Kinetic Theory

Making Connections: Things Great and Small—Atomic and Molecular Origin of Pressure in a Gas

The figure below shows a box filled with a gas. We know from our previous discussions that putting more gas into the box produces greater pressure, and that increasing the temperature of the gas also produces a greater pressure. But why should increasing the temperature of the gas increase the pressure in the box? A look at the atomic and molecular scale gives us some answers, and an alternative expression for the ideal gas law.

The figure shows an expanded view of an elastic collision of a gas molecule with the wall of a container. Calculating the average force exerted by such molecules will lead us to the ideal gas law, and to the connection between temperature and molecular kinetic energy. We assume that a molecule is small compared with the separation of molecules in the gas, and that its interaction with other molecules can be ignored. We also assume the wall is rigid and that the molecule’s direction changes, but that its speed remains constant (and hence its kinetic energy and the magnitude of its momentum remain constant as well). This assumption is not always valid, but the same result is obtained with a more detailed description of the molecule’s exchange of energy and momentum with the wall.

If the molecule’s velocity changes in the \(x\)-direction, its momentum changes from \(–{\text{mv}}_{x}\) to \(+{\text{mv}}_{x}\). Thus, its change in momentum is \(\text{Δ}\text{mv}\phantom{\rule{0.20em}{0ex}}\text{= +}{\text{mv}}_{x}–(–{\text{mv}}_{x})=2{\text{mv}}_{x}\). The force exerted on the molecule is given by

\(F=\cfrac{\text{Δ}p}{\text{Δ}t}=\cfrac{2{\text{mv}}_{x}}{\text{Δ}t}\text{.}\)

There is no force between the wall and the molecule until the molecule hits the wall. During the short time of the collision, the force between the molecule and wall is relatively large. We are looking for an average force; we take \(\text{Δ}t\) to be the average time between collisions of the molecule with this wall. It is the time it would take the molecule to go across the box and back (a distance \(2l)\) at a speed of \({v}_{x}\). Thus \(\text{Δ}t=2l/{v}_{x}\), and the expression for the force becomes

\(F=\cfrac{2{\text{mv}}_{x}}{2l/{v}_{x}}=\cfrac{{\text{mv}}_{x}^{2}}{l}\text{.}\)

This force is due to one molecule. We multiply by the number of molecules \(N\) and use their average squared velocity to find the force

\(F=N\cfrac{m\overline{{v}_{x}^{2}}}{l},\)

where the bar over a quantity means its average value. We would like to have the force in terms of the speed \(v\), rather than the \(x\)-component of the velocity. We note that the total velocity squared is the sum of the squares of its components, so that

\(\overline{{v}^{2}}=\overline{{v}_{x}^{2}}+\overline{{v}_{y}^{2}}+\overline{{v}_{z}^{2}}\text{.}\)

Because the velocities are random, their average components in all directions are the same:

\(\overline{{v}_{x}^{2}}=\overline{{v}_{y}^{2}}=\overline{{v}_{z}^{2}}\text{.}\)

Thus,

\(\overline{{v}^{2}}=3\overline{{v}_{x}^{2}},\)

or

\(\overline{{v}_{x}^{2}}=\cfrac{1}{3}\overline{{v}^{2}}.\)

Substituting \(\cfrac{1}{3}\overline{{v}^{2}}\) into the expression for \(F\) gives

\(F=N\cfrac{m\overline{{v}^{2}}}{3l}\text{.}\)

The pressure is \(F/A,\) so that we obtain

\(P=\cfrac{F}{A}=N\cfrac{m\overline{{v}^{2}}}{3\text{Al}}=\cfrac{1}{3}\cfrac{\text{Nm}\overline{{v}^{2}}}{V},\)

where we used \(V=\text{Al}\) for the volume. This gives the important result.

\(\text{PV}=\cfrac{1}{3}\text{Nm}\overline{{v}^{2}}\)

This equation is another expression of the ideal gas law.