# 2.2 Pressure, temperature, and rms speed  (Page 8/18)

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$\lambda =\frac{V}{4\sqrt{2}\pi {r}^{2}N}.$

In an ideal gas, we can substitute $V\text{/}N={k}_{\text{B}}T\text{/}p$ to obtain

$\lambda =\frac{{k}_{\text{B}}T}{4\sqrt{2}\pi {r}^{2}p}.$

The mean free time     $\tau$ is simply the mean free path divided by a typical speed, and the usual choice is the rms speed. Then

$\tau =\frac{{k}_{\text{B}}T}{4\sqrt{2}\pi {r}^{2}p{v}_{\text{rms}}}.$

## Calculating mean free time

Find the mean free time for argon atoms $\left(M=39.9\phantom{\rule{0.2em}{0ex}}\text{g/mol}\right)$ at a temperature of $0\phantom{\rule{0.2em}{0ex}}\text{°C}$ and a pressure of 1.00 atm. Take the radius of an argon atom to be $1.70\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}$

## Solution

1. Identify the knowns and convert into SI units. We know the molar mass is 0.0399 kg/mol, the temperature is 273 K, the pressure is $1.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{Pa},$ and the radius is $1.70\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}\phantom{\rule{0.2em}{0ex}}\text{m}.$
2. Find the rms speed: ${v}_{\text{rms}}=\sqrt{\frac{3RT}{M}}=413\phantom{\rule{0.2em}{0ex}}\frac{\text{m}}{\text{s}}$ .
3. Substitute into the equation for the mean free time:
$\tau =\frac{{k}_{\text{B}}T}{4\sqrt{2}\pi {r}^{2}p{v}_{\text{rms}}}=\frac{\left(1.38\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-23}\phantom{\rule{0.2em}{0ex}}\text{J/K}\right)\phantom{\rule{0.2em}{0ex}}\left(273\phantom{\rule{0.2em}{0ex}}\text{K}\right)}{4\sqrt{2}\pi {\left(1.70\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{2}\left(1.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{Pa}\right)\left(413\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)}=1.76\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}\phantom{\rule{0.2em}{0ex}}\text{s}\text{.}$

## Significance

We can hardly compare this result with our intuition about gas molecules, but it gives us a picture of molecules colliding with extremely high frequency.

Check Your Understanding Which has a longer mean free path, liquid water or water vapor in the air?

In a liquid, the molecules are very close together, constantly colliding with one another. For a gas to be nearly ideal, as air is under ordinary conditions, the molecules must be very far apart. Therefore the mean free path is much longer in the air.

## Summary

• Kinetic theory is the atomic description of gases as well as liquids and solids. It models the properties of matter in terms of continuous random motion of molecules.
• The ideal gas law can be expressed in terms of the mass of the gas’s molecules and $\stackrel{\text{–}}{{v}^{2}},$ the average of the molecular speed squared, instead of the temperature.
• The temperature of gases is proportional to the average translational kinetic energy of molecules. Hence, the typical speed of gas molecules ${v}_{\text{rms}}$ is proportional to the square root of the temperature and inversely proportional to the square root of the molecular mass.
• In a mixture of gases, each gas exerts a pressure equal to the total pressure times the fraction of the mixture that the gas makes up.
• The mean free path (the average distance between collisions) and the mean free time of gas molecules are proportional to the temperature and inversely proportional to the molar density and the molecules’ cross-sectional area.

## Conceptual questions

How is momentum related to the pressure exerted by a gas? Explain on the molecular level, considering the behavior of molecules.

If one kind of molecule has double the radius of another and eight times the mass, how do their mean free paths under the same conditions compare? How do their mean free times compare?

The mean free path is inversely proportional to the square of the radius, so it decreases by a factor of 4. The mean free time is proportional to the mean free path and inversely proportional to the rms speed, which in turn is inversely proportional to the square root of the mass. That gives a factor of $\sqrt{8}$ in the numerator, so the mean free time decreases by a factor of $\sqrt{2}.$

What is the average velocity of the air molecules in the room where you are right now?

Why do the atmospheres of Jupiter, Saturn, Uranus, and Neptune, which are much more massive and farther from the Sun than Earth is, contain large amounts of hydrogen and helium?

Since they’re more massive, their gravity is stronger, so the escape velocity from them is higher. Since they’re farther from the Sun, they’re colder, so the speeds of atmospheric molecules including hydrogen and helium are lower. The combination of those facts means that relatively few hydrogen and helium molecules have escaped from the outer planets.

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