Question 3.9: Given that the vapor pressure of n-butane at 350 K is 9.4573...

Values of T_{c }  and  P_{c} for n-butane from App. B yield:

T_{r} = \frac{350}{425.1} = 0.8233          and          P_{r}  = \frac{9.4573}{37.96} = 0.2491

Parameter q is given by Eq. (3.51) with Ω, Ψ, and α(T_{r}) for the RK equation from Table 3.1:

q = \frac{\Psi \alpha (T_{r};\omega )}{\Omega T_{r} }        (3.51)

Table 3.1: Parameter Assignments for Equations of State

 q = \frac{\Psi T^{-1/2}_{r} }{\Omega T_{r} } = \frac{\Psi }{\Omega }T^{-3/2}_{r} = \frac{0.42748}{0.08664} ( 0.8233 )^{-3/2}  = 6.6048

Parameter β is found from Eq. (3.50):

\beta = \Omega \frac{P_{r} }{T_{r} } = \frac{ ( 0.08664 ) ( 0.2491 )}{0.8233} = = 0.026214

(a) For the saturated vapor, we write the RK form of Eq. (3.48) that results upon substitution of appropriate values for ε and σ from Table 3.1:

Z = 1 +\beta – q\beta \frac{ Z- \beta }{(Z +εβ) (Z+ \sigma \beta )}         (3.48)

 Z = 1 + β  − qβ\frac{(Z-\beta )}{Z(Z+\beta )}

or

 Z = 1 + 0.026214  −  ​​( 6.6048 ) ( 0.026214)V\frac{Z  −  0.026214 )}{Z(Z + 0.026214)}

Solution by iteration starting from Z = 1 yields Z = 0.8305, and

V^{V} = \frac{ZRT}{P}= \frac{( 0.8305 ) ( 83.14 ) ( 350 )}{9.4573} = 2555  cm^{3}· mol^{−1}

An experimental value is 2482  cm^{3}· mol^{−1}

(b) For the saturated liquid, we apply Eq. (3.49) in its RK form:

Z = β + ( Z + εβ ) ( Z + σβ )\left(\frac{ 1+\beta -Z}{q\beta } \right)      (3.49)

Z = β + Z ( Z + β )\left(\frac{1 + \beta -Z}{q\beta } \right)

or

 Z = 0.026214 + Z ( Z + 0.026214 )\frac{()1.026214  −  Z )}{(6.6048)(0.026214)}

Solution by iteration yields Z = 0.04331, and

V^{l} = \frac{ZRT}{P} = \frac{ ( 0.04331 ) ( 83.14 ) ( 350 )}{9.4573} = 133.3  cm^{3}· mol^{−1}

An experimental value is 115.0  cm^{3}· mol^{−1}