A new and relatively unstudied compound is being investigated for its potential use as a refrigerant. The compound has a critical temperature { T }_{ c }=500 K and { P }_{ c }=30 bar. The vapor pressure at T=350 K is { P }^{ sat }=4 bar. Estimate the molar volume of the compound at each of the following conditions:
A) In the liquid phase at T=350 K and P=6 bar
B) In the vapor phase at T=350 K and P=3 bar
C) In the vapor phase at T=400 K and P=6 bar
D) In the liquid phase at T=300 K and P=1 bar
A) We will be using the Peng Robinson Equation of State
P=\frac{R T}{\underline{V}-b}-\frac{a}{\underline{V}(\underline{V}+b)+b(\underline{V}-b)}
Determine reduced properties
\mathrm{T}_{\mathrm{r}}=\frac{\mathrm{T}}{\mathrm{T}_{\mathrm{c}}}=\frac{350 \mathrm{~K}}{500 \mathrm{~K}}=0.7
\mathrm{P}_{\mathrm{r}}=\frac{\mathrm{P}}{\mathrm{P}_{\mathrm{c}}}=\frac{6~ \mathrm{bar}}{30 ~\mathrm{bar}}=0.2
Find the acentric factor
\omega=-1-\log _{10} \frac{P^{\text {sat }}}{P_c}=-1-\log _{10} \frac{4 \text { bar }}{30~ \mathrm{bar}}=-0.124
NOTE: Very few compounds have negative acentric factors, with hydrogen and helium being the prime examples that do. However, conceptually and mathematically, the fact that the acentric factor is negative presents no barrier in the solution of the problem.
Find k
\mathrm{k}=0.3746+1.54226 \omega-0.2699 \,\omega^2=0.179
Find α
\alpha=\left(1+\mathrm{k}\left(1-\mathrm{T}_{\mathrm{r}}^{0.5}\right)\right)^2=1.059
Find \mathrm{a}_{\mathrm{c}}
\mathrm{a}_{\mathrm{c}}=\frac{0.45724 \mathrm{R}^2 \mathrm{~T}_{\mathrm{c}}^2}{\mathrm{P}_{\mathrm{c}}}=2.63 * 10^{-5} \frac{\mathrm{m}^6 ~\mathrm{bar}}{\mathrm{mol}^2}
Find a
\mathrm{a}=\mathrm{a}_{\mathrm{c}} \alpha=2.785 \times 10^{-5} \frac{\mathrm{m}^6~ \mathrm{bar}}{\mathrm{mol}^2}
Find b
\mathrm{b}=\frac{0.07780 \mathrm{RT}_{\mathrm{c}}}{\mathrm{P}_{\mathrm{c}}}=1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}
Plugging in to the P.R equation
\begin{aligned}& 6~ \mathrm{bar}\\&=\frac{\left(8.314 \times 10^{-5} \frac{\mathrm{m}^3 ~\mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right) 350 \mathrm{~K}}{\underline{\mathrm{V}}-1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}} \\& -\frac{2.785 \times 10^{-5} \frac{\mathrm{m}^6~ \mathrm{bar}}{\mathrm{mol}^2}}{\underline{\mathrm{V}}\left(\underline{\mathrm{V}}+\left(1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)+1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}\left(\underline{\mathrm{V}}-\left(1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)}\\& \bf \underline{V}^{\mathrm{L}}=0.000155 \frac{\mathbf{m}^3}{\mathbf{~mol}}\end{aligned}
B) At constant temperatures, nothing changes in the P.R equation except for Pressure input. Therefore, we may use the values obtained in part A)
\begin{aligned}& 3 \text { bar } \\& =\frac{\left(8.314 \times 10^{-5} \frac{\mathrm{m}^3~ \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right) 350 \mathrm{~K}}{\underline{\mathrm{V}}-1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}} \\& -\frac{2.785 \times 10^{-5} \frac{\mathrm{m}^6 ~\mathrm{bar}}{\mathrm{mol}^2}}{\underline{\mathrm{V}}\left(\underline{\mathrm{V}}+\left(1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)+1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}\left(\underline{\mathrm{V}}-\left(1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)} \\& \bf\underline{V}^{{V}}=0.008789 \frac{{m}^3}{{~mol}} \end{aligned}
C) Determine reduced properties
