Chapter 19, Problem 19.73E

Interpretation Introduction

(a)

Interpretation:

The percentage of atoms having velocity within $1\%$ of the root-mean-square speeds from the graph of $G\left(v\right)$ is to be stated.

Concept introduction:

The probability distribution function of the velocities of the gas molecules in each dimension is given by $g_x\left(v_x\right)$, $g_y\left(v_y\right)$ and $g_z\left(v_z\right)$. This is a vector quantity. This probability function in the range of $-\infty$ to $+\infty$ is represented by,

${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{g}_x\left(v_x\right)d{v}_x}=1$

The Maxwell-Boltzmann distribution is shown below.

$G\left(v\right)dv=4\pi {\left(\frac{m}{2\pi kT}\right)}^{3/2}{v}^2\cdot e^{-m{v}^2/2kT}dv$

This distribution depends on the mass of the particle and absolute temperature.

Answer to Problem 19.73E

The percentage of atoms having a velocity within $1\%$ of the root-mean-square speeds from the graph of $G\left(v\right)$ is approximately $0.068$.

Explanation of Solution

The given temperature is $298\text{K}$.

The molar mass of the helium gas is $4\times {10}^{-3}\text{kg}/\text{mol}$.

The graph of $G\left(v\right)$ verses $v$ is shown in figure 1.

Figure 1

The formula to calculate the root-mean-square speed is given below.

$v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$ …(1)

Where,

• $R$ represents the gas constant with a value of $8.314\text{J}/\text{mol}\text{K}$.

• $T$ represents the temperature.

• $M$ represents the molar mass of the gas.

Substitute the values the value of the molar mass of helium, $R$ and $T$ in the equation (1).

$\begin{array}{rll}{v}_{\text{rms}}& =& \sqrt{\frac{3RT}{M}}\\ & =& \sqrt{\frac{\left(3\right)\left(8.314\text{J}/\text{mol}\text{K}\right)\left(298\text{K}\right)\left(\frac{1{\text{kg·m}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{J}}\right)}{4\times {10}^{-3}\text{kg}/\text{mol}}}\\ & =& 1363.15\text{m}/\text{s}\\ & \approx & 1363\text{m}/\text{s}\end{array}$

The corresponding value of the percentage of atoms having $1363\text{m}/\text{s}$ velocity is represented as shown below.

Figure 2

Therefore, the percentage of atoms having velocity within $1\%$ of the root-mean-square speeds from the graph of $G\left(v\right)$ is approximately $0.068$.

Conclusion

The percentage of atoms having velocity within $1\%$ of the root-mean-square speeds from the graph of $G\left(v\right)$ is approximately $0.068$.

Interpretation Introduction

(b)

Interpretation:

The percentage of atoms having velocity within $1\%$ of the most probable speed from the graph of $G\left(v\right)$ is to be stated.

Concept introduction:

The probability distribution function of the velocities of the gas molecules in each dimension is given by $g_x\left(v_x\right)$, $g_y\left(v_y\right)$ and $g_z\left(v_z\right)$. This is a vector quantity. This probability function in the range of $-\infty$ to $+\infty$ is represented by,

${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{g}_x\left(v_x\right)d{v}_x}=1$

The Maxwell-Boltzmann distribution is given by,

$G\left(v\right)dv=4\pi {\left(\frac{m}{2\pi kT}\right)}^{3/2}{v}^2\cdot e^{-m{v}^2/2kT}dv$

This distribution depends on the mass of the particle and absolute temperature.

Answer to Problem 19.73E

The percentage of atoms having velocity within $1\%$ of the most probable speed from the graph of $G\left(v\right)$ is approximately $0.072$.

Explanation of Solution

The given temperature is $298\text{K}$.

The molar mass of the helium gas is $4\times {10}^{-3}\text{kg}/\text{mol}$.

The graph of $G\left(v\right)$ verses $v$ is shown in figure 1.

Figure 1

The formula to calculate the most probable speed is given below as,

${\nu }_{\text{most}\text{prob}}=\sqrt{\frac{2RT}{M}}$

Where,

• $R$ represents the gas constant with a value of $8.314\text{J}/\text{mol}\text{K}$.

