A) Using Peng-Robinson Equation at its Critical Point
\begin{aligned}& \mathrm{k}=0.3746+1.54226 \omega-0.2699 \omega^2 \\& \mathrm{k}=0.3746+1.54226(0.216)-0.2699(0.216)^2=0.6866\end{aligned}
\begin{aligned}& \alpha=\left(1+\mathrm{k}\left(1-\mathrm{T}_{\mathrm{r}}^{0.5}\right)\right)^2 \\& \alpha=\left(1+0.6866\left(1-(1)^2\right)\right)^2=1 \\& \mathrm{a}_{\mathrm{c}}=\frac{0.45724 \mathrm{R}^2 \mathrm{~T}_{\mathrm{c}}^2}{\mathrm{P}_{\mathrm{c}}} \\& \mathrm{a}_{\mathrm{c}}=\frac{0.45724\left(8.314 \times 10^{-5} \frac{\mathrm{m}^3 \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right)^2(562.05 \mathrm{~K})^2}{(48.95 ~\mathrm{bar})} \\& \mathrm{a}_{\mathrm{c}}=2.03 \times 10^{-5} \frac{\mathrm{m}^6 \mathrm{bar}}{\mathrm{mol}^2} \\& \mathrm{a}=\mathrm{a}_{\mathrm{c}} \alpha\end{aligned}
\begin{aligned}& \mathrm{a}=2.03 \times 10^{-5} \frac{\mathrm{m}^6 \mathrm{bar}}{\mathrm{mol}^2} \times 1=2.03 \times 10^{-5} \frac{\mathrm{m}^6 \mathrm{bar}}{\mathrm{mol}^2} \\& \mathrm{~b}=\frac{0.07780 \mathrm{RT}_{\mathrm{c}}}{\mathrm{P}_{\mathrm{c}}} \\& \mathrm{b}=\frac{0.07780\left(8.314 \times 10^{-5} \frac{\mathrm{m}^3 \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right)(562.05 \mathrm{~K})}{(48.95~ \mathrm{bar})}=7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\end{aligned}
Plug a and b into the PR Equation
\begin{aligned}& P=\frac{R T}{\underline{V}-b}-\frac{a}{\underline{V}(\underline{V}+b)+b(\underline{V}-b)} \\& 48.95~ \mathrm{bar}=\frac{\left(8.314 \times 10^{-5} \frac{\mathrm{m}^3 \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right)(562.05 \mathrm{~K})}{\underline{\mathrm{V}}-\left(7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)} \\& -\frac{2.03 \times 10^{-5} \frac{\mathrm{m}^6 \mathrm{bar}}{\mathrm{mol}^2}}{\underline{\mathrm{V}}\left(\underline{\mathrm{V}}+\left(7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)+\left(7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\left(\underline{\mathrm{V}}-\left(7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)} \\& \underline{\mathrm{V}}^{\mathrm{L}}=\underline{\mathrm{V}}^{\mathrm{L}}=2.836 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}} \end{aligned}
B) Using the same calculations as in part A), we find
\mathrm{a}=2.96 \times 10^{-5} \frac{\mathrm{m}^6 \mathrm{bar}}{\mathrm{mol}^2}
Of course, b does not need to be recalculated because it is not temperature dependent
\mathrm{b}=7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}
Converting, we find
\begin{aligned}& \mathrm{P}=1 \mathrm{~atm}=1.01325~ \mathrm{bar} \\& \mathrm{T}=500^{\circ} \quad \mathrm{R}=277.8 \mathrm{~K}\end{aligned}
Plugging into the PR Equation