\begin{aligned}& \mathrm{T}_{\mathrm{r}}=\frac{\mathrm{T}}{\mathrm{T}_{\mathrm{c}}}=\frac{400 \,K}{500 \,K}=0.8 \\& \mathrm{P}_{\mathrm{r}}=\frac{\mathrm{P}}{\mathrm{P}_{\mathrm{c}}}=\frac{6~ \mathrm{bar}}{30~ \mathrm{bar}}=0.2\end{aligned}
Find the accentric factor
\omega=-1-\log _{10} \frac{\mathrm{P}^{\text {sat }}}{\mathrm{P}_{\mathrm{c}}}=-1-\log _{10} \frac{4 ~\mathrm{bar}}{30~ \mathrm{bar}}=-0.124
Find k
\mathrm{k}=0.3746+1.54226 \omega-0.2699 \omega^2=0.179
Find α
\alpha=\left(1+\mathrm{k}\left(1-\mathrm{T}_{\mathrm{r}}^{0.5}\right)\right)^2=1.038
Find a_c
\mathrm{a}_{\mathrm{c}}=\frac{0.45724 \mathrm{R}^2 \mathrm{~T}_{\mathrm{c}}^2}{\mathrm{P}_{\mathrm{c}}}=2.63 \times 10^{-5} \frac{\mathrm{m}^6 ~\mathrm{bar}}{\mathrm{mol}^2}
Find a
\mathrm{a}=\mathrm{a}_{\mathrm{c}} \alpha=2.729 \times 10^{-5} \frac{\mathrm{m}^6~ \mathrm{bar}}{\mathrm{mol}^2}
Find b
\mathrm{b}=\frac{0.07780 \mathrm{RT}_{\mathrm{c}}}{\mathrm{P}_{\mathrm{c}}}=1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}
Plugging in to the P.R equation
\begin{aligned}& 6 \text { bar } \\& =\frac{\left(8.314 \times 10^{-5} \frac{\mathrm{m}^3~ \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right) 400 \mathrm{~K}}{\underline{\mathrm{V}}-1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}} \\& -\frac{2.729 \times 10^{-5} \frac{\mathrm{m}^6~ \mathrm{bar}}{\mathrm{mol}^2}}{\underline{\mathrm{V}}\left(\underline{\mathrm{V}}+\left(1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)+1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}^3}\left(\underline{\mathrm{V}}-\left(1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)} \\& \underline{\mathbf{V}}^{\mathbf{V}}=\mathbf{0 . 0 0 4 7 5 6 \frac { \mathbf { m } ^ { 3 } } { \mathbf { m o l } ^ { 3 } }}\end{aligned}
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{300 \mathrm{~K}}{500 \mathrm{~K}}=0.6 \\& \mathrm{P}_{\mathrm{r}}=\frac{\mathrm{P}}{\mathrm{P}_{\mathrm{c}}}=\frac{1~ \mathrm{bar}}{30~ \mathrm{bar}}=0.033\end{aligned}
Find the accentric factor
\omega=-1-\log _{10} \frac{P^{\mathrm{sat}}}{\mathrm{P}_{\mathrm{c}}}=-1-\log _{10} \frac{4~ \mathrm{bar}}{30~ \mathrm{bar}}=-0.124
Find k
\mathrm{k}=0.3746+1.54226 \omega-0.2699 \omega^2=0.179
Find α
\alpha=\left(1+\mathrm{k}\left(1-\mathrm{T}_{\mathrm{r}}^{0.5}\right)\right)^2=1.08
Find \mathrm{a}_{\mathrm{c}}
\mathrm{a}_{\mathrm{c}}=\frac{0.45724 \mathrm{R}^2 \mathrm{~T}_{\mathrm{c}}^2}{\mathrm{P}_{\mathrm{c}}}=2.63 \times 10^{-5} \frac{\mathrm{m}^6 ~\mathrm{bar}}{\mathrm{mol}^2}
Find a
\mathrm{a}=\mathrm{a}_{\mathrm{c}} \alpha=2.840 \times 10^{-5} \frac{\mathrm{m}^6~ \mathrm{bar}}{\mathrm{mol}^2}
Find b
\mathrm{b}=\frac{0.07780 \mathrm{RT}_{\mathrm{c}}}{\mathrm{P}_{\mathrm{c}}}=1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}
Plugging in to the P.R equation
\begin{aligned}& \rm 6~bar\\&=\frac{\left(8.314 \times 10^{-5} \frac{\mathrm{m}^3 ~\mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right) 300 \mathrm{~K}}{\underline{\mathrm{V}}-1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}} \\& -\frac{2.840 \times 10^{-5} \frac{\mathrm{m}^6 ~\mathrm{bar}}{\mathrm{mol}^2}}{\underline{\mathrm{V}}\left(\underline{\mathrm{V}}+\left(1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)+1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}\left(\underline{\mathrm{V}}-\left(1.078 \times 10^{-4} \frac{\mathrm{m}^3}{\mathrm{~mol}}\right)\right)} \\& \bf\underline{V}^{{L}}=0.000142 \frac{{m}^3}{{~mol}}\end{aligned}