Example

Diffusion of water vapor in a narrow tube is occurring as in the previous example under the same conditions. However, at a given time $t,$ the level is $zm$ from the top. As diffusion proceeds, the level drops slowly. Derive the equation for the time $t_F$ for the level to drop from a starting point of $z_{0}m$ at $t=0$ to $z_F$ at $t=t_{F}s.$

$N_A=\frac{D_{AB}P}{RTz{p}_{BM}}(p_{A1}-p_{A2})$

Assuming a cross$-$sectional area of $1m^2,$ the level drops $dzm$ in $dts$ and ${}_A\frac{dz\mathrm{.}1}{M_A}$ is the mol of A that have left and diffused. Then,

$N_A*1=\frac{{}_A(dz\mathrm{.}1)}{M_{A}dt}$