Question 3.8: Reported values for the virial coefficients of isopropanol v...
Reported values for the virial coefficients of isopropanol vapor at 200°C are:
B = − 388 cm^{3} · mol^{−1} C = − 26,000 cm^{6} · mol^{−2}
Calculate V and Z for isopropanol vapor at 200°C and 10 bar:
(a) For the ideal-gas state; (b) By Eq. (3.36); (c) By Eq. (3.38).
Z =\frac{PV}{RT} = 1 + \frac{BP}{RT} (3.36)
Z = \frac{PV}{RT} = 1 + \frac{B}{V} +\frac{C}{V^{2} } (3.38)
The absolute temperature is T = 473.15 K, and the appropriate value of the gas constant for pressure in bar and volume in cm^{3}·mol^{−1} is R = 83.14 bar·cm^{3}·mol^{−1}·K^{−1}.
(a) For the ideal-gas state, Z = 1, and
V^{ig} = \frac{RT}{P}=\frac{(83.14 ) ( 473.15 )}{10} = 3934 cm^{3} · mol^{-1}
(b) From the second equality of Eq. (3.36), we have
V = \frac{RT}{P} + B = 3934 − 388 = 3546 cm^{3} · mol^{-1}
and
Z = \frac{PV}{RT} = \frac{V}{RT/P} = \frac{V}{V^{ig} } = \frac{3546}{3934} = 0.9014
(c) For convenient solution by iteration, Eq. (3.38) may be written:
V_{i+1} = \frac{RT}{P} \left(1+\frac{B}{V_{i} }+\frac{C}{V^{2}_{i} } \right)
where i is the iteration number. Iteration is initiated with the ideal-gas state valueV^{ig}. Iteration to convergence yields:
V = 3488 cm^{3} ·mol^{−1}
from which Z = 0.8866. In comparison with this result, the ideal-gas-state value is 13% too high, and the two-term virial equation of state, Eq. (3.36), gives a value 1.7% too high.