Chapter 19, Problem 19.62E

Interpretation Introduction

(a)

Interpretation:

The diffusion constant for $\text{He}$ is to be calculated.

Concept introduction:

The diffusion of any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=\frac{3}{8d^2\rho }\cdot \sqrt{\frac{RT}{\pi M}}$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $\rho$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Answer to Problem 19.62E

The diffusion constant of helium is $0.844{\text{cm}}^2\cdot {\text{s}}^{-1}$.

Explanation of Solution

It is given that the helium is at $1.00\text{atm}$, that is, standard pressure and $273\text{K}$. The molar mass of helium is $4.00\text{g}/\text{mol}$ and the diameter of helium molecules is $2.65\stackrel{\circ }{\text{A}}$.

The diffusion constant can be calculated by the formula,

$D=\frac{3}{8d^2\rho }\cdot \sqrt{\frac{RT}{\pi M}}$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $\rho$ is the particle density.

• $R$ is the gas constant and its value is $\text{8}\text{.314}\text{J}\cdot {\text{mol}}^{-1}\cdot {\text{K}}^{-1}$ or $\text{8}\text{.314}\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-2}\cdot {\text{mol}}^{-1}\cdot {\text{K}}^{-1}$.

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=\frac{3kT}{8d^{2}p}\cdot \sqrt{\frac{RT}{\pi M}}$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=\frac{3\times 1.381\times {10}^{-23}{\text{m}}^2\cdot \text{kg}\cdot {\text{s}}^{-2}\cdot {\text{K}}^{-1}\times 273\text{K}}{8{\left(2.65\stackrel{\circ }{\text{A}}\right)}^2\times 1.00\text{atm}}\cdot \sqrt{\frac{8.314{\text{m}}^2\cdot \text{kg}\cdot {\text{s}}^{-2}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\times 273\text{K}}{\pi \times 4.00\times {10}^{-3}\text{kg}\cdot {\text{mol}}^{-1}}}$

The unit of pressure is converted from $\text{atm}$ to $\text{Pa}$. The conversion factor is $\text{1}\text{atm}=\text{101325}\text{Pa}$.

The conversion factor of $\text{Pa}$ is,

$\begin{array}{l}\text{1}\text{Pa}=\text{1}\text{N}/{\text{m}}^2\\ 1\text{N}=\text{kg}\cdot \text{m}\cdot {\text{s}}^{-2}\end{array}$

Substitute the units of pressure and solve the square root as shown below.

$\begin{array}{l}D=\frac{3\times 1.381\times {10}^{-23}{\text{m}}^2\cdot \text{kg}\cdot {\text{s}}^{-2}\cdot {\text{K}}^{-1}\times 273\text{K}}{8{\left(2.65\times {\text{10}}^{-10}\text{m}\right)}^2\times \text{101325}\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-2}}\times 424.993\text{m}\cdot {\text{s}}^{-1}\\ D=8.444\times {10}^{-5}{\text{m}}^2\cdot {\text{s}}^{-1}\end{array}$

The units of diffusion constant are usually expressed in $\text{cm}$. The conversion factor is $\text{1}\text{cm}={10}^{-2}\text{m}$. The diffusion constant of helium is $0.844{\text{cm}}^2\cdot {\text{s}}^{-1}$.

Conclusion

The diffusion constant of helium is $0.844{\text{cm}}^2\cdot {\text{s}}^{-1}$.

Interpretation Introduction

(b)

Interpretation:

The diffusion constant for $\text{Xe}$ is to be calculated.

Concept introduction:

The diffusion any gas particle among its own gas particles is known as self-diffusion. The diffusion constant is directly proportional to the product of mean free path and average velocity. The self-diffusion of a particle is given as,

$D=\frac{3}{8d^2\rho }\cdot \sqrt{\frac{RT}{\pi M}}$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $\rho$ is the particle density.

• $R$ is the gas constant.

• $T$ is the temperature.

Answer to Problem 19.62E

The diffusion constant of xenon is $0.0647{\text{cm}}^2\cdot {\text{s}}^{-1}$.

Explanation of Solution

It is given that the xenon is at $1.00\text{atm}$, that is, standard pressure and $273\text{K}$. The molar mass of xenon is $131.29\text{g}/\text{mol}$ and the diameter of xenon molecules is $4.00\stackrel{\circ }{\text{A}}$.

The diffusion constant can be calculated by the formula,

$D=\frac{3}{8d^2\rho }\cdot \sqrt{\frac{RT}{\pi M}}$

Where,

• $D$ is the self-diffusion constant.

• $M$ is the molar mass.

• $d$ is the diameter of given particles.

• $\rho$ is the particle density.

• $R$ is the gas constant and its value is $\text{8}\text{.314}\text{J}\cdot {\text{mol}}^{-1}\cdot {\text{K}}^{-1}$ or $\text{8}\text{.314}\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-2}\cdot {\text{mol}}^{-1}\cdot {\text{K}}^{-1}$.

• $T$ is the temperature.

The diffusion constant can be represented in the terms of pressure as,

$D=\frac{3kT}{8d^{2}p}\cdot \sqrt{\frac{RT}{\pi M}}$

Substitute the values of molar mass, diameter, pressure, gas constant and temperature.

$D=\frac{3\times 1.381\times {10}^{-23}{\text{m}}^2\cdot \text{kg}\cdot {\text{s}}^{-2}\cdot {\text{K}}^{-1}\times 273\text{K}}{8{\left(4.00\stackrel{\circ }{\text{A}}\right)}^2\times 1.00\text{atm}}\cdot \sqrt{\frac{8.314{\text{m}}^2\cdot \text{kg}\cdot {\text{s}}^{-2}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\times 273\text{K}}{\pi \times 131.29\times {10}^{-3}\text{kg}\cdot {\text{mol}}^{-1}}}$

The units of pressure are converted from $\text{atm}$ to $\text{Pa}$. The conversion factor is $\text{1}\text{atm}=\text{101325}\text{Pa}$.

The conversion factor of $\text{Pa}$ is,

$\begin{array}{l}\text{1}\text{Pa}=\text{1}\text{N}/{\text{m}}^2\\ 1\text{N}=\text{kg}\cdot \text{m}\cdot {\text{s}}^{-2}\end{array}$

Substitute the units of pressure and solve the square root as shown below.

$\begin{array}{l}D=\frac{3\times 1.381\times {10}^{-23}{\text{m}}^2\cdot \text{kg}\cdot {\text{s}}^{-2}\cdot {\text{K}}^{-1}\times 273\text{K}}{8{\left(4.00\times {\text{10}}^{-10}\text{m}\right)}^2\times \text{101325}\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-2}}\times 74.181\text{m}\cdot {\text{s}}^{-1}\\ D=6.47\times {10}^{-6}{\text{m}}^2\cdot {\text{s}}^{-1}\end{array}$

The units of diffusion constant are usually expressed in $\text{cm}$. The conversion factor is $\text{1}\text{cm}={10}^{-2}\text{m}$. The diffusion constant of xenon is $0.0647{\text{cm}}^2\cdot {\text{s}}^{-1}$.

Conclusion

The diffusion constant of xenon is $0.0647{\text{cm}}^2\cdot {\text{s}}^{-1}$.