Chapter 16, Problem 67P

(a)

To determine

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{\text{mp}}\right)}$ for the value of $v=\frac{v_{mp}}{50.0}$.

Answer to Problem 67P

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=\frac{v_{mp}}{50.0}$ is $1.09\times {10}^{-3}$.

Explanation of Solution

Given information:Value of average speed is $\frac{v_{mp}}{50.0}$.

Write the expression for the Maxwell-Boltzmann speed distribution function,

$N_v=4\pi N{\left(\frac{m_0}{2\pi k_{B}T}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}^2{e}^{\left(-\frac{m_0{v}^2}{2k_{B}T}\right)}$ (1)

Here,

$N_v$ is the Maxwell-Boltzmann speed distribution function.

$N$ is the total number of molecules of gas.

$T$ is the absolute temperature of gas.

$v$ is the speed of the fraction of molecules of gas.

$k_B$ is the Boltzmann constant.

$m_0$ is the mass of the gas molecule.

Write the expression for the average speed of a gas molecule.

$v=\sqrt{\frac{8k_{B}T}{\pi m_0}}$

Here,

$v$ is the average speed of a gas molecule.

Write the expression for the most probable speed of a gas molecule.

$v_{mp}=\sqrt{\frac{2k_{B}T}{m_0}}$

Here,

$v_{mp}$ is the most probable speed of a gas molecule.

Formula to calculate the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ using equation(1).

$\begin{array}{c}\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}=\frac{4\pi N{\left(\frac{m_0}{2\pi k_{B}T}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}^2{e}^{\left(-\frac{m_0{v}^2}{2k_{B}T}\right)}}{4\pi N{\left(\frac{m_0}{2\pi k_{B}T}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}_{mp}^2{e}^{\left(-\frac{m_0{v}_{mp}^2}{2k_{B}T}\right)}}\\ ={\left(\frac{v}{v_{mp}}\right)}^2{e}^{\left(\frac{m_0{v}_{mp}^2}{2k_{B}T}-\frac{m_0{v}^2}{2k_{B}T}\right)}\\ ={\left(\frac{v}{v_{mp}}\right)}^2{e}^{\frac{m_0{v}_{mp}^2}{2k_{B}T}\left(1-{\left(\frac{v}{v_{mp}}\right)}^2\right)}\end{array}$ (2)

Substitute $\sqrt{\frac{2k_{B}T}{m_0}}$ for $v_{mp}$ in equation (2) to find $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$.

$\begin{array}{c}\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{v}{v_{mp}}\right)}^2{e}^{\frac{m_0{\left(\sqrt{\frac{2k_{B}T}{m_0}}\right)}^2}{2k_{B}T}\left(1-{\left(\frac{v}{v_{mp}}\right)}^2\right)}\\ ={\left(\frac{v}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{v}{v_{mp}}\right)}^2\right)}\end{array}$ (3)

Substitute $\frac{v_{mp}}{50.0}$ for $v$ in equation (3) to find $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$,

$\begin{array}{c}\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{\frac{v_{mp}}{50.0}}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{\frac{v_{mp}}{50.0}}{v_{mp}}\right)}^2\right)}\\ =1.0868\times {10}^{-3}\\ \approx 1.09\times {10}^{-3}\end{array}$

Thus, the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=\frac{v_{mp}}{50.0}$ is $1.09\times {10}^{-3}$.

Conclusion:

Therefore, the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=\frac{v_{mp}}{50.0}$ is $1.09\times {10}^{-3}$.

(b)

To determine

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=\frac{v_{mp}}{10.0}$.

Answer to Problem 67P

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=\frac{v_{mp}}{10.0}$ is $2.69\times {10}^{-2}$.

Explanation of Solution

Given information: Value of average speed is $\frac{v_{mp}}{10.0}$.

From equation (3), formula to calculate the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ is,

$\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{v}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{v}{v_{mp}}\right)}^2\right)}$

Substitute $\frac{v_{mp}}{10.0}$ for $v$ in above expression to find $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$.

$\begin{array}{c}\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{\frac{v_{mp}}{10.0}}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{\frac{v_{mp}}{10.0}}{v_{mp}}\right)}^2\right)}\\ =2.69\times {10}^{-2}\end{array}$

Thus, the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=\frac{v_{mp}}{10.0}$ is $2.69\times {10}^{-2}$.

