A 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:

Flux of benzene in air $(NA)=mo\frac{l}{m}in.c{m}^2$

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

b$)2\mathrm{.}86{10}^{-5}$

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

d$)2\mathrm{.}76{10}^{-5}$

If the diameter of beaker is reduced to $5cm,$ the flux of benzne 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 diameter of beaker is doubled to $20cm,$ the rate of evaporation benzne will be $($rate$)=\frac{g}{m}in$

a$)0\mathrm{.}745$

b$)0\mathrm{.}382$

c$)4\mathrm{.}78{10}^{-2}$

d$)9\mathrm{.}56{10}^{-2}$