calculate the enthalpy and entropy of saturated isobutene vapor at 360 k from the following...

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Abellar, Rodgie John Assignment Aries, Allen Jerry ChE Thermodynamics 2 Belnas, Katrina Lyn January 27,2014 BSChE-4 Problem: Calculate the enthalpy and entropy of saturated isobutene vapor at 360 K from the following information: 1. compressibility-factor data for isobutene vapor. 2. The vapor pressure of isobutane at 360 K is 15.41 bar. 3. H 0 ig =18,115.0 Jmol 1 and S 0 ig =295.976 Jmol 1 K 1 for the ideal gas reference state at 300 K and 1 bar. 4. The ideal-gas heat capacity of interest is: C P ig R =1.7765+33.037 × 10 3 T ( T / K ) Solution: Compressibility Factors Z for Isobutane P (bar) 340 K 350 K 360 K 370 K 380 K 0.1 0.997 7 0.997 19 0.997 37 0.997 53 0.997 67 0.5 0.987 45 0.988 3 0.989 07 0.989 77 0.990 4 2 0.958 95 0.962 06 0.964 83 0.967 3 0.969 53 4 0.924 22 0.930 69 0.936 35 0.941 32 0.945 74 6 0.887 42 0.898 16 0.907 34 0.915 29 0.922 23 8 0.845 75 0.862 18 0.875 86 0.887 45 0.897 43 10 0.796 59 0.821 17 0.840 77 0.856 95 0.870 61 12 0.773 1 0.801 03 0.823 15 0.841 34 14 0.755 06 0.785 31 0.809 23

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Enthalpy and Entropy by Graphical Evaluation of Residual Properties

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Problem:

Calculate the enthalpy and entropy of saturated isobutene vapor at 360 K from the following information:1. compressibility-factor data for isobutene vapor.2. The vapor pressure of isobutane at 360 K is 15.41 bar.3. and for the ideal gas reference state at 300 K and 1 bar.4. The ideal-gas heat capacity of interest is:

Solution:

Compressibility Factors Z for Isobutane

P (bar)340 K350 K360 K370 K380 K

0.10.99770.997190.997370.997530.99767

0.50.987450.98830.989070.989770.9904

20.958950.962060.964830.96730.96953

40.924220.930690.936350.941320.94574

60.887420.898160.907340.915290.92223

80.845750.862180.875860.887450.89743

100.796590.821170.840770.856950.87061

120.77310.801030.823150.84134

140.755060.785310.80923

15.410.71727

Solving for in every pressure at 360 K:

P (bar) (bar-1)

0(0.031500)

0.10.026300

0.50.021860

20.017585

40.015913

60.015443

80.015518

100.015923

120.016581

140.017496

15.410.018347

*Value(s) in parenthesis are found by extrapolation

Plotting at each pressure:

At bar,

Graphing from 350 K to remove decreasing side:

At bar,

At bar,

At bar,

At bar,

At bar,

At bar,

At bar,

At bar,

Differentiating each trend line equation at and dividing each by corresponding pressure gives:

P (bar) (K-1bar-1)

0(1.757 x 10-4)

0.11.700 x 10-4

0.51.474 x 10-4

21.320 x 10-4

41.342 x 10-4

61.446 x 10-4

81.608 x 10-4

101.838 x 10-4

122.093 x 10-4

142.387 x 10-4

15.41(2.594 x 10-4)

*Value(s) in parenthesis are found by extrapolation

Plotting vs. : ,

Graphing : and

Abellar, Rodgie JohnAssignmentAries, Allen JerryChE Thermodynamics 2Belnas, Katrina LynJanuary 27,2014BSChE-4

20.0175850.03517

20.0160.032

20.01550.031

20.01550.031

20.0160.032

20.0165

20.01750.035

1.410.0183470.025869

Summary: