# A (2 mathrm{~cm})-diameter, (19 mathrm{~cm})-long tube is placed touching a pool of liquid. The end away from

## Question:

A $$2 \mathrm{~cm}$$-diameter, $$19 \mathrm{~cm}$$-long tube is placed touching a pool of liquid. The end away from the liquid pool $$(\mathrm{z}=0.19 \mathrm{~m})$$ is in an air stream (component C) so that it is pure air, $$\mathrm{y}_{\mathrm{C}}(\mathrm{z}=0.19 \mathrm{~m})=1.0$$. The liquid is $$80 \mathrm{~mol} \%$$ component $$\mathrm{A}$$ and $$20 \mathrm{~mol} \%$$ component $$\mathrm{B}$$. The ratio of these two components in the gas at $$\mathrm{z}=0$$ is the same as in the liquid, $$\mathrm{y}_{\mathrm{A}, \mathrm{i}} / \mathrm{y}_{\mathrm{B}, \mathrm{i}}=4$$. The total vapor pressure of the liquid is $$13,005 \mathrm{~Pa}$$. Thus $$13,005 / \mathrm{P}_{\text {total }}=\left(\mathrm{y}_{\mathrm{A}, \mathrm{i}}+\mathrm{y}_{\mathrm{B}, i}\right)$$. Total pressure is $$107,404 \mathrm{~Pa}$$, and temperature is $$40^{\circ} \mathrm{C}$$. Assume ideal gas. At $$40^{\circ} \mathrm{C}$$ and 107,404 Pa, $$D_{\mathrm{AB}}=2.45 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}(\mathrm{C}=$$ air $$), D_{\mathrm{BC}}=$$ $$2.69 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}$$, and $$D_{\mathrm{AC}}=1.01 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s} . \mathrm{MW} \mathrm{B}=18, \mathrm{MW}$$ air $$=28.9, \mathrm{MW} \mathrm{A}=$$ 60.1. A partial solution is $$\mathrm{N}_{\mathrm{A}}=0.000228074 \mathrm{~mol} /\left(\mathrm{m}^{2} \mathrm{~s}\right)$$. What are the molar fluxes, $$\mathrm{mol} /\left(\mathrm{s}^{2}\right)$$, of B and C (air) at steady state?

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