On a clear night the Moon shows sharp black shadows and no haze at all. The astronauts who walked there in 1969 needed sealed suits, because the Moon has no air. The Earth, less than four times wider, holds an ocean of air a hundred kilometres deep. Yet even the Earth's air holds almost no hydrogen, the most common element in the universe.
Temperature is molecular energy
The last lesson found two expressions for the same product \(pV\). Kinetic theory gives \(pV = \tfrac13 Nm\overline{v^2}\), from the hits on the walls. Experiment gives \(pV = NkT\), the ideal gas equation. Setting them equal tells us what temperature is. This lesson uses \(k = 1.38 \times 10^{-23}\,\text{J/K}\) and \(R = 8.31\,\text{J/(mol K)}\).
\[ pV = \tfrac13 Nm\,\overline{v^2} \]
Why. the kinetic pressure \(p = \tfrac13 (Nm/V)\,\overline{v^2}\), multiplied by \(V\).
\[ pV = \tfrac23 N\left(\tfrac12 m\,\overline{v^2}\right) \]
Why. write \(\tfrac13\) as \(\tfrac23 \times \tfrac12\), so the average kinetic energy of one molecule appears in the bracket.
\[ pV = NkT \]
Why. the ideal gas equation, written per molecule.
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Your turn. equate the two expressions for \(pV\) and divide by \(\tfrac23 N\). Which line is it?
The absolute temperature of an ideal gas is a measure of the average translational kinetic energy of its molecules: \(\tfrac12 m\overline{v^2} = \tfrac32 kT\). At the same temperature, the molecules of every gas have the same average translational kinetic energy.
The result is striking in two ways. First, the energy depends on \(T\) alone. A molecule of hydrogen and a molecule of oxygen at the same temperature carry the same average energy. The oxygen molecule is \(16\) times heavier, so it moves \(\sqrt{16} = 4\) times slower. Second, absolute zero now has a meaning: it is the temperature at which, in this classical picture, the molecules would stop.
At \(300\,\text{K}\), each molecule of any gas carries on average \(\tfrac32 kT = 1.5 \times 1.38 \times 10^{-23} \times 300\), which is \(6.21 \times 10^{-21}\,\text{J}\). That is a tiny energy, but a mole holds \(6.02 \times 10^{23}\) molecules. One mole at \(300\,\text{K}\) therefore holds \(\tfrac32 RT = 1.5 \times 8.31 \times 300 = 3740\,\text{J}\) of translational energy. That is enough to lift a \(10\,\text{kg}\) bag of rice through \(37\,\text{m}\).
The formula also shows how speed grows with temperature. From \(\tfrac12 m v_{\text{rms}}^2 = \tfrac32 kT\), \(v_{\text{rms}} = \sqrt{3kT/m}\). This matches \(\sqrt{3RT/M}\) from the last lesson, because \(R = N_Ak\) and \(M = N_Am\). So \(v_{\text{rms}}\) grows as \(\sqrt T\). To double the molecular speed, a gas must be heated from \(300\,\text{K}\) to \(1200\,\text{K}\).
Hydrogen and oxygen are in two vessels at the same temperature. Compare their molecules.
At what temperature is the average translational kinetic energy of a gas molecule \(1.0 \times 10^{-20}\,\text{J}\)? Take \(k = 1.38 \times 10^{-23}\,\text{J/K}\). Give the answer in kelvin.
A crowd of different speeds
The molecules of a gas do not share one speed. Collisions keep changing each speed: one molecule gains and another loses. At any instant a few molecules are almost at rest, most move at a few hundred metres per second, and a few move very fast. In 1860 James Clerk Maxwell found how the speeds are shared in equilibrium. The result is the Maxwell distribution.
The distribution is drawn as a curve \(f(v)\) against the speed \(v\). The height of the curve at a speed tells how many molecules have speeds near it. The area under the curve between two speeds is the number of molecules with speeds in that range. So the area under the whole curve is the total number of molecules. The curve starts at zero, rises to a peak and then falls in a long tail towards high speeds.
Nitrogen is heated from \(300\,\text{K}\) to \(1200\,\text{K}\) in a closed vessel. What happens to the peak of its speed distribution?
Choose the chips in the figure above. At \(300\,\text{K}\) the peak is at \(422\,\text{m/s}\). At \(1200\,\text{K}\) it is at \(844\,\text{m/s}\), twice as fast for four times the temperature. The curve is also lower and wider, because the shaded area, the number of molecules, never changes. Heating does not make every molecule fast. It shifts the whole crowd towards higher speeds and spreads it out.
Maxwell's formula is
\[ f(v) = 4\pi N\left(\dfrac{m}{2\pi kT}\right)^{3/2} v^2 e^{-mv^2/2kT} \]
Here \(f(v)\,dv\) molecules have speeds between \(v\) and \(v + dv\). The factor \(v^2\) makes the curve rise from zero: there are more ways to move fast than slowly. The exponential lowers it again, because a large kinetic energy is rare.
