Chapter 13, Problem 72P

(a)

To determine

To Prove: The concentration of oxygen is about $8.7{\text{ mol / m}}^{\text{3}}$.

Answer to Problem 72P

From the ideal gas law equation we can prove that the concentration of oxygen at ${20}^0\text{C}$ is about $8.7{\text{ mol / m}}^{\text{3}}$ .

Explanation of Solution

Given:

Distance $\Delta x=2\text{ mm }=2\times {10}^{-3}\text{ m}$ .

Temperature $T={20}^{\circ }\text{C }=293\text{ K}$ .

Area $A=2\times {10}^{-9}{\text{ m}}^{\text{2}}$ .

Pressure $P=1\text{ atm }=1.013\times {10}^5{\text{ N/m}}^{\text{2}}$ .

Formula used:

Concentration of oxygen $C_0$ is expressed by the following formula:

Concentration of oxygen $C_O=\frac{P}{R\times T}$ .

Ideal gas equation:

  $PV=nRT$

Here, $P$ is the pressure,

  $V$ is the volume,

  $T$ is the temperature,

  $R$ is the universal gas constant,

  $n$ is the no of moles

Calculation:

By using ideal gas law theory, Pressure of $1\text{atm}$ means pressure caused by the oxygen is the percent of oxygen time pressure which is equal to $21\%ofP\text{ is }\left(0.21\times 1\text{ atm}\right)$

  $\begin{array}{l}P\times V=n\times R\times T\\ \frac{n}{V}=\frac{P}{R\times T}=\frac{0.21\times 1.01\times {10}^5}{8.314\times 293}=8.7{\text{ mol / m}}^{\text{3}}\end{array}$

Conclusion:

Hence, the concentration of oxygen in the air at ${20}^0\text{C}$ is about $8.7{\text{ mol / m}}^{\text{3}}$ .

(b)

To determine

To Find: The diffusion rate.

Answer to Problem 72P

The diffusion rate $J$ is $4.35\times {10}^{-11}\text{ mol / s}$ .

Explanation of Solution

Given:

Distance $\Delta x=2\text{ mm }=2\times {10}^{-3}\text{ m}$ .

Area $A=2\times {10}^{-9}{\text{ m}}^{\text{2}}$ .

Diffusion constant $D=1\times {10}^{-5}{\text{ m}}^{\text{2}}\text{/s}$ .

Concentration at point $C_1=8.7{\text{ mol/m}}^{\text{3}}$ .

Concentration at point $C_2=\frac{C_1}{2}=4.35{\text{ mol/m}}^{\text{3}}$ .

Formula used:

  $\begin{array}{l}\text{Diffusion rate }J=D\times A\times \frac{\Delta C}{\Delta x}\\ \Delta C=C_1-C_2\end{array}$

Here, $J$ is the diffusion rate,

  $D$ is the diffusion constant,

  $A$ is the area,

  $\Delta x$ is the diffusion distance,

  $\Delta C$ is the change in concentration.

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}\Delta C=8.7-4.35=4.35{\text{ mol / m}}^{\text{3}}\\ \text{Diffusion rate }J=1\times {10}^{-5}\times 2\times {10}^{-9}\times \frac{4.35}{2\times {10}^{-3}}=4.35\times {10}^{-11}\text{ mol/s}\\ \end{array}$

Conclusion:

Thus the diffusion rate $J$ is $4.35\times {10}^{-11}\text{ mol/s}$ .

(c)

To determine

To Find: The average time for diffusion.

Answer to Problem 72P

The average time of diffusion is $0.6\text{ s}$ .

Explanation of Solution

Given:

Distance $\Delta x=2\text{ mm }=2\times {10}^{-3}\text{ m}$ .

Area $A=2\times {10}^{-9}{\text{ m}}^{\text{2}}$ .

Diffusion constant $D=1\times {10}^{-5}{\text{ m}}^{\text{2}}\text{/s}$ .

Concentration at point $C_1=8.7{\text{ mol/m}}^{\text{3}}$ .

Concentration at point $C_2=\frac{C_1}{2}=4.35{\text{ mol/m}}^{\text{3}}$ .

Average concentration $\overline{C}=\frac{C_1+C_2}{2}=6.525{\text{ mol/m}}^{\text{3}}$

  $\Delta C=C_1-C_2=4.35{\text{ mol/m}}^{\text{3}}$ .

Formula used:

  $\text{Time }t=\frac{\overline{C}\times {\left(\Delta x\right)}^2}{\Delta C\times D}$

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

  $\Delta x$ is the diffusion distance,

  $\Delta C$ is the difference in concentration,

  $D$ is the diffusion constant.

Calculation:

Substitute the given values in the above equation,

  $\text{Time }t=\frac{6.525\times {\left(2\times {10}^{-3}\right)}^2}{4.35\times 1\times {10}^{-5}}=0.6\text{ s}$ .

Conclusion:

Thus, the average time for a molecule to diffuse is $0.6\text{ s}$ .