Chapter 19, Problem 19.48E

Interpretation Introduction

(a)

Interpretation:

The mean free paths for nitrogen and oxygen atoms are to be calculated.

Concept introduction:

The mean free path of collisions between gaseous atoms is given by the formula given below.

$\lambda =\frac{kT}{\sqrt{2}\pi d^{2}p}$

The average collision frequency of one atom is given by the formula,

$z=\frac{4\pi \rho d^2\sqrt{kT}}{\sqrt{\pi m}}$

The total number of collisions is given by the formula,

$Z=\frac{2\pi {\rho }_1{\rho }_2{\left[\frac{d_1+d_2}{2}\right]}^2\sqrt{kT}}{\sqrt{\pi m}}$

Answer to Problem 19.48E

The mean free paths for nitrogen and oxygen atoms are $8.44\times {10}^{-8}\text{m}$ and $9.43\times {10}^{-8}\text{m}$ respectively.

Explanation of Solution

The mean free path of collisions between gaseous atoms is given by the formula given below.

$\lambda =\frac{kT}{\sqrt{2}\pi d^{2}p}$ …(1)

Where,

• $p$ is the pressure.

• $T$ is the temperature.

• $d$ is the collision diameter.

• $k$ is the Boltzmann constant.

Substitute the values in the equation (1) for nitrogen atom as given below.

$\begin{array}{rll}\lambda & =& \frac{kT}{\sqrt{2}\pi d^{2}p}\\ & =& \frac{\left(1.381\times {10}^{-23}\text{J}/\text{K}\right)\left(273\text{K}\right)}{\sqrt{2}\times 3.14\times {\left(3.15\stackrel{\circ }{\text{A}}\times \frac{{10}^{-10}\text{m}}{1\stackrel{\circ }{\text{A}}}\right)}^2\times 1\text{atm}}\times \frac{\text{1}\text{L}\text{atm}}{\text{101}\text{.33}\text{J}}\times \frac{{\left(\text{0}\text{.1}\text{m}\right)}^3}{1\text{L}}\\ & =& 8.44\times {10}^{-8}\text{m}\end{array}$

Substitute the values in the equation (1) for oxygen atom as given below.

$\begin{array}{rll}\lambda & =& \frac{kT}{\sqrt{2}\pi d^{2}p}\\ & =& \frac{\left(1.381\times {10}^{-23}\text{J}/\text{K}\right)\left(273\text{K}\right)}{\sqrt{2}\times 3.14\times {\left(2.98\stackrel{\circ }{\text{A}}\times \frac{{10}^{-10}\text{m}}{1\stackrel{\circ }{\text{A}}}\right)}^2\times 1\text{atm}}\times \frac{\text{1}\text{L}\text{atm}}{\text{101}\text{.33}\text{J}}\times \frac{{\left(\text{0}\text{.1}\text{m}\right)}^3}{1\text{L}}\\ & =& 9.43\times {10}^{-8}\text{m}\end{array}$

Conclusion

The mean free paths for nitrogen and oxygen atoms are $8.44\times {10}^{-8}\text{m}$ and $9.43\times {10}^{-8}\text{m}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The average collision frequencies for nitrogen and oxygen atoms are to be calculated.

Concept introduction:

The mean free path of collisions between gaseous atoms is given by the formula given below.

$\lambda =\frac{kT}{\sqrt{2}\pi d^{2}p}$

The average collision frequency of one atom is given by the formula,

$z=\frac{4\pi \rho d^2\sqrt{kT}}{\sqrt{\pi m}}$

The total number of collisions is given by the formula,

$Z=\frac{2\pi {\rho }_1{\rho }_2{\left[\frac{d_1+d_2}{2}\right]}^2\sqrt{kT}}{\sqrt{\pi m}}$.

Answer to Problem 19.48E

The average collision frequencies for nitrogen and oxygen atoms are $96.18\times {10}^8{\text{s}}^{-\text{1}}$ and $317.55\times {10}^8{\text{s}}^{-\text{1}}$.

