$10\mathrm{.}31.$ Repeat Example $10\mathrm{.}14$ using water as the absorbing liquid.

$($a$)$ Show that the solubility of toluene in water at $25C,$ listed in Table $10\mathrm{.}3,$ corresponds to a Henry's law constant of ~~$9800$atm. Assume that this value at $25C=77F$ is applicable also at $100F($only a fair assumption!$).$

$($b$)$ Then compute $x_{\text{t}\text{o}\text{l}\text{u}\text{e}\text{n}\text{e}\text{,}\text{b}\text{o}\text{t}\text{t}\text{o}\text{m}}$ and from it $\frac{L}{G}.$

$($c$)$ Comment on the feasibility o

f using water

$$ this is the solution of example $10\mathrm{.}14$

in this example

$($Here we do not write this as $57000ppm$ because in liquids ppm always means ppm by mass, and this is a mol fraction!$)$ Then we may write

$\frac{L}{G}=\frac{Y_{\text{i}\text{b}\text{o}\text{t}\text{t}\text{o}\text{m}}-Y_{\text{i}\text{t}\text{o}\text{p}}}{x_{\text{i}\text{b}\text{o}\text{t}\text{t}\text{o}\text{m}}-x_{\text{i}\text{t}\text{o}\text{p}}}$~~$\frac{y_{\text{i}\text{b}\text{o}\text{t}\text{t}\text{o}\text{m}}-y_{\text{i}\text{t}\text{o}\text{p}}}{x_{\text{i}\text{b}\text{o}\text{t}\text{t}\text{o}\text{m}}-x_i\text{t}\text{o}\text{p}}=\frac{5000ppm-50ppm}{0\mathrm{.}057-0}=0\mathrm{.}087$

The molar flow rate of gas is

$G=1000\frac{scf}{\text{m}\text{i}\text{n}}*\frac{\text{l}\text{b}\text{m}\text{o}\text{l}}{385\mathrm{.}3scf}=2\mathrm{.}6\frac{\text{l}\text{b}\text{m}\text{o}\text{l}}{\text{m}\text{i}\text{n}}=1\mathrm{.}25\frac{lb}{s}=1800\frac{\text{m}\text{o}\text{l}}{\text{m}\text{i}\text{n}}$

so the required liquid flow rate is

$L=0\mathrm{.}087*2\mathrm{.}6\frac{\text{l}\text{b}\text{m}\text{o}\text{l}}{\text{m}\text{i}\text{n}}=0\mathrm{.}23\frac{\text{l}\text{b}\text{m}\text{o}\text{l}}{\text{m}\text{i}\text{n}}=44\frac{lb}{\text{m}\text{i}\text{n}}=20\frac{kg}{\text{m}\text{i}\text{n}}$

We can also use the Henry's law expression to estimate how thoroughly we must strip the solvent before reusing it$.$ At the top of the column we also arbitrarily specify that

$y_{\text{t}\text{o}\text{l}\text{u}\text{e}\text{n}\text{e}}^{**}=0\mathrm{.}8y_{\text{t}\text{o}\text{l}\text{u}\text{e}\text{n}\text{e}}=0\mathrm{.}8*50ppm=40ppm$

$x_{\text{t}\text{o}\text{l}\text{u}\text{e}\text{n}\text{e}}=\frac{1(atm)*40*{10}^{-6}}{0\mathrm{.}070(atm)}=5\mathrm{.}71*{10}^{-4}=0\mathrm{.}000571$

This is a difficult but not impossible stripping requirement. If we substituted this value of $x_i$ top into the above calculation in the place of the assumed value of zero, it would increase the computed value of $\frac{L}{G}$ by $1\%,$ which we ignore.

$.$