Chapter 1, Problem 1B.7P

Interpretation Introduction

Interpretation: The escape velocity from the surface of the Earth and Mars of radius, $R$ has to be calculated.  The temperatures at which ${\text{H}}_{\text{2}}\text{,}\text{He}$ and ${\text{O}}_{\text{2}}$ have mean speeds equal to the escape speeds have to be calculated.  The proportions of molecules that have enough speed to escape at temperatures, $240\text{K}$ and $1500\text{K}$ have to be calculated.

Concept introduction: The minimum speed required for the body to escape the influence of the gravitational force of the large body is called escape velocity.  The value of escape velocity is less for the objects which are at a greater distance from the body and for less massive bodies.

Answer to Problem 1B.7P

(i) The escape velocity for the Earth is $\underset{\_}{15.8km{s}^{-1}}$.  The escape velocity for the Mars is $\underset{\_}{7.13km{s}^{-1}}$.  (ii) The temperature at which the mean speed corresponds to the escape velocity at the surface of the Earth for hydrogen molecule is $\underset{\_}{23.7\times 10^{-3}K}$.  The temperature at which the mean speed corresponds to the escape velocity at the surface of the Earth for $\text{He}$ is $\underset{\_}{47.19\times 10^{-3}K}$.  The temperature at which the mean speed corresponds to the escape velocity at the surface of the Earth for ${\text{O}}_{\text{2}}$ is $\underset{\_}{377.3\times 10^{-3}K}$.  The temperature at which the mean speed corresponds to the escape velocity at the surface of the Mars for hydrogen molecule is $\underset{\_}{4.84\times 10^{-3}K}$.  The temperature at which the mean speed corresponds to the escape velocity at the surface of the Mars for $\text{He}$ is $\underset{\_}{9.61\times 10^{-3}K}$.  The temperature at which the mean speed corresponds to the escape velocity at the surface of the Mars for ${\text{O}}_{\text{2}}$ is $\underset{\_}{76.83\times 10^{-3}K}$.

(i) The proportion of molecules that have enough speed to escape at temperature $240\text{K}$ is shown below.

Escape velocity	${\text{H}}_{\text{2}}$	$\text{He}$	${\text{O}}_{\text{2}}$
$v_{\text{esc}}=15.8\text{km}{\text{s}}^{-1}$	$2.5\times {10}^{-27}$	$3.2\times {10}^{-54}$	$0$
$v_{\text{esc}}=7.13\text{km}{\text{s}}^{-1}$	$1.1\times {10}^{-5}$	$5.1\times {10}^{-11}$	$5.6\times {10}^{-88}$

(ii) The proportion of molecules that have enough speed to escape at temperature $1500\text{K}$ is shown below.

 

Escape velocity	${\text{H}}_{\text{2}}$	$\text{He}$	${\text{O}}_{\text{2}}$
$v_{\text{esc}}=15.8\text{km}{\text{s}}^{-1}$	$1.5\times {10}^{-4}$	$9.5\times {10}^{-9}$	$1.9\times {10}^{-69}$
$v_{\text{esc}}=7.13\text{km}{\text{s}}^{-1}$	$2.5\times {10}^{-1}$	$4.3\times {10}^{-2}$	$4.2\times {10}^{-14}$

Explanation of Solution

(i)

The expression for the gravitational force, $F$ is given by Newton’s gravitational force law.

    $F=\frac{Gmm\text{'}}{r^2}$

Where,

• • $G$ is the gravitational constant.
• • $m$ and $m\text{'}$ are the masses of the objects.
• • $r$ is the distance between the centres of the two masses.

The minimum work, $w$ required to bring the object of mass, $m_{\text{planet}}$ from infinity to the surface of the planet of radius, $R_{\text{planet}}$ is,

    $\begin{array}{c}w={\displaystyle \underset{0}{\overset{\infty }{\int }}Fdr}\\ =Gm{m}_{\text{planet}}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{r^2}dr}\\ =-Gm{m}_{\text{planet}}{\left[\frac{1}{r^2}\right]}_{R_{\text{planet}}}^{\infty }\\ =\left(\frac{G{m}_{\text{planet}}}{R_{\text{planet}}}\right)m\end{array}$

    $w=\left(g_{\text{planet}}{R}_{\text{planet}}\right)m$

Where,

• • $g_{\text{planet}}$ is the gravitational constant of the planet and

  $g_{\text{planet}}=\frac{G{m}_{\text{planet}}}{R_{\text{planet}}^2}$

Convert the work into kinetic energy to calculate the escape velocity, $v_{\text{esc}}$.

