Chapter 19, Problem 19.27E

Interpretation Introduction

(a)

Interpretation:

The percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $10\text{m}/\text{s}$ and $20\text{m}/\text{s}$ is to be calculated using Maxwell-Boltzmann distribution function.

Concept introduction:

The Maxwell-Boltzmann distribution is given by,

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

Where,

• $m$ is the mass of the system.

• $v$ is the average velocity.

• $k$ is the Boltzmann constant.

• $T$ is the temperature.

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

This probability distribution gives the distribution of velocities of any gas at a certain temperature.

Answer to Problem 19.27E

The value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $10\text{m}/\text{s}$ and $20\text{m}/\text{s}$ is $8.24\times {10}^{-5}$.

Explanation of Solution

It is given that the ${\text{O}}_2$ molecules at $300\text{K}$ move between $10\text{m}/\text{s}$ and $20\text{m}/\text{s}$. The average velocity is $15\text{m}/\text{s}$ and $dv$ is $10\text{m}/\text{s}$. The mass of ${\text{O}}_2$ molecule is $32\text{g}/\text{mol}$.

The Maxwell-Boltzmann distribution function is,

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

Where,

• $m$ is the mass of the system.

• $v$ is the average velocity.

• $k$ is the Boltzmann constant.

• $T$ is the temperature.

Substitute the values of mass of the system, average velocity, Boltzmann constant and temperature in the given formula.

$\begin{array}{l}G\left(v\right)dv=\left[\begin{array}{l}4\pi {\left(\frac{32\times {10}^{-3}\text{kg}/\text{mol}}{2\pi \left(8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300\text{K}\right)}\right)}^{3/2}{\left(15\text{m}/\text{s}\right)}^2\times \\ \exp \left\{\frac{-32\times {10}^{-3}\text{kg}/\text{mol}{\left(15\text{m}/\text{s}\right)}^2}{2\left(8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300\text{K}\right)}\right\}\times 10\text{m}/\text{s}\end{array}\right]\\ G\left(v\right)dv=\left[4\pi \times 2.9178{\left(15\text{m}/\text{s}\right)}^2\times \exp \left\{-0.001443\right\}\times 10\text{m}/\text{s}\right]\\ G\left(v\right)dv=8.24\times {10}^{-5}\end{array}$

Thus, the value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $10\text{m}/\text{s}$ and $20\text{m}/\text{s}$ is $8.24\times {10}^{-5}$.

Conclusion

The value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $10\text{m}/\text{s}$ and $20\text{m}/\text{s}$ is $8.28\times {10}^{-5}$.

Interpretation Introduction

(b)

Interpretation:

The percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $100\text{m}/\text{s}$ and $110\text{m}/\text{s}$ is to be calculated using Maxwell-Boltzmann distribution function.

Concept introduction:

The Maxwell-Boltzmann distribution is given by,

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

Where,

• $m$ is the mass of the system.

• $v$ is the average velocity.

• $k$ is the Boltzmann constant.

• $T$ is the temperature.

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

This probability distribution gives the distribution of velocities of any gas at a certain temperature.

Answer to Problem 19.27E

The value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $100\text{m}/\text{s}$ and $110\text{m}/\text{s}$ is $3.77\times {10}^{-3}$.

Explanation of Solution

It is given that the ${\text{O}}_2$ molecules at $300\text{K}$ move between $100\text{m}/\text{s}$ and $110\text{m}/\text{s}$. The average velocity is $105\text{m}/\text{s}$ and $dv$ is $10\text{m}/\text{s}$. The mass of ${\text{O}}_2$ molecule is $32\text{g}/\text{mol}$.

The Maxwell-Boltzmann distribution function is,

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

Where,

• $m$ is the mass of the system.

• $v$ is the average velocity.

• $k$ is the Boltzmann constant.

• $T$ is the temperature.

Substitute the values of mass of the system, average velocity, Boltzmann constant and temperature in the given formula.

