Question 24.4: Compare estimates the diffusion coefficient of an ethanol-wa......

The Wilke–Chang correlation will be used for estimation of the diffusion coefficients. This correlation will require the molar volume of each species at its normal boiling point. From Table 24.4, the molecular volume of water is 18.9 cm³/gmole. From Table 24.5, using the group contribution method, the molar volume of ethanol (EtOH) is estimated as

V_{\mathrm{EtOH}}=2\cdot V_{C}+1\cdot V_{O}+6\cdot V_{H}=2(14.8)+1(7.4)=59.2\,\mathrm{cm}^{2}/\mathrm{gmole}

For the ethanol-water system, let A ethanol and B water. First, if ethanol is the solute, V_{A}=59.2\,\mathrm{cm}^{3}/\mathrm{gmole}, and if water is the solvent, \Phi_{B}=2.6\mathrm{~and~}\mu_{B}=1.394\mathrm{~cP} Consequently, by equation (24-52)

\frac{D_{A B}\,\mu_{B}}{T}=\frac{7.4\times10^{-8}(\Phi_{B}M_{B})^{1/2}}{V_{A}^{0.6}}            (24-52)

D_{A B}={\frac{T}{\mu_{B}}}{\frac{7.4\times10^{-8}(\Phi_{B}M_{B})^{1/2}}{V_{A}^{0.6}}}={\frac{(283)}{(1.394)}}{\frac{7.4\times10^{-8}(2.6\cdot18)^{1/2}}{(59.2)^{0.6}}}=8.9\times10^{-6}\,\mathrm{cm^{2}/s}

For comparison, from Appendix J, Table J.2, D_{A B}=8.3\times10^{-6}\,\mathrm{cm}^{2}/s, which is within 10% of the estimated value.

Now consider that water is the solute with V_{B}=18.9\,\mathrm{cm}^{3}/\mathrm{gmole}, and ethanol is the solvent with \Phi_{A}=1.5\;\mathrm{and}\;\mu_{A}=1.306\ cP.  In context to equation (24-52), D_{A B} is now given by

D_{B A}=\frac{T}{\mu_{A}}\frac{7.4\times10^{-8}(\Phi_{A}M_{A})^{1/2}}{V_{B}^{0.6}}=\frac{(283)}{(1.306)}\frac{7.4\times10^{-8}(1.5\cdot46)^{1/2}}{(18.9)^{0.6}}=2.28\times10^{-5}\,\mathrm{cm^{2}/s}

This result shows that D_{A B}\neq D_{B}, for liquids.