# 13.4 Kinetic theory: atomic and molecular explanation of pressure  (Page 4/5)

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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?

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.

## Section summary

• Kinetic theory is the atomistic description of gases as well as liquids and solids.
• Kinetic theory models the properties of matter in terms of continuous random motion of atoms and molecules.
• The ideal gas law can also be expressed as
$\text{PV}=\frac{1}{3}\text{Nm}\overline{{v}^{2}},$
where $P$ is the pressure (average force per unit area), $V$ is the volume of gas in the container, $N$ is the number of molecules in the container, $m$ is the mass of a molecule, and $\overline{{v}^{2}}$ is the average of the molecular speed squared.
• Thermal energy is defined to be the average translational kinetic energy $\overline{\text{KE}}$ of an atom or molecule.
• The temperature of gases is proportional to the average translational kinetic energy of atoms and molecules.
$\overline{\text{KE}}=\frac{1}{2}m\overline{{v}^{2}}=\frac{3}{2}\text{kT}$

or

$\sqrt{\overline{{v}^{2}}}={v}_{\text{rms}}=\sqrt{\frac{3\text{kT}}{m}}\text{.}$
• The motion of individual molecules in a gas is random in magnitude and direction. However, a gas of many molecules has a predictable distribution of molecular speeds, known as the Maxwell-Boltzmann distribution .

## Conceptual questions

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

## Problems&Exercises

Some incandescent light bulbs are filled with argon gas. What is ${v}_{\text{rms}}$ for argon atoms near the filament, assuming their temperature is 2500 K?

$1\text{.}\text{25}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s}$

Average atomic and molecular speeds $\left({v}_{\text{rms}}\right)$ are large, even at low temperatures. What is ${v}_{\text{rms}}$ for helium atoms at 5.00 K, just one degree above helium’s liquefaction temperature?

(a) What is the average kinetic energy in joules of hydrogen atoms on the $\text{5500}\text{º}\text{C}$ surface of the Sun? (b) What is the average kinetic energy of helium atoms in a region of the solar corona where the temperature is $6\text{.}\text{00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{K}$ ?

(a) $1\text{.}\text{20}×{\text{10}}^{-\text{19}}\phantom{\rule{0.25em}{0ex}}\text{J}$

(b) $1\text{.}\text{24}×{\text{10}}^{-\text{17}}\phantom{\rule{0.25em}{0ex}}\text{J}$

The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would oxygen molecules (molecular mass is equal to 32.0 g/mol) have an average velocity ${v}_{\text{rms}}$ equal to Earth’s escape velocity of 11.1 km/s?

The escape velocity from the Moon is much smaller than from Earth and is only 2.38 km/s. At what temperature would hydrogen molecules (molecular mass is equal to 2.016 g/mol) have an average velocity ${v}_{\text{rms}}$ equal to the Moon’s escape velocity?

$\text{458}\phantom{\rule{0.25em}{0ex}}\text{K}$

Nuclear fusion, the energy source of the Sun, hydrogen bombs, and fusion reactors, occurs much more readily when the average kinetic energy of the atoms is high—that is, at high temperatures. Suppose you want the atoms in your fusion experiment to have average kinetic energies of $6\text{.}\text{40}×{\text{10}}^{–\text{14}}\phantom{\rule{0.25em}{0ex}}\text{J}$ . What temperature is needed?

Suppose that the average velocity $\left({v}_{\text{rms}}\right)$ of carbon dioxide molecules (molecular mass is equal to 44.0 g/mol) in a flame is found to be $1\text{.}\text{05}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ . What temperature does this represent?

$1\text{.}\text{95}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{K}$

Hydrogen molecules (molecular mass is equal to 2.016 g/mol) have an average velocity ${v}_{\text{rms}}$ equal to 193 m/s. What is the temperature?

Much of the gas near the Sun is atomic hydrogen. Its temperature would have to be $1\text{.}5×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{K}$ for the average velocity ${v}_{\text{rms}}$ to equal the escape velocity from the Sun. What is that velocity?

$6\text{.}\text{09}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{m/s}$

There are two important isotopes of uranium— ${}^{\text{235}}\text{U}$ and ${}^{\text{238}}\text{U}$ ; these isotopes are nearly identical chemically but have different atomic masses. Only ${}^{\text{235}}\text{U}$ is very useful in nuclear reactors. One of the techniques for separating them (gas diffusion) is based on the different average velocities ${v}_{\text{rms}}$ of uranium hexafluoride gas, ${\text{UF}}_{6}$ . (a) The molecular masses for ${}^{\text{235}}\text{U}\phantom{\rule{0.25em}{0ex}}$ ${\text{UF}}_{6}$ and ${}^{\text{238}}\text{U}$ $\phantom{\rule{0.25em}{0ex}}{\text{UF}}_{6}$ are 349.0 g/mol and 352.0 g/mol, respectively. What is the ratio of their average velocities? (b) At what temperature would their average velocities differ by 1.00 m/s? (c) Do your answers in this problem imply that this technique may be difficult?

#### Questions & Answers

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