Useful in several parts of the problem are the number of moles:
n=\frac{M}{M W}=\frac{120~ \mathrm{lb}_{\mathrm{m}}}{16 \frac{\mathrm{lb}_{\mathrm{m}}}{\mathrm{lb~mol}}}=7.5 ~\mathrm{lb~mol}
And the molar volume:
\underline{V}=\frac{6 \mathrm{ft}^3}{7.5 \mathrm{lb~mol}}=0.8 \frac{\mathrm{ft}^3}{\mathrm{lb~mol}}
Or equivalently:
\underline{V}=\left(0.8 \frac{\mathrm{ft}^3}{\mathrm{lb~mol}}\right)\left(\frac{12 \mathrm{~in}}{1 \mathrm{~ft}}\right)^3\left(\frac{2.54 \mathrm{~cm}}{1 \mathrm{~in}}\right)^3\left(\frac{1 \mathrm{~lb~mol}}{453.6 \mathrm{~gmol}}\right)=49.94 \frac{\mathrm{cm}^3}{\mathrm{~mol}}
The temperature -100°F is equivalent to (-100+459.6)=359.6°R
Which is also equivalent to (5/9)(359.6)=199.8 K.
A) Applying the ideal gas law:
P=\frac{R T}{\widehat{V}}=\frac{\left(10.731 \frac{\mathrm{psia} \,\cdot \,\mathrm{ft}^3}{\mathrm{lb~mol} \,\cdot\,{ }^{\circ} \mathrm{R}}\right)\left(359.6^{\circ} \mathrm{R}\right)}{0.8 \frac{\mathrm{ft}^3}{\mathrm{lb~mol}}}=(4824 \,\mathrm{psia})\left(\frac{1 \mathrm{~atm}}{14.696 \,\mathrm{psia}}\right)=\mathbf{3 2 8} \mathbf{~ a t m}
B) For the van der Waals equation of state, from section 7.2.7:
a=\frac{27}{64} \frac{R^2 T_c^2}{P_c} \quad \text { and } \quad b=\frac{R T_c}{8 P_c}
Here using data from Appendix C for methane:
\begin{gathered}a=\frac{27}{64} \frac{R^2 T_c^2}{P_c} \quad \text { and } \quad b=\frac{R T_c}{8 P_c} \\ a=\frac{27}{64} \frac{\left(83.14 \frac{\mathrm{bar~cm}^3}{\mathrm{~mol} \mathrm{~K}}\right)^2(190.56 \mathrm{~K})^2}{(45.99 \mathrm{~bar})} \quad \text { and } \quad b=\frac{\left(83.14 \frac{\mathrm{bar~cm}^3}{\mathrm{~mol} \mathrm{~K}}\right)(190.56 \mathrm{~K})}{8(45.99 \mathrm{~bar})} \\ a=2.303 \times 10^6 \frac{\mathrm{bar} \mathrm{cm}^6}{\mathrm{~mol}^2} \quad \text { and } \quad b=43.06 \frac{\mathrm{cm}^3}{\mathrm{~mol}^3} \\ P=\frac{\left(83.14 \frac{\mathrm{bar~cm}^3}{\mathrm{~mol} \mathrm{~K}}\right)(199.8 \mathrm{~K})}{49.94-43.06 \frac{\mathrm{cm}^3}{\mathrm{~mol}}}-\frac{2.303 \times 10^6 \frac{\mathrm{bar} \mathrm{~cm}^6}{\mathrm{~mol}^2}}{\left(49.94 \frac{\mathrm{cm}^3}{\mathrm{~mol}^2}\right)^2}=(1372 \mathrm{~bar})\left(\frac{1 \mathrm{~atm}}{1.013 \mathrm{~bar}}\right)\\ =\bf 1355 ~atm\end{gathered}
C) Applying the expressions from section 7.2.7 and data from Appendix C to find the Soave a and b:
