Chapter 19, Problem 19.49E

Interpretation Introduction

(a)

Interpretation:

The mean free paths for argon and xenon 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 19.49E

The mean free paths for argon and xenon atoms are $1.24\times {10}^{16}\text{m}$ and $5.23\times {10}^{15}\text{m}$ 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 argon atom as given below.

$\begin{array}{rll}\lambda & =& \frac{kT}{\sqrt{2}\pi d^{2}p}\\ & =& \frac{\left(1.381\times {10}^{-23}\text{J}/\text{K}\right)\left(273\text{K}\right)}{\sqrt{2}\times 3.14\times {\left(2.60\stackrel{\circ }{\text{A}}\times \frac{{10}^{-10}\text{m}}{1\stackrel{\circ }{\text{A}}}\right)}^2\times 1\text{atm}}\times \frac{\text{1}\text{L}\text{atm}}{\text{101}\text{.33}\text{J}}\times \frac{{\left(\text{0}\text{.1}\text{m}\right)}^3}{1\text{L}}\\ & =& 1.24\times {10}^{16}\text{m}\end{array}$

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

$\begin{array}{rll}\lambda & =& \frac{kT}{\sqrt{2}\pi d^{2}p}\\ & =& \frac{\left(1.381\times {10}^{-23}\text{J}/\text{K}\right)\left(273\text{K}\right)}{\sqrt{2}\times 3.14\times {\left(4.00\stackrel{\circ }{\text{A}}\times \frac{{10}^{-10}\text{m}}{1\stackrel{\circ }{\text{A}}}\right)}^2\times 1\text{atm}}\times \frac{\text{1}\text{L}\text{atm}}{\text{101}\text{.33}\text{J}}\times \frac{{\left(\text{0}\text{.1}\text{m}\right)}^3}{1\text{L}}\\ & =& 5.23\times {10}^{15}\text{m}\end{array}$

Conclusion

The mean free paths for argon and xenon atoms are $1.24\times {10}^{16}\text{m}$ and $5.23\times {10}^{15}\text{m}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The average collision frequencies for argon and xenon 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 19.49E

The average collision frequencies for argon and xenon atoms are $15.311\times {10}^8{\text{s}}^{\text{-1}}$ and $199.86\times {10}^7{\text{s}}^{\text{-1}}$.

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 mixture given is a $50:50$ mixture. Thus, the number of moles of argon and xenon in the mixture is $0.5$ each. This gives the density of argon and xenon at STP as follows.

$\begin{array}{rll}{\rho }_{\text{xenon}}& =& \frac{6.022\times {10}^{23}}{0.5\times 0.001{\text{m}}^{\text{3}}\times 22.4}\\ & =& 0.134\times {10}^{26}{\text{m}}^{-\text{3}}\end{array}$

$\begin{array}{rll}{\rho }_{\text{argon}}& =& \frac{6.022\times {10}^{23}}{0.5\times 0.001{\text{m}}^{\text{3}}\times 22.4}\\ & =& 0.134\times {10}^{26}{\text{m}}^{-\text{3}}\end{array}$

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

$\begin{array}{rll}z& =& \frac{4\pi \rho d^2\sqrt{kT}}{\sqrt{\pi m}}\\ & =& \frac{4\times 3.14\times 0.134\times {10}^{26}{\text{m}}^{-\text{3}}\times {\left(2.60\stackrel{\circ }{\text{A}}\times \frac{{10}^{-10}\text{m}}{1\stackrel{\circ }{\text{A}}}\right)}^2\sqrt{\left(1.381\times {10}^{-23}\text{J}/\text{K}\right)\left(273\text{K}\right)}}{\sqrt{\left(6.63\times {10}^{-26}\text{kg}\right)\left(3.14\right)}}\\ & =& 15.311\times {10}^8{\text{s}}^{-\text{1}}\end{array}$

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

$\begin{array}{rll}z& =& \frac{4\pi \rho d^2\sqrt{kT}}{\sqrt{\pi m}}\\ & =& \frac{4\times 3.14\times 0.134\times {10}^{26}{\text{m}}^{-3}\times {\left(4.00\stackrel{\circ }{\text{A}}\times \frac{{10}^{-10}\text{m}}{1\stackrel{\circ }{\text{A}}}\right)}^2\sqrt{\left(1.381\times {10}^{-23}\text{J}/\text{K}\right)\left(273\text{K}\right)}}{\sqrt{\left(2.18\times {10}^{-25}\text{kg}\right)\left(3.14\right)}}\\ & =& 199.86\times {10}^7{\text{s}}^{\text{-1}}\end{array}$

Conclusion

The average collision frequencies for argon and xenon atoms are $15.311\times {10}^8{\text{s}}^{-\text{1}}$ and $199.86\times {10}^7{\text{s}}^{-\text{1}}$.

Interpretation Introduction

(c)

Interpretation:

The total number of collisions between argon and xenon 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 19.49E

The total number of collisions between argon and xenon atoms is $37.79\times {10}^{33}{\text{m}}^{-\text{3}}{\text{s}}^{-1}$.

Explanation of Solution

The total number of collisions is given by the formula,

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

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 mixture given is a $50:50$ mixture. Thus, the number of moles of argon and xenon in the mixture is $0.5$ each. This gives the density of argon and xenon at STP as follows.

$\begin{array}{rll}{\rho }_{\text{xenon}}& =& \frac{6.022\times {10}^{23}}{0.5\times 0.001{\text{m}}^{\text{3}}\times 22.4}\\ & =& 0.134\times {10}^{26}{\text{m}}^{-\text{3}}\end{array}$

$\begin{array}{rll}{\rho }_{\text{argon}}& =& \frac{6.022\times {10}^{23}}{0.5\times 0.001{\text{m}}^{\text{3}}\times 22.4}\\ & =& 0.134\times {10}^{26}{\text{m}}^{-\text{3}}\end{array}$

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

$\begin{array}{rll}Z& =& \frac{2\pi {\rho }_1{\rho }_2{\left[\frac{d_1+d_2}{2}\right]}^2\sqrt{kT}}{\sqrt{\pi m}}\\ & =& \frac{\left(\begin{array}{l}4\times 3.14\times 0.134\times {10}^{26}{\text{m}}^{-\text{3}}\times 0.134\times {10}^{26}{\text{m}}^{-\text{3}}\times {\left(\frac{\left(4.00\stackrel{\circ }{\text{A}}\times \frac{{10}^{-10}\text{m}}{1\stackrel{\circ }{\text{A}}}\right)+\left(2.60\stackrel{\circ }{\text{A}}\times \frac{{10}^{-10}\text{m}}{1\stackrel{\circ }{\text{A}}}\right)}{2}\right)}^2\times \\ \sqrt{\left(1.381\times {10}^{-23}\text{J}/\text{K}\right)\left(273\text{K}\right)}\end{array}\right)}{\sqrt{\left(\text{0}\text{.508}\times {10}^{-25}\text{kg}\right)\left(3.14\right)}}\\ & =& 37.79\times {10}^{33}{\text{m}}^{-\text{3}}{\text{s}}^{-1}\end{array}$

Conclusion

The total number of collisions between argon and xenon atoms is $37.79\times {10}^{33}{\text{m}}^{-\text{3}}{\text{s}}^{-1}$.