Explanation of Solution

The average collision frequency of one atom is given by the formula,

$z=\frac{4\pi \rho d^2\sqrt{kT}}{\sqrt{\pi m}}$ …(2)

Where,

• $\rho$ is the density.

• $T$ is the temperature.

• $m$ is the mass of an atom.

• $d$ is the collision diameter.

• $k$ is the Boltzmann constant.

The ratio of nitrogen to oxygen in air is $4:1$. Thus, the number of moles of nitrogen and oxygen in the mixture is $0.8$ and $0.2$ respectively. This gives the density of argon and xenon at STP as follows.

$\begin{array}{rll}{\rho }_{\text{nitrogen}}& =& \frac{6.022\times {10}^{23}}{0.8\times 0.001{\text{m}}^{\text{3}}\times 22.4}\\ & =& 0.34\times {10}^{26}{\text{m}}^{-3}\end{array}$

$\begin{array}{rll}{\rho }_{\text{oxygen}}& =& \frac{6.022\times {10}^{23}}{0.2\times 0.001{\text{m}}^{\text{3}}\times 22.4}\\ & =& 1.34\times {10}^{26}{\text{m}}^{-\text{3}}\end{array}$

The mass of nitrogen and oxygen is $2.33\times {10}^{-26}\text{kg}$ and $2.66\times {10}^{-26}\text{kg}$ respectively.

Substitute the values in the equation (2) for nitrogen atom as given below.

$\begin{array}{rll}z& =& \frac{4\pi \rho d^2\sqrt{kT}}{\sqrt{\pi m}}\\ & =& \frac{4\times 3.14\times 0.34\times {10}^{26}{\text{m}}^{-\text{3}}\times {\left(3.15\stackrel{\circ }{\text{A}}\times \frac{{10}^{-10}\text{m}}{1\stackrel{\circ }{\text{A}}}\right)}^2\sqrt{\left(1.381\times {10}^{-23}\text{J}/\text{K}\right)\left(273\text{K}\right)}}{\sqrt{\left(2.33\times {10}^{-26}\text{kg}\right)\left(3.14\right)}}\\ & =& \frac{260.176\times {10}^{-5}}{2.705\times {10}^{-13}}{\text{s}}^{-\text{1}}\\ & =& 96.18\times {10}^8{\text{s}}^{-\text{1}}\end{array}$

Substitute the values in the equation (2) for oxygen atom as given below.

$\begin{array}{rll}z& =& \frac{4\pi \rho d^2\sqrt{kT}}{\sqrt{\pi m}}\\ & =& \frac{4\times 3.14\times 1.34\times {10}^{26}{\text{m}}^{-3}\times {\left(2.98\stackrel{\circ }{\text{A}}\times \frac{{10}^{-10}\text{m}}{1\stackrel{\circ }{\text{A}}}\right)}^2\sqrt{\left(1.381\times {10}^{-23}\text{J}/\text{K}\right)\left(273\text{K}\right)}}{\sqrt{\left(2.66\times {10}^{-26}\text{kg}\right)\left(3.14\right)}}\\ & =& \frac{917.709\times {10}^{-5}}{2.890\times {10}^{-13}}\\ & =& 317.55\times {10}^8{\text{s}}^{-\text{1}}\end{array}$

Conclusion

The average collision frequencies for nitrogen and oxygen atoms are $96.18\times {10}^8{\text{s}}^{-\text{1}}$ and $317.55\times {10}^8{\text{s}}^{-\text{1}}$.

Interpretation Introduction

(c)

Interpretation:

The total number of collisions between nitrogen and oxygen atoms is to be calculated.

Concept introduction:

The mean free path of collisions between gaseous atoms is given by the formula given below.

$\lambda =\frac{kT}{\sqrt{2}\pi d^{2}p}$

The average collision frequency of one atom is given by the formula,

$z=\frac{4\pi \rho d^2\sqrt{kT}}{\sqrt{\pi m}}$

The total number of collisions is given by the formula,

$Z=\frac{2\pi {\rho }_1{\rho }_2{\left[\frac{d_1+d_2}{2}\right]}^2\sqrt{kT}}{\sqrt{\pi m}}$.

