A)
\begin{aligned} & \mathrm{T}_{\mathrm{c}}=\mathrm{T}_{\mathrm{b}}\left[0.584+0.965 \sum \mathrm{T}_{\mathrm{b}}-\left(\sum \mathrm{T}_{\mathrm{b}}\right)^2\right]^{-1} \\ & \mathrm{P}_{\mathrm{c}}=\left(0.113+0.0032 \mathrm{n}_{\mathrm{a}}-\sum \mathrm{P}_{\mathrm{c}}\right)^{-2} \\ & \mathrm{~T}_{\mathrm{b}}=198.2+\sum \mathrm{T}_{\mathrm{b}}\end{aligned}
| Group |
Occurrence |
Tb |
Tc |
Pc |
| -CH3 |
2 |
23.58 |
0.0141 |
-0.0012 |
| -CH2- |
2 |
22.88 |
0.0189 |
0 |
| >CH- |
2 |
21.74 |
0.0164 |
0.002 |
| =CH- |
2 |
24.96 |
0.0129 |
-0.0006 |
| F |
1 |
-0.03 |
0.011 |
-0.0057 |
| Cl |
1 |
38.13 |
0.0105 |
-0.0049 |
\begin{array}{|c|c|} \hline\sum{\rm T_b} & 224.42\\ \hline \sum{\rm T_c} & 0.1461\\ \hline \sum{\rm P_c} & -0.0102\\ \hline \sum{\rm \Delta H_{form}} & -381.93\\ \hline \rm n_a & 24\\ \hline \end{array}
\begin{aligned} & \bf T_c=600.61 K \\ & \bf P_c=25~ \mathbf{bar}\end{aligned}
B) The compound is an octene with two halogen substitutions. Estimate the acentric factor to be about 0.39. This is based on octane having an acentric factor of 0.399, and the fact that butene has a slightly smaller (~0.01) acentric factor than butane, as propene has a slightly smaller acentric factor than propane.
C) While the analogy described above makes sense based upon the limited selection of compounds available in Appendix C, it does not account for the halogen substitutions in any way, and so can’t be considered very meaningful.
D)
P=\frac{R T}{\underline{V}-b}-\frac{a}{\underline{V}(\underline{V}+b)+b(\underline{V}-b)}
Determine reduced properties
\begin{aligned}& \mathrm{T}_{\mathrm{r}}=\frac{\mathrm{T}}{\mathrm{T}_{\mathrm{c}}}=\frac{500 \mathrm{~K}}{600.61 \mathrm{~K}}=0.8324 \\& \mathrm{P}_{\mathrm{r}}=\frac{\mathrm{P}}{\mathrm{P}_{\mathrm{c}}}=\frac{5~ \mathrm{bar}}{25~ \mathrm{bar}}=0.2\end{aligned}
We had found the accentric factor previously
ω = 0.39
Find k
k = 0.3746 + 1.54226ω − 0.2699 ω² = 0.935
Find α
\alpha=\left(1+\mathrm{k}\left(1-\mathrm{T}_{\mathrm{r}}^{0.5}\right)\right)^2=1.171
Find a_c
\mathrm{a}_{\mathrm{c}}=\frac{0.45724 \mathrm{R}^2 \mathrm{~T}_{\mathrm{c}}^2}{\mathrm{P}_{\mathrm{c}}}=4.56 \times 10^7 \frac{\mathrm{cm}^6 ~\mathrm{bar}}{\mathrm{mol}^2}
Find a
\mathrm{a}=\mathrm{a}_{\mathrm{c}} \alpha=5.34 \times 10^7 \frac{\mathrm{cm}^6~ \mathrm{bar}}{\mathrm{mol}^2}
Find b
\mathrm{b}=\frac{0.07780 \mathrm{RT}_{\mathrm{c}}}{\mathrm{P}_{\mathrm{c}}}=155.39 \frac{\mathrm{cm}^3}{\mathrm{~mol}}
\begin{aligned}& 5~ \mathrm{bar}=\frac{\left(83.14 \frac{\mathrm{cm}^3 ~\mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right) 500 \mathrm{~K}}{\underline{\mathrm{V}}-155.39 \frac{\mathrm{cm}^3}{\mathrm{~mol}}} \\& \quad-\frac{5.34 \times 10^7 \frac{\mathrm{cm}^6 ~\mathrm{bar}}{\mathrm{mol}^2}}{\underline{\mathrm{V}}\left(\underline{\mathrm{V}}+\left(155.39 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\right)\right)+155.39 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\left(\underline{\mathrm{V}}-\left(155.39 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\right)\right)} \\& \underline{V}^{\mathbf{L}}=\mathbf{2 3 6 . 7} \frac{\mathbf{c m}^3}{\mathbf{m o l}^{\mathbf{m o l}}} \\& \underline{\mathbf{V}}^{\mathbf{V}}=\mathbf{7 0 5 0} \frac{\mathbf{c m}^3}{\mathbf{m o l}^3}\end{aligned}
Unclear whether the compound is a liquid or a vapor at this temperature and pressure.
