Mean Free Path, Real Gases and the Atmosphere

How far between collisions?

Light an agarbatti at one end of a room. Its molecules move at about \(500\,\text{m/s}\), yet the smell takes a minute or more to cross the room. The reason is collisions. A molecule moves in a straight line only until it strikes another molecule, then leaves in a new direction. Its path is a zigzag, and its progress across the room is slow. The average length of a straight stretch between collisions is the mean free path \(\lambda\).

Treat the molecules as balls of diameter \(d\). Two molecules collide if their centres come within \(d\) of each other. So a moving molecule strikes every molecule whose centre lies inside a cylinder of radius \(d\) around its path. In a time \(t\) it moves a distance \(vt\) and sweeps a cylinder of volume \(\pi d^2 vt\). If there are \(n\) molecules per unit volume, it meets \(n\pi d^2 vt\) of them.

Dividing the distance \(vt\) by the number of collisions gives \(\lambda = 1/(\pi d^2 n)\). This treats the other molecules as standing still. They move too. On average, one molecule moves relative to another at \(\sqrt2\) times its own speed. So collisions come \(\sqrt2\) times as often, and the exact result has \(\sqrt2\) in the denominator. With \(n = p/kT\) from the ideal gas equation, \(\lambda\) can be written with the pressure and the temperature.

For air at \(300\,\text{K}\) and \(1\,\text{atm}\), take \(d = 3 \times 10^{-10}\,\text{m}\) and \(n = 2.45 \times 10^{25}\,\text{m}^{-3}\). Then \(\pi d^2 = 2.83 \times 10^{-19}\,\text{m}^2\) and \(\sqrt2\,\pi d^2 n = 9.8 \times 10^6\,\text{m}^{-1}\). So \(\lambda = 1.0 \times 10^{-7}\,\text{m}\), about \(0.1\,\mu\text{m}\). That is \(340\) molecular diameters. A molecule makes \(\bar v/\lambda\) collisions each second, the collision frequency. With an average speed of \(468\,\text{m/s}\), that is about \(4.6 \times 10^9\) collisions every second.

Watch the counter in Fig 6.1. The first few free paths give an average that swings from \(21\,\text{nm}\) to over \(100\,\text{nm}\). As more paths are added, the average steadies near \(100\,\text{nm}\). No single free path equals \(\lambda\). The mean free path is a property of the crowd, like the average height of a class.

The formula also shows how \(\lambda\) changes. At a fixed temperature, \(\lambda\) is inversely proportional to the pressure. In a vacuum chamber at \(10^{-3}\,\text{atm}\) it grows to \(0.1\,\text{mm}\). At a fixed pressure, \(\lambda\) is proportional to \(T\), because a hotter gas at the same pressure is less crowded. Bigger molecules have shorter free paths, as \(1/d^2\).

When a gas is not ideal

Two assumptions of the ideal gas fail when a gas is squeezed hard or cooled. The molecules do have a size, and they do pull on one another when close. At ordinary pressures neither matters, because the molecules fill less than a thousandth of the space and are far apart. At a hundred atmospheres they fill a few percent of the space and are close neighbours. A gas that departs from \(pV = nRT\) is a real gas.

The departure is measured by the compressibility factor \(Z = pV/nRT\). For an ideal gas \(Z = 1\) at every pressure.

Fig 6.2 shows both effects. At moderate pressure, A, the attraction between molecules wins. A molecule about to strike the wall is pulled backwards by the others behind it, so it strikes more gently. The pressure is lower than the ideal value and \(Z < 1\). At high pressure, C, the size of the molecules wins. They cannot enter the space the others fill, so the gas resists squeezing and \(Z > 1\).

Johannes van der Waals made two corrections in 1873. The molecules cannot move into the space the others occupy, so the free volume is \(V - nb\). The constant \(b\) is about four times the volume of the molecules in one mole. The attraction lowers the measured pressure by an amount proportional to \((n/V)^2\). So \(an^2/V^2\) is added back to \(p\). The units follow: \(an^2/V^2\) is a pressure, so \(a\) is in Pa m⁶/mol², and \(nb\) is a volume, so \(b\) is in m³/mol.

A real gas is nearly ideal under two conditions. At low pressure the molecules are far apart, so their size and their attraction hardly matter. At high temperature the molecules move so fast that the weak attraction barely bends their paths. Below a certain temperature, the critical temperature \(T_c = 8a/27Rb\), squeezing turns a gas into a liquid. For carbon dioxide \(T_c = 304\,\text{K}\), just \(31\,^\circ\text{C}\). Oxygen and nitrogen have \(T_c\) near \(-119\,^\circ\text{C}\) and \(-147\,^\circ\text{C}\), so they could not be liquefied until they were first cooled below these temperatures.

The weight of the air

The air near the ground is squeezed by the weight of all the air above it. Climb a little, and there is less air above you, so the pressure falls. In the unit on fluids you met \(dp = -\rho g\,dh\) for a liquid, whose density is fixed. Air is different. Its density is proportional to its pressure, \(\rho = pM/RT\), so the thin air higher up weighs less and the pressure falls more and more slowly. This part uses \(g = 9.8\,\text{m/s}^2\), as the atmosphere needs, and \(M = 0.029\,\text{kg/mol}\) for air.

The length \(H = RT/Mg\) is the scale height. For air at \(300\,\text{K}\), \(H = 8.31 \times 300/(0.029 \times 9.8) = 8772\,\text{m}\), about \(8.8\,\text{km}\). Each climb of \(8.8\,\text{km}\) divides the pressure by \(e = 2.72\). Half the air lies below \(H\ln2 = 6.1\,\text{km}\). The scale height is also the height the whole atmosphere would have if its density stayed at the sea-level value. At a fixed temperature the density is proportional to the pressure, so it falls with height in exactly the same way.

Move the slider to \(2.2\,\text{km}\), the height of Shimla: \(p/p_0 = e^{-2.2/8.77} = 0.78\). At Leh, \(3.5\,\text{km}\), it is \(e^{-3.5/8.77} = 0.67\), or \(68\,\text{kPa}\). Each breath at Leh draws in only two thirds of the air, and two thirds of the oxygen, of a breath at the sea. At the top of Everest, \(8.85\,\text{km}\), the formula gives \(0.36\). The measured value is about \(0.33\), because the real air there is far colder than \(300\,\text{K}\).

This pays off the hook. At \(68\,\text{kPa}\) water boils at about \(89\,^\circ\text{C}\) instead of \(100\,^\circ\text{C}\). Dal simmering at \(89\,^\circ\text{C}\) cooks slowly, so kitchens in Leh use a pressure cooker to raise the pressure and the boiling point again. The breathlessness has the same cause: the partial pressure of oxygen falls from \(0.21\,\text{atm}\) at the sea to about \(0.14\,\text{atm}\) at Leh.