Chapter 21, Problem 21.62AP

A vessel contains 1.00 × 10⁴ oxygen molecules at 500 K. (a) Make an accurate graph of the Maxwell speed distribution function versus speed with points at speed intervals of 100 m/s. (b) Determine the most probable speed from this graph. (c) Calculate the average and rms speeds for the molecules and label these points on your graph. (d) From the graph, estimate the fraction of molecules with speeds in the range 300 m/s to 600 m/s.

To determine

The graph of the Maxwell speed distribution function versus speed with points at speed intervals of $100\text{m}/\text{s}$.

Answer to Problem 21.62AP

The of the Maxwell speed distribution function versus speed with points at speed intervals of $100\text{m}/\text{s}$ is shown below;

Explanation of Solution

The Maxwell distribution curve is the graph between the distribution of speed and the change in speed or speed interval.

The number of molecules of oxygen in vessel is $1.00\times {10}^4$ and the temperature is $500\text{K}$. The interval of speed is $100\text{m}/\text{s}$. The range of speed is $300\text{m}/\text{s}$ to $600\text{m}/\text{s}$.

Write the expression of Maxwell’s speed distribution function.

    $N_{\text{v}}=4\pi N\left(\frac{m_{\text{0}}}{2\pi k_{\text{B}}T}\right)v^2{e}^{-m_0{v}^2/2k_{\text{B}}T}$                                             (1)

Here, $N$ is the total number of molecules of oxygen in the vessel, $m_{\text{0}}$ is the mass of molecule of oxygen, $k_{\text{B}}$ is the Boltzmann’s constant, $T$ is the absolute temperature and $e^{-m_0{v}^2/2k_{\text{B}}T}$ is the Boltzmann factor.

The mass of the molecules of oxygen

    $m_{\text{0}}=\frac{M}{N_{\text{A}}}$

Here, $M$ is the molecular mass of the oxygen in $\text{kg}$ and $N_{\text{A}}$ is the Avogadro number.

The molecular mass of the oxygen molecules in $\text{kg}$ is $0.032$.

Substitute $\frac{M}{N_{\text{A}}}$ for $m_{\text{0}}$ in equation (1).

    $N_{\text{v}}=4\pi N\left(\frac{\left(\frac{M}{N_{\text{A}}}\right)}{2\pi k_{\text{B}}T}\right)v^2{e}^{-\left(\frac{M}{N_{\text{A}}}\right)v^2/2k_{\text{B}}T}$

Substitute $1.00\times {10}^4$ for $N$, $0.032\text{kg}$ for $M$, $6.02\times {10}^{23}$ for $N_{\text{A}}$, $500\text{K}$ for $T$ and $1.38\times {10}^{-23}\text{J}/\text{molecules}\cdot \text{K}$ for $k_{\text{B}}$ in above equation.

    $\begin{array}{c}{N}_{\text{v}}=4\pi \left(1.00\times {10}^4\right)\left[\begin{array}{l}\left(\frac{\left(\frac{0.032\text{kg}}{6.02\times {10}^{23}}\right)}{2\pi \left(1.38\times {10}^{-23}\text{J}/\text{molecules}\cdot \text{K}\right)\left(500\text{K}\right)}\right)\\ v^2{e}^{-\left(\frac{0.032\text{kg}}{6.02\times {10}^{23}}\right)v^2/2\left(1.38\times {10}^{-23}\text{J}/\text{molecules}\cdot \text{K}\right)\left(500\text{K}\right)}\end{array}\right]\\ =\left(1.71\times {10}^{-4}\right)v^2{e}^{-\left(3.85\times {10}^{-6}\right)v^2}\end{array}$

Substitute the values of $v$ and form a table.

$v\left(\text{m}/\text{s}\right)$	$N_{\text{V}}$
0	0
100	1.64
200	5.86
300	10.88
400	14.78
500	16.33
600	15.39
700	12.7
800	9.31
900	6.13
1000	3.64
1100	1.961
1200	0.96
1300	0.43
1400	0.18
1500	0.07

On the basis of the table, a graph is plotted below;

To determine

The most probable speed from the graph.

Answer to Problem 21.62AP

The most probable speed is $510\text{m}/\text{s}$.

Explanation of Solution

The most probable speed occurs where $N_{\text{V}}$ is maximum. From the above graph the value of speed when $N_{\text{v}}$ is maximum is $510\text{m}/\text{s}$.

