076
0.9503

 TP
B 0  0.083 
0.422
B  0.139 
0.172
1
Tr1.6

  exp  B 0   B 1




r
r
Tr4.2
Pale blue boxes are input fields, pink boxes are the final output.
T (K)
P (bar)
423.15
30
Tc (K)

Pc (bar)
365.6
46.65
0
0.14
1
Tr
Pr
B
B

1.1574
0.6431
-0.2510
0.0459
0.8729

 TP
B 0  0.083 
0.422
B1  0.139 
0.172
Tr1.6

  exp  B 0   B 1

r
r



Tr4.2
Pale blue boxes are input fields, pink boxes are the final output.
id
fˆethylene
 yethyleneethylene P  0.35* 0.9503*30 bar  9.98 bar
id
fˆpropylene
 y propylene propylene P  0.65* 0.8729 *30 bar  17.02 bar
φ
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10

P 
1 N N
ln ˆk 
B

yi y j  2 ik   ij  
 kk

RT 
2 i 1 j 1

 ik   ki  2 Bik  Bii  Bkk
 ij   ji  2 Bij  Bii  B jj
 ii   jj   kk  0
P
ln ˆ1 
B11  y2212  y3213  y2 y3 12  13   23  

RT
P
ln ˆ2 
B22  y32 23  y1212  y3 y1  23  12  13  

RT
P
ln ˆ3 
B33  y1213  y22 23  y1 y2 13   23  12  

RT
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
Spreadsheet for computing fugacity coefficients in a binary mixture (problem 11.27 in SVA)
T (K)
P (bar)
373.15
35
Species 1
Species 2
Species 3
Mole
Fraction
0.21
0.43
0.36
Name
Methane
Ethane
Propane
Cross-Parameters
1,2
1,3
2,3
kij
Tc (K)
Pc (bar)
0 241.22641
47.01
0
265.488
43.26
0 336.00586
45.26
3
2nd Virial Coefficients (cm /mol)
B11
-21
B22
-113
B33
-244
B12
-52
B13
-78
B23
-167
12 (cm /mol)
3
13 (cm /mol)
3
23 (cm /mol)
3
 ij 
Pcij 
i   j
0.286
0.279
0.276
Z cij 
Vcij
ˆ1
ˆ2
ˆ3
Z ci  Z cj
0
B
0.1288
0.0650
-0.0266
0
B
0.1115
0.0978
0.0283
0.012
0.1
0.152
Tr
1.9578
1.2222
1.0091
Pr
0.7610
0.7184
0.8239
B
-0.0610
-0.2231
-0.3330
0.0560
0.0820
0.1260
Tr
1.5469
1.4055
1.1105
Pr
0.7446
0.8090
0.7734
B
-0.1270
-0.1618
-0.2738

Zc
0.2825
0.2810
0.2775
B 0  0.083 
0.422
B1  0.139 
0.172
2
1
1
Tr1.6
Tr4.2
3
Bij 
RTcij
Pcij
B
0
  ij B 1 
P
 B11  y2212  y3213  y2 y3 12  13   23  
RT
P
ln ˆ2 
 B22  y32 23  y1212  y3 y1  23  12  13  
RT
P
ln ˆ3 
 B33  y1213  y22 23  y1 y2 13   23  12  
RT
ln ˆ1 
0.01895
-0.12320
-0.25478
ln ˆ3

Zc
Tcij  TciTcj 1  kij 
2
Z cij RTcij
 V 1/ 3  Vcj1/ 3 
Vcij   ci


2


30.33
107.52
23.10
ln ˆ1
ln ˆ2
Vc
3
(cm /mol)
98.6
145.5
200
Vc
(cm3/mol)
120.5
143.4
171.3
Tc (K)
Pc (bar)
190.6
45.99
305.3
48.72
369.8
42.48
1.019
0.884
0.775
f1 (bar)
f2 (bar)
f3 (bar)
7.49
13.31
9.77
fˆiid  yi fi  yii P
T (K)
P (bar)
373.15
35

Tc (K) Pc (bar)
190.6
45.99
0
0.012
1
Tr
Pr
B
B

1.9578
0.7610
-0.0610
0.1288
0.9771

 TP
B 0  0.083 
B1  0.139 
0.422
Tr1.6
0.172


  exp  B 0   B1
r
r



Tr4.2
Pale blue boxes are input fields, pink boxes are the final output.
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
T (K)
Chapter 10
P (bar)
373.15

