A $15$ cm tall tube has liquid acetone $($A$)$ at $298$K$.$ Dry air $($B$)$ at $298$K and $1$ atm is blown across the

mouth of the tube so that evaporated acetone is carried away by convection after it transfers through an air layer

within the beaker. At a particular time $($t$1),$ the liquid level is located at $6$ cm from the top. The air layer above the

liquid is assumed stagnant. The vapor pressure of acetone at $298$K is $0\mathrm{.}291$ atm. The mole fraction of acetone in

the air at the top of the beaker is zero and is at equilibrium at the gas$-$liquid interface. The diffusion coefficient for

acetone in air at $298$K and $1$ atm is $0\mathrm{.}124$ cm$2/$sec$.$ At time t$1,$ compute:

a$.$Rate of evaporation of acetone as a molar flux in mol$/$cm$2/$sec$.$

b$.$Concentration profiles $($for both A and B$)$ in the stagnant air layer. Make a plot of the

concentrations.

c$.$ Molar fluxes $($for both A and B$)$ due to diffusion only, at various locations. Discuss the

direction of the fluxes.

d$.\backslash$Diffusion velocities, the species velocities $($relative to stationary coordinates$),$ and average

velocity at various locations in the stagnant layer.

e$.$Time for the acetone level in the beaker to drop $3$ cm if the density of liquid acetone is $0\mathrm{.}76$

gm$/$cm$3.$ The molecular weight of acetone is $58\mathrm{.}08.$

f$.$ How much acetone level in the beaker would drop in a day?