\begin{aligned}& 1.01325~ \mathrm{bar}=\frac{\left(8.314 \times 10^{-5} \frac{\mathrm{m}^3 \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right)(277.8 \mathrm{~K})}{\underline{\mathrm{V}}-\left(7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)} \\& -\frac{2.96 \times 10^{-5} \frac{\mathrm{m}^6 \mathrm{bar}}{\mathrm{mol}^2}}{\underline{\mathrm{V}}\left(\underline{\mathrm{V}}+\left(7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)+\left(7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\left(\underline{\mathrm{V}}-\left(7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)} \\& \underline{\mathrm{V}}^{\mathrm{L}}=8.676 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\end{aligned}
C) Using the same calculations as in part B), we find
\mathrm{a}=2.36 \times 10^{-5} \frac{\mathrm{m}^6 \mathrm{bar}}{\mathrm{mol}^2}
Of course, b does not need to be recalculated because it is not temperature dependent
\mathrm{b}=7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}
Converting, we find
\begin{aligned}& \mathrm{P}=4 \mathrm{~atm}=4.053 ~\mathrm{bar} \\& \mathrm{T}=800^{\circ} \quad \mathrm{R}=444.4 \mathrm{~K}\end{aligned}
Plugging into the PR Equation
\begin{aligned}& \text { 4.053 bar } \\& =\frac{\left(8.314 \times 10^{-5} \frac{\mathrm{m}^3 \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right)(444.4 \mathrm{~K})}{\underline{\mathrm{V}}-\left(7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)} \\& -\frac{2.96 \times 10^{-5} \frac{\mathrm{m}^6 \mathrm{bar}}{\mathrm{mol}^2}}{\underline{\mathrm{V}}\left(\underline{\mathrm{V}}+\left(7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)+\left(7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\left(\underline{\mathrm{V}}-\left(7.427 \times 10^{-5} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)} \\& \underline{\mathrm{V}}^{\mathrm{V}}=8.637 \times 10^{-3} \frac{\mathrm{m}^3}{\mathrm{~mol}^3}\end{aligned}
D) The change in molar enthalpy when benzene is heated and compressed from T=500°R and P=1 atm to T=800°R and P=4 atm
State 1:
\begin{aligned}& \mathrm{T}_1=500^{\circ} \mathrm{R}=277.8 \mathrm{~K} \\& \mathrm{P}_1=1 \mathrm{~atm}=1.01325 ~\mathrm{bar} \\& \mathrm{T}_{\mathrm{r} 1}=\frac{277.8 \mathrm{~K}}{562.05 \mathrm{~K}}=0.4943 \\& \mathrm{P}_{\mathrm{r} 1}=\frac{1.013 \,\mathrm{bar}}{48.95~ \mathrm{bar}}=0.0207\end{aligned}
State 2:
\begin{aligned}& \mathrm{T}_2=800^{\circ} \mathrm{R}=444.4 \mathrm{~K} \\& \mathrm{P}_2=4 \mathrm{~atm}=4.053 \mathrm{bar} \\& \mathrm{T}_{\mathrm{r} 2}=\frac{444.4 \mathrm{~K}}{562.05 \mathrm{~K}}=0.7907 \\& \mathrm{P}_{\mathrm{r} 2}=\frac{4.053~ \mathrm{bar}}{48.95~ \mathrm{bar}}=0.0828\end{aligned}
The equation we will be using to find the difference of molar enthalpies
