Kinetic Theory of Gases

The ideal gas equation $pV = nRT$ was discovered by measuring pressures and volumes. Kinetic theory explains it: a gas is a vast number of tiny molecules flying about at random, and the pressure on a wall is nothing more than the average force of molecules bouncing off it. This note states the assumptions of the model, derives $pV = \tfrac{1}{3}Nm\langle c^2\rangle$ step by step, and compares it with $pV = NkT$ to show that temperature measures the mean kinetic energy of the molecules. The derivation and the assumptions are asked in almost every series of Paper 4, usually for four to six marks.

The kinetic model of a gas

Picture a box of air at room temperature. It contains around $10^{25}$ molecules per cubic metre, each moving at several hundred metres per second in a random direction, colliding with each other and with the walls billions of times a second. Every collision with a wall pushes on the wall. Individually these pushes are tiny and irregular, but there are so many of them that their average is a perfectly steady force: the pressure of the gas.

To turn this picture into an equation, we need a simplified model with clear assumptions. A gas that behaves exactly according to this model is an ideal gas.

Some consequences follow directly. With no forces between collisions, molecules travel in straight lines at constant speed between collisions. With no intermolecular forces, the molecules have no potential energy, so all the internal energy of an ideal gas is kinetic. That fact is central to internal energy.

When the model fails

The assumptions explain why real gases deviate from ideal behaviour at high pressure and low temperature:

• At high pressure the molecules are packed close together, so their own volume is no longer negligible compared with the container.
• At low temperature the molecules move slowly, so the weak attractive forces between them have time to act during close approaches; the forces are no longer negligible. Near the boiling point these forces cause the gas to condense.

Pressure from molecular collisions

The derivation is examined in steps, and each step usually earns a mark. Learn it as a sequence of physical ideas, not as algebra to memorise.

One molecule, one direction

Consider a cube of side $L$ containing a single molecule of mass $m$. Let its velocity have a component $c_x$ perpendicular to one face, called face A.

Step 1: change in momentum at one collision. The molecule hits face A with momentum $+mc_x$ (towards the face). The collision is elastic and the wall is fixed, so it rebounds with $-mc_x$. The change in momentum of the molecule is

$$\Delta p = (-mc_x) - (+mc_x) = -2mc_x$$

so its magnitude is $2mc_x$.

Step 2: time between collisions with face A. After rebounding, the molecule must travel to the opposite face and back, a distance $2L$, at speed $c_x$ in the $x$-direction, before it hits face A again. The other components of velocity do not affect this time.

$$\Delta t = \frac{2L}{c_x}$$

Step 3: average force. By Newton's second law, force equals rate of change of momentum. The average force exerted on the molecule by the wall is $2mc_x / \Delta t$; by Newton's third law, the molecule exerts an equal and opposite force on the wall:

$$F = \frac{2mc_x}{2L/c_x} = \frac{mc_x^{2}}{L}$$

Step 4: pressure. Face A has area $L^2$:

$$p = \frac{F}{L^2} = \frac{mc_x^{2}}{L^3} = \frac{mc_x^{2}}{V}$$

where $V = L^3$ is the volume of the container.

Many molecules

Now put in $N$ molecules, each with its own $c_x$. The total pressure on face A is the sum of their contributions:

$$p = \frac{m}{V}\left(c_{x1}^{2} + c_{x2}^{2} + \dots + c_{xN}^{2}\right) = \frac{Nm\langle c_x^{2}\rangle}{V}$$

where $\langle c_x^{2}\rangle$ is the mean of the squares of the $x$-components.

Collisions between molecules do not change this result: in an elastic collision the total momentum and kinetic energy are shared differently, but on average the molecules' $x$-velocities are unaffected.

Extending to three dimensions

Each molecule's speed $c$ is related to its components by Pythagoras in three dimensions:

$$c^{2} = c_x^{2} + c_y^{2} + c_z^{2}$$

Averaging over all molecules, $\langle c^{2}\rangle = \langle c_x^{2}\rangle + \langle c_y^{2}\rangle + \langle c_z^{2}\rangle$. Because the motion is random, no direction is special, so the three averages are equal:

$$\langle c_x^{2}\rangle = \langle c_y^{2}\rangle = \langle c_z^{2}\rangle = \tfrac{1}{3}\langle c^{2}\rangle$$

Substituting into the pressure equation:

Since $Nm$ is the total mass of gas and $Nm/V$ is its density $\rho$, an equivalent form is

$$p = \tfrac{1}{3}\rho\langle c^{2}\rangle$$

This is convenient when a question gives the density instead of the number of molecules.

Mean-square and root-mean-square speed

The molecules in a gas have a wide spread of speeds. The derivation produces the mean-square speed $\langle c^{2}\rangle$: square every speed, then take the mean. Its square root is a speed in $\text{m s}^{-1}$:

The order of operations is exactly as the name says, read backwards: square the speeds, take the mean, then take the root. The r.m.s. speed is not the same as the mean speed. For speeds $200$, $400$, $600$ and $800\ \text{m s}^{-1}$:

• mean speed $= 2000/4 = 500\ \text{m s}^{-1}$
• mean-square speed $= (4 + 16 + 36 + 64) \times 10^{4}/4 = 3.0 \times 10^{5}\ \text{m}^2\ \text{s}^{-2}$
• r.m.s. speed $= \sqrt{3.0 \times 10^{5}} = 548\ \text{m s}^{-1}$

Squaring gives extra weight to the faster molecules, so $c_{\text{r.m.s.}}$ is always at least as large as the mean speed.