  $\frac{1}{2}m{v}_{\text{esc}}^2=\left(g_{\text{planet}}{R}_{\text{planet}}\right)m$

Or,

  $v_{\text{esc}}=2{\left(g_{\text{planet}}{R}_{\text{planet}}\right)}^{1/2}$                                                                                   (1)

Substitute the values in the above equation to calculate the escape velocity for the Earth.

    $\begin{array}{c}{v}_{\text{esc}}=2{\left(g_{\text{earth}}{R}_{\text{earth}}\right)}^{1/2}\\ =2{\left(9.81\text{m}{\text{s}}^{-2}\times 6.37\times {10}^6\text{m}\right)}^{1/2}\\ =15.8\times {10}^3\text{m}{\text{s}}^{-1}\times \frac{1\text{km}}{{10}^3\text{m}}\\ =\underset{\_}{15.8km{s}^{-1}}\end{array}$

Hence, the escape velocity for the Earth is $\underset{\_}{15.8km{s}^{-1}}$.

(ii)

To determine the gravitational constant of Mars $g_{\text{Mars}}$, the gravitational constant of the earth, radius of the earth and the radius of Mars is used.  The fraction of the masses of the planet Mars and Earth is,

    $\begin{array}{c}\frac{m_{\text{Mars}}}{m_{\text{Earth}}}=0.108\\ \frac{g_{\text{Mars}}}{g_{\text{Earth}}}=\frac{\frac{G{m}_{\text{Mars}}}{R_{\text{Mars}}^2}}{\frac{G{m}_{\text{Earth}}}{R_{\text{Earth}}^2}}\\ g_{\text{Mars}}=g_{\text{Earth}}\times \frac{m_{\text{Mars}}}{m_{\text{Earth}}}\times \frac{R_{\text{Earth}}^2}{R_{\text{Mars}}^2}\end{array}$

Substitute the value of $g_{\text{Earth}}$, $\frac{m_{\text{Mars}}}{m_{\text{Earth}}}$, $R_{\text{Earth}}$ and $R_{\text{Mars}}$ in the above equation, to calculate $g_{\text{Mars}}$.

    $\begin{array}{c}{g}_{\text{Mars}}=9.81\text{m}{\text{s}}^{-2}\times \left(0.108\right)\times \left[\frac{{\left(6.37\times {10}^6\text{m}\right)}^2}{{\left(3.38\times {10}^6\text{m}\right)}^2}\right]\\ =1.059\times \frac{40.5769\times {10}^{12}}{11.4244\times {10}^{12}}\\ =1.059\times 3.552\\ =3.761\text{m}{\text{s}}^{-2}\end{array}$

Substitute the values of $g_{\text{Mars}}$ and $R_{\text{Mars}}$ in the equation (1) to calculate the escape velocity for the Mars.

    $\begin{array}{c}{v}_{\text{esc}}=2{\left(g_{\text{Mars}}{R}_{\text{Mars}}\right)}^{1/2}\\ =2{\left(3.761\text{m}{\text{s}}^{-2}\times 3.38\times {10}^6\text{m}\right)}^{1/2}\\ =2\times 3.565\times {10}^3\text{m}{\text{s}}^{-1}\\ =7.13\times {10}^3\text{m}{\text{s}}^{-1}\times \frac{1\text{km}}{{10}^3\text{m}}=\underset{\_}{7.13km{s}^{-1}}\end{array}$

Hence, the escape velocity for the Mars is $\underset{\_}{7.13km{s}^{-1}}$.

The expression to calculate the temperature at which the mean speed corresponds to the escape velocity is,

    $T=\frac{\pi M{v}_{\text{esc}}}{8R}$                                                                                                   (2)

Where,

• • $R$ is the gas constant.
• • $T$ is the temperature.
• • $M$ is the molar mass.
• • $v_{\text{esc}}$ is the escape velocity.

The molar mass of ${\text{H}}_{\text{2}}=\text{2}\text{.016}\times {10}^{-3}\text{kg}{\text{mol}}^{-1}$, $\text{He}\text{=}4.003\times {10}^{-3}\text{kg}{\text{mol}}^{-1}$ and ${\text{O}}_{\text{2}}=32\times {10}^{-3}\text{kg}{\text{mol}}^{-1}$.

Substitute the values of $M$, $R$, $T$ in the above equation to calculate the temperature at which the mean speed corresponds to the escape velocity at the surface of the Earth for hydrogen molecule.

