Chapter 15, Problem 15.19P

Interpretation Introduction

Interpretation:

The solubility of naphthalene (1) in carbon dioxide (2) at a given temperature and pressure should be estimated and compare the results with given graph and comment on them and differences should be discussed at P₁^sat =0.0102 bar at 80^oC.

Concept Introduction:

The solubility of solid in the solvent carbon dioxide is calculated by following formula which is equation (15.28)

   $\begin{array}{l}{y}_1=\frac{P_1{}^{sat}}{P}{F}_1\\ \text{where}\\ F_1\equiv \frac{{\varphi }_1{}^{sat}}{\widehat{{\varphi }_1}}\exp \frac{V_1{}^s\left(P-P_1{}^{sat}\right)}{RT}\mathrm{.....}\left(1\right)\end{array}$

And for naphthalene at infinite dilution in CO₂,

   $\ln {\widehat{{\varphi }_1}}^{\infty }=\frac{b_1}{b_2}\left(Z_2-1\right)-\ln \left(Z_2-{\beta }_2\right)-q_2\left[2\left(1-l_{12}\right){\left(\frac{a_1}{a_2}\right)}^{1/2}-\frac{b_1}{b_2}\right]I_2$ .....(2)

Answer to Problem 15.19P

The solubility of naphthalene increases then after some time remains constant. Solubility is affected by the temperature.

Explanation of Solution

Given information:

It is given that the operating conditions are

   $\begin{array}{l}T={80}^{\circ }\text{C+273}\text{.15=353}\text{.15}\text{K}\\ \text{and }P=300\text{bar}\end{array}$

   $P_1{}^{sat}=0.0102$

SVE is given with $l_{12}=0.088$

Solubility graph of naphthalene (1) in carbon dioxide (2) is given as

For simplicity, considering for naphthalene at infinite dilution in carbon dioxide, the fugacity coefficient in equation (1) is at infinite dilution, hence from equation (1) function F₁ is

   $F_1\equiv \frac{{\varphi }_1{}^{sat}}{{\widehat{{\varphi }_1}}^{\infty }}\exp \frac{V_1{}^s\left(P-P_1{}^{sat}\right)}{RT}$

Since it is given that vapor pressure is very small and the saturated vapor is for practical purposes an ideal gas, hence at this condition ${\varphi }_1{}^{sat}=1$ and for very small vapor pressure term $P-P_1{}^{sat}\approx P$

Hence equation (1) becomes

   $F_1\equiv \frac{1}{{\widehat{{\varphi }_1}}^{\infty }}\exp \frac{V_1{}^s\times P}{RT}$

Hence solubility is

   $y_1=\frac{P_1{}^{sat}}{P}\times \frac{1}{{\widehat{{\varphi }_1}}^{\infty }}\exp \frac{V_1{}^s\times P}{RT}$

Where, 1 is used for naphthalene and 2 will use for carbon dioxide.

From equation (2)

   $\ln {\widehat{{\varphi }_1}}^{\infty }=\frac{b_1}{b_2}\left(Z_2-1\right)-\ln \left(Z_2-{\beta }_2\right)-q_2\left[2\left(1-l_{12}\right){\left(\frac{a_1}{a_2}\right)}^{1/2}-\frac{b_1}{b_2}\right]I_2$

Where

$a_i\left(T\right)=\frac{\psi \alpha \left(T_{ri}\right)R^2{T}_{ci}{}^2}{P_{ci}}$ and $b_i=\frac{\Omega R{T}_{ci}}{P_{ci}}$

${\beta }_i=\frac{b_{i}P}{RT}$ and $q_i=\frac{a_i\left(T\right)}{b_{i}RT}$

For vapors

   $I_i{}^V=I_2=\frac{1}{\sigma -\epsilon }\ln \frac{Z_i{}^V+\sigma {\beta }_i}{Z_i{}^V+\epsilon {\beta }_i}$

And it is given that $l_{12}=0.088$

For the calculation of $Z_2$, for vapors we are using general form of cubic equation of state

   $Z_i{}^V=Z_2=1+{\beta }_i-q_i{\beta }_i\left(\frac{Z_i{}^V-{\beta }_i}{\left(Z_i{}^V+\epsilon {\beta }_i\right)\left(Z_i{}^V+\sigma {\beta }_i\right)}\right)$

From SRK equation for the calculation of parameters assigned to equation of state for vapors is

   $\begin{array}{l}\sigma =1\\ \epsilon =0\\ \Omega =0.08664\\ \psi =0.42748\\ Z_C=\frac{1}{3}\\ \alpha \left(T_r,\omega \right)={\left[1+\left(0.480+1.574\omega -0.176{\omega }^2\right)\left(1-T_r{}^{1/2}\right)\right]}^2\end{array}$