A gas in a rigid vessel is heated. Which statements about its Maxwell distribution are true?
Three speeds that describe the crowd
Three speeds describe the distribution. The most probable speed \(v_{\text{mp}}\) sits at the peak: more molecules have speeds near it than near any other. The average speed \(\bar v\) is the ordinary mean. The root mean square speed \(v_{\text{rms}}\) is the one that fixes the pressure and the energy. The long tail pulls the mean above the peak, and squaring pulls \(v_{\text{rms}}\) higher still.
All three have the common factor \(\sqrt{RT/M}\). Their ratio is \(\sqrt2 : \sqrt{8/\pi} : \sqrt3 = 1.414 : 1.596 : 1.732\). Dividing by \(1.414\) gives \(1 : 1.13 : 1.22\). For nitrogen at \(300\,\text{K}\) they are \(422\), \(476\) and \(517\,\text{m/s}\). The order never changes: \(v_{\text{mp}} < \bar v < v_{\text{rms}}\), for every gas at every temperature.
In these formulas \(M\) is the molar mass, the mass of one mole, in kg/mol. It is never the mass of the gas sample. The form \(v_{\text{rms}} = \sqrt{3pV/m}\) is also true, but there \(m\) is the total mass of the sample. Mixing the two meanings gives wrong answers. Two vessels of hydrogen at \(1\,\text{atm}\) and \(2\,\text{atm}\), at one temperature, have the same \(v_{\text{rms}}\).
The root mean square speed of the molecules of a gas is \(600\,\text{m/s}\). Find their average speed and their most probable speed.
\(\bar v = v_{\text{rms}}\sqrt{8/3\pi} = 600 \times 0.921 = 553\,\text{m/s}\).
\(v_{\text{mp}} = v_{\text{rms}}\sqrt{2/3} = 600 \times 0.816 = 490\,\text{m/s}\).
As always, \(v_{\text{mp}} < \bar v < v_{\text{rms}}\).
The rms speed of the molecules of a gas is \(500\,\text{m/s}\). What is their most probable speed, in m/s?
Why light gases escape
A molecule leaves a planet for good if it moves upward faster than the escape speed. The chapter on gravitational energy and orbits found \(11.2\,\text{km/s}\) for the Earth and \(2.4\,\text{km/s}\) for the Moon. At \(300\,\text{K}\), hydrogen molecules have \(v_{\text{rms}} = 1934\,\text{m/s}\) and oxygen molecules \(483\,\text{m/s}\). Both are well below even the Moon's escape speed. So how does any gas escape?
The answer is the tail of the distribution. Fig 3.1 shows that about \(21\,\%\) of hydrogen molecules at \(300\,\text{K}\) move faster than \(2.4\,\text{km/s}\). For helium it is \(2.8\,\%\). For nitrogen it is one molecule in \(10^{13}\), and for oxygen fewer still. The fast molecules near the top of an atmosphere escape, collisions refill the tail, and the leak continues for billions of years.
So the Moon, with its small escape speed and a sunlit surface near \(390\,\text{K}\), lost every gas long ago. The Earth holds nitrogen and oxygen easily, because their tails beyond \(11.2\,\text{km/s}\) are empty for all practical purposes. Hydrogen and helium, the lightest gases, still escape slowly. That is why the Earth's air holds almost no free hydrogen, although hydrogen fills the universe.
The rms speed of hydrogen in the upper air is far below the Earth's escape speed of \(11.2\,\text{km/s}\). Why does hydrogen escape all the same?
In 1859 Maxwell, then 28, read Clausius's papers on the motion of gas molecules. Within a year he asked a question nobody had asked. The molecules collide millions of times a second, so they cannot share one speed; how are the speeds shared out? His answer, published in 1860, was the first law of physics built on probability. It was tested directly only in the twentieth century, with beams of molecules passed through slotted spinning discs.
A helium balloon bought at a fair floats on the ceiling at night and sags by morning. A balloon of the same rubber filled with air stays firm for days. Helium atoms, at \(v_{\text{rms}} = 1367\,\text{m/s}\), are small and move nearly three times as fast as air molecules. So they strike the rubber more often and pass through its tiny pores faster.
- Thinking that a heavier gas at the same temperature has more kinetic energy per molecule.\(\bar K = \tfrac32 kT\) depends on \(T\) only. A heavier molecule has the same energy and a smaller speed.
- Taking the peak of the Maxwell curve to rise on heating.The area is fixed by the number of molecules. The curve widens, so its peak falls.
- Putting the mass of the sample for \(M\) in \(\sqrt{3RT/M}\).\(M\) is the molar mass in kg/mol. The amount of gas does not change any of the three speeds.