Acetone is being absorbed by water in a packed tower having a cross$-$sectional area of $0\mathrm{.}186m2$ at $293K$ and $101\mathrm{.}32$kPa$(1$atm$).$ The inlet air contains $2\mathrm{.}6$mol$\%$ acetone and the outlet contains $0\mathrm{.}5$mol$\%.$ The gas flow is $13\mathrm{.}65kg$ mol inert air$/$h$.$ The pure$-$water inlet flow is $45\mathrm{.}36$ kgmol water$/$h$.$ The following are the mass transfer film coefficients and the corresponding values at the interface. The equilibrium data is given by $y=1\mathrm{.}186x.$

$\backslash$table$[[$Film mass transfer coefficient,Value,Unit$],[k_y^{\text{'}}a,3\mathrm{.}78{10}^{-2},$Kgmo$\frac{l}{s}*m^3*mol$ frac$],[k_x^{\text{'}}a,6\mathrm{.}16{10}^{-2},$kgmo$\frac{l}{s}*m^3*mol$ frac$]]$

$H_G=\frac{V}{k_y^{\text{'}}aS}=\frac{V}{k_{y}a(1-y{)}_{MM}S}$

$H_L=\frac{L}{k_x^{\text{'}}aS}=\frac{L}{k_{x}a(1-x{)}_{M_M}S}$

$N_G=\frac{y_1-y_2}{(y_1-y_i{)}_M}$

$N_L=\frac{x_1-x_2}{(x_i-x{)}_M}$

Use the average values for $L$ and $V$ in this formula.

Where

$(y-y_i{)}_M=\frac{(y_1-y_{i1})-(y_2-y_{i2})}{ln[\frac{y_1-y_{i1}}{y_2-y_{i2}}]}$