As noted in the text, the speed distribution of molecules in the Earth's atmosphere has a significant impact on its composition.
a) What is the average speed of a nitrogen molecule in the atmosphere, at a temperature of 18.0 °C and a (partial) pressure of 78.8 kPa?
b) What is the average speed of a hydrogen molecule at the same temperature and pressure?
THINK
Average speed of a molecule can be found from the expression using kinetic theory of gases.
Average speed is independent of the pressure of the gas. It depends only upon the temperature of the gas and on its molecular or molar mass.
RESEARCH
The average speed of a given molecule is given by
\(v_{\mathrm{ave}}=\sqrt{\frac{8 k_{\mathrm{B}} T}{\pi m}} \ldots \ldots\) (1)
Here \(T\) is the absolute temperature of the gas, and \(m\) is the mass of the each molecule, \(k_{\mathrm{B}}\) is
the Boltzmann constant.
From the equation of average speed, it can be noticed that average speed is independent of pressure of the gas.
CALCULATE
a) The temperature of the nitrogen molecule is \(T=18.0^{\circ} \mathrm{C}\)
\(=(18+273) \mathrm{K}\)
\(=291 \mathrm{~K}\)
The Boltzmann constant, \(k_{\mathrm{B}}=1.381 \times 10^{-23} \mathrm{~J} / \mathrm{K}\)
The mass of \(\mathrm{N}_{2}\) molecule, \(m=28.01 \mathrm{amu}\)
\(=28.01 \mathrm{amu}\left(\frac{1.661 \times 10^{-27} \mathrm{~kg}}{1 \mathrm{amu}}\right)\)
\(=46.5246 \times 10^{-27} \mathrm{~kg}\)
The average speed of the nitrogen molecule is given by
$$ \begin{aligned} v_{\text {ave }} &=\sqrt{\frac{8\left(1.381 \times 10^{-23} \mathrm{~J} / \mathrm{K}\right)(291 \mathrm{~K})}{(3.142)\left(46.5246 \times 10^{-27} \mathrm{~kg}\right)}} \\ &=\sqrt{21.9931 \times 10^{4}} \\ &=468.968 \mathrm{~m} / \mathrm{s} \\ &=469 \mathrm{~m} / \mathrm{s} \end{aligned} $$
(b) The temperature of the hydrogen molecule is \(T=18.0^{\circ} \mathrm{C}\)
\(=(18+273) \mathrm{K}\)
\(=291 \mathrm{~K}\)
The Boltzmann constant, \(k_{\mathrm{B}}=1.381 \times 10^{-23} \mathrm{~J} / \mathrm{K}\)
The mass of \(\mathrm{H}_{2}\) molecule, \(m=2.016 \mathrm{amu}\)
\(=2.016 \mathrm{amu}\left(\frac{1.661 \times 10^{-27} \mathrm{~kg}}{1 \mathrm{amu}}\right)\)
\(=3.3485 \times 10^{-27} \mathrm{~kg}\)
The average speed of the hydrogen molecule is given by
$$ \begin{aligned} v_{\text {ave }} &=\sqrt{\frac{8\left(1.381 \times 10^{-23} \mathrm{~J} / \mathrm{K}\right)(291 \mathrm{~K})}{(3.142)\left(3.3485 \times 10^{-27} \mathrm{~kg}\right)}} \\ &=\sqrt{305.5766 \times 10^{4}} \\ &=1748.07 \mathrm{~m} / \mathrm{s} \\ &=1748 \mathrm{~m} / \mathrm{s} \end{aligned} $$