Tc (K) Pc (bar)
35
305.3
48.72
0
0.1
1
Tr
Pr
B
B

1.2222
0.7184
-0.2231
0.0650
0.8805

 TP
B 0  0.083 
B1  0.139 
0.422


Tr1.6
  exp  B 0   B1
0.172



r
r
Tr4.2
Pale blue boxes are input fields, pink boxes are the final output.
T (K)
P (bar)
373.15

Tc (K) Pc (bar)
35
369.8
42.48
0
0.152
1
Tr
Pr
B
B

1.0091
0.8239
-0.3330
-0.0266
0.7594

 TP
B 0  0.083 
0.422
B1  0.139 
0.172
Tr1.6

  exp  B 0   B1

r
r



Tr4.2
Pale blue boxes are input fields, pink boxes are the final output.
f1 (bar)
f2 (bar)
f3 (bar)
Updated 4/5/2017
7.18
13.25
9.57
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SVNAS 8th Edition Annotated Solutions
Chapter 10
,
Part
a
b
c
d
e
f
g
and
I
II
G1E
H1E
S1E
CPE S2E
H2E
G2E
S2E
H2E
G2E
-622.0 -1920.0 -4.354
4.2 -3.951 -1794 -497.4 -4.354 -1920.0 -491.4
1095.0 1595.0 1.677
3.3 1.993 1694 1039.9 1.677 1595.0 1044.7
407.0
984.0 1.935 -2.7 1.677
903
352.8 1.935
984.0
348.9
632.0
-208.0 -2.817 23.0 -0.614
482
683.5 -2.817
-208.0
716.5
1445.0
605.0 -2.817 11.0 -1.764
935 1513.7 -2.817
605.0 1529.5
734.0
-416.0 -3.857 11.0 -2.803
-86
833.9 -3.857
-416.0
849.7
759.0 1465.0 2.368 -8.0 1.602 1225
699.5 2.368 1465.0
688.0
x1
0.02715
0.09329
0.1749
0.3276
0.40244
0.56689
0.63128
0.66233
0.69984
0.72792
0.77514
VE
87.5
265.6
417.4
534.5
531.7
421.1
347.1
321.7
276.4
252.9
190.7
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
0.79243
0.82954
0.86835
0.93287
0.98233
Chapter 10
178.1
138.4
98.4
37.6
10
600
500
VE
400
VE actual
300
200
100
0
0
0.2
0.4
0.6
0.8
1
x1
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
3500
3000
2500
VE1
VE2
VE
2000
1500
1000
500
0
-500 0
0.2
0.4
0.6
0.8
1
x1
B  y12 B11  2 y1 y2 B12  y22 B22
BP
504.25* 2
1
 0.965
RT
83.14*348.15
H R BP P dB
504.25* 2
2




*3.55  0.1202
RT RT R dT
83.14*348.15 83.14
SR
P dB
2


*3.55  0.08540
R
R dT
83.14
Z 1
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
1
0.95
0.9
ɸhat1 ɸhat2
0.85
0.8
0.75
ɸhat1
0.7
ɸhat2
0.65
0.6
0.55
0.5
0
0.2
0.4
0.6
0.8
1
y1
≡
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
isa function of T only:
,
; substitution yields:
x1
HE
0.0426
-23.3
0.0817
-45.7
0.1177
-66.5
0.151
-86.6
0.2107
-118.2
0.2624
-144.6
0.3472
-176.6
0.4158
-195.7
0.5163
-204.2
0.6156
-191.7
0.681
-174.1
0.7621
-141
0.8181
-116.8
0.865
-85.6
0.9276
-43.5
0.9624
-22.6
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
0
0
0.2
0.4
0.6
0.8
1
-50
HE
-100
-150
HE actual
-200
equation
-250
Updated 4/5/2017
x1
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SVNAS 8th Edition Annotated Solutions
Chapter 10
100
0
-100
0
0.2
0.4
0.6
1
HEbar1
-200
HEbar
0.8
HEbar2
-300
-400
-500
-600
-700
x1
N
N
B   yi y j Bij
i 1 j 1
B

