91影视

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

$V=A\mathrm{螖}x$

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:

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

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

$J_x=\frac{N}{A\mathrm{螖}t}$

$J_x=鈭�D\frac{dn}{dx}$

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

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

$\begin{array}{r}鈫�\mathrm{螖}t=\frac{(\mathrm{螖}x{)}^2}{D}\\ 鈫�\mathrm{螖}x=\sqrt{D\mathrm{螖}t}\end{array}$

substitute with,$螖t=60$s and $D=5脳{10}^{-10}{\mathrm{m}}^2路{\mathrm{s}}^{-1}$, so:

$\mathrm{螖}x=\sqrt{5脳{10}^{鈭�10}脳60}=1.732脳{10}^{鈭�4}\mathrm{m}$

$鈫�\mathrm{螖}x=1.732脳{10}^{鈭�4}\mathrm{m}$