Blow up a balloon for a festival and tie it. Nothing solid holds its rubber skin out against the pull of the rubber. There is no frame inside, only air. Yet the skin stays tight for days. The air inside must push on every square centimetre of the rubber, all the time and equally in every direction.
A model made of molecules
The gas laws of the last lesson describe what a gas does. They do not say why. The kinetic theory answers the why. It pictures a gas as an enormous number of molecules in ceaseless motion, and it explains pressure and temperature with Newton's laws alone. Daniel Bernoulli first tried this in 1738. Clausius, Maxwell and Boltzmann completed it in the nineteenth century.
A gas is made of a very large number of identical molecules. They move at random, with no preferred direction. The molecules are points: their own volume is negligible beside the volume of the vessel. They exert no forces on one another except during collisions. Their collisions, with each other and with the walls, are perfectly elastic and last a negligible time.
These assumptions describe an ideal gas, and they are close to the truth for air. At \(300\,\text{K}\) and \(1\,\text{atm}\) a cubic metre holds \(2.45 \times 10^{25}\) molecules. So each molecule has a cube of side \(3.4\,\text{nm}\) to itself, more than ten times the size of a molecule, about \(0.3\,\text{nm}\). The molecules themselves fill less than a thousandth of the space. A gas is mostly empty space, crossed at high speed by tiny points.
Which of these are assumptions of the kinetic theory of an ideal gas?
One molecule and one wall
Follow a single molecule of mass \(m\) in a cubical box of side \(L\). Let its velocity have components \(v_x\), \(v_y\) and \(v_z\). It strikes the right wall, which is perpendicular to the \(x\) axis. The collision is elastic and the wall is smooth, so \(v_x\) reverses while \(v_y\) and \(v_z\) do not change. The momentum of the molecule changes from \(+mv_x\) to \(-mv_x\), a change of \(2mv_x\). By Newton's third law the wall receives \(2mv_x\) in the opposite direction.
After a hit the molecule must cross to the far wall and return before it strikes the right wall again. That is a distance \(2L\) along \(x\) at the speed \(v_x\), taking a time \(2L/v_x\). Collisions with the side walls change \(v_y\) or \(v_z\), not \(v_x\), so they do not delay the return. In Fig 1.3 the hits come every \(3.75\,\text{s}\), exactly \(2L/v_x\).
From one hit to a pressure
Now add the pushes together. The wall receives \(2mv_x\) every \(2L/v_x\). The average force from this one molecule is the momentum it delivers per second, which is \(mv_x^2/L\). Every molecule does the same with its own \(v_x\). Their pushes add to a steady force, and the force divided by the area is the pressure.
\[ \Delta p = mv_x - (-mv_x) = 2mv_x \]
Why. the collision is elastic, so the molecule leaves the smooth wall with \(v_x\) reversed and \(v_y\), \(v_z\) unchanged. The wall receives this momentum.
\[ \Delta t = \dfrac{2L}{v_x} \]
Why. between two hits on the same wall the molecule travels \(2L\) along \(x\) at the constant speed \(v_x\).
\[ F_1 = \dfrac{2mv_x}{2L/v_x} = \dfrac{mv_x^2}{L} \]
Why. an average force is the momentum delivered per unit time (Newton's second law).
\[ F = \dfrac{m}{L}\left(v_{x1}^2 + v_{x2}^2 + \dots + v_{xN}^2\right) = \dfrac{Nm\,\overline{v_x^2}}{L} \]
Why. the \(N\) molecules push independently, so their forces add; the bar means the average over all the molecules.
\[ \overline{v_x^2} = \tfrac13\,\overline{v^2} \]
Why. \(v^2 = v_x^2 + v_y^2 + v_z^2\) for every molecule, and random motion favours no direction, so the three averages are equal.
\[ p = \dfrac{F}{L^2} = \dfrac{Nm\,\overline{v^2}}{3L^3} = \dfrac{1}{3}\,\dfrac{Nm}{V}\,\overline{v^2} = \dfrac{1}{3}\rho\,\overline{v^2} \]
Read it. the wall has area \(L^2\), the box has volume \(V = L^3\), and the density of the gas is \(\rho = Nm/V\).
The derivation ignored collisions between molecules. They do not matter. Two molecules that collide elastically share their momentum and energy in a new way. The totals stay the same, so the average \(\overline{v^2}\) is unchanged. The shape of the vessel does not matter either: any wall can be divided into small flat pieces, and each piece feels the same pressure.
Why a third? A molecule that moves along \(x\) hits the right wall; one that moves along \(y\) never does. Random motion shares the square of the speed equally among the three directions. Only the \(x\) share, one third, pushes on a wall facing \(x\).
A molecule of mass \(m\) strikes a smooth wall with velocity components \(v_x\) towards the wall and \(v_y\) along it, and rebounds elastically. What momentum does the wall receive?
Set the speed to \(1000\,\text{m/s}\) in the sim above. Each collision now delivers twice the momentum, and the molecules also reach the piston twice as often. The two factors multiply, and the pressure reads \(400\,\text{kPa}\), four times as much. Now return to \(500\,\text{m/s}\) and halve the length to \(10\,\text{cm}\). The push of each hit is unchanged, but the trip across and back is half as long. The hits come twice as often, and the pressure doubles to \(200\,\text{kPa}\).