$\begin{array}{l}G\left(v\right)dv=\left[\begin{array}{l}4\pi {\left(\frac{32\times {10}^{-3}\text{kg}/\text{mol}}{2\pi \left(8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300\text{K}\right)}\right)}^{3/2}{\left(105\text{m}/\text{s}\right)}^2\times \\ \exp \left\{\frac{-32\times {10}^{-3}\text{kg}/\text{mol}{\left(105\text{m}/\text{s}\right)}^2}{2\left(8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300\text{K}\right)}\right\}\times 10\text{m}/\text{s}\end{array}\right]\\ G\left(v\right)dv=\left[4\pi \times 2.9178{\left(105\text{m}/\text{s}\right)}^2\times \exp \left\{-0.0707\right\}\times 10\text{m}/\text{s}\right]\\ G\left(v\right)dv=3.77\times {10}^{-3}\end{array}$

Thus, the value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $100\text{m}/\text{s}$ and $110\text{m}/\text{s}$ is $3.77\times {10}^{-3}$.

Conclusion

The value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $100\text{m}/\text{s}$ and $110\text{m}/\text{s}$ is $3.77\times {10}^{-3}$.

Interpretation Introduction

(c)

Interpretation:

The percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $1000\text{m}/\text{s}$ and $1010\text{m}/\text{s}$ is to be calculated using Maxwell-Boltzmann distribution function.

Concept introduction:

The Maxwell-Boltzmann distribution is given by,

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

Where,

• $m$ is the mass of the system.

• $v$ is the average velocity.

• $k$ is the Boltzmann constant.

• $T$ is the temperature.

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

This probability distribution gives the distribution of velocities of any gas at a certain temperature.

Answer to Problem 19.27E

The value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $1000\text{m}/\text{s}$ and $1010\text{m}/\text{s}$ is $0.368$.

Explanation of Solution

It is given that the ${\text{O}}_2$ molecules at $300\text{K}$ move between $1000\text{m}/\text{s}$ and $1010\text{m}/\text{s}$. The average velocity is $1005\text{m}/\text{s}$ and $dv$ is $10\text{m}/\text{s}$. The mass of ${\text{O}}_2$ molecule is $32\text{g}/\text{mol}$.

The Maxwell-Boltzmann distribution function is,

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

Where,

• $m$ is the mass of the system.

• $v$ is the average velocity.

• $k$ is the Boltzmann constant.

• $T$ is the temperature.

Substitute the values of mass of the system, average velocity, Boltzmann constant and temperature in the given formula.

$\begin{array}{l}G\left(v\right)dv=\left[\begin{array}{l}4\pi {\left(\frac{32\times {10}^{-3}\text{kg}/\text{mol}}{2\pi \left(8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300\text{K}\right)}\right)}^{3/2}{\left(1005\text{m}/\text{s}\right)}^2\times \\ \exp \left\{\frac{-32\times {10}^{-3}\text{kg}/\text{mol}{\left(1005\text{m}/\text{s}\right)}^2}{2\left(8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300\text{K}\right)}\right\}\times 10\text{m}/\text{s}\end{array}\right]\\ G\left(v\right)dv=\left[4\pi \times 2.9178{\left(1005\text{m}/\text{s}\right)}^2\times \exp \left\{-0.00645\right\}\times 10\text{m}/\text{s}\right]\\ G\left(v\right)dv=0.368\end{array}$

Thus, the value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $1000\text{m}/\text{s}$ and $1010\text{m}/\text{s}$ is $0.368$.

Conclusion

The value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $1000\text{m}/\text{s}$ and $1010\text{m}/\text{s}$ is $0.368$.

Interpretation Introduction

(d)

Interpretation:

The percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $5000\text{m}/\text{s}$ and $5010\text{m}/\text{s}$ is to be calculated using Maxwell-Boltzmann distribution function.

Concept introduction:

The Maxwell-Boltzmann distribution is given by,

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

Where,

• $m$ is the mass of the system.

• $v$ is the average velocity.

• $k$ is the Boltzmann constant.

• $T$ is the temperature.

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

This probability distribution gives the distribution of velocities of any gas at a certain temperature.

Answer to Problem 19.27E

The value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $5000\text{m}/\text{s}$ and $5010\text{m}/\text{s}$ is $0$.

Explanation of Solution

It is given that the ${\text{O}}_2$ molecules at $300\text{K}$ move between $5000\text{m}/\text{s}$ and $5010\text{m}/\text{s}$. The average velocity is $5005\text{m}/\text{s}$ and $dv$ is $10\text{m}/\text{s}$. The mass of ${\text{O}}_2$ molecule is $32\text{g}/\text{mol}$.

