through stagnant B

The figure shown below show a cylinder in which benzene is diffusing into the external pure air. The cylinder tube is $10cm$ in diameter and $20cm$ tall from which $4cm$ is filled with benzene only. If all the benzene is evaporated after $48hr,$ determine the following :

Data : $T=30c,P=1$atm,$R=82\mathrm{.}05$atm$*c\frac{m^3}{m}ol.K,$ benzene : $=0\mathrm{.}876\frac{g}{c}{m}^3,MW=78,P^{**}=119mmHg,$DAB$=0\mathrm{.}095c\frac{m^2}{s}$

Solution:

If the height of beaker is reduced to $10cm,$ the flux of benzene will be $(NA)=mo\frac{l}{m}in.c{m}^2$

a$)1\mathrm{.}38{10}^{-5}$

b$)5\mathrm{.}52{10}^{-5}$

c$)1\mathrm{.}56{10}^{-5}$

d$){7\mathrm{.}76810}^{-5}$

If the height of beaker is reduced to $10cm,$ the rate of evaporation benzne will be $($rate$)=mo\frac{l}{m}in$

a$)4\mathrm{.}08{10}^{-5}$

b$)3\mathrm{.}12{10}^{-3}$

c$)1\mathrm{.}22{10}^{-3}$

d$)1\mathrm{.}56{10}^{-5}$

If the height of benzne is double to $8cm,$ the flux of benzene will be $(NA)=mo\frac{l}{m}in.c{m}^2$

a$)1\mathrm{.}76{10}^{-5}$

b$)1\mathrm{.}56{10}^{-5}$

c$)3\mathrm{.}12{10}^{-5}$

d$)3\mathrm{.}5{10}^{-5}$

If the height of benzne is double to $8cm,$ the time of evaporation will be

a$)48hr$

b$)72hr$

c$)24hr$

d$)96hr$