91Ó°ÊÓ

D is the diffusion constant for sucrose (table sugar) in water at room temperature.$D=5×{10}^{-10}{\mathrm{m}}^2·{\mathrm{s}}^{-1}$We can estimate how far a sucrose molecule will diffuse by taking $∆x$the distance over which diffusion has occurred. This region's volume is:.

$V=A\mathrm{Δ}x$

where A denotes the container's cross sectional area If N is the total number of molecules in this region, the particle density is as follows:

$n=\frac{N}{V}=\frac{N}{A\mathrm{Δ}x}$

The flux can be calculated by taking $∆x$the time required for the volume$A∆x$ to acquire the N molecules, yielding:

$J_x=\frac{N}{A\mathrm{Δ}t}$

$J_x=−D\frac{dn}{dx}$

substitute from equations $J_x=\frac{N}{A\mathrm{Δ}t}$and$n=\frac{N}{V}=\frac{N}{A\mathrm{Δ}x}$into equation $J_x=−D\frac{dn}{dx}$, so:

$$\begin{gathered} \frac{N}{A\mathrm{Δ}t}=−D\frac{d}{dx}\left[\frac{N}{A\mathrm{Δ}x}\right] \\ =D\left[\frac{N}{A(\mathrm{Δ}x{)}^2}\right] \end{gathered}$$

$\begin{array}{r}→\mathrm{Δ}t=\frac{(\mathrm{Δ}x{)}^2}{D}\\ →\mathrm{Δ}x=\sqrt{D\mathrm{Δ}t}\end{array}$

substitute with,$\mathrm{Δ}t=60$s and $D=5×{10}^{-10}{\mathrm{m}}^2·{\mathrm{s}}^{-1}$, so:

$\mathrm{Δ}x=\sqrt{5×{10}^{−10}×60}=1.732×{10}^{−4}\mathrm{m}$

$→\mathrm{Δ}x=1.732×{10}^{−4}\mathrm{m}$