Question

At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere? (Given: Mass of oxygen molecule(m)=$2.76 \times {10^{ - 26}}\;kg $
Boltzmann’s constant ${k_B} = 1.38 \times {10^{ - 23}}J{K^{ - 1}}$)
$A.$ $5.016 \times {10^4}\;K$
$B.$ $2.508 \times {10^4}\;K$
$C.$ $1.254 \times {10^4}\;K$
$D.$ $8.360 \times {10^4}\;K$

Answer 1

Hint: - The root-mean-square speed takes into account both molecular weight and temperature, two factors that directly affect the kinetic energy of a material.
Formula Used: -
${V_{rms}}$=$\sqrt {\dfrac{{3KT}}{m}} $
Because both oxygen and hydrogen gas is diatomic, oxygen is 16 times heavier than hydrogen as measured per atom or molecule.

Complete step-by-step solution -
The root-mean-square rate is the calculation of particle velocity in a gas.
The ${V_{rms}}$ is the square root of the average of the squares of the speeds of the molecules is ${\left({v}^{2}\right)}^\dfrac{1}{2}$ which can also be mentioned as ${V_{rms}} = $$\sqrt {\dfrac{{3{R}T}}{m}} $ of the molecules in a gas.
RMS speed is inversely proportional to the square root of mass (molecular or molar).
The root-mean-square speed takes into account molecular weight and temperature, two factors which directly affect a material's kinetic energy.
The Boltzmann constant is the proportionality factor which relates the average relative kinetic energy of particles in a gas to the gas's thermodynamic temperature.
We have, mass of oxygen molecule is $2.76 \times {10^{ - 26}}\;kg$ and ${k_B} = 1.38 \times {10^{ - 23}}J{K^{ - 1}}$
Escape Velocity is referred to as the minimum velocity that someone or object has to predict to transcend the planet earth's gravitational pull. In other words, the minimum speed needed to escape the gravitational field is speed escape,
Escape velocity is given by
${V_{escape}} = \sqrt {2gR} $
Where, $g$ is the acceleration due to gravity and $R$ is the radius of earth
Therefore, $g = 9.8 \dfrac{m}{{s}^2}$
$R = 6.4 \times {10^6}$
Therefore, \[{V_{escape}} = \sqrt {2 \times 9.8 \times 6.4 \times {{10}^6}} \]
\[ \to {V_{escape}} = \sqrt {2 \times 9.8 \times 6.4 \times {{10}^6}} \]
\[ \to {V_{escape}} = \sqrt {125.44 \times {{10}^6}} \]
\[ \to {V_{escape}} = 11.2 \times {10^3} \dfrac{m}{s} \]
Now let the temperature of the molecule be $T$ when it attains velocity, \[{V_{escape}}\]
As per the question, rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere i.e.,
${V_{esape}} = {V_{rms}}$,
where ${V_{rms}}$ is the speed of an oxygen molecule.
Hence as per root mean square of the velocity,
${V_{rms}} = $$\sqrt {\dfrac{{3{K_b}T}}{m}} $, where m is the molar mass of the gas in kilograms per mole, $T$ is the temperature, $K$ is molar mass gas constant.
$ \to $${V_{rms}} = $$\sqrt {\dfrac{{3{K_b}T}}{m}} $
$ \to T = \dfrac{{{{\left( {11.2 \times {{10}^3}} \right)}^2} \times m}}{{3{K_b}}}$ since, ${V_{esape}} = {V_{rms}}$ and \[{V_{escape}} = 11.2 \times {10^3} \dfrac{m}{s}\]
Putting the value of ${K_B} = 1.38 \times {10^{ - 23}}J{K^{ - 1}}$ and m =$2.76 \times {10^{ - 26}}\;kg$
$ \to T = \dfrac{{{{\left( {11.2 \times {{10}^3}} \right)}^2} \times \;\left( {2.76 \times {{10}^{ - 26}}} \right)}}{{3 \times \left( {1.38 \times {{10}^{ - 23}}} \right)}}$
$ \to T = 8.3626 \times {10^4}\;K$
Therefore, option D is the correct answer and the temperature at which the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere is $8.360 \times {10^4}\;K$.

Note: - Root mean square speed, ${V_{rms}} = $$\sqrt {\dfrac{{3{K_b}T}}{m}} $ and escape velocity ${V_{escape}} = \sqrt {2gR} $ are very important for solving such type of numerical. Also we must remember the properties and nature like oxygen and hydrogen gas are diatomic and oxygen is 16 times heavier than hydrogen as measured per atom or molecule.