The second result is Boyle's law, now explained. At a fixed temperature the molecules keep their speeds. Halve the volume and they strike each square metre of wall twice as often. So \(p\) doubles and \(pV\) stays constant. The kinetic theory has recovered a law found by experiment, from nothing but moving points.
In a rigid vessel, the molecules of a gas are made to move with twice their root mean square speed. By what factor does the pressure change?
The root mean square speed
The pressure depends on \(\overline{v^2}\), the mean of the squared speeds. Its square root is a speed, the root mean square speed, \(v_{\text{rms}} = \sqrt{\overline{v^2}}\). The name gives the recipe backwards: square every speed, take the mean, then take the root. It is not the same as the average speed.
Take five molecules with speeds \(200\), \(400\), \(500\), \(600\) and \(800\,\text{m/s}\). Their average speed is \(2500/5 = 500\,\text{m/s}\). Their squares add to \((4 + 16 + 25 + 36 + 64) \times 10^4\), and the mean square is \(2.9 \times 10^5\,\text{m}^2/\text{s}^2\). So \(v_{\text{rms}} = 539\,\text{m/s}\). Squaring gives the fast molecules more weight. So \(v_{\text{rms}}\) is larger than the average speed unless all the speeds are equal.
The root mean square speed of \(N\) molecules with speeds \(v_1, v_2, \dots, v_N\) is \(v_{\text{rms}} = \sqrt{(v_1^2 + v_2^2 + \dots + v_N^2)/N}\). A molecule moving at \(v_{\text{rms}}\) has the average kinetic energy of the gas.
Now solve the pressure formula for the speed. From \(p = \tfrac13\rho v_{\text{rms}}^2\), we get \(v_{\text{rms}} = \sqrt{3p/\rho}\). Both \(p\) and \(\rho\) can be measured. And the ideal gas equation gives \(p/\rho = RT/M\), so \(v_{\text{rms}} = \sqrt{3RT/M}\). Here \(M\) is the molar mass in kg/mol, not the mass of the sample.
Here is the balloon's answer. Air at \(1\,\text{atm}\) has a density of \(1.18\,\text{kg/m}^3\), so \(v_{\text{rms}} = \sqrt{3 \times 1.013 \times 10^5/1.18}\), which is \(508\,\text{m/s}\). The molecules that hold the skin out move at half a kilometre per second, twice as fast as a passenger jet. For nitrogen at \(300\,\text{K}\), \(v_{\text{rms}} = \sqrt{3 \times 8.31 \times 300/0.028}\), which is \(517\,\text{m/s}\).
Hydrogen at STP has a density of \(0.090\,\text{kg/m}^3\). Find the root mean square speed of its molecules. Check the answer with \(v_{\text{rms}} = \sqrt{3RT/M}\), taking \(M = 2\,\text{g/mol}\) and \(T = 273\,\text{K}\).
From the pressure: \(v_{\text{rms}} = \sqrt{3 \times 1.013 \times 10^5/0.090}\), which is \(\sqrt{3.38 \times 10^6} = 1838\,\text{m/s}\).
From the temperature: \(v_{\text{rms}} = \sqrt{3 \times 8.31 \times 273/0.002}\), which is \(\sqrt{3.40 \times 10^6} = 1845\,\text{m/s}\).
The two agree within half a percent. Hydrogen molecules move at nearly \(2\,\text{km/s}\), almost four times as fast as air molecules.
Find the root mean square speed of oxygen molecules (\(M = 32\,\text{g/mol}\)) at \(300\,\text{K}\), in m/s. Take \(R = 8.31\,\text{J/(mol K)}\).
Four gases are at the same temperature. Rank them by the root mean square speed of their molecules, fastest first.
In his book Hydrodynamica of 1738, Daniel Bernoulli pictured air as tiny particles flying in all directions and striking the walls of a cylinder. He showed that squeezing the air into a smaller space makes the hits more frequent, and so he explained Boyle's law. He also saw that faster particles would press harder, which links pressure to heat. Almost nobody believed him. The idea waited more than a hundred years for Clausius and Maxwell.
Sound travels through air at about \(343\,\text{m/s}\), while the molecules of air move at about \(500\,\text{m/s}\). This is no accident. A sound wave is passed from molecule to molecule by collisions, so it cannot outrun the molecules that carry it. The chapter on sound in the unit on waves finds its speed exactly.
- Taking the momentum change in a wall collision as \(mv\).The molecule reverses, so the change is \(2mv_x\): only the component towards the wall, doubled.
- Using \(M\) in g/mol in \(v_{\text{rms}} = \sqrt{3RT/M}\).With \(R = 8.31\,\text{J/(mol K)}\), \(M\) must be in kg/mol: \(32\,\text{g/mol} = 0.032\,\text{kg/mol}\).
- Treating the root mean square speed as the average speed.Square first, average, then take the root. \(v_{\text{rms}}\) is larger than the average speed.