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

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

\frac{H^R}{R T}=-T \int_0^P\left(\frac{\partial Z}{\partial T}\right)_P \frac{d P}{P} \quad \text { (const } T \text { ) }      (6.46)

\frac{S^R}{R}=-T \int_0^P\left(\frac{\partial Z}{\partial T}\right)_P \frac{d P}{P}-\int_0^P(Z-1) \frac{d P}{P} \quad \text { (const } T \text { ) }      (6.48)

\int_0^P\left(\frac{\partial Z}{\partial T}\right)_P \frac{d P}{P} \quad \int_0^P(Z-1) \frac{d P}{P}

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.

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

\int_0^P\left(\frac{\partial Z}{\partial T}\right)_P \frac{d P}{P}=26.37 \times 10^{-4} K ^{-1}            \quad \int_0^P(Z-1) \frac{d P}{P}=-0.2596

By Eq. (6.46),

\frac{H^R}{R T}=-(360)\left(26.37 \times 10^{-4}\right)=-0.9493

By Eq. (6.48),

\frac{S^R}{R}=-0.9493-(-0.2596)=-0.6897

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

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

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

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

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    (6.51)

8.314 \times ICPH \left(300,360 ; 1.7765,33.037 \times 10^{-3}, 0.0,0.0\right)=6324.8  J \cdot mol ^{-1}

8.314 \times \operatorname{ICPS}\left(300,360 ; 1.7765,33.037 \times 10^{-3}, 0.0,0.0\right)=19.174  J \cdot mol ^{-1} \cdot K ^{-1}

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

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

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

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 H_0^{i g}  and  S_0^{i g} 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.