In the diffusion system shown in the figure, pure liquid A evaporates into gas B at constant pressure and temperature. During an experiment under steady state conditions, the position of liquid$-$gas interface is kept fixed at $z=Z_1,$ and the mole fraction of $A$ is $y_{Al},$ which corresponds to equilibrium with the liquid at the interface. The solubility of $B$ in $A$ is negligible. At the tube mouth $)=(z_2$ mole fraction of A remains constant at $y_{A2}.$

a$)$ Derive the following equation for the total molar flux of $A$ at any $z,$

$N_A=\frac{C{D}_{AB}}{z-z_1}ln(\frac{1-y_A}{1-y_{A1}})$

b$)$ Derive the following expression for the concentration profile of $B$ in the gas phase,

$\frac{y_B}{y_{B1}}=(\frac{y_{B2}}{y_{B1}}{)}^{\frac{z-z_1}{z_2-z_1}}$

For the case $T=23C,P=\frac{1}{b}ar,D_{AB}=8\mathrm{.}35{10}^{-6}\frac{m^2}{s},z_2-Z_1=150mm,$ and vapor pressure of $A$ at $23C100mmHg,$ calculate,

c$)$ The total molar flux of $A,$

d$)$ The distance where the partial pressure of $A$ is the half of that at the liquid$-$gas interface.

Due to the large flow rate of gas $B,$ the partial pressure of $A$ at the tube mouth may be taken zero.