Chapter 17, Problem 28SP

To determine

The temperature of ${\text{H}}_2$ gas and ${\text{N}}_2$ gas at which the rms speed is equal to escape speed from the Earth’s gravitational field.

Answer to Problem 28SP

Solution:

$1.0\times {10}^4\text{ K}$, $1.4\times {10}^5\text{ K}$

Explanation of Solution

Given data:

The escape speed from the Earth’s gravitational field is $11.2\text{ km/s}$.

The molecular mass of ${\text{H}}_2$ is $2.0\text{ kg/kmol}$.

The molecular mass of ${\text{N}}_2$ is $28.0\text{ kg/kmol}$.

Formula used:

Write the expression of mass of gas molecule:

$m_o=\frac{M}{N_A}$

Here, $m_o$ is the mass of gas molecule, $M$ is the molecular mass, and $N_A$ is the Avogadro’s number.

Write the expression of root mean square speed of gas molecule:

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

Here, $v_{rms}$ is the root mean square speed of gas molecule, $k_B$ is the Boltzmann constant, $T$ is the temperature, and $m_o$ is the mass of the gas molecule.

Explanation:

Recall the expression of mass of gas molecule:

$m_o=\frac{M}{N_A}$

For ${\text{H}}_2$ molecule, substitute $2.0\text{ kg/kmol}$ for $M$ and $6.02\times {10}^{26}{\text{ kmol}}^{-1}$ for $N_A$

$\begin{array}{c}{m}_o=\frac{2.0\text{ kg/kmol}}{6.02\times {10}^{26}{\text{ kmol}}^{-1}}\\ =3.3\times {10}^{-27}\text{ kg}\end{array}$

Recall the expression of root mean square speed of ${\text{H}}_2$ gas molecule:

$v_{rms}=\sqrt{\frac{3k_B{T}_{{\text{H}}_2}}{m_o}}$

Here, $T_{{\text{H}}_2}$ is the temperature of ${\text{H}}_2$ gas molecule.

For ${\text{H}}_2$ molecule, substitute $11.2\text{ km/s}$ for $v_{rms}$, $1.38\times {10}^{-23}\text{ J/K}$ for $k_B$, and $3.3\times {10}^{-27}\text{ kg}$ for $m_o$

$\begin{array}{l}11.2\text{ km/s}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{ J/K}\right)T_{{\text{H}}_2}}{\left(3.3\times {10}^{-27}\text{ kg}\right)}}\\ {\left(11.2\text{ km/s}\left(\frac{1000\text{ m/s}}{1\text{ km/s}}\right)\right)}^2=\frac{3\left(1.38\times {10}^{-23}\text{ J/K}\right)T_{{\text{H}}_2}}{\left(3.3\times {10}^{-27}\text{ kg}\right)}\end{array}$

Solve for $T_{{\text{H}}_2}$

$\begin{array}{c}{T}_{{\text{H}}_2}=\frac{{\left(11.2\text{ km/s}\left(\frac{1000\text{ m/s}}{1\text{ km/s}}\right)\right)}^2\left(3.3\times {10}^{-27}\text{ kg}\right)}{3\left(1.38\times {10}^{-23}\text{ J/K}\right)}\\ =\frac{\left(4.139\times {10}^{-19}\right)}{\left(4.14\times {10}^{-23}\right)}\text{ K}\\ =0.99\times {10}^4\text{ K}\\ =\text{1}\times {10}^4\text{ K}\end{array}$

Recall the expression of mass of gas molecule:

$m_o=\frac{M}{N_A}$

For ${\text{N}}_2$ molecule, substitute $28.0\text{ kg/kmol}$ for $M$ and $6.02\times {10}^{26}{\text{ kmol}}^{-1}$ for $N_A$

$\begin{array}{c}{m}_o=\frac{28.0\text{ kg/kmol}}{6.02\times {10}^{26}{\text{ kmol}}^{-1}}\\ =4.6\times {10}^{-26}\text{ kg}\end{array}$

Recall the expression of root mean square speed of ${\text{N}}_2$ gas molecule:

$v_{rms}=\sqrt{\frac{3k_B{T}_{{\text{N}}_2}}{m_o}}$

Here, $T_{{\text{N}}_2}$ is the temperature of ${\text{N}}_2$ gas molecule.

For ${\text{N}}_2$ molecule, substitute $11.2\text{ km/s}$ for $v_{rms}$, $1.38\times {10}^{-23}\text{ J/K}$ for $k_B$, and $4.6\times {10}^{-26}\text{ kg}$ for $m_o$

$\begin{array}{l}11.2\text{ km/s}=\sqrt{\frac{3\left(1.38\times {10}^{-23}\text{ J/K}\right)T_{{\text{N}}_2}}{\left(4.6\times {10}^{-26}\text{ kg}\right)}}\\ {\left(11.2\text{ km/s}\left(\frac{1000\text{ m/s}}{1\text{ km/s}}\right)\right)}^2=\frac{3\left(1.38\times {10}^{-23}\text{ J/K}\right)T_{{\text{N}}_2}}{\left(4.6\times {10}^{-26}\text{ kg}\right)}\end{array}$

Solve for $T_{{\text{N}}_2}$

$\begin{array}{c}{T}_{{\text{N}}_2}=\frac{{\left(11.2\text{ km/s}\left(\frac{1000\text{ m/s}}{1\text{ km/s}}\right)\right)}^2\left(4.6\times {10}^{-26}\text{ kg}\right)}{3\left(1.38\times {10}^{-23}\text{ J/K}\right)}\\ =\frac{\left(5.77\times {10}^{-18}\right)}{\left(4.14\times {10}^{-23}\right)}\text{ K}\\ =1.39\times {10}^5\text{ K}\\ =\text{1}\text{.4}\times {10}^5\text{ K}\end{array}$

Conclusion:

Hence, the temperature of ${\text{H}}_2$ gas and N₂ gas at which the rms speed is equal to escape speed is $1\times {10}^4\text{ K}$ and $1.4\times {10}^5\text{ K}$ , respectively.