B  y12 B11  2 y1 y2 B12  y22 B22

RTc 0
0.422
0.172
B   B1 , with B0  0.083  1.6 and B1  0.139  4.2
Pc
Tr
Tr
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
i   j
ij 
2
Tcij  TciTcj 1  kij 
Pcij 
Z cij RTcij
Z cij 
Vcij
Z ci  Z cj
2
V
Vcij  


1/ 3
ci
 Vcj1/ 3 

2

3
dB
dB
dB
dB
 y12 11  2 y1 y2 12  y22 22
dT
dT
dT
dT
dBij
dT
V

RTc
Pc
 dBij0
dBij1


 dT
dT

RT
B,
P
Updated 4/5/2017
 R  dBij0
dBij1
 

 Pc  dTr
dTr


G R  BP ,

dBij0 0.675
 , with
 2.6 and

dT
Tr

H R  BP  TP
dB
,
dT
and S R   P
dBij1
dT

0.722
Tr5.2
dB
dT
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SVNAS 8th Edition Annotated Solutions
Chapter 10
Spreadsheet for computing residual properties of a binary mixture, for SVA problem 11.37(a)
T (K)
P (bar)
333.15
1.7
Vc
Mole
3
Fraction
Name
Tc (K)
Pc (bar) (cm /mol)
Zc

Species 1 Acetone
0.28
508.2
47.01
209
0.233
0.307
Species 2 1,3-butadiene
0.72
425.2
42.77
220.4
0.267
0.190
Cross-Parameters
Vc
Tc (K)
Pc (bar) (cm3/mol)
464.8512
45.02
214.6
kij
0
3
2nd Virial Coefficients (cm /mol)
B11
-912
B22
-500
B12
-665
i   j
ij 
Pcij 
3
Derivatives of 2nd Virial Coefficients (cm /((mol K))
dB11/dT
7.10
dB22/dT
3.42
dB12/dT
4.84
Zc
0.2500

0.2485
Zcij 
Vcij
B
-0.8744
-0.3402
0
B
-0.5579
Pr
0.0362
0.0397
B
-0.7464
-0.5405
Tr
0.7167
Pr
0.0378
B
-0.6361
Tcij  TciTcj 1  kij 
2
Zcij RTcij
0
Tr
0.6555
0.7835
Zci  Zcj
2
1
1
B 0  0.083 
0.422
B  0.139 
0.172
1
0
dB
dT
6.4892
2.5674
0
dB
dT
4.0817
dB
dT
2.0236
1.2729
dB
dT
1.6049
1
1
Tr1 .6
Tr4 .2
3
 Vci1/ 3  Vcj1/ 3 
Vcij  


2


Bij 
RTcij
Pcij
B
0
  ij B 1 
dBij RTc  dBij0
dBij1 
dBij0 0.675
dBij1 0.722

 , with


 2.6 and
 5.2
3
Mixture 2nd Virial Coefficient (cm /mol)
dT
Pc  dT
dT 
dT
dT
T
Tr
r


B
-599
3
2
2
Derivative of Mixture 2nd Virial Coefficient (cm /((mol K))
B  y1 B11  2 y1 y2 B12  y2 B22
dB/dT
4.28
Mixture Properties:
V (cm3/mol)
Z
R
G (J/mol)
R
H (J/mol)
R
S (J/(mol K))
dB
dB
dB
dB
 y12 11  2 y1 y2 12  y22 22
dT
dT
dT
dT
15695
0.9632
-101.8
-344.3
-0.728
N
V
N
B   yi y j Bij
i 1 j 1
B

RT
B,
P
GR  BP ,
H R  BP  TP
dB
,
dT
and S R  P
dB
dT
B  y12 B11  2 y1 y2 B12  y22 B22

RTc 0
0.422
0.172
B   B1 , with B0  0.083  1.6 and B1  0.139  4.2
Pc
Tr
Tr
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
i   j
ij 
2
Tcij  TciTcj 1  kij 
Pcij 
Z cij RTcij
Z cij 
Vcij
Z ci  Z cj
2
V
Vcij  


