For flow inside tubes at constant volumetric flow rates$(Q,$ the mass transfer rate of a component A $($either from the fluid stream to the tube inside surface or from the tube inside surface to the fluid stream$)$ can be calculated using the following relations:

$Q(C_{A,\text{o}\text{u}\text{t}}-C_{A,in})=N_{A}S=S{k}_c$ubrace$(\frac{(C_{Ai}-C_{A,in})-(C_{Ai}-C_{A,\text{o}\text{u}\text{t}})}{ln(\frac{C_{Ai}-C_{A,in}}{C_{Ai}-C_{A,\text{o}\text{u}\text{t}}})}$ubrace${)}_{?}=(C_{Ai}-C_A{)}_{\text{m}\text{e}\text{a}\text{n}}$ where the final term is the $log$ mean driving force at the inlet and outlet of the tube, $C_{A|}$ :

concentration of $A$ at the inside surface of the tube, $C_{A,in}$ : concentration of $A$ in the inlet fluid stream,

$C_{A,\text{o}\text{u}\text{t}}$ : concentration of $A$ in the outlet fluid stream, and $S$ : the inside surface area of the tube through which component $A$ is transported.

a$)$ Derive this relation starting from a steady state component mass balance over a differential

element on the tube as given below.

$N_{A}dS=k_c(C_{Ai}-C_A)dS=Qd{C}_A$

$Q$: volumetric flow rate $of$ the fluid,$\frac{m^3}{s}$

b$)$ Benzene$-$free air at $25C$ and $101\mathrm{.}3$kPa enters a $5-cm-$diameter tube at an average velocity of

$5\frac{m}{s}.$ The inner surface of the $6-m-$long tube is coated with a thin film of pure benzene at $25C.$ The vapor pressure of benzene $(C6H6)$ at $25C$ is $13$kPa, and the solubility of air in benzene is assumed to be negligible. Calculate the molar concentration of benzene in the

outlet air, and the evaporation rate of benzene in $k\frac{g}{h}$ given that the diffusivity of benzene in air at $25C$ is ${0\mathrm{.}88}^{**}{10}^{-5}\frac{m^2}{s}.($ Can you solve it according to mass transfer and writing assumption? $)$