Conclusion:

Therefore, the most probable speed is $510\text{m}/\text{s}$.

To determine

The average and rms speeds for the molecules and label these points on the graph.

Answer to Problem 21.62AP

The average and rms speeds for the molecules is $575\text{m}/\text{s}$ and $624\text{m}/\text{s}$ respectively and the graph is shown below,

Explanation of Solution

Write the expression of average velocity.

    $v_{\text{avg}}=\sqrt{\frac{8k_{\text{B}}T}{\pi m_0}}$

The mass of the molecules of oxygen

    $m_{\text{0}}=\frac{M}{N_{\text{A}}}$

Substitute $\frac{M}{N_{\text{A}}}$ for $m_{\text{0}}$ in above equation.

    $v_{\text{avg}}=\sqrt{\frac{8k_{\text{B}}T}{\pi \left(\frac{M}{N_{\text{A}}}\right)}}$

The molecular mass of the oxygen molecules in $\text{kg}$ is $0.032$.

Substitute $1.38\times {10}^{-23}\text{J}/\text{molecule}\cdot \text{K}$ for $k_{\text{B}}$, $500\text{K}$ for $T$, $0.032\text{kg}$ for $M$ and $6.02\times {10}^{23}$ for $N_{\text{A}}$ in above equation.

    $\begin{array}{c}{v}_{\text{avg}}=\sqrt{\frac{8\left(1.38\times {10}^{-23}\text{J}/\text{molecule}\cdot \text{K}\right)500\text{K}}{\pi \left(\frac{0.032\text{kg}}{6.02\times {10}^{23}}\right)}}\\ =575\text{m}/\text{s}\end{array}$

Thus, the average speed is $575\text{m}/\text{s}$.

Write the expression of rms velocity.

    $v_{\text{rms}}=\sqrt{\frac{3k_{\text{B}}T}{m_0}}$

Substitute $\frac{M}{N_{\text{A}}}$ for $m_{\text{0}}$ in above equation.

    $v_{\text{rm}}=\sqrt{\frac{3k_{\text{B}}T}{\left(\frac{M}{N_{\text{A}}}\right)}}$

Substitute $1.38\times {10}^{-23}\text{J}/\text{molecule}\cdot \text{K}$ for $k_{\text{B}}$, $500\text{K}$ for $T$, $0.032\text{kg}$ for $M$ and $6.02\times {10}^{23}$ for $N_{\text{A}}$ in above equation.

    $\begin{array}{c}{v}_{\text{rms}}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{J}/\text{molecule}\cdot \text{K}\right)500\text{K}}{\left(\frac{0.032\text{kg}}{6.02\times {10}^{23}}\right)}}\\ =624\text{m}/\text{s}\end{array}$

Thus, the rms velocity of the oxygen molecules is $624\text{m}/\text{s}$.

The graph of Maxwell’s curve is shown below;

The point $\text{A}$ on the graph is the average velocity of the oxygen molecules and the point $\text{B}$ is the average velocity of the oxygen molecules.

Conclusion:

Therefore, the average and rms speeds for the molecules is $575\text{m}/\text{s}$ and $624\text{m}/\text{s}$ respectively and the graph is shown below,

To determine

The fraction of molecules with the speed in the range of $300\text{m}/\text{s}$ to $600\text{m}/\text{s}$.

Answer to Problem 21.62AP

The fraction of molecules with the speed in the range of $300\text{m}/\text{s}$ to $600\text{m}/\text{s}$ is $44\%$.

Explanation of Solution

The figure given below shows the Maxwell’s curve,

The area under the distribution curve in the range $300\text{m/s}$ to $600\text{m/s}$ gives the fraction of molecules with speed in the range $300\text{m/s}$ to $600\text{m/s}$.

Conclusion:

Write the area under the curve in the range $300\text{m/s}$ to $600\text{m/s}$

    $\begin{array}{l}F\approx 11\times 300+\left(100\times 5.5\right)+\frac{1}{2}\left(200\times 5.5\right)\\ =4400\\ =4400\times \frac{1}{100}\%\\ =44\%\end{array}$

Therefore, the fraction of molecules with the speed in the range of $300\text{m}/\text{s}$ to $600\text{m}/\text{s}$ is $44\%$.