1/ 3
ci
 Vcj1/ 3 

2

3
dB
dB
dB
dB
 y12 11  2 y1 y2 12  y22 22
dT
dT
dT
dT
dBij
dT
V

RTc
Pc
 dBij0
dBij1


 dT
dT

RT
B,
P
 R  dBij0
dBij1
 

 Pc  dTr
dTr


G R  BP ,
N
N
B   yi y j Bij
i 1 j 1
B


dBij0 0.675
 , with
 2.6 and

dT
Tr

H R  BP  TP
dB
,
dT
and S R   P
dBij1
dT

0.722
Tr5.2
dB
dT
B  y12 B11  2 y1 y2 B12  y22 B22

RTc 0
0.422
0.172
B   B1 , with B0  0.083  1.6 and B1  0.139  4.2
Pc
Tr
Tr
Updated 4/5/2017
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SVNAS 8th Edition Annotated Solutions
Chapter 10
i   j
ij 
2
Tcij  TciTcj 1  kij 
Pcij 
Z cij RTcij
Z cij 
Vcij
Z ci  Z cj
2
V
Vcij  


1/ 3
ci
 Vcj1/ 3 

2

3
dB
dB
dB
dB
 y12 11  2 y1 y2 12  y22 22
dT
dT
dT
dT
dBij
dT
V

RTc
Pc
 dBij0
dBij1


 dT
dT

RT
B,
P
Updated 4/5/2017

dBij0 0.675
 , with
 2.6 and

dT
Tr

G R  BP ,
H R  BP  TP
dB
,
dT
dBij1
dT

0.722
Tr5.2
and S R   P
dB
dT
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SVNAS 8th Edition Annotated Solutions
Chapter 10
Spreadsheet for computing residual properties of a binary mixture, for SVA problem 11.37(c)
T (K)
P (bar)
298.15
1
Vc
Mole
3
Fraction
Name
Tc (K)
Pc (bar) (cm /mol)
Zc

Species 1
Methyl chloride
0.45
416.3
66.8
143
0.276
0.153
Species 2
Ethyl chloride
0.55
460.4
52.7
200
0.275
0.190
Cross-Parameters kij
Tc (K)
437.7951
0
Vc
Pc (bar) (cm3/mol)
59.02
169.9
3
2nd Virial Coefficients (cm /mol)
B11
-374
B22
-682
B12
-507
i   j
ij 
Pcij 
Derivatives of 2nd Virial Coefficients (cm3/((mol K))
dB11/dT
2.78
dB22/dT
5.37
dB12/dT
3.87
Pr
0.0150
0.0190
B0
-0.6369
-0.7627
B1
-0.5599
-0.9278
dB0
dT
1.6078
2.0889
dB1
dT
4.0963
6.9148
Tr
0.6810
Pr
0.0169
B0
-0.6973
B1
-0.7245
dB0
dT
1.8326
dB1
dT
5.3221
Tcij  TciTcj 1  kij 
2
Z cij RTcij
Z cij 
Vcij
 V 1/ 3  Vcj1/ 3 
Vcij   ci


2



0.1715
Zc
0.2755
Tr
0.7162
0.6476
Z ci  Z cj
2
B1 0.139
0.172
Tr1.6
Tr4.2
Bij 
dBij1
dT

RTcij
Pcij
B
0
 ij B1 
0.722
Tr5.2
dB
dB
dB
dB
 y12 11  2 y1 y2 12  y22 22
dT
dT
dT
dT
24257
0.9785
-53.3
-175.6
-0.410
V
N
N
RT
B,
P
B   yi y j Bij
i 1 j 1
B
0.422
3
dBij RTc  dBij0
dBij1 
dBij0 0.675

 , with


 2.6 and
Mixture 2nd Virial Coefficient (cm3/mol)
dT
Pc  dT
dT 
dT
Tr


B
-533
3
2
2
Derivative of Mixture 2nd Virial Coefficient (cm /((mol K))
B  y1 B11  2 y1 y2 B12  y2 B22
dB/dT
4.10
Mixture Properties:
V (cm3/mol)
Z
GR (J/mol)
HR (J/mol)
SR (J/(mol K))
B 0 0.083

G R  BP ,
H R  BP  TP
dB
,
dT
and S R   P
dB
dT
B  y12 B11  2 y1 y2 B12  y22 B22

RTc 0
0.422
0.172
B   B1 , with B0  0.083  1.6 and B1  0.139  4.2
Pc
Tr
Tr
ij 
i   j
2
Tcij  TciTcj 1  kij 
Pcij 
Z cij 
Z cij RTcij
Vcij
Z ci  Z cj
2
 V 1/ 3  Vcj1/ 3 
Vcij   ci