And the characteristics properties of pure naphthalene and carbon dioxide is given in Appendix B, Table B.1

For naphthalene

   $\omega =0.302,T_{C1}/\text{K=748}\text{.4,}{P}_{C1}=40.51$

For pure carbon dioxide

$\omega =0.224,T_{C2}/\text{K=304}\text{.2,}{P}_{C2}=73.83$

One by one solving each quantity

For carbon dioxide

   $T_{r2}=\frac{T}{T_{C2}}=\frac{353.15}{304.2}=1.16$

   $\begin{array}{l}\alpha \left(T_{r2},\omega \right)={\left[1+\left(0.480+1.574\omega -0.176{\omega }^2\right)\left(1-T_r{}^{1/2}\right)\right]}^2\\ \Rightarrow \alpha \left(T_{r2},\omega \right)={\left[1+\left(0.480+1.574\times 0.224-0.176\times {0.224}^2\right)\left(1-{1.16}^{1/2}\right)\right]}^2\\ \alpha \left(T_{r2},\omega \right)=0.877\end{array}$

   $\begin{array}{l}{a}_2\left(T\right)=\frac{\psi \alpha \left(T_{r2}\right)R^2{T}_{c2}{}^2}{P_{c2}}\\ a_2\left(T\right)=\frac{0.42748\times 0.877\times {83.14}^2\times {304.2}^2}{73.83}\text{bar}{\text{cm}}^6{\text{mol}}^{-2}\\ a_2\left(T\right)=3248037.131\text{bar}{\text{cm}}^6{\text{mol}}^{-2}\end{array}$

   $\begin{array}{l}{b}_2=\frac{\Omega R{T}_{c2}}{P_{c2}}\\ b_2=\frac{0.08664\times 83.14\times 304.2}{73.83}{\text{cm}}^3{\text{mol}}^{-1}\\ b_2=29.68{\text{cm}}^3{\text{mol}}^{-1}\end{array}$

   $\begin{array}{l}{\beta }_2=\frac{b_{2}P}{RT}=\frac{29.68{\text{cm}}^3{\text{mol}}^{-1}\times P}{83.14\frac{{\text{bar cm}}^3}{\text{mol K}}\times 353.15\text{K}}\\ {\beta }_2=1.01\times {10}^{-3}P\end{array}$

And

   $\begin{array}{l}{q}_2=\frac{a_2\left(T\right)}{b_{2}RT}\\ \frac{3248037.131\text{bar}{\text{cm}}^6{\text{mol}}^{-2}}{29.68{\text{cm}}^3{\text{mol}}^{-1}\times 83.14\frac{{\text{bar cm}}^3}{\text{mol K}}\times 353.15\text{K}}\\ \\ q_2=3.727\end{array}$

Therefore,

   $\begin{array}{l}{Z}_2=1+{\beta }_2-q_2{\beta }_2\left(\frac{Z_2-{\beta }_2}{\left(Z_2+\epsilon {\beta }_2\right)\left(Z_2+\sigma {\beta }_2\right)}\right)\\ \Rightarrow Z_2=1+1.01\times {10}^{-3}P-3.727\times 1.01\times {10}^{-3}P\left(\frac{Z_2-1.01\times {10}^{-3}P}{\left(Z_2+0\times 1.01\times {10}^{-3}P\right)\left(Z_2+1\times 1.01\times {10}^{-3}P\right)}\right)\\ \Rightarrow Z_2=1-2.75\times {10}^{-3}P\left(\frac{Z_2-1.01\times {10}^{-3}P}{Z_2\left(Z_2+1.01\times {10}^{-3}P\right)}\right)\end{array}$

And

   $\begin{array}{l}{I}_2=\frac{1}{\sigma -\epsilon }\ln \frac{Z_2+\sigma {\beta }_2}{Z_2+\epsilon {\beta }_2}\\ I_2=\frac{1}{1-0}\times \ln \left(\frac{Z_2+\sigma {\beta }_2}{Z_2+\epsilon {\beta }_2}\right)\\ \Rightarrow I_2=\ln \frac{Z_2+{\beta }_2}{Z_2}\end{array}$

For naphthalene

   $T_{r1}=\frac{T}{T_{C1}}=\frac{353.15}{748.4}=0.472$

   $\begin{array}{l}\alpha \left(T_{r1},\omega \right)={\left[1+\left(0.480+1.574\omega -0.176{\omega }^2\right)\left(1-T_{r1}{}^{1/2}\right)\right]}^2\\ \Rightarrow \alpha \left(T_{r1},\omega \right)={\left[1+\left(0.480+1.574\times 0.302-0.176\times {0.302}^2\right)\left(1-{0.472}^{1/2}\right)\right]}^2\\ \alpha \left(T_{r1},\omega \right)=1.674\end{array}$

