CENG 221 Homework 1 Solution
Q1:
Part I
Antoine Equation:
Let liquid() and vapour() be two phases of pure component at equilibrium.
Then dG = dG
dG = VdP SdT
V dP* S dT = V dP* S dT
dP* S S S
=
=
dT V V V
(1)
As G = 0 , therefore, H = T S
dP* S
H
=
=
dT V T V
For P =
(2)
RT
V
Then Eq.(2) can be changed as following:
dP* H P*
=
dT
RT 2
dP* H dT
=
P*
RT 2
d (ln P* ) =
H
1
d( )
R
T
if H is a constant at a small range of temperatures, Then
Let B=
H
R
ln( P* ) = A
B
T
(3)
For the temperature in Eq.3 is K, and H is depend on temperature.then Eq.3 is modified as,
ln( PA* ) = A
B
T ( C) + C
o
where A, B, and C are "Antoine coefficients" that vary from substance to substance.
Sublimations and vaporizations of the same substance have separate sets of Antoine
coefficients, as do components in mixtures. The Antoine equation is accurate to a few percent
for most volatile substances (with vapor pressures over 10 Torr).
Assumptions:
1. The vapour is an idea gas of negligible molecular size and there is no interaction between
gas molecules.
2. The solution is ideal solution, which is defined as a solution formed with no accompanying
energy change, when the intermolecular attractive forces between the molecules of the solvent
are the same as those between the molecules in the separate components.
3. The system is a close system and in equilibrium. The volume of liquid is negligible
compared to volume of vapour.
Part II
log P* = A
B
(P:mmHg, T oC)
T +C
P*
P
=
760 101.3
P* =
log
760* P
101.3
760
B
P = A
101.3
T +C
ln10*log
ln
B
760
P = [A
]*ln(10)
T +C
101.3
B
760
+ ln P = [ A
]*ln(10)
T +C
101.3
ln P = [ A *ln(10) ln
B *ln(10)
760
]
T +C
101.3
then new A= [ A *ln(10) ln
760
]
101.3
new B= B *ln(10)
new C=C
Table 1. The Antoine coefficients of different compounds
Compound
Reference
Cyclohexane
13.297
2802.08
224.53
Gaw and Swinton, 1968
1-Hexene.
13.794
2654.81
225.85
Forziati, Camin, et al., 1950
Methol
13.88
2921.56
216.95
Ambrose and Sprake, 1970
1-Heptene
13.881
2921.14
216.95
Bent, Cuthbertson, et al., 1936
Decane
13.996
3477.521
194.48
Williamham, Taylor, et al., 1945
Heptane
14.285
3641.17
239.65
Williamham, Taylor, et al., 1945
Acetone
14.793
2021.574
240.705
Ambrose, Sprake, et al., 1974
Toluene
15.069
3094.51
219.377
Williamham, Taylor, et al., 1945
Methane
13.36
928.18
267.67
Cutler and Morrison, 1965
Ethane
14.983
1828.943
279.57
Carruth and Kobayashi, 1973
Pentane
13.791
2465.187
232.696
Osborn and Douslin, 1974
Note: The attachment is Excel File which I got from Angela. There are thousand of the A,B and
C parameters.