• $T$ represents the temperature.

• $M$ represents the molar mass of the gas.

Substitute the values the value of the molar mass of helium, $R$ and $T$ in the equation (1).

$\begin{array}{rll}{\nu }_{\text{most}\text{prob}}& =& \sqrt{\frac{2RT}{M}}\\ & =& \sqrt{\frac{\left(2\right)\left(8.314\text{J}/\text{mol}\text{K}\right)\left(298\text{K}\right)\left(\frac{1{\text{kg·m}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{J}}\right)}{4\times {10}^{-3}\text{kg}/\text{mol}}}\\ & =& 1113.00\text{m}/\text{s}\end{array}$

The corresponding value of the percentage of atoms having $1113.00\text{m}/\text{s}$ velocity is represented as shown below.

Figure 3

Therefore, the percentage of atoms having velocity within $1\%$ of the most probable speed from the graph of $G\left(v\right)$ is approximately $0.072$.

Conclusion

The percentage of atoms having velocity within $1\%$ of the most probable speed from the graph of $G\left(v\right)$ is approximately $0.072$.

Interpretation Introduction

(c)

Interpretation:

The percentage of atoms having velocity within $1\%$ of the mean speed from the graph of $G\left(v\right)$ is to be stated. Whether the corresponding percentages are similar or not is to be stated.

Concept introduction:

The probability distribution function of the velocities of the gas molecules in each dimension is given by $g_x\left(v_x\right)$, $g_y\left(v_y\right)$ and $g_z\left(v_z\right)$. This is a vector quantity. This probability function in the range of $-\infty$ to $+\infty$ is represented by,

${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{g}_x\left(v_x\right)d{v}_x}=1$

The Maxwell-Boltzmann distribution is given by,

$G\left(v\right)dv=4\pi {\left(\frac{m}{2\pi kT}\right)}^{3/2}{v}^2\cdot e^{-m{v}^2/2kT}dv$

This distribution depends on the mass of the particle and absolute temperature.

Answer to Problem 19.73E

The percentage of atoms having velocity within $1\%$ of the mean speed from the graph of $G\left(v\right)$ is approximately $0.069$. All the percentages have a relative same value.

Explanation of Solution

The given temperature is $298\text{K}$.

The molar mass of the helium gas is $4\times {10}^{-3}\text{kg}/\text{mol}$.

The graph of $G\left(v\right)$ verses $v$ is shown in figure 1.

Figure 1

The formula to calculate the mean speed is given below as,

$\overline{\nu }=\sqrt{\frac{8RT}{\pi M}}$

Where,

• $R$ represents the gas constant with a value of $8.314\text{J}/\text{mol}\text{K}$.

• $T$ represents the temperature.

• $M$ represents the molar mass of the gas.

Substitute the values the value of molar mass of helium, $R$ and $T$ in the equation (1).

$\begin{array}{rll}\overline{\nu }& =& \sqrt{\frac{8RT}{\pi M}}\\ & =& \sqrt{\frac{\left(8\right)\left(8.314\text{J}/\text{mol}\text{K}\right)\left(298\text{K}\right)\left(\frac{1{\text{kg·m}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{J}}\right)}{\left(3.14\right)\left(4\times {10}^{-3}\text{kg}/\text{mol}\right)}}\\ & =& 1256.21\text{m}/\text{s}\\ & \approx & 1256\text{m}/\text{s}\end{array}$

The corresponding value of percentage of atoms having $1256\text{m}/\text{s}$ velocity is represented as shown below.

Figure 4

Therefore, the percentage of atoms having velocity within $1\%$ of the mean speed from the graph of $G\left(v\right)$ is approximately $0.069$.

The percentage of atoms having velocity within $1\%$ of the root-mean-square speeds from the graph of $G\left(v\right)$ is approximately $0.068$.

The percentage of atoms having velocity within $1\%$ of the most probable speed from the graph of $G\left(v\right)$ is approximately $0.072$.

Therefore, all the percentages have a relative same value.

Conclusion

The percentage of atoms having velocity within $1\%$ of the mean speed from the graph of $G\left(v\right)$ is approximately $0.069$. All the percentages have a relative same value.