Question 6.6: Estimate V, H^R, and S^R for an equimolar mixture of carbon ...

The pseudocritical parameters are found by Eqs. (6.78) through (6.80) with critical constants from Table B.1 of App. B:

$\omega \equiv \sum\limits_{i}{y_{i}\omega _{i } }$     (6.78)

$P_{pc} \equiv \sum\limits_{i}{y_{i} P_{c_{i} } }$    (6.80)

$ω = y_1 ω_1 + y_2 ω_2 = ( 0.5 ) ( 0.224 ) + ( 0.5 ) ( 0.152 ) = 0.188$

$T_{pc } = y _1 T_{c 1} + y _2 T_{c 2} = ( 0.5 ) ( 304.2 ) + ( 0.5 ) ( 369.8 ) = 337.0 K$

$P pc = y_1 P c_1 + y_2 P c _2 = ( 0.5 ) ( 73.83 ) + ( 0.5 ) ( 42.48 ) = 58.15  bar$

Then,

$T_{pr} =\frac{450}{337.0} = 1.335           P_{pr} = \frac{140}{58.15} =2041$

Values of $Z^0 and Z^1$ from Tables D.3 and D.4 at these reduced conditions are:

$Z^0 = 0.697 and Z^1 = 0.205$

By Eq. (3.57),

$Z = 1 + \frac{BP}{RT} = 1+ \left(\frac{BP_{c} }{R T_{c} } \right) \frac{P_{r} }{T_R} = 1 + \widehat{B} \frac{P_{r} }{T_R}$      (3.57)

$Z = Z^0 + ω Z^1 = 0.697 + ( 0.188 ) ( 0.205 ) = 0.736$

Thus,

$V = \frac{ZRT}{P} = \frac{( 0.736 ) ( 83.14 ) ( 450 )}{140} = 196.7 cm^3 ⋅mol^−1$

Similarly, from Tables D.7 and D.8, with substitution into Eq. (6.66):

$\frac{H^{r} }{RT_c} = \frac{( H^{R} )^{0} }{R t_c} + ω \frac{( H^{R} )^{1} }{R t_c}$      (6.66)

$(\frac{H^{R} }{RT_{pc} } )^{0} = − 1.730 (\frac{H^{R} }{RT_{pc} } )^{1} = − 0.169$

$\frac{H^{R} }{R T_{pc} } = = − 1.730 + ( 0.188 ) ( − 0.169 ) = − 1.762$

and

$H^R = ( 8.314 ) ( 337.0 ) ( −1.762 ) = − 4937  J ⋅mol^−1$

From Tables D.11 and D.12 and substitution into Eq. (6.67),

and