SVA Example 6.3
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Transcript of SVA Example 6.3
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Table 6.1 Compressibility Factors Z for Isobutane
P (bar) 340 350 360 370 380 T (K)
0.1 0.997 0.99719 0.99737 0.99753 0.99767
0.5 0.98745 0.9883 0.98909 0.98977 0.9904
2 0.95895 0.96206 0.96483 0.9673 0.96953
4 0.92422 0.93069 0.93635 0.94132 0.94574
6 0.88742 0.89816 0.90734 0.91529 0.92223
8 0.84575 0.86218 0.87586 0.88745 0.89743
10 0.79659 0.82117 0.84077 0.85695 0.87061
12 0.7731 0.80103 0.82315 0.84134
14 0.75506 0.78531 0.80923
15.41 0.71727
For the enthalpy of isobutane:
SVA Example 6.3
= ∗ −
0 ( )
= 0 +
+
0 =18115.0
= 18115.0 +
0+ ∗ −
0
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To evaluate
a. Calculate by estimating the derivative of the graph Z vs T
P
0.1 1.7E-05
0.5 7.35E-05
2 0.000262
4 0.0005315
6 0.0008565
to estimate the slope of the tangent line 8 0.0012635
10 0.001789
12 0.0025025
14 0.003025
P (dZ/dT)/P
0.1 0.00017
0.5 0.000147
2 0.000131
4 0.000132875
6 0.00014275
8 0.000157938
10 0.0001789
12 0.000208542
14 0.000216071
:
′ ≈ + ∆ − ∆ (ℎ − ) ′ ≈ − − ∆∆ ( − )
′ ≈ + ∆ − − ∆2∆ ( − )
y = -1.32283E-07x3 + 3.63387E-06x2 - 2.11857E-05x + 0.000164219
R² = 0.974094729
0
0.00005
0.0001
0.00015
0.0002
0.00025
0 2 4 6 8 10 12 14 16
(dZ/dT)/P vs P
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b. Compute for P = 0 and P = 15.41 by using the best fit for the curve
P (dZ/dT)/P
0 0.000164219
0.1 0.00017
0.5 0.0001472 0.000131
4 0.000132875
6 0.00014275
8 0.000157938
10 0.0001789
12 0.000208542
14 0.000216071
15.41 0.000216601
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c. Compute for by Trapezoidal method
The value of the integral is 0.00258364
Hres/RT -0.9301104
Hres -2783.85763 J/mol
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(J/mol)
Ho 18115 To 300
Hig 24439.6477 T 360
Hres -2783.857632 A 1.7765
H 21655.79006 B 0.033037tau 1.2
ICPH 760.7226
dH 6324.648