E)
P=\frac{\mathrm{RT}}{\underline{\mathrm{V}}-\mathrm{b}}-\frac{\mathrm{a}}{\underline{\mathrm{V}}(\underline{\mathrm{V}}+\mathrm{b})+\mathrm{b}(\underline{\mathrm{V}}-\mathrm{b})}
Determine reduced properties
\begin{aligned}& \mathrm{T}_{\mathrm{r}}=\frac{\mathrm{T}}{\mathrm{T}_{\mathrm{c}}}=\frac{800 \mathrm{~K}}{600.61 \mathrm{~K}}=1.33 \\& \mathrm{P}_{\mathrm{r}}=\frac{\mathrm{P}}{\mathrm{P}_{\mathrm{c}}}=\frac{5~ \mathrm{bar}}{25~ \mathrm{bar}}=0.2\end{aligned}
We had found the accentric factor previously
ω = 0.38
Find k
k = 0.3746 + 1.54226ω − 0.2699 ω² = 0.935
Find α
\alpha=\left(1+\mathrm{k}\left(1-\mathrm{T}_{\mathrm{r}}^{0.5}\right)\right)^2=0.734
Find a_c
\mathrm{a}_{\mathrm{c}}=\frac{0.45724 \mathrm{R}^2 \mathrm{~T}_{\mathrm{c}}^2}{\mathrm{P}_{\mathrm{c}}}=4.56 \times 10^7 \frac{\mathrm{cm}^6 ~\mathrm{bar}}{\mathrm{mol}^2}
Find a
\mathrm{a}=\mathrm{a}_{\mathrm{c}} \alpha=3.35 \times 10^7 \frac{\mathrm{cm}^6~ \mathrm{bar}}{\mathrm{mol}^2}
Find b
\begin{aligned}& \mathrm{b}=\frac{0.07780 \mathrm{RT}_{\mathrm{c}}}{\mathrm{P}_{\mathrm{c}}}=155.39 \frac{\mathrm{cm}^3}{\mathrm{~mol}} \\& 5~ \mathrm{bar}=\frac{\left(83.14 \frac{\mathrm{cm}^3 ~\mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right) 800 \mathrm{~K}}{\underline{\mathrm{V}}-155.39 \frac{\mathrm{cm}^3}{\mathrm{~mol}}} \\& -\frac{15.35 \times 10^7 \frac{\mathrm{cm}^6~ \mathrm{bar}}{\mathrm{mol}^2}}{\underline{\mathrm{V}}\left(\underline{\mathrm{V}}+\left(155.39 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\right)\right)+155.39 \frac{\mathrm{cm}^3}{\mathrm{~mol}^3}\left(\underline{\mathrm{V}}-\left(155.39 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\right)\right)} \\& \underline{\mathrm{V}}=\mathbf{1 2 9 6 0} \frac{\mathbf{c m}^3}{\mathbf{m o l}^3}\end{aligned}