\underline{\mathrm{H}}_2-\underline{\mathrm{H}}_1=\left(\underline{\mathrm{H}}_2-\underline{\mathrm{H}}_2^{\mathrm{ig}}\right)+\left(\underline{\mathrm{H}}_2^{\mathrm{ig}}-\underline{\mathrm{H}}_1^{\mathrm{ig}}\right)-\left(\underline{\mathrm{H}}_1-\underline{\mathrm{H}}_1^{\mathrm{ig}}\right)
Using the Lee Kesler Correlation Residuals Enthalpy Values
\begin{aligned}& \frac{\left(\underline{\mathrm{H}}_1-\underline{\mathrm{H}}_1^{\mathrm{ig}}\right)}{\mathrm{RT}_{\mathrm{c}}}=\left[\frac{\underline{\mathrm{H}}-\underline{\mathrm{H}}_{\mathrm{ig}}}{\mathrm{RT}_{\mathrm{c}}}\right]^0+\omega\left[\frac{\underline{\mathrm{H}}-\underline{\mathrm{H}}_{\mathrm{ig}}}{\mathrm{RT}}\right]_{\mathrm{c}}^1=-5.482+(0.216)(-8.942)=-7.413 \\& \left(\underline{\mathrm{H}}_1-\underline{\mathrm{H}}_1^{\mathrm{ig}}\right)=(-7.413)\left(8.314 \frac{\mathrm{J}}{\mathrm{mol~K}}\right)(562.05 \mathrm{~K})=-34.64 \frac{\mathrm{kJ}}{\mathrm{mol}} \\& \frac{\left(\underline{\mathrm{H}}_2-\underline{\mathrm{H}}_2^{\mathrm{ig}}\right)}{\mathrm{RT}_{\mathrm{c}}}=\left[\frac{\underline{\mathrm{H}}-\underline{\mathrm{H}}_{\mathrm{ig}}}{\mathrm{RT}_{\mathrm{c}}}\right]^0+\omega\left[\frac{\underline{\mathrm{H}}-\underline{\mathrm{H}}_{\mathrm{ig}}}{\mathrm{RT}}\right]_{\mathrm{c}}^1=-0.134+(0.216)(-0.199)=-0.177 \\& \left(\underline{\mathrm{H}}_2-\underline{\mathrm{H}}_2^{\mathrm{ig}}\right)=(-0.177)\left(8.314 \frac{\mathrm{J}}{\mathrm{mol~K}}\right)(562.05 \mathrm{~K})=-0.827 \frac{\mathrm{kJ}}{\mathrm{mol}}\end{aligned}
Find the \left(\underline{H}_2^{\mathrm{ig}}-\underline{H}_1^{\mathrm{ig}}\right)
\left(\underline{\mathrm{H}}_2^{\mathrm{ig}}-\underline{\mathrm{H}}_1^{\mathrm{ig}}\right)=\mathrm{dH}=\mathrm{C}_{\mathrm{P}}^* \mathrm{dT}
Using Appendix D
\frac{C_P^*}{\mathrm{R}}=\mathrm{A}+\mathrm{BT}+\mathrm{CT}^2+\mathrm{DT}^3+\mathrm{ET}^4
\begin{array}{|c|c|c|c|c|c|c|}\hline \bf{Name} & \bf {Formula} & \bf A & \bf B\times10^3 & \bf C\times10^5 & \bf D\times10^8 & \bf E\times10^{11}\\ \hline \rm{Benzene} & \rm C_6H_6 & 3.551 & -6.184 & 14.365 & -19.807 & 8.234\\ \hline\end{array}
\begin{aligned}&\mathrm{C}_{\mathrm{P}}^*=\mathrm{R}\left(3.551-6.184 \times 10^{-3} \mathrm{~T}+1.437 \times 10^{-4} \mathrm{~T}^2-1.981 \times 10^{-7} \mathrm{~T}^3+8.234\right. \\& \qquad\times 10^{-11}\rm T^4)\\&\mathrm{dH}=\int_{277.8 \mathrm{~K}}^{444.4 \mathrm{~K}} \mathrm{x}\left(3.551-6.184 \times 10^{-3} \mathrm{~T}+1.437 * 10^{-4} \mathrm{~T}^2-1.981 \times 10^{-7} \mathrm{~T}^3+8.234\right.\\ & \qquad \times 10^{-11}\rm T^4)\rm dT \\\\ & \mathrm{dH}=16.78 \frac{\mathrm{kJ}}{\mathrm{mol}}\\\\ & \underline{\mathrm{H}}_2-\underline{\mathrm{H}}_1=\left(-0.827 \frac{\mathrm{kJ}}{\mathrm{mol}}\right)+\left(16.78 \frac{\mathrm{kJ}}{\mathrm{mol}}\right)-\left(-34.64 \frac{\mathrm{kJ}}{\mathrm{mol}}\right)=\mathbf{5 0 . 5 9} \frac{\mathbf{k J}}{\mathbf{m o l}}\end{aligned}