   $\begin{array}{l}{a}_1\left(T\right)=\frac{\psi \alpha \left(T_{r1}\right)R^2{T}_{c1}{}^2}{P_{c1}}\\ a_1\left(T\right)=\frac{0.42748\times 1.674\times {83.14}^2\times {748.4}^2}{40.51}\text{bar}{\text{cm}}^6{\text{mol}}^{-2}\\ a_1\left(T\right)=6.839\times {10}^7\text{bar}{\text{cm}}^6{\text{mol}}^{-2}\end{array}$

   $\begin{array}{l}{b}_1=\frac{\Omega R{T}_{c1}}{P_{c2}}\\ b_1=\frac{0.08664\times 83.14\times 748.4}{40.51}{\text{cm}}^3{\text{mol}}^{-1}\\ b_1=133.076{\text{cm}}^3{\text{mol}}^{-1}\end{array}$

And,

   $V_1{}^s=124.5\frac{{\text{cm}}^3}{\text{mol}}$

Put the values in equation (2)

   $\begin{array}{l}\ln {\widehat{{\varphi }_1}}^{\infty }=\frac{b_1}{b_2}\left(Z_2-1\right)-\ln \left(Z_2-{\beta }_2\right)-q_2\left[2\left(1-l_{12}\right){\left(\frac{a_1}{a_2}\right)}^{1/2}-\frac{b_1}{b_2}\right]I_2\\ \\ \ln {\widehat{{\varphi }_1}}^{\infty }=\frac{133.076}{29.68}\times \left(Z_2-1\right)-\ln \left(Z_2-{\beta }_2\right)-3.727\times \left[2\left(1-0.088\right){\left(\frac{6.839\times {10}^7}{3248037.131}\right)}^{1/2}-\frac{133.076}{29.68}\right]\times I_2\\ \\ \Rightarrow \ln {\widehat{{\varphi }_1}}^{\infty }=4.484\left(Z_2-1\right)-\ln \left(Z_2-{\beta }_2\right)-14.4832\times I_2\end{array}$

Therefore, solubilities at different pressure $P=20\text{bar, 40 bar}\mathrm{.......}\text{ 300 bar}$ are

   $y_1=\frac{0.0102}{P}\times \frac{1}{{\widehat{{\varphi }_1}}^{\infty }}\exp \frac{124.5\times P}{83.14\times 353.15}$

Put the values of pressure and hence solubilities are

P1	β2	Z2	I2	ln(f1)	f1	y1
20	0.0202	0.944	0.021173	-0.47849	0.619718	0.000895792
40	0.0404	0.887	0.04454	-0.98525	0.373347	0.000809263
60	0.0606	0.828	0.070634	-1.52951	0.216642	0.001012043
80	0.0808	0.768	0.100034	-2.11397	0.120758	0.001482243
100	0.101	0.709	0.133179	-2.73612	0.064822	0.002404564
120	0.1212	0.653	0.170253	-3.39027	0.0337	0.004195473
140	0.1414	0.605	0.210033	-4.0444	0.01752	0.007529222
160	0.1616	0.569	0.249986	-4.65524	0.009512	0.013209077
180	0.1818	0.546	0.287407	-5.18826	0.005582	0.021779138
200	0.202	0.535	0.320321	-5.62472	0.003608	0.033011622
220	0.2222	0.533	0.348461	-5.97226	0.002548	0.046242035
240	0.2424	0.536	0.373106	-6.25881	0.001914	0.061451133
260	0.2626	0.542	0.395079	-6.50057	0.001503	0.07863101
280	0.2828	0.551	0.414259	-6.69709	0.001235	0.096735253
300	0.303	0.561	0.431852	-6.86828	0.00104	0.116627234

The graph between pressure and the solubilities is

On comparison of graph from the given graph

From the found graph, one can clearly conclude that the solubility of the naphthalene is constant initially at low pressure but as pressure increases its solubility also increases reaches up to 0.12 but after very high pressure above 300 bar it remains constant. The given graph of solubility v/s pressure shows that at $l_{12}=0.088$, the maximum solubility goes upto 0.01 at pressure of around 300 bar and it is nearly constant after 100 bar pressure and at low pressure it decreases as pressure increases upto 50-60 bar, after this temperature it achieves its peak value and then there are negligible changes in solubility at higher pressure.

The temperature of graph found is $T={80}^{\circ }\text{C}$ while the graph given is at temperature $T={35}^{\circ }\text{C}$, so one can say temperature is affecting the solubility.

Conclusion

The solubility of naphthalene increases then after some time remains constant. Solubility is affected by the temperature.