The escape speed from the Earth is \({\bf{1}}{\bf{.12 \times 1}}{{\bf{0}}^{\bf{4}}}\;{\bf{m/s}}\),that is, a gas molecule traveling away from Earth near the outer boundary of the Earth’s atmosphere would, at this speed, be able to escape from the Earth’s gravitational field and be lost in the atmosphere. At what temperature is the RMS speed of (a) oxygen molecules and (b) helium atoms equal to \({\bf{1}}{\bf{.12 \times 1}}{{\bf{0}}^{\bf{4}}}\;{\bf{m/s}}\)? (c) Can you explain why our atmosphere contains oxygen but not helium?

The square root of the mean of the square of the molecules’ speedin a gas at a certain temperature Tis termed as the root-mean square speed of that gas.

The RMS speed is given as

\({v_{{\rm{rms}}}} = \sqrt {\frac{{3kT}}{m}} \). …… (i)

Here, \(k\)is the Boltzmann’s constant, T is the temperature,and m is the mass of each molecule.

The RMS speed of oxygen molecules and helium atoms is\({v_{{\rm{rms}}}} = 1.12 \times {10^4}\;{\rm{m/s}}\).

The molecular mass of oxygen is 32 u.

The atomic mass of the helium atom is 4 u.

The mass of the oxygen molecule can be calculated in the following manner:

\(\begin{aligned}m &= 32\;{\rm{u}} \times \frac{{1.66 \times {{10}^{ - 27}}\;{\rm{kg}}}}{{1\;{\rm{u}}}}\\m &= 53.12 \times {10^{ - 27}}\;{\rm{kg}}\end{aligned}\)

From equation (i), the temperature is given by

\(T = \frac{{mv_{{\rm{rms}}}^2}}{{3k}}\). …… (ii)

Substitute the values into the above equation.

\(\begin{aligned}T &= \frac{{\left( {53.12\; \times {{10}^{ - 27}}\;{\rm{kg}}} \right){{\left( {1.12 \times {{10}^4}\;{\rm{m/s}}} \right)}^2}}}{{3\left( {1.38 \times {{10}^{ - 23}}\;{\rm{J/K}}} \right)}}\\ &= 1.61\; \times {10^5}\;{\rm{K}}\end{aligned}\)

Thus, the RMS speed of oxygen molecules is equal to \(1.12 \times {10^4}\;{\rm{m/s}}\)at the temperature of\(1.61\; \times {10^5}\;{\rm{K}}\).

The mass of the helium atom can be calculated in the following manner:

\(\begin{aligned}m &= 4\;{\rm{u}} \times \frac{{1.66 \times {{10}^{ - 27}}\;{\rm{kg}}}}{{1\;{\rm{u}}}}\\m &= 6.64 \times {10^{ - 27}}\;{\rm{kg}}\end{aligned}\)

From equation (ii), the temperature of helium atoms can be calculated in the following manner:

\(\begin{aligned}T &= \frac{{\left( {6.64\; \times {{10}^{ - 27}}\;{\rm{kg}}} \right){{\left( {1.12 \times {{10}^4}\;{\rm{m/s}}} \right)}^2}}}{{3\left( {1.38 \times {{10}^{ - 23}}\;{\rm{J/K}}} \right)}}\\ &= 2.01\; \times {10^4}\;{\rm{K}}\end{aligned}\)

Thus, the RMS speed of helium atoms is equal to \(1.12 \times {10^4}\;{\rm{m/s}}\)at the temperature of\(2.01\; \times {10^4}\;{\rm{K}}\).

The temperature at which oxygen molecules can escape the Earth’s atmosphere is \(1.61 \times {10^5}\;{\rm{m/s}}\) while the temperature at which helium atoms can escape the Earth’s atmosphere is \(2.01 \times {10^4}\;{\rm{m/s}}\). It can be seen that oxygen molecules require a much higher temperature in order to escape the Earth’s surface as compared to helium atoms. So, it is very difficult for oxygen molecules to leave the Earth’s atmosphere.

However, the temperature of the Earth during its formation was much higher, which is comparable to the temperature at which helium atoms can escape.In that case, the helium atoms being at a much lower temperature than the oxygen molecules might have escaped from the atmosphere.

Therefore, our atmosphere contains oxygen but not helium.