Calculate the enthalpy and entropy of saturated isobutane vapor at 360 K from the following information: 1. Table 6.1 gives compressibility-factor data (values of Z) for isobutane vapor. 2. The vapor pressure of isobutane at 360 K is 15.41 bar.3. Set H_{0)^{ig} = 18,115.0 J . mol^−1 and

Question

Calculate the enthalpy and entropy of saturated isobutane vapor at 360 K from the following information:

1. Table 6.1 gives compressibility-factor data (values of Z) for isobutane vapor.

2. The vapor pressure of isobutane at 360 K is 15.41 bar.

3. Set [latex] H_0^{\prime g}=18,115.0 J \cdot mol ^{-1} ext { and } S_0^{\prime g}=295.976 J \cdot mol ^{-1} \cdot K ^{-1} [/latex] for the reference state at 300 K and 1 bar. [These values are in accord with the bases adopted by R. D. Goodwin and W. M. Haynes, Nat. Bur. Stand. (U.S.), Tech. Note 1051, 1982.]

4. The ideal-gas-state heat capacity of isobutane vapor at temperatures of interest is:

[latex] C_P^{\prime g} / R=1.7765+33.037 imes 10^{-3} T \quad(T K ) [/latex]

Table 6.1: Compressibility Factor Z for Isobutane

P bar	340 K	350 K	360 K	370 K	380 K
0.100	0.99700	0.99719	0.99737	0.99753	0.99767
0.500	0.98745	0.98830	0.98907	0.98977	0.99040
2.00	0.95895	0.96206	0.96483	0.96730	0.96953
4.00	0.92422	0.93069	0.93635	0.94132	0.94574
6.00	0.88742	0.89816	0.90734	0.91529	0.92223
8.00	0.84575	0.86218	0.87586	0.88745	0.89743
10.0	0.79659	0.82117	0.84077	0.85695	0.87061
12.0	. . . . . .	0.77310	0.80103	0.82315	0.84134
14.0	. . . . . .	. . . . . .	0.75506	0.78531	0.80923
15.41	. . . . . .	. . . . . .	0.71727

Accepted Answer

Calculating [latex]H^{R} and S^{R}[/latex] at 360 K and 15.41 bar by application of Eqs. (6.46) and (6.48) requires evaluation of two integrals:

[latex] \frac{H^R}{R T}=-T \int_0^P\left(\frac{\partial Z}{\partial T} ight)_P \frac{d P}{P} \quad ext { (const } T ext { ) } [/latex] (6.46)

[latex] \frac{S^R}{R}=-T \int_0^P\left(\frac{\partial Z}{\partial T} ight)_P \frac{d P}{P}-\int_0^P(Z-1) \frac{d P}{P} \quad ext { (const } T ext { ) } [/latex] (6.48)

[latex] \int_0^P\left(\frac{\partial Z}{\partial T} ight)_P \frac{d P}{P} \quad \int_0^P(Z-1) \frac{d P}{P} [/latex]

Graphical integration requires simple plots of (∂Z/∂T )P/P and (Z −1)/P vs. P. Values of (Z−1)/P are found from the compressibility-factor data at 360 K. The quantity (∂Z/∂T )P/P requires evaluation of the partial derivative (∂Z/∂T)P, given by the slope of a plot of Z vs. T at constant pressure. For this purpose, separate plots are made of Z vs. T for each pressure at which compressibility-factor data are given, and a slope is determined at 360 K for each curve (for example, by construction of a tangent line at 360 K or by fitting the data points to a simple function and evaluating its slope at 360 K). Data for the required plots are shown in Table 6.2.

Table 6.2: Values of the Integrands Required in Ex. 6.3
Values in parentheses are by extrapolation.

P bar	[latex] [(\partial Z / \partial T) p / P] imes 10^4 K ^{-1} \cdot bar ^{-1} [/latex]	[latex] [-(Z-1) / P] imes 10^2 bar ^{-1} [/latex]
0.00	(1.780)	(2.590)
0.10	1.700	2.470
0.50	1.514	2.186
2.00	1.293	1.759
4.00	1.290	1.591
6.00	1.395	1.544
8.00	1.560	1.552
10.00	1.777	1.592
12.00	2.073	1.658
14.00	2.432	1.750
15.41	(2.720)	(1.835)

The values of the two integrals, as determined from the plots, are:

[latex] \int_0^P\left(\frac{\partial Z}{\partial T} ight)_P \frac{d P}{P}=26.37 imes 10^{-4} K ^{-1} \quad \int_0^P(Z-1) \frac{d P}{P}=-0.2596 [/latex]

By Eq. (6.46),

[latex] \frac{H^R}{R T}=-(360)\left(26.37 imes 10^{-4} ight)=-0.9493 [/latex]

By Eq. (6.48),

[latex] \frac{S^R}{R}=-0.9493-(-0.2596)=-0.6897 [/latex]

[latex] ext { For } R=8.314 J \cdot mol ^{-1} \cdot K ^{-1} [/latex]

[latex] H^R=(-0.9493)(8.314)(360)=-2841.3 J \cdot mol ^{-1} [/latex]

[latex] S^R=(-0.6897)(8.314)=-5.734 J \cdot mol ^{-1} \cdot K ^{-1} [/latex]

Values of the integrals in Eqs. (6.50) and (6.51), with parameters from the given equation for [latex]C_P^{i g} / R [/latex] , are:

[latex] H=H_0^{i g}+\int_{T_0}^T C_P^{i g} d T+H^R [/latex] (6.50)

[latex] S=S_0^{i g}+\int_{T_0}^T C_P^{i g} \frac{d T}{T}-R \ln \frac{P}{P_0}+S^R [/latex] (6.51)

[latex] 8.314 imes ICPH \left(300,360 ; 1.7765,33.037 imes 10^{-3}, 0.0,0.0 ight)=6324.8 J \cdot mol ^{-1} [/latex]

[latex] 8.314 imes \operatorname{ICPS}\left(300,360 ; 1.7765,33.037 imes 10^{-3}, 0.0,0.0 ight)=19.174 J \cdot mol ^{-1} \cdot K ^{-1} [/latex]

Substitution of numerical values into Eqs. (6.50) and (6.51) yields:

[latex]H = 18,115.0 + 6324.8 − 2841.3 = − 21,598.5 J ⋅mol^{-1} [/latex]

[latex]S = 295.976 + 19.174 − 8.314 ln 15.41 − 5.734 = 286.676 J ⋅mol^{-1} ⋅K^{-1} [/latex]

Although calculations are here carried out for just one state, enthalpies and entropies can be evaluated for any number of states, given adequate data. All of these processes, described here as graphical operations, are readily automated in a spreadsheet or simple computer program. After having completed a set of calculations, one is not irrevocably committed to the particular values of [latex] H_0^{i g} and S_0^{i g}[/latex] initially assigned. The scale of values for either the enthalpy or the entropy can be shifted by addition of a constant to all values. In this way one can give arbitrary values to H and S for some particular state so as to make the scales convenient for some particular purpose.

Question 6.3: Calculate the enthalpy and entropy of saturated isobutane va...

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