Question 7.18: You are designing a process in which toluene is used as a so...

A) According to Appendix C, the critical properties of toluene are \rm T_c=591.75 K, \rm P_c=41.08 bar and ω=0.264. The Peng-Robinson equation is:

P=\frac{R T}{\underline{V}-b}-\frac{a}{\underline{V}(\underline{V}+b)+b(\underline{V}-b)}

The expressions in Section 7.2.7 are used to find a and b.When \rm  T=T_c, the values are a=2.35 x 10^7\,\rm bar~cm^6/mol² and b=93.2 cm³/mol.

When \rm T=T_c and \rm P=P_c, the solution is V=385 cm³/mol.

B) When P=1 bar and T=300 K, a=4.01 x 10^7\,\rm bar~cm^6/mol²

There are solutions for V at 1404 cm³/mol and ~23,320 cm³/mol, but the smallest solution is the one corresponding to the liquid phase. This is V=107.4 cm³/mol.

C) When P=3 bar and T=500 K, the three solutions for V are 150.8, 414 and 13,200 cm³/mol. The largest is the vapor solution.

D) The change in molar enthalpy determined using residual properties is:

\underline{H}_2-\underline{H}_1=\left(\underline{H}_2-\underline{H}_2^{i g}\right)+\left(\underline{H}_2^{i g}-\underline{H}_1^{i g}\right)-\left(\underline{H}_1-\underline{H}_1^{i g}\right)

We know from the problem statement that state (1) is a liquid and state (2) is a vapor, and we will compute the residual molar enthalpy for each.

The expression for residual molar enthalpy as given in Section 7.2.8 is:

\frac{\underline{H}^R}{R T}=(Z-1)-\left\{\left(\frac{A}{B \sqrt{8}}\right)\left(1+\frac{\kappa \sqrt{T_r}}{\sqrt{\alpha}}\right) \ln \left[\frac{Z+(1+\sqrt{2}) B}{Z+(1-\sqrt{2}) B}\right]\right\}

With

\begin{gathered}A=\frac{a P}{R^2 T^2} \\B=\frac{b P}{R T}\end{gathered}

Start with the vapor (2) at 500 K and 3 bar. Appendix C gives \rm T_c=591.75 K, \rm P_c=41.08 bar and ω= 0.264. Use the method summarized in Section 7.2.7 to find a and b. The results are:

\begin{aligned}& \kappa=0.763 \\& \mathrm{~T}_{\mathrm{r}}=0.845 \\& \alpha=1.127 \\& \mathrm{a}_{\mathrm{c}}=2.69 \times 10^7 \mathrm{bar~cm}^6 / \mathrm{mol}^2 \\& \mathrm{a}=3.036 \times 10^7 \mathrm{bar~cm} / \mathrm{mol}^2 \\& \mathrm{~b}=93.17 \mathrm{~cm}^3 / \mathrm{mol}\end{aligned}

Inserting a and b to determine A and B at this temperature gives:

\begin{gathered}A=\frac{\left(3.036 \times 10^7 \frac{\mathrm{bar~cm}^6}{\mathrm{~mol}^2}\right)(3 \mathrm{~bar})}{\left(83.14 \frac{\mathrm{bar~cm}^3}{\mathrm{mol~K}}\right)^2(500 \mathrm{~K})^2}=0.0527 \\B=\frac{\left(93.17 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\right)(3 \mathrm{~bar})}{\left(83.14 \frac{\mathrm{bar~cm}^3}{\mathrm{mol~K}}\right)(500 \mathrm{~K})}=0.00672\end{gathered}

There are three solutions to the Peng-Robinson equation at T=500 K and P=3 bar. The largest, determined using the Solver feature in Microsoft EXCEL, corresponds to the vapor phase and this is V=13,200 cm³/mol. At this condition:

Z=\frac{P \underline{V}}{R T}=\frac{(3 \text { bar })\left(13,200 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\right)}{\left(83.14 \frac{\mathrm{bar~cm}^3}{\mathrm{mol~K}}\right)(500 \mathrm{~K})}=0.953

Plugging all of the above results into the equation:

\frac{\underline{H}^R}{R T}=(Z-1)-\left\{\left(\frac{A}{B \sqrt{8}}\right)\left(1+\frac{\kappa \sqrt{T_r}}{\sqrt{\alpha}}\right) \ln \left[\frac{Z+(1+\sqrt{2}) B}{Z+(1-\sqrt{2}) B}\right]\right\}

Gives

\begin{gathered}\frac{\mathrm{H}_2^R}{R T}=-0.139 \\\underline{H}_2^R=-5767 \frac{\text { bar~cm}^3}{\mathrm{~mol}}=-577 \frac{\mathrm{J}}{\mathrm{mol}}\end{gathered}

Now we repeat the calculation for the initial state (1), 300 K and 1 bar. The results are:

\begin{aligned}& \kappa=0.763 \\& T_r=0.507 \\& \alpha=1.488 \\& a_c=2.69 \times 10^7 b a r~ c m^6 / \mathrm{mol}^2 \\& a=4.01 \times 10^7 b a r~c m^6 / \mathrm{mol}^2 \\& b=93.17 \mathrm{~cm}^3 / \mathrm{mol}\end{aligned}

Inserting a and b to determine A and B at this temperature gives:

\begin{gathered}A=\frac{\left(4.01 \times 10^7 \frac{\mathrm{bar~cm}^6}{\mathrm{~mol}^2}\right)(1 \,\mathrm{bar})}{\left(83.14 \frac{\mathrm{bar~cm}^3}{\mathrm{mol~K}}\right)^2(300 \mathrm{~K})^2}=0.0644 \\B=\frac{\left(93.17 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\right)(1 \,\mathrm{bar})}{\left(83.14 \frac{\mathrm{bar~cm}^3}{\mathrm{mol~K}}\right)(300 \mathrm{~K})}=0.00374\end{gathered}

Again there are three solutions to the Peng-Robinson equation at T=300 K and P=1 bar. This time we are concerned with the smallest of the three, the liquid solution, which is V=107.4 cm³/mol. At this condition:

Z=\frac{P \underline{V}}{R T}=\frac{(1 \,\mathrm{bar})\left(107.4 \frac{\mathrm{cm}^3}{\mathrm{~mol}}\right)}{\left(83.14 \frac{\mathrm{bar~cm}^3}{\mathrm{mol~K}}\right)(300 \mathrm{~K})}=0.00430

Plugging all of the above results into the equation:

\frac{\underline{H}^R}{R T}=(Z-1)-\left\{\left(\frac{A}{B \sqrt{8}}\right)\left(1+\frac{\kappa \sqrt{T_r}}{\sqrt{\alpha}}\right) \ln \left[\frac{Z+(1+\sqrt{2}) B}{Z+(1-\sqrt{2}) B}\right]\right\}

Gives:

\begin{gathered}\frac{\underline{H}_1^R}{R T}=-14.89 \\\underline{H}_1^R=-371,350 \frac{\text { bar~cm}^3}{\mathrm{~mol}}=-37,135 \frac{\mathrm{J}}{\mathrm{mol}}\end{gathered}

This leaves the ideal gas enthalpy change:

\underline{H}_2^{i g}-\underline{H}_1^{i g}=\int_{T_1=300 K}^{T_2=500 K} C_P^*

Using C_P^* from Appendix D:

\underline{H}_2^{i g}-\underline{H}_1^{i g}=27,656 \frac{\mathrm{J}}{\mathrm{mol}}

Returning to the enthalpy equation