Ideal gases and kinetic theoryEdexcel A-Level Physics: Revision notes

An ideal gas is a model in which the molecules obey these assumptions:

• the molecules are in constant, random motion
• the volume of the molecules is negligible compared with the volume of the container
• there are no intermolecular forces except during collisions
• collisions between molecules, and with the walls, are perfectly elastic
• the time of a collision is negligible compared with the time between collisions

Pressure arises because molecules hitting a wall change momentum. By Newton's second law the wall exerts a force on each molecule, and by Newton's third law the molecule exerts an equal and opposite force on the wall. The sum of these forces per unit area is the pressure.

Take a cubic box of side $l$ containing $N$ molecules of mass $m$.

1. A molecule moving at $c_x$ towards a wall rebounds elastically, so its momentum changes by $2mc_x$.
2. It returns to the same wall after a time $2l/c_x$.
3. Force from one molecule = rate of change of momentum = $mc_x^2/l$.
4. For $N$ molecules, $F = Nm\langle c_x^2\rangle/l$, so $p = F/l^2 = Nm\langle c_x^2\rangle/V$.
5. Random motion means $\langle c^2\rangle = 3\langle c_x^2\rangle$.

Therefore $pV = \frac{1}{3}Nm\langle c^2\rangle$.

$\langle c^2\rangle$ is the mean square speed. Its square root is the root mean square (rms) speed $c_{rms}$.

For an ideal gas, $pV = NkT$, where $N$ is the number of molecules, $T$ is the thermodynamic temperature in kelvin and $k$ is the Boltzmann constant, $1.38 \times 10^{-23}$ J K⁻¹.

Worked example. Find the pressure when $4.8 \times 10^{24}$ helium atoms occupy $0.020$ m³ at $300$ K.

$p = NkT/V = (4.8 \times 10^{24} \times 1.38 \times 10^{-23} \times 300)/0.020 = 9.9 \times 10^{5}$ Pa.

For a fixed number of molecules at constant $T$, $pV$ is constant. This is Boyle's law: $p \propto 1/V$.

Equating $\frac{1}{3}Nm\langle c^2\rangle = NkT$ and cancelling $N$ gives $m\langle c^2\rangle = 3kT$, so

$\frac{1}{2}m\langle c^2\rangle = \frac{3}{2}kT$

The mean translational kinetic energy of a molecule depends only on the thermodynamic temperature, not on its mass. At the same temperature, lighter molecules therefore have a greater rms speed, $c_{rms} = \sqrt{3kT/m}$.

Worked example. An argon atom (mass $6.63 \times 10^{-26}$ kg) at $300$ K has mean kinetic energy $\frac{3}{2} \times 1.38 \times 10^{-23} \times 300 = 6.2 \times 10^{-21}$ J and $c_{rms} = \sqrt{3 \times 1.38 \times 10^{-23} \times 300 / 6.63 \times 10^{-26}} = 433$ m s⁻¹.

Method. Trap a fixed mass of air above oil in a uniform sealed tube. Use a foot pump to raise the pressure in steps, reading $p$ from a Bourdon gauge and the air column length $l$ from a scale. Because the tube is uniform, $V \propto l$.

Keeping the temperature constant. Increase the pressure slowly and wait before each reading, because compressing the gas does work on it and warms it. Keep the tube away from heat sources, or use a water bath.

Analysis. Plot $p$ against $1/l$. A straight line through the origin shows $p \propto 1/V$, so $pV$ is constant. Alternatively, calculate $pl$ for each reading and check that it is constant within uncertainty.