Distribution of Molecular Speeds
Distribution of Molecular Speeds
The motion of molecules in a gas is random in magnitude and direction for individual molecules, but a gas of many molecules has a predictable distribution of molecular speeds. This distribution is called the Maxwell-Boltzmann distribution, after its originators, who calculated it based on kinetic theory, and has since been confirmed experimentally. (See the figure below.) The distribution has a long tail, because a few molecules may go several times the rms speed. The most probable speed \({v}_{\text{p}}\) is less than the rms speed \({v}_{\text{rms}}\). The second figure below shows that the curve is shifted to higher speeds at higher temperatures, with a broader range of speeds.
The Maxwell-Boltzmann distribution of molecular speeds in an ideal gas. The most likely speed \({v}_{\text{p}}\) is less than the rms speed \({v}_{\text{rms}}\). Although very high speeds are possible, only a tiny fraction of the molecules have speeds that are an order of magnitude greater than \({v}_{\text{rms}}\).
The distribution of thermal speeds depends strongly on temperature. As temperature increases, the speeds are shifted to higher values and the distribution is broadened.
The Maxwell-Boltzmann distribution is shifted to higher speeds and is broadened at higher temperatures.
What is the implication of the change in distribution with temperature shown in the figure above for humans? All other things being equal, if a person has a fever, he or she is likely to lose more water molecules, particularly from linings along moist cavities such as the lungs and mouth, creating a dry sensation in the mouth.
Example: Calculating Temperature: Escape Velocity of Helium Atoms
In order 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 solve for mass \(m\) of the helium atom, we can use information from the periodic table:
\(m=\cfrac{\text{molar mass}}{\text{number of atoms per mole}}.\)
- To solve for temperature \(T\), we can rearrange either
\(\overline{\text{KE}}=\cfrac{1}{2}m\overline{{v}^{2}}=\cfrac{3}{2}\text{kT}\)
or
\(\sqrt{\overline{{v}^{2}}}={v}_{\text{rms}}=\sqrt{\cfrac{3\text{kT}}{m}}\)
to yield
\(T=\cfrac{m\overline{{v}^{2}}}{\text{3}k},\)
where \(k\) is the Boltzmann constant and \(m\) is the mass of a helium atom.
4. Plug the known values into the equations and solve for the unknowns.
\(m=\cfrac{\text{molar mass}}{\text{number of atoms per mole}}=\cfrac{4\text{.}\text{0026}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{kg/mol}}{6\text{.}\text{02}×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}\text{mol}}=6\text{.}\text{65}×{\text{10}}^{-\text{27}}\phantom{\rule{0.25em}{0ex}}\text{kg}\)
\(T=\cfrac{(6\text{.}\text{65}×{\text{10}}^{-\text{27}}\phantom{\rule{0.25em}{0ex}}\text{kg}){(\text{11}\text{.}1×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s})}^{2}}{3(1\text{.}\text{38}×{\text{10}}^{-\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K})}=1\text{.}\text{98}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{K}\)
Discussion
This temperature is much higher than atmospheric temperature, which is approximately 250 K \((–\text{25}\text{º}\text{C}\) or \(–\text{10}\text{º}\text{F})\) at high altitude. Very few helium atoms are left in the atmosphere, but there were many when the atmosphere was formed. The reason for the loss of helium atoms is that there are a small number of helium atoms with speeds higher than Earth’s escape velocity even at normal temperatures. The speed of a helium atom changes from one instant to the next, so that at any instant, there is a small, but nonzero chance that the speed is greater than the escape speed and the molecule escapes from Earth’s gravitational pull. Heavier molecules, such as oxygen, nitrogen, and water (very little of which reach a very high altitude), 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, so few have speeds above the escape velocity that billions of years are required to lose significant amounts of the atmosphere. the figure below shows the impact 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 comparison between Earth and the Moon is discussed in this tutorial’s Problems and Exercises.
This photograph of Apollo 17 Commander Eugene Cernan driving the lunar rover on the Moon in 1972 looks as though it was taken at night with a large spotlight. In fact, the light is coming from the Sun. Because the acceleration due to gravity on the Moon is so low (about 1/6 that of Earth), the Moon’s escape velocity is much smaller. As a result, gas molecules escape very easily from the Moon, leaving it with virtually no atmosphere. Even during the daytime, the sky is black because there is no gas to scatter sunlight. (credit: Harrison H. Schmitt/NASA)
Check Your Understanding
If you consider a very small object such as a grain of pollen, in a gas, then the number of atoms and molecules striking its surface would also be relatively small. Would the grain of pollen experience any fluctuations in pressure due to statistical fluctuations in the number of gas atoms and molecules striking it in a given amount of time?
Solution
Yes. Such fluctuations actually occur for a body of any size in a gas, but since the numbers of atoms and molecules are immense for macroscopic bodies, the fluctuations are a tiny percentage of the number of collisions, and the averages spoken of in this section vary imperceptibly. Roughly speaking the fluctuations are proportional to the inverse square root of the number of collisions, so for small bodies they can become significant. This was actually observed in the 19th century for pollen grains in water, and is known as the Brownian effect.
PhET Explorations: Gas Properties
Pump gas molecules into a box and see what happens as you change the volume, add or remove heat, change gravity, and more. Measure the temperature and pressure, and discover how the properties of the gas vary in relation to each other.
This lesson is part of:
Temperature, Kinetic Theory, and Gas Laws