Chapter 13, Problem 71P

To determine

To Find: The time required for a glycine molecule.

The comparison of diffuse speed to rms speed.

Answer to Problem 71P

The time required for a glycine molecule to diffuse the distance of $15\text{μm}$in water at $20°\text{C}$ is$0.28\text{ s}$.

The comparison of diffuse speed to rms speed is $\text{1}\text{.74}\times {\text{10}}^{\text{-7}}$.

Explanation of Solution

Given:

Distance $\Delta x=15\text{μm }=15\times {10}^{-6}\text{ m}$

Temperature $T={20}^{\circ }\text{C}$.

Concentration distance distance ${\text{d}}_1=1{\text{ mol/m}}^{\text{3}}$.

Concentration distance ${\text{d}}_2=0.4{\text{ mol/m}}^{\text{3}}$.

Diffusion constant $D=95\times {10}^{-11}{\text{ m}}^{\text{2}}\text{/s}$.

Formula used:

Equation for diffusion time t is expressed by the following formula:

Diffusion time $t=\frac{\overline{c}\times {\left(\Delta x\right)}^2}{\Delta c\times D}$

Here, $\overline{c}$is the average concentration,

  $\Delta x$is the diffusion distance,

  $\Delta c$is the difference in concentration distance,

  $D$is the diffusion constant.

Average concentration $\overline{c}=\frac{1}{2}\times \left(\text{Concentration}\text{distance }(d_1+d_2)\right)$

  $\Delta c\text{ = difference in }\left(\text{Concentration}\text{distance }(d_1-d_2)\right)$.is the concentration distance.

Here, $d_1,d_2$is the concentration distance.

  $\begin{array}{l}\text{Diffusion speed }{v}_{diffuse}=\frac{\Delta x}{\text{diffusion time }(t)}\\ \text{rms speed }{v}_{rms}=\sqrt{\frac{3kT}{m}}\end{array}$

Here,

  $\Delta x$is the diffusion distance,

  $k$ is Boltzmann constant,

  $T$is temperature in$\text{K}$,

  $m$ is mass of molecule.

Calculation:

Substitute the given values in the above equation:

Average concentration $\overline{c}=\frac{1}{2}\times \left(1+0.4\right)=0.7{\text{ mol / m}}^{\text{3}}$.

  $\begin{array}{l}\Delta c=1-0.4=0.6{\text{ mol / m}}^{\text{3}}\\ \end{array}$

Diffusion time $t=\frac{0.7\times {\left(15\times {10}^{-6}\right)}^2}{0.6\times \left(95\times {10}^{-11}\right)}=0.28\text{ s}$.

  $\begin{array}{l}\text{Diffusion speed }{v}_{\text{diffuse}}=\frac{15\times {10}^{-6}}{0.28}=5.35\times {10}^{-5}\text{ m/s}\\ \text{rms speed }{v}_{rms}=\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 293}{75\times 1.66\times {10}^{-27}}}=312.13\text{ m/s }\\ \frac{v_{\text{diffuse}}}{v_{rms}}=\frac{5.35\times {10}^{-5}\text{ m/s}}{312.13\text{ m/s}}=1.71\times {10}^{-7}\end{array}$

It’s about 7^th order of magnitude difference.

Conclusion:

Thus, the time required for a glycine molecule to diffuse the distance of $15\text{μm}$in water at ${20}^0\text{C}$ is$0.28\text{ s}$. The comparison of diffuse speed to rms speed is$\text{1}\text{.74}\times {\text{10}}^{\text{-7}}$.