Temperature and molecular kinetic energy

We now have two equations for the same quantity $pV$: one from experiment and one from the molecular model.

$$pV = NkT \qquad\text{and}\qquad pV = \tfrac{1}{3}Nm\langle c^{2}\rangle$$

Setting them equal:

$$\tfrac{1}{3}Nm\langle c^{2}\rangle = NkT \quad\Rightarrow\quad m\langle c^{2}\rangle = 3kT$$

Multiplying both sides by $\tfrac{1}{2}$ gives the mean translational kinetic energy of a molecule:

This is one of the most important results in A Level physics. It says:

• The mean kinetic energy of the molecules of an ideal gas is directly proportional to the thermodynamic temperature. Temperature is a measure of the average random kinetic energy of the molecules.
• At $T = 0\ \text{K}$ the mean kinetic energy would be zero. This gives absolute zero a physical meaning.
• The mean kinetic energy depends only on temperature, not on the mass of the molecule. At the same temperature, a light helium atom and a heavy xenon atom have the same mean kinetic energy; the helium atom moves faster.

The r.m.s. speed and temperature

Rearranging $\tfrac{1}{2}m\langle c^{2}\rangle = \tfrac{3}{2}kT$:

$$c_{\text{r.m.s.}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}$$

where the second form uses $k/m = R/M$ with $M$ the molar mass in $\text{kg mol}^{-1}$. Two consequences are asked often:

• $c_{\text{r.m.s.}} \propto \sqrt{T}$: doubling the r.m.s. speed requires four times the thermodynamic temperature.
• At the same temperature, $c_{\text{r.m.s.}} \propto 1/\sqrt{m}$: lighter molecules move faster.

Total kinetic energy of a gas

For $N$ molecules the total translational kinetic energy is $N \times \tfrac{3}{2}kT = \tfrac{3}{2}NkT$. For $n$ moles this is $\tfrac{3}{2}nRT$, and since $pV = NkT$ it also equals $\tfrac{3}{2}pV$. For a monatomic ideal gas (such as helium or argon) this kinetic energy is the whole of the internal energy.

Worked examples

Practice questions

Answers

1. Any four of: a large number of molecules in random motion; volume of molecules negligible compared with the volume of the container; no intermolecular forces except during collisions; collisions are elastic; time of collisions negligible compared with time between collisions.
2. Mean $= 2500/5 = 500\ \text{m s}^{-1}$. Mean square $= (9 + 16 + 25 + 36 + 49) \times 10^{4}/5 = 2.7 \times 10^{5}\ \text{m}^2\ \text{s}^{-2}$; r.m.s. $= \sqrt{2.7 \times 10^{5}} = 520\ \text{m s}^{-1}$.
3. $c_{\text{r.m.s.}} = \sqrt{3RT/M} = \sqrt{3 \times 8.31 \times 273/0.0020} = 1.84 \times 10^{3}\ \text{m s}^{-1}$.
4. $\langle E_K\rangle = 1.5 \times 1.38 \times 10^{-23} \times 400 = 8.28 \times 10^{-21}\ \text{J}$. $c_{\text{r.m.s.}} = \sqrt{3 \times 8.31 \times 400/0.040} = 499\ \text{m s}^{-1}$ (about $500\ \text{m s}^{-1}$).
5. (a) $p = nRT/V = 3.0 \times 8.31 \times 320/0.50 = 1.60 \times 10^{4}\ \text{Pa}$. (b) $E_K = \tfrac{3}{2}nRT = 1.5 \times 3.0 \times 8.31 \times 320 = 1.20 \times 10^{4}\ \text{J}$.
6. $T = Mc^2/3R = 0.028 \times 1000^2/(3 \times 8.31) = 1123\ \text{K}$, which is $850\ ^\circ\text{C}$.
7. The mean kinetic energy of the molecules is proportional to $T$, so the molecules move faster. Each collision with a wall produces a larger change in momentum, and collisions with the walls are more frequent. The rate of change of momentum at the walls increases, so the force per unit area, the pressure, increases. The volume is fixed, so the number of molecules per unit volume is unchanged.
8. Total $E_K = N \times \tfrac{3}{2}kT = \tfrac{3}{2}NkT = \tfrac{3}{2}pV = 1.5 \times 1.0 \times 10^{5} \times 50 = 7.5 \times 10^{6}\ \text{J}$. If the room warms at constant pressure, air escapes: $N$ falls in proportion as $T$ rises, so $NT$, and hence the total kinetic energy, stays the same.
9. Change in momentum at one wall $= 2mc_x$ (elastic collisions, so the molecule rebounds with the same speed). Time between collisions with that wall $= 2L/c_x$ (no intermolecular forces, so constant velocity between walls; molecular volume negligible, so the distance travelled is $2L$; collision time negligible). Force $= mc_x^2/L$; pressure $= mc_x^2/L^3 = mc_x^2/V$. For $N$ molecules (a large number, so the force is steady), $p = Nm\langle c_x^2\rangle/V$. Random motion gives $\langle c_x^2\rangle = \langle c_y^2\rangle = \langle c_z^2\rangle$ and $\langle c^2\rangle = 3\langle c_x^2\rangle$, so $pV = \tfrac{1}{3}Nm\langle c^2\rangle$.
10. (a) $1:1$: the mean kinetic energy depends only on temperature, which is the same for both. (b) $c_{\text{He}}/c_{\text{Ne}} = \sqrt{20/4.0} = 2.24$. (c) At the same $V$ and $T$, $N \propto p$, so $N_{\text{He}}/N_{\text{Ne}} = 3.0/5.0 = 0.60$. Mass ratio $= 0.60 \times 4.0 : 1 \times 20 = 2.4 : 20$, so the helium fraction of the mass is $2.4/22.4 = 0.107$, about $11\%$.