The Maxwell-Boltzmann distribution function is,

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

Where,

• $m$ is the mass of the system.

• $v$ is the average velocity.

• $k$ is the Boltzmann constant.

• $T$ is the temperature.

Substitute the values of mass of the system, average velocity, Boltzmann constant and temperature in the given formula.

$G\left(v\right)dv=\left[\begin{array}{l}4\pi {\left(\frac{32\times {10}^{-3}\text{kg}/\text{mol}}{2\pi \left(8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300\text{K}\right)}\right)}^{3/2}{\left(5005\text{m}/\text{s}\right)}^2\times \\ \exp \left\{\frac{-32\times {10}^{-3}\text{kg}/\text{mol}{\left(5005\text{m}/\text{s}\right)}^2}{2\left(8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300\text{K}\right)}\right\}\times 10\text{m}/\text{s}\end{array}\right]$

The value of exponential is negligible and approximately equal to zero thus, the value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $5000\text{m}/\text{s}$ and $5010\text{m}/\text{s}$ is $0$.

Conclusion

The value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $5000\text{m}/\text{s}$ and $5010\text{m}/\text{s}$ is $0$.

Interpretation Introduction

(e)

Interpretation:

The percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $10000\text{m}/\text{s}$ and $10010\text{m}/\text{s}$ is to be calculated using Maxwell-Boltzmann distribution function. The explanation regarding distribution of velocities is to be stated.

Concept introduction:

The Maxwell-Boltzmann distribution is given by,

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

Where,

• $m$ is the mass of the system.

• $v$ is the average velocity.

• $k$ is the Boltzmann constant.

• $T$ is the temperature.

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

This probability distribution gives the distribution of velocities of any gas at a certain temperature.

Answer to Problem 19.27E

The value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $10000\text{m}/\text{s}$ and $10010\text{m}/\text{s}$ is $0$.

Explanation of Solution

It is given that the ${\text{O}}_2$ molecules at $300\text{K}$ move between $10000\text{m}/\text{s}$ and $10010\text{m}/\text{s}$. The average velocity is $10005\text{m}/\text{s}$ and $dv$ is $10\text{m}/\text{s}$. The mass of ${\text{O}}_2$ molecule is $32\text{g}/\text{mol}$.

The Maxwell-Boltzmann distribution function is,

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

Where,

• $m$ is the mass of the system.

• $v$ is the average velocity.

• $k$ is the Boltzmann constant.

• $T$ is the temperature.

Substitute the values of mass of the system, average velocity, Boltzmann constant and temperature in the given formula.

$G\left(v\right)dv=\left[\begin{array}{l}4\pi {\left(\frac{32\times {10}^{-3}\text{kg}/\text{mol}}{2\pi \left(8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300\text{K}\right)}\right)}^{3/2}{\left(10005\text{m}/\text{s}\right)}^2\times \\ \exp \left\{\frac{-32\times {10}^{-3}\text{kg}/\text{mol}{\left(10005\text{m}/\text{s}\right)}^2}{2\left(8.314\text{J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300\text{K}\right)}\right\}\times 10\text{m}/\text{s}\end{array}\right]$

The value of exponential is negligible and approximately equal to zero thus, the value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $10000\text{m}/\text{s}$ and $10010\text{m}/\text{s}$ is $0$.

On comparison of the percentage of molecules moving with the given velocities it is observed that as the velocity increases, the percentage of molecules moving with that velocity increases till the interval of $1000\text{m}/\text{s}$ and $1100\text{m}/\text{s}$. After this range of velocities, the percentage of molecules moving decreases with the increase in velocity.

Conclusion

The value of percentage of ${\text{O}}_2$ molecules at $300\text{K}$ moving between $10000\text{m}/\text{s}$ and $10010\text{m}/\text{s}$ is $0$.

As the velocity increases the percentage of molecules moving with that velocity increases till the interval of $1000\text{m}/\text{s}$ and $1100\text{m}/\text{s}$. After this range of velocities, the percentage of molecules moving decreases with the increase in velocity.