2


Updated 4/5/2017
3
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SVNAS 8th Edition Annotated Solutions
Chapter 10
dB
dB
dB
dB
 y12 11  2 y1 y2 12  y22 22
dT
dT
dT
dT
0
dBij1
RTc  dBij



dT
Pc  dT
dT

dBij
V
RT
B,
P

dBij0 0.675
 , with
 2.6 and

dT
Tr

G R  BP ,
H R  BP  TP
0
Vc
Tc (K)
Pc (bar) (cm3/mol)
226.2727
62.00
80.6
i   j
2
Z RT
Pcij  cij cij
Vcij
2nd Virial Coefficients (cm 3/mol)
B11
-7
B22
-228
B12
-56
Derivatives of 2nd Virial Coefficients (cm /((mol K))
dB11/dT
0.19
dB22/dT
1.89
dB12/dT
V
Vcij  


1/ 3
ci
Z cij 
V
1/ 3
cj
2
dT



0.722
Tr5.2
and S R   P
dB
dT
Tr
2.3229
0.7226
Pr
0.0882
0.0266
B0
-0.0266
-0.6267
B1
0.1340
-0.5343
dB0
dT
0.0754
1.5711
dB1
dT
0.0090
3.9114
Tr
1.2956
Pr
0.0484
B0
-0.1959
B1
0.0810
dB0
dT
0.3443
dB1
dT
0.1878
Tcij  TciTcj 1  kij 
ij 
3

0.1455
Zc
0.2655

dB
,
dT
Spreadsheet for computing residual properties of a binary mixture, for SVA problem 11.37(d)
T (K)
P (bar)
293.15
3
Vc
Mole
3
Fraction
Name
Tc (K)
Pc (bar) (cm /mol)
Zc

Species 1
Nitrogen
0.83
126.2
34
89.2
0.289
0.038
Species 2
Ammonia
0.17
405.7
112.8
72.5
0.242
0.253
Cross-Parameters kij
dBij1
Z ci  Z cj
2
B 0 0.083
0.422
B 0.139
0.172
1
Tr1.6
Tr4.2
3
Bij 
RTcij
Pcij
B
0
 ij B1 
0.50
dBij RTc  dBij0
dBij1 
dB0 0.675
dBij1 0.722

 , with ij 


and
 5.2
2.6
Mixture 2nd Virial Coefficient (cm 3/mol)
dT
Pc  dT
dT 
dT
dT
T
Tr
r


B
-27
3
Derivative of Mixture 2nd Virial Coefficient (cm /((mol K))
B  y12 B11  2 y1 y2 B12  y22 B22
dB/dT
0.32
Mixture Properties:
V (cm3/mol)
Z
GR (J/mol)
HR (J/mol)
SR (J/(mol K))
Updated 4/5/2017
8098
0.9967
-8.1
-36.5
-0.097
dB
dB
dB 2 dB11
 y1
 2 y1 y2 12  y22 22
dT
dT
dT
dT
V
RT
B,
P
G R  BP ,
H R  BP  TP
dB
,
dT
and S R   P
dB
dT
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SVNAS 8th Edition Annotated Solutions
N
Chapter 10
N
B   yi y j Bij
B  y12 B11  2 y1 y2 B12  y22 B22
i 1 j 1
B


RTc 0
0.422
0.172
B   B1 , with B0  0.083  1.6 and B1  0.139  4.2
Pc
Tr
Tr
ij 
i   j
2
Tcij  TciTcj 1  kij 
Pcij 
Z cij 
Z cij RTcij
Vcij
Z ci  Z cj
2
 V 1/ 3  Vcj1/ 3 
Vcij   ci


2


3
dB
dB
dB
dB
 y12 11  2 y1 y2 12  y22 22
dT
dT
dT
dT
0
dBij1
RTc  dBij



dT
Pc  dT
dT

dBij
V
RT
B,
P
Updated 4/5/2017

dBij0 0.675
 , with
 2.6 and

dT
Tr

G R  BP ,
H R  BP  TP
dB
,
dT
dBij1
dT

0.722
Tr5.2
and S R   P
dB
dT
p. 45 of 53
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SVNAS 8th Edition Annotated Solutions
Chapter 10
Spreadsheet for computing residual properties of a binary mixture, for SVA problem 11.37(e)
T (K)
P (bar)
293.15
4.2
Vc
Mole
3
Fraction
Name
Tc (K)
Pc (bar) (cm /mol)
Zc