    $\begin{array}{c}T=\frac{3.14\times 2.016\times {10}^{-3}\text{kg}{\text{mol}}^{-1}\times {\left(15.8\text{km}{\text{s}}^{-1}\right)}^2}{8\times 8.31\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}}\\ =\underset{\_}{23.7\times 10^{-3}K}\end{array}$

Hence, the temperature at which the mean speed corresponds to the escape velocity at the surface of the Earth for hydrogen molecule is $\underset{\_}{23.7\times 10^{-3}K}$.

Substitute the values of $M$, $R$, $T$ in the equation (2) to calculate the temperature at which the mean speed corresponds to the escape velocity at the surface of the Earth for $\text{He}$.

  $\begin{array}{c}T=\frac{3.14\times 4.003\times {10}^{-3}\text{kg}{\text{mol}}^{-1}\times {\left(15.8\text{km}{\text{s}}^{-1}\right)}^2}{8\times 8.31\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}}\\ =\underset{\_}{47.19\times 10^{-3}K}\end{array}$

Hence, the temperature at which the mean speed corresponds to the escape velocity at the surface of the Earth for $\text{He}$ is $\underset{\_}{47.19\times 10^{-3}K}$.

Substitute the values of $M$, $R$, $T$ in the equation (2) to calculate the temperature at which the mean speed corresponds to the escape velocity at the surface of the Earth for ${\text{O}}_{\text{2}}$.

  $\begin{array}{c}T=\frac{3.14\times 32\times {10}^{-3}\text{kg}{\text{mol}}^{-1}\times {\left(15.8\text{km}{\text{s}}^{-1}\right)}^2}{8\times 8.31\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}}\\ =\underset{\_}{377.3\times 10^{-3}K}\end{array}$

Hence, the temperature at which the mean speed corresponds to the escape velocity at the surface of the Earth for ${\text{O}}_{\text{2}}$ is $\underset{\_}{377.3\times 10^{-3}K}$.

Substitute the values of $M$, $R$, $T$ in the above equation to calculate the temperature at which the mean speed corresponds to the escape velocity at the surface of the Mars for hydrogen molecule.

    $\begin{array}{c}T=\frac{3.14\times 2.016\times {10}^{-3}\text{kg}{\text{mol}}^{-1}\times {\left(7.13\text{km}{\text{s}}^{-1}\right)}^2}{8\times 8.31\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}}\\ =\underset{\_}{4.84\times 10^{-3}K}\end{array}$

Hence, the temperature at which the mean speed corresponds to the escape velocity at the surface of the Mars for hydrogen molecule is $\underset{\_}{4.84\times 10^{-3}K}$.

Substitute the values of $M$, $R$, $T$ in the equation (2) to calculate the temperature at which the mean speed corresponds to the escape velocity at the surface of the Mars for $\text{He}$.

  $\begin{array}{c}T=\frac{3.14\times 4.003\times {10}^{-3}\text{kg}{\text{mol}}^{-1}\times {\left(7.13\text{km}{\text{s}}^{-1}\right)}^2}{8\times 8.31\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}}\\ =\underset{\_}{9.61\times 10^{-3}K}\end{array}$

Hence, the temperature at which the mean speed corresponds to the escape velocity at the surface of the Mars for $\text{He}$ is $\underset{\_}{9.61\times 10^{-3}K}$.

Substitute the values of $M$, $R$, $T$ in the equation (2) to calculate the temperature at which the mean speed corresponds to the escape velocity at the surface of the Mars for ${\text{O}}_{\text{2}}$.

  $\begin{array}{c}T=\frac{3.14\times 32\times {10}^{-3}\text{kg}{\text{mol}}^{-1}\times {\left(7.13\text{km}{\text{s}}^{-1}\right)}^2}{8\times 8.31\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}}\\ =\underset{\_}{76.79\times 10^{-3}K}\end{array}$

Hence, the temperature at which the mean speed corresponds to the escape velocity at the surface of the Mars for ${\text{O}}_{\text{2}}$ is $\underset{\_}{76.79\times 10^{-3}K}$.

(i)

The expression for the distribution of speeds at a temperature $T$ is,

    $f\left(v\right)=4\pi {\left(\frac{M}{2\pi RT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}^2{e}^{\raisebox{1ex}{$-M{v}^2$}\!\left/ \!\raisebox{-1ex}{$2RT$}\right.}$

Where,

• • $M$ is the molar mass.
• • $R$ is the gas constant.
• • $T$ is the temperature.
• • $v$ is the velocity.