\begin{gathered}m=0.480+1.574 \omega-0.176 \omega^2 \\ m=0.480+1.574(0.011)-0.176(0.011)^2=0.497 \\ \alpha=\left[1+m\left(1-T_r^{0.5}\right)\right]^2 \\ \alpha=\left[1+(0.497)\left(1-\left(\frac{199.8 \mathrm{~K}}{190.56 \mathrm{~K}}\right)^{0.5}\right)\right]^2 \\ \alpha=0.9763 \\ a_c=0.42747 R^2 \frac{T_C^2}{P_c} \\ a=0.42747 \frac{\left(83.14 \frac{\rm{ bar~cm }^3}{\text { mol } K}\right)^2(190.56 \mathrm{~K})^2}{(45.99 \text { bar })}=2.333 \times 10^6 \frac{\mathrm{bar} \mathrm{~cm}^6}{\mathrm{~mol}^2} \\ a_c \alpha=\left(2.333 \times 10^6 \frac{\mathrm{bar} \mathrm{~cm}^6}{\mathrm{~mol}^2}\right)(0.9763)=2.278 \times 10^6 \frac{\mathrm{bar} ~\mathrm{cm}^6}{\mathrm{~mol}^2} \\ b=0.08664 R \frac{T_c}{P_c}\\b=\frac{(0.08664)\left(83.14 \frac{\mathrm{bar~cm}^3}{\mathrm{~mol} \mathrm{~K}^3}\right)(190.56 \mathrm{~K})}{(45.99 \mathrm{~bar})}=29.85 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\end{gathered}
The Soave equation is:
\begin{gathered}P=\frac{R T}{\underline{V}-b}-\frac{a}{\underline{V}(\underline{V}+b)} \\ P=\frac{\left(83.14 \frac{\mathrm{bar~cm}^3}{\mathrm{~mol} \mathrm{~K}^3}\right)(199.8 \mathrm{~K})}{49.94-29.85 \frac{\mathrm{cm}^3}{\mathrm{~mol}}}-\frac{2.278 \times 10^6 \frac{\mathrm{bar} \mathrm{~cm}^6}{\mathrm{~mol}^2}}{\left(49.94 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\right)\left(49.94+29.85 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\right)}=255.2 \mathrm{~bar}\end{gathered}
This is equivalent to ~252 atm.
D) When using the Lee-Kessler approach, we must know P and T to find Z. Since P is the unknown, there is no clear way to proceed other than iterating. For example, if the answer from part C were the correct P, then by the definition of Z:
Z=\frac{P \underline{V}}{R T}=\frac{(255 \,\mathrm{bar})\left(49.94 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\right)}{\left(83.14 \frac{\mathrm{bar~cm}^3}{\mathrm{~mol} \mathrm{~K}}\right)(199.8 \mathrm{~K})}=0.767
But is this the Z we get from the Lee Kessler approach?
\begin{gathered}P_r=\frac{P}{P_c}=\frac{255}{45.99}=5.54 \\T_r=\frac{T}{T_c}=\frac{199.8}{190.56}=1.05\end{gathered}
From the figures Z^0 \sim 0.75 and Z^1 \sim-0.18
\begin{gathered}Z=Z^0+\omega Z^1 \\Z=0.75+(0.011)(-0.13) \sim 0.75\end{gathered}
This is slightly smaller than the value we obtained from the definition of Z, but is probably within the accuracy of the graphs. To check, we could insert this Z into the definition of Z and compute a new P :
P=\frac{Z R T}{\underline{V}}=\frac{0.75\left(83.14 \frac{\mathrm{bar~cm}^3}{\mathrm{~mol} \mathrm{~K}}\right)(199.8 \mathrm{~K})}{\left(49.94 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\right)}=249 \,\mathrm{bar}
P_r=\frac{249}{45.99}=5.42
And when we use this \rm P_r(T_r is still 1.05) in the figure, we get Z=0.75, so this is a correct answer. But due to the accuracy of the figures we can’t realistically say it has more than two significant figures:
P~250 bar~250 atm
E)
\rho=\frac{120 \,l b_m}{6 \,f t^3}=20 \frac{l b_m}{f t^3}
From the figure, with T = -100°F and ρ=20 \rm lb_m/ft³, P~3300 psia. This is equivalent to ~225 atm.