Q2
Given:
Antoine equation
P
2.3 atm
A mixture of toluene and n-heptane
Find: calculate T, X,Y ,x, y, Kh,Kt and
Soln:
For pure compound:
Pt * = Ph* = PT = 2.3atm = 233.45kPa
ln Pt* = 14.0098 3103.01/(T + 219.79)
ln Ph* = 13.8587 2991.32 /(T + 216.64)
Ttb=142.854
Thb=139.23
T
Tmin
Tmax
Ph
Pt
Xh
Xt
Yh
Yt
xh
xt
yh
yt
Kh
0
Kt
alpha
139.227 233.45 214.16
139.727 236.22 216.75
0.8578
0.1422
0.8679
0.1321
0.8677
0.1323
0.8773
0.142 1.0119 0.9285 1.0898
140.227 239.01 219.36
0.7169
0.2831
0.7339
0.2661
0.7336
0.2664
0.75
0.2829 1.0238 0.9397 1.0896
140.727 241.83
222
0.5773
0.4227
0.598
0.402
0.5976
0.4024
0.618
0.4225 1.0359
141.227 244.68 224.66
0.439
0.561
0.4601
0.5399
0.4598
0.5402
0.481
0.5608 1.0481 0.9624 1.0891
141.727 247.55 227.35
0.302
0.698
0.3203
0.6797
0.32
0.68
0.3388
0.6979 1.0604 0.9739 1.0888
142.227 250.44 230.06
0.1663
0.8337
0.1784
0.8216
0.1782
0.8218
0.191
0.8337 1.0728 0.9855 1.0886
142.727 253.37
232.8
0.0318
0.9682
0.0345
0.9655
0.0344
0.9656
0.0374
0.9682 1.0853 0.9972 1.0884
142.845 254.06 233.45
0.951 1.0893
144
143
T(oC)
142
141
140
139
138
0
0.2
0.4
0.6
0.8
1.2
x, y
1.2
1
yh
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
xh
0.8
1.09
1.0898
1.0896
Alpha
1.0894
1.0892
1.089
1.0888
1.0886
1.0884
1.0882
139.5
140
140.5
141
141.5
T(oC)
142
142.5
143
Q3:
We wished to use a distillation system with a still pot to batch
distill a mixture of methanol and water. A total condenser is used. The
feed is 57 mole% methanol (use VLE data in Table 3-3) Please find the
Wfinal, D, and XD using 1 Kg as basis of your calculation. The final
concentration in the still is 15 mole% methanol.
Solution:
Given: F = 1kg;
Find: Xd, Wfinal, D
x
y
0
2
4
6
8
10
15
20
30
40
50
60
70
80
90
95
100
Xf = 0.57;
y-x
0
13.4
23
30.4
36.5
41.8
51.7
57.9
66.5
72.9
77.9
82.5
87
91.5
95.8
97.9
100
Xw,final = 0.15
1/(y-x)
11.4
19
24.4
28.5
31.8
36.7
37.9
36.5
32.9
27.9
22.5
17
11.5
5.8
2.9
0
8.77193
5.263158
4.098361
3.508772
3.144654
2.724796
2.638522
2.739726
3.039514
3.584229
4.444444
5.882353
8.695652
17.24138
34.48276
10
9
8
7
6
Series1
5
4
3
2
1
0
0
20
40
.57 *32
= .702kg = 21.94mol
.57 *32 + .43*18
Mass of water=1-.702=.298=16.56mol
F=21.94+16.56=38.5mol
Mass of methaol=
By Rayleigh Equation,
dX w
dW
=
W
Yd X w
W final
ln
F
By Simpsons Rule,
0.15
X W , final
dX w
dX w
=
= X f
0.57 X X
Xd Xw
d
w
60
80
100
Area =
W final
ln
F
( X f X w, final )[ F ( X w, final ) + 4 F ( X m ) + F ( X f )]
6
( X f X w, final )[ F ( X w, final ) + 4 F ( X m ) + F ( X f )]
=
6
(0.57 0.15)(2.725+4*2.9+4.18)
=
= 1.295
6
Wfinal = F * e1.295 = 11.84mol
By overall material balance,
F =W + D
FX f = WX w + DX d
Xd =
FX f WX w
F W
38.5 0.57 11.84 0.15
= 0.756
38.5 11.84
1/(y-x)
D=38.5-11.84=26.66mol
4.6
4.5
4.4
4.3
4.2
4.1
4
3.9
3.8
3.7
3.6
3.5
3.4
3.3
3.2
3.1
3
2.9
2.8
2.7
2.6
2.5
2.4
2.3
2.2
2.1
2
Series1
10
20
30
40
x
50
60
70
