Question 3.10: Determine the molar volume of n-butane at 510 K and 25 bar b...
Determine the molar volume of n-butane at 510 K and 25 bar based on each of the following:
(a) The ideal-gas state.
(b) The generalized compressibility-factor correlation.
(c) Equation (3.57), with the generalized correlation for \widehat{B} .
Z = 1 + \frac{BP}{RT}= 1 + \left(\frac{BP_{c} }{RT_{C} } \right) \frac{P_{r} }{T_{r} }= 1 + \widehat{B} \frac{P_{r} }{T_{r} } (3.57)
(d) Equation (3.63), with the generalized correlations for \widehat{B} and \widehat{C} .
Z = 1 + \widehat{B}\frac{P_{r} }{T_{r}Z } + \widehat{C} (\frac{P_{r} }{T_{r}Z })^{2} (3.63)
(a) For the ideal-gas state,
V = \frac{RT}{P} = \frac{ ( 83.14 ) ( 510 )}{25} = 1696.1 cm^{3}· mo^{-1}
(b) With values of T_{c} and P_{c} given in Table B.1 of App. B,
T_{r} = \frac{510}{425.1} = 1200 P_{r} = \frac{25}{37.96}= 0.659
Interpolation in Tables D.1 and D.2 then provides:
Z^{0} = 0.865 Z^{1} = 0.038
By Eq. (3.53) with ω = 0.200,
Z = Z^{0} + ω Z^{1} = 0.865 + ( 0.200 ) ( 0.038 ) = 0.873
V = \frac{ZRT}{P} = \frac{ ( 0.873 ) ( 83.14 ) ( 510 )}{25} = 1480.7 cm^{3}· mol^{−1}
If Z^{1}, the secondary term, is neglected, Z = Z^{0} = 0.865. This two-parameter corresponding- states correlation yields V = 1467.1 cm^{3}· mol^{−1} , which is less than 1% lower than the value given by the three-parameter correlation.
(c) Values of B^{0} and B^{1} are given by Eqs. (3.61) and (3.62):
B^{0}= 0.083 − \frac{0.422}{T^{1.6}_{r} } (3.61)
B^{1} = 0.139 – \frac{0.1722}{T^{4.2}_{r}} (3.62)
B^{0}= −0.232 B^{1} = 0.059Equations (3.59) and (3.57) then yield:
\widehat{B} = B^{0}+ ω B^{1}= −0.232 + ( 0.200 ) ( 0.059 ) = −0.220
Z = 1+ ( −0.220 )\frac{0.659}{1.200}= 0.879
from which V = 1489.1 cm^{3}· mol^{−1} , a value less than 1% higher than that given by the full Lee/Kesler compressibility-factor correlation.
(d) Values of C^{0} and C^{1} are given by Eqs. (3.66) and (3.67):
C^{0} = 0.01407 + \frac{ 0.02432}{T_{r}} – \frac{0.00313}{T^{10.5}_{r} } (3.66)
C^{1} = −0.02676 + \frac{ 0.05539}{T^{2.7}_{r}} – \frac{0.00242}{T^{10.5}_{r}} (3.67)
C^{0}= 0.0339 C^{1} = 0.0067Equation (3.65) then yields:
\widehat{C} =C^{0} + ωC^{1} = 0.0339 + ( 0.200 ) ( 0.0067 ) = 0.0352With this value of \widehat{C} and the value of \widehat{B} from part (c), Eq. (3.63) becomes,
Z = 1 + ( −0.220 )\left(\frac{0.659}{1.200Z} \right)+( 0.0352 ) \left(\frac{0.659}{1.200 Z} \right)^{2}
Solution for Z yields Z = 0.876 and V = 1485.8 cm^{3}· mol^{−1} . The value of V differs from that of part (c) by about 0.2%. An experimental value for V is 1480.7 cm^3·mol^–1. Significantly, the results of parts (b), (c), and (d) are in excellent agreement. Mutual agreement at these conditions is suggested by Fig. 3.13.
