Question

A gas molecule of mass $M$ at the surface of the Earth has kinetic energy equivalent to $0^\circ \,C$. If it were to go up straight without colliding with any other molecules, how high would it rise? Assume that the height attained is much less than the radius of the earth. ( ${k_B}$ is the Boltzmann constant)
A) $0$
B) $\dfrac{{273{k_B}}}{{2Mg}}$
C) $\dfrac{{564{k_B}}}{{2Mg}}$
D) $\dfrac{{819{k_B}}}{{2Mg}}$

Answer 1

Hint
The temperature of a gas is defined as the average kinetic energy associated with a gas at a given temperature. Use the law of conservation of energy and equalize the kinetic energy of the gas molecule at the surface with the potential energy of the gas at its highest point.
Formula used:
- Average energy of a gas: ${E_{avg}} = \dfrac{3}{2}{k_B}T$ where ${k_B}$ is the Boltzmann constant and $T$ is its temperature in Kelvin
- Potential energy of a molecule: $U = mgh$ where $m$ is its mass, g is the gravitational acceleration on the surface of Earth, $h$ is its height

Complete step by step answer
At the surface of the Earth, the molecule will only have kinetic energy associated with its thermal energy. Since the gas has temperature $0^\circ C = 237\,K$, we can calculate the average kinetic energy of the gas using the relation
$\Rightarrow {E_{avg}} = \dfrac{3}{2}{k_B}T$
Substituting the value of $T = 273\,K$, we get
$\Rightarrow {E_{avg}} = \dfrac{{819}}{2} \times {k_B}$
When the gas molecule reaches its highest point, all its energy will have converted to potential energy and it will have no kinetic energy. So we can equalize the kinetic energy of the gas molecule with the potential energy of the gas molecule at height $h$ as
$\Rightarrow \dfrac{{819}}{2} \times {k_B} = Mgh$ where $g$ is the gravitational acceleration on the surface of the Earth.
Solving for $h$, we get
$h = \dfrac{{819{k_B}}}{{2Mg}}$ which corresponds to option (D).

Note
Here we have assumed that the gas molecule will rise straight up however in reality gas molecules continuously interact with other molecules and get deflected in random directions. As a result, no gas molecule in reality would move completely straight in any direction for more than a few nanometres. While calculating the average kinetic energy of the molecule, we should take the temperature in the Kelvin scale and not in the Centigrade scale.