Answer to Problem 19.48E

The total number of collisions between nitrogen and oxygen atoms is $32.93\times {10}^{34}{\text{m}}^{-\text{3}}{\text{s}}^{-1}$.

Explanation of Solution

The total number of collisions is given by the formula,

$Z=\frac{2\pi {\rho }_1{\rho }_2{\left[\frac{d_1+d_2}{2}\right]}^2\sqrt{kT}}{\sqrt{\pi \mu }}$ …(3)

Where,

• $\rho$ is the density.

• $T$ is the temperature.

• $\mu$ is the reduced mass.

• $d$ is the collision diameter.

• $k$ is the Boltzmann constant.

The ratio of nitrogen to oxygen in air is $4:1$. Thus, the number of moles of nitrogen and oxygen in the mixture is $0.8$ and $0.2$ respectively. This gives the density of argon and xenon at STP as follows.

$\begin{array}{rll}{\rho }_{\text{nitrogen}}& =& \frac{6.022\times {10}^{23}}{0.8\times 0.001{\text{m}}^{\text{3}}\times 22.4}\\ & =& 0.34\times {10}^{26}{\text{m}}^{-3}\end{array}$

$\begin{array}{rll}{\rho }_{\text{oxygen}}& =& \frac{6.022\times {10}^{23}}{0.2\times 0.001{\text{m}}^{\text{3}}\times 22.4}\\ & =& 1.34\times {10}^{26}{\text{m}}^{-\text{3}}\end{array}$

The mass of nitrogen and oxygen is $2.33\times {10}^{-26}\text{kg}$ and $2.66\times {10}^{-26}\text{kg}$ respectively.

The reduced mass is calculated as follows:

$\mu =\frac{m_{{\text{N}}_{\text{2}}}{m}_{{\text{O}}_{\text{2}}}}{m_{{\text{N}}_{\text{2}}}+m_{{\text{O}}_{\text{2}}}}$

Substitute the mass of nitrogen and oxygen in above formula.

$\begin{array}{l}\mu =\frac{\left(2.33\times {10}^{-26}\text{kg}\right)\left(2.66\times {10}^{-26}\text{kg}\right)}{2.33\times {10}^{-26}\text{kg}+2.66\times {10}^{-26}\text{kg}}\\ \mu =\frac{6.1978\times {10}^{-52}{\text{kg}}^2}{4.99\times {10}^{-26}\text{kg}}\\ \mu =1.242\times {10}^{-26}\text{kg}\end{array}$

Thus, the reduced mass of nitrogen and oxygen is $1.242\times {10}^{-26}\text{kg}$.

Substitute the values in the equation (3) as given below.

$\begin{array}{rll}Z& =& \frac{\left(\begin{array}{l}2\times 3.14\times 1.34\times {10}^{26}{\text{m}}^{-\text{3}}\times 0.134\times {10}^{26}{\text{m}}^{-\text{3}}\times \\ {\left(\frac{\left(3.15\stackrel{\circ }{\text{A}}\times \frac{{10}^{-10}\text{m}}{1\stackrel{\circ }{\text{A}}}\right)+\left(2.98\stackrel{\circ }{\text{A}}\times \frac{{10}^{-10}\text{m}}{1\stackrel{\circ }{\text{A}}}\right)}{2}\right)}^2\times \\ \sqrt{\left(1.381\times {10}^{-23}\text{J}/\text{K}\right)\left(273\text{K}\right)}\end{array}\right)}{\sqrt{\left(1.242\times {10}^{-26}\text{kg}\right)\left(3.14\right)}}\\ & =& \frac{65.042\times {10}^{21}}{1.975\times {10}^{-13}}{\text{m}}^{-\text{3}}{\text{s}}^{-1}\\ & =& 32.93\times {10}^{34}{\text{m}}^{-\text{3}}{\text{s}}^{-1}\end{array}$

Conclusion

The total number of collisions between nitrogen and oxygen atoms is $32.93\times {10}^{34}{\text{m}}^{-\text{3}}{\text{s}}^{-1}$.