Question

23-43. An infinitely sharp zone of solute is placed at the center of a column at time $\mathbf{t}\mathbf{=}\mathbf{0}$. After diffusion for time${\mathbf{t}}_{\mathbf{1}}$ , the standard deviation of the Gaussian band is $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{mm}$. After$\mathbf{20}\mathbf{min}$ more, at time ${\mathbf{t}}_{\mathbf{2}}$, the standard deviation is$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{mm}$ . What will be the width after another$\mathbf{20}\mathbf{min}$ , at time ${\mathbf{t}}_{\mathbf{3}}$?

Short answer

The solution is

${\mathrm{p}}_3=2.646$

Step-by-step solution

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An indefinitely sharp zone of solute was positioned at the center of a column at time$\mathrm{t}=0$ . After diffusion, the Gaussian band's standard deviation wasrole="math" localid="1654840869606" $1.0\mathrm{mm}$ for time ${\mathrm{t}}_1$. The standard deviation was $2.0\mathrm{mm}$at time after ${\mathrm{t}}_2$after $20\min$. At time${\mathrm{t}}_3$ , we'll calculate the width after another $20\min$.

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Standard deviation is proportional to $\sqrt{\mathrm{t}}$and it is given:

$$\begin{gathered} {\mathrm{p}}_1=1\mathrm{mm} \\ {\mathrm{p}}_2=2\mathrm{mm} \\ {\mathrm{t}}_2={\mathrm{t}}_1+20 \\ {\mathrm{t}}_3={\mathrm{t}}_1+40{\mathrm{t}}_{1^{\text{'}}} \\ {\mathrm{p}}_3=? \end{gathered}$$

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To find the value of ${\mathrm{t}}_1$, we'll first compare the following expressions for ${\mathrm{p}}_1$and${\mathrm{p}}_2$:

$$\begin{gathered} \frac{{\mathrm{p}}_1}{{\mathrm{p}}_2}=\sqrt{\frac{{\mathrm{t}}_1}{{\mathrm{t}}_1+20}} \\ \frac{1}{2}=\sqrt{\frac{{\mathrm{t}}_1}{{\mathrm{t}}_1+20}} \end{gathered}$$

We'll now subtract the root from the equation and find ${\mathrm{t}}_1$

$$\begin{gathered} {\left(\frac{1}{2}\right)}^2=\frac{{\mathrm{t}}_1}{{\mathrm{t}}_1+20} \\ \frac{1}{4}{\mathrm{t}}_1+5={\mathrm{t}}_1 \\ 5={\mathrm{t}}_1-\frac{1}{4}{\mathrm{t}}_1 \\ {\mathrm{t}}_1=\frac{20}{3} \end{gathered}$$

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The value of ${\mathrm{p}}_3$is then calculated by comparing the values of ${\mathrm{p}}_2$and ${\mathrm{p}}_3$:

$$\begin{gathered} \frac{{\mathrm{p}}_1}{{\mathrm{p}}_3}=\sqrt{\frac{{\mathrm{t}}_1}{{\mathrm{t}}_1+40}} \\ \frac{1}{{\mathrm{p}}_3}=\sqrt{\frac{{\displaystyle \frac{20}{3}}}{{\displaystyle \frac{20}{3}}+40}} \\ \frac{1}{{\mathrm{p}}_3}=0.378 \\ {\mathrm{p}}_3=\frac{1}{0.378} \\ {\mathrm{p}}_3=2.646 \end{gathered}$$