Species 1
sulfur dioxide
0.32
430.8
78.84
122
0.269
0.245
Species 2
ethylene
0.68
282.3
50.4
131
0.281
0.087
Cross-Parameters kij
0
Vc
Tc (K)
Pc (bar) (cm3/mol)
348.7332
63.06
126.4
i   j
2
Z cij RTcij
Pcij 
Vcij
3
2nd Virial Coefficients (cm /mol)
B11
-398
B22
-147
B12
-235
dB12/dT
dBij
Mixture 2nd Virial Coefficient (cm 3/mol)
dT
B
-211
3
Derivative of Mixture 2nd Virial Coefficient (cm /((mol K))
dB/dT
1.62
Mixture Properties:
V (cm3/mol)
Z
GR (J/mol)
HR (J/mol)
SR (J/(mol K))
5593
0.9637
-88.5
-288.4
-0.682
B1
-0.7274
-0.0078
dB0
dTr
1.8365
0.6120
dB1
dTr
5.3445
0.5934
Tr
0.8406
Pr
0.0666
B0
-0.4741
B1
-0.2176
dB0
dTr
1.0601
dB1
dTr
1.7809
Z ci  Z cj
2
B 0 0.083
0.422
B1 0.139
0.172
Tr1.6
Tr4.2
3
Bij 
dBij
R  dBij


Pc  dTr
dTr

0

B0
-0.6983
-0.3143
1
RTcij
Pcij
B
0
 ij B1 

dB0 0.675
dBij1 0.722
 , with ij 
and
 5.2
2.6

dTr
dTr
Tr
Tr

B  y12 B11  2 y1 y2 B12  y22 B22
dB
dB
dB 2 dB11
 y1
 2 y1 y2 12  y22 22
dT
dT
dT
dT
V
and
Updated 4/5/2017
Z cij 
 V 1/ 3  Vcj1/ 3 
Vcij   ci


2


1.79
Pr
0.0533
0.0833
Tcij  TciTcj 1  kij 
ij 
Derivatives of 2nd Virial Coefficients (cm 3/((mol K))
dB11/dT
3.32
dB22/dT
1.09

0.1660
Zc
0.2750
Tr
0.6805
1.0384
RT
B,
P
G R  BP ,
H R  BP  TP
then
dB
,
dT
and S R   P
dB
dT
and
p. 46 of 53
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written consent of McGraw-Hill Education.
SVNAS 8th Edition Annotated Solutions
Chapter 10
Updated 4/5/2017
p. 47 of 53
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SVNAS 8th Edition Annotated Solutions
has the same sign over the whole composition range, both
Updated 4/5/2017
Chapter 10
and
p. 48 of 53
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SVNAS 8th Edition Annotated Solutions
Chapter 10
2
ME
MEbar1
ME values
1.5
MEbar2
1
0.5
0
-0.5
0
0.2
0.4
0.6
0.8
1
xi
35
30
ME values
25
ME
20
MEbar1
15
MEbar2
10
5
0
0
0.2
0.4
0.6
0.8
1
xi
Updated 4/5/2017
p. 49 of 53
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SVNAS 8th Edition Annotated Solutions
Chapter 10
)
Updated 4/5/2017
p. 50 of 53
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written consent of McGraw-Hill Education.
SVNAS 8th Edition Annotated Solutions
Chapter 10
Updated 4/5/2017
p. 51 of 53
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SVNAS 8th Edition Annotated Solutions
. Therefore the sign of
Chapter 10
is the same as the sign of
, and by the same argument the sign of
T (K)
283.15
303.15
323.15
Updated 4/5/2017
. Similarly, at
is of opposite sign as the
GE (J/mol)
544
513
494.2
HE (J/mol)
932.1
893.4
845.9
p. 52 of 53
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SVNAS 8th Edition Annotated Solutions
Chapter 10
Updated 4/5/2017
p. 53 of 53
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