# 2.2 Pressure, temperature, and rms speed  (Page 5/18)

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## Calculating temperature: escape velocity of helium atoms

To escape Earth’s gravity, an object near the top of the atmosphere (at an altitude of 100 km) must travel away from Earth at 11.1 km/s. This speed is called the escape velocity . At what temperature would helium atoms have an rms speed equal to the escape velocity?

## Strategy

Identify the knowns and unknowns and determine which equations to use to solve the problem.

## Solution

1. Identify the knowns: v is the escape velocity, 11.1 km/s.
2. Identify the unknowns: We need to solve for temperature, T . We also need to solve for the mass m of the helium atom.
3. Determine which equations are needed.
• To get the mass m of the helium atom, we can use information from the periodic table:
$m=\frac{M}{{N}_{A}}.$
• To solve for temperature T , we can rearrange
$\frac{1}{2}\phantom{\rule{0.2em}{0ex}}m\stackrel{\text{–}}{{v}^{2}}=\frac{3}{2}\phantom{\rule{0.2em}{0ex}}{k}_{\text{B}}T$

to yield
$T=\frac{m\stackrel{\text{–}}{{v}^{2}}}{3{k}_{\text{B}}}.$
4. Substitute the known values into the equations and solve for the unknowns,
$m=\frac{M}{{N}_{A}}=\frac{4.0026\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{kg/mol}}{6.02\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{23}\phantom{\rule{0.2em}{0ex}}\text{mol}}=6.65\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-27}\phantom{\rule{0.2em}{0ex}}\text{kg}$

and
$T=\frac{{\left(6.65\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-27}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\phantom{\rule{0.2em}{0ex}}\left(11.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)}^{2}}{3\phantom{\rule{0.2em}{0ex}}\left(1.38\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\phantom{\rule{0.2em}{0ex}}\text{J/K}\right)}=1.98\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{K}\text{.}$

## Significance

This temperature is much higher than atmospheric temperature, which is approximately 250 K $\left(-25\phantom{\rule{0.2em}{0ex}}\text{°}\text{C or}\phantom{\rule{0.2em}{0ex}}-10\phantom{\rule{0.2em}{0ex}}\text{°}\text{F}\right)$ at high elevation. Very few helium atoms are left in the atmosphere, but many were present when the atmosphere was formed, and more are always being created by radioactive decay (see the chapter on nuclear physics). The reason for the loss of helium atoms is that a small number of helium atoms have speeds higher than Earth’s escape velocity even at normal temperatures. The speed of a helium atom changes from one collision to the next, so that at any instant, there is a small but nonzero chance that the atom’s speed is greater than the escape velocity. The chance is high enough that over the lifetime of Earth, almost all the helium atoms that have been in the atmosphere have reached escape velocity at high altitudes and escaped from Earth’s gravitational pull. Heavier molecules, such as oxygen, nitrogen, and water, have smaller rms speeds, and so it is much less likely that any of them will have speeds greater than the escape velocity. In fact, the likelihood is so small that billions of years are required to lose significant amounts of heavier molecules from the atmosphere. [link] shows the effect of a lack of an atmosphere on the Moon. Because the gravitational pull of the Moon is much weaker, it has lost almost its entire atmosphere. The atmospheres of Earth and other bodies are compared in this chapter’s exercises.

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