12.2 Kinetic Theory of an Ideal Gas: Pressure and Temperature

Stand under a tin roof in a hailstorm and you hear the pressure of the hail as a roar of separate impacts. Each hailstone gives the roof a small kick as it bounces off; there are so many kicks every second that they add up to a steady downward force. The air around you does exactly the same to every surface it touches. About $3\times10^{23}$ molecules hit each square centimetre of your skin every second, each bouncing off with a tiny change of momentum, and together they push with $10\ \text{N}$ on every square centimetre — the weight of a $1\ \text{kg}$ mass. That steady push is atmospheric pressure.

In Section 12.1 the gas equation $pV=Nk_BT$ was an experimental fact. In this section you will derive the pressure of a gas from Newton's laws, using a simple model of molecules as tiny elastic balls — the kinetic theory of gases. The result, $p=\tfrac13\rho\overline{v^2}$, links a quantity you measure with a gauge to the speeds of molecules you cannot see. Comparing it with the gas equation then tells you what temperature is: a measure of the average kinetic energy of the molecules. From that you will find how fast molecules move, why light gases escape from planets, and how uranium isotopes are separated.

The assumptions of kinetic theory

These are exactly the conditions of Definition 12.2, written as rules of motion. They describe dilute gases well: in air at room conditions the molecules occupy less than $0.1\%$ of the volume (Exercise I.6 of Section 12.1), and a molecule spends almost all its time flying freely.

The pressure of an ideal gas

To find the pressure, we need averages over the molecules' velocities. Because the speeds of different molecules differ, two different averages of speed appear, and it is important to keep them apart.

Now place $N$ molecules, each of mass $m$, in a cubical box of side $l$, so $V=l^3$. Take a molecule with velocity components $(v_x,v_y,v_z)$ and follow it as it bounces between the two faces perpendicular to the $x$-axis.

• One collision. The collision with the right-hand face is elastic and the face is smooth, so $v_x$ reverses and $v_y$, $v_z$ are unchanged. The molecule's momentum changes from $mv_x$ to $-mv_x$; by Newton's third law the wall receives an impulse $2mv_x$ (Section 4.1).
• Collision rate. Between hits on the same face, the molecule crosses the box and back, a distance $2l$ in the $x$-direction, at speed $v_x$: it hits that face $\dfrac{v_x}{2l}$ times per second.
• Average force from one molecule. Impulse per second $=2mv_x\times\dfrac{v_x}{2l}=\dfrac{mv_x^2}{l}$. Collisions with the other faces do not change $v_x$.
• All the molecules. Adding over the $N$ molecules, the force on the face is $F=\dfrac ml\left(v_{x1}^2+v_{x2}^2+\dots\right)=\dfrac{Nm}{l}\overline{v_x^2}$.

The pressure is $p=\dfrac{F}{l^2}=\dfrac{Nm}{l^3}\overline{v_x^2}=\dfrac{Nm}{V}\cdot\dfrac13\overline{v^2}$. Collisions between molecules do not spoil this result: they conserve momentum, so in a gas in equilibrium they simply hand a velocity component from one molecule to another, and on average the distribution of velocities — and so $\overline{v_x^2}$ — stays the same. Nor does the shape of the box matter, since pressure is the same on every face of a gas in equilibrium.

The kinetic interpretation of temperature

We now have two equations for the same gas. Kinetic theory gives $pV=\tfrac23N\left(\tfrac12m\overline{v^2}\right)$; experiment gives $pV=Nk_BT$. They agree only if the bracket — the mean translational kinetic energy of one molecule — is proportional to the absolute temperature.

This is one of the great results of physics: the thermometer reading measures how vigorously the molecules are moving. Several consequences follow at once.

• Same temperature, same mean kinetic energy. In a mixture of hydrogen and oxygen at the same temperature, the light hydrogen molecules and the heavy oxygen molecules have the same average kinetic energy, $\tfrac32k_BT$; so the hydrogen molecules must move faster.
• Dalton's law explained. In a mixture, each kind of molecule contributes $\tfrac13n_{V1}m_1\overline{v_1^2}=n_{V1}k_BT$ to the pressure; adding the contributions gives $p=(n_{V1}+n_{V2}+\dots)k_BT=p_1+p_2+\dots$ (Law 12.4).
• Absolute zero. In this model $T=0$ is the temperature at which the translational motion of the molecules would stop. (Quantum physics shows that some motion always remains, but the classical picture is right at ordinary temperatures.)
• Order of magnitude. At room temperature ($300\ \text{K}$), $\tfrac32k_BT=$ $6.2\times10^{-21}\ \text{J}$ — tiny, but there are $10^{25}$ molecules in every cubic metre.

Exercises

Take $R=8.31\ \text{J mol}^{-1}\text{K}^{-1}$, $k_B=1.38\times10^{-23}\ \text{J/K}$, $N_A=6.02\times10^{23}\ \text{mol}^{-1}$ and $1\ \text{atm}=1.013\times10^5\ \text{Pa}$. Treat all gases as ideal.