The integral of the above equation can be represented as,

$f\left(v\right)=4\pi {\left(\frac{M}{2\pi RT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\displaystyle \underset{v_{\text{esc}}}{\overset{\infty }{\int }}{v}^2{e}^{\raisebox{1ex}{$-M{v}^2$}\!\left/ \!\raisebox{-1ex}{$2RT$}\right.}}$ (3)

At $v_{\text{esc}}=15.8\text{km}{\text{s}}^{-1}$ and temperature, $T=240\text{K}$, the proportion of molecules that have enough speed to escape is shown below.

For ${\text{H}}_{\text{2}}$ substitute the value of $M=2.015\text{g}{\text{mol}}^{-1}$, $R=8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $T=240\text{K}$ in equation (3)

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{\text{2}\text{.015}\text{g}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 240\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {\displaystyle \underset{15.8}{\overset{\infty }{\int }}{(15.8)}^2{e}^{\raisebox{1ex}{$-2.015\times {\left(15.8\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 240\text{K}$}\right.}}\\ =\underset{\_}{2.5\times 10^{-27}}\end{array}$

For $\text{He}$ substitute the value of $M=4.003\text{g}{\text{mol}}^{-1}$, $R=8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $T=240\text{K}$ in equation (3)

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{4.003\text{g}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 240\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {\displaystyle \underset{15.8}{\overset{\infty }{\int }}{(15.8)}^2{e}^{\raisebox{1ex}{$-4.003\text{g}{\text{mol}}^{-1}\times {\left(15.8\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 240\text{K}$}\right.}}\\ =\underset{\_}{3.2\times 10^{-54}}\end{array}$

For ${\text{O}}_{\text{2}}$ substitute the value of $M=32\text{g}{\text{mol}}^{-1}$, $R=8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $T=240\text{K}$ in equation (3)

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{32\text{g}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 240\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {\displaystyle \underset{15.8}{\overset{\infty }{\int }}{(15.8)}^2{e}^{\raisebox{1ex}{$-32\text{g}{\text{mol}}^{-1}\times {\left(15.8\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 240\text{K}$}\right.}}\\ =\underset{\_}{0}\end{array}$

At $v_{\text{esc}}=7.13\text{km}{\text{s}}^{-1}$ and temperature, $T=240\text{K}$, the proportion of molecules that have enough speed to escape is shown below

For ${\text{H}}_{\text{2}}$ substitute the value of $M=2.015\text{g}{\text{mol}}^{-1}$, $R=8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $T=240\text{K}$ in equation (3)

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{\text{2}\text{.015}\text{g}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 240\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {\displaystyle \underset{7.13}{\overset{\infty }{\int }}{(7.13)}^2{e}^{\raisebox{1ex}{$-2.015\times {\left(7.13\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 240\text{K}$}\right.}}\\ =\underset{\_}{1.1\times 10^{-5}}\end{array}$

For $\text{He}$ substitute the value of $M=4.003\text{g}{\text{mol}}^{-1}$, $R=8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $T=240\text{K}$ in equation (3)

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{4.003\text{g}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 240\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {\displaystyle \underset{7.13}{\overset{\infty }{\int }}{(7.13)}^2{e}^{\raisebox{1ex}{$-4.003\text{g}{\text{mol}}^{-1}\times {\left(7.13\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 240\text{K}$}\right.}}\\ =\underset{\_}{5.1\times 10^{-11}}\end{array}$

For ${\text{O}}_{\text{2}}$ substitute the value of $M=32\text{g}{\text{mol}}^{-1}$, $R=8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $T=240\text{K}$ in equation (3)

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{32\text{g}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 240\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {\displaystyle \underset{7.13}{\overset{\infty }{\int }}{(7.13)}^2{e}^{\raisebox{1ex}{$-32\text{g}{\text{mol}}^{-1}\times {\left(7.13\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 240\text{K}$}\right.}}\\ =\underset{\_}{5.6\times 10^{-88}}\end{array}$

The proportion of molecules that have enough speed to escape at temperature $240\text{K}$  is shown below.