Conclusion:

Therefore, the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=\frac{v_{mp}}{10.0}$ is $2.69\times {10}^{-2}$.

(c)

To determine

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=\frac{v_{mp}}{2.00}$.

Answer to Problem 67P

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=\frac{v_{mp}}{2.00}$ is $0.529$.

Explanation of Solution

Given information: Value of average speed is $\frac{v_{mp}}{2.00}$.

From equation (3), formula to calculate the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ is,

$\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{v}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{v}{v_{mp}}\right)}^2\right)}$

Substitute $\frac{v_{mp}}{2.00}$ for $v$ in equation (3) to find $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$.

$\begin{array}{c}\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{\frac{v_{mp}}{2.00}}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{\frac{v_{mp}}{2.00}}{v_{mp}}\right)}^2\right)}\\ =0.529\end{array}$

Thus, the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=\frac{v_{mp}}{2.00}$ is $0.529$.

Conclusion:

Therefore, the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=\frac{v_{mp}}{2.00}$ is $0.529$.

(d)

To determine

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=v_{mp}$.

Answer to Problem 67P

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=v_{mp}$ is $1.00$.

Explanation of Solution

Given information: Value of average speed is $v_{mp}$.

From equation (3), formula to calculate the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$.

$\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{v}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{v}{v_{mp}}\right)}^2\right)}$

Substitute $v_{mp}$ for $v$ in above expression to find $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$.

$\begin{array}{c}\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{v_{mp}}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{v_{mp}}{v_{mp}}\right)}^2\right)}\\ =1.00\end{array}$

Thus, the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=v_{mp}$ is $1.00$.

Conclusion:

Therefore, the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=v_{mp}$ is $1.00$.

(e)

To determine

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=2.00v_{mp}$.

Answer to Problem 67P

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=2.00v_{mp}$ is $0.199$.

Explanation of Solution

Given information: Value of average speed is $2.00v_{mp}$.

From equation (3), formula to calculate the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ is,

$\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{v}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{v}{v_{mp}}\right)}^2\right)}$

Substitute $2.00v_{mp}$ for $v$ in above expression to find $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$.

$\begin{array}{c}\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{2.00v_{mp}}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{2.00v_{mp}}{v_{mp}}\right)}^2\right)}\\ =0.199\end{array}$

Thus, the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=2.00v_{mp}$ is $0.199$.

Conclusion:

Therefore, the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=2.00v_{mp}$ is $0.199$.

(f)

To determine

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=10.0v_{mp}$.

Answer to Problem 67P

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=10.0v_{mp}$ is $1.01\times {10}^{-41}$.

Explanation of Solution

Given information: Value of average speed is $10.0v_{mp}$.

From equation (3), formula to calculate the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ is,

$\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{v}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{v}{v_{mp}}\right)}^2\right)}$

Substitute $10.0v_{mp}$ for $v$ in above expression to find $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$.

$\begin{array}{c}\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{10.0v_{mp}}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{10.0v_{mp}}{v_{mp}}\right)}^2\right)}\\ =1.01\times {10}^{-41}\end{array}$

Thus, the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=10.0v_{mp}$ is $1.01\times {10}^{-41}$.

Conclusion:

Therefore, thenumerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=10.0v_{mp}$ is $1.01\times {10}^{-41}$.

(g)

To determine

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=50.0v_{mp}$.

Answer to Problem 67P

The numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=50.0v_{mp}$ is $1.25\times {10}^{-1082}$

Explanation of Solution

Given information: Value of average speed is $50.0v_{mp}$.

From equation (3), formula to calculate the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ is,

$\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{v}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{v}{v_{mp}}\right)}^2\right)}$

Substitute $50.0v_{mp}$ for $v$ in above equation to find $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$.

$\begin{array}{c}\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}={\left(\frac{50.0v_{mp}}{v_{mp}}\right)}^2{e}^{\left(1-{\left(\frac{50.0v_{mp}}{v_{mp}}\right)}^2\right)}\\ =1.25\times {10}^{-1082}\end{array}$

Thus, the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=50.0v_{mp}$ is $1.25\times {10}^{-1082}$.

Conclusion:

Therefore, the numerical value of the $\frac{N_v\left(v\right)}{N_v\left(v_{mp}\right)}$ for the value of $v=50.0v_{mp}$ is $1.25\times {10}^{-1082}$.