Escape velocity	${\text{H}}_{\text{2}}$	$\text{He}$	${\text{O}}_{\text{2}}$
$v_{\text{esc}}=15.8\text{km}{\text{s}}^{-1}$	$2.5\times {10}^{-27}$	$3.2\times {10}^{-54}$	$0$
$v_{\text{esc}}=7.13\text{km}{\text{s}}^{-1}$	$1.1\times {10}^{-5}$	$5.1\times {10}^{-11}$	$5.6\times {10}^{-88}$

(ii)

At $v_{\text{esc}}=15.8\text{km}{\text{s}}^{-1}$ and temperature, $T=1500\text{K}$, the proportion of molecules that have enough speed to escape is shown below.

For ${\text{H}}_{\text{2}}$ substitute the value of $M=2.015\text{g}{\text{mol}}^{-1}$, $R=8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $T=1500\text{K}$ in equation (3)

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{\text{2}\text{.015}\text{g}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1500\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {\displaystyle \underset{15.8}{\overset{\infty }{\int }}{(15.8)}^2{e}^{\raisebox{1ex}{$-2.015\times {\left(15.8\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1500\text{K}$}\right.}}\\ =\underset{\_}{1.5\times 10^{-4}}\end{array}$

For $\text{He}$ substitute the value of $M=4.003\text{g}{\text{mol}}^{-1}$, $R=8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $T=1500\text{K}$ in equation (3)

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{4.003\text{g}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1500\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {\displaystyle \underset{15.8}{\overset{\infty }{\int }}{(15.8)}^2{e}^{\raisebox{1ex}{$-4.003\text{g}{\text{mol}}^{-1}\times {\left(15.8\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1500\text{K}$}\right.}}\\ =\underset{\_}{9.5\times 10^{-9}}\end{array}$

For ${\text{O}}_{\text{2}}$ substitute the value of $M=32\text{g}{\text{mol}}^{-1}$, $R=8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $T=1500\text{K}$ in equation (3)

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{32\text{g}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1500\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {\displaystyle \underset{15.8}{\overset{\infty }{\int }}{(15.8)}^2{e}^{\raisebox{1ex}{$-32\text{g}{\text{mol}}^{-1}\times {\left(15.8\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1500\text{K}$}\right.}}\\ =\underset{\_}{1.9\times 10^{-69}}\end{array}$

At $v_{\text{esc}}=7.13\text{km}{\text{s}}^{-1}$ and temperature, $T=1500\text{K}$, the proportion of molecules that have enough speed to escape is shown below

For ${\text{H}}_{\text{2}}$ substitute the value of $M=2.015\text{g}{\text{mol}}^{-1}$, $R=8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $T=1500\text{K}$ in equation (3)

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{\text{2}\text{.015}\text{g}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1500\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {\displaystyle \underset{7.13}{\overset{\infty }{\int }}{(7.13)}^2{e}^{\raisebox{1ex}{$-2.015\times {\left(7.13\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1500\text{K}$}\right.}}\\ =\underset{\_}{2.5\times 10^{-1}}\end{array}$

For $\text{He}$ substitute the value of $M=4.003\text{g}{\text{mol}}^{-1}$, $R=8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $T=1500\text{K}$ in equation (3)

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{4.003\text{g}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1500\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {\displaystyle \underset{7.13}{\overset{\infty }{\int }}{(7.13)}^2{e}^{\raisebox{1ex}{$-4.003\text{g}{\text{mol}}^{-1}\times {\left(7.13\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1500\text{K}$}\right.}}\\ =\underset{\_}{4.3\times 10^{-2}}\end{array}$

For ${\text{O}}_{\text{2}}$ substitute the value of $M=32\text{g}{\text{mol}}^{-1}$, $R=8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $T=1500\text{K}$ in equation (3)

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{32\text{g}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1500\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {\displaystyle \underset{7.13}{\overset{\infty }{\int }}{(7.13)}^2{e}^{\raisebox{1ex}{$-32\text{g}{\text{mol}}^{-1}\times {\left(7.13\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1500\text{K}$}\right.}}\\ =\underset{\_}{4.2\times 10^{-14}}\end{array}$

The proportion of molecules that have enough speed to escape at temperature $1500\text{K}$  is shown below.

Escape velocity	${\text{H}}_{\text{2}}$	$\text{He}$	${\text{O}}_{\text{2}}$
$v_{\text{esc}}=15.8\text{km}{\text{s}}^{-1}$	$1.5\times {10}^{-4}$	$9.5\times {10}^{-9}$	$1.9\times {10}^{-69}$
$v_{\text{esc}}=7.13\text{km}{\text{s}}^{-1}$	$2.5\times {10}^{-1}$	$4.3\times {10}^{-2}$	$4.2\times {10}^{-14}$