State the assumptions of the kinetic theory of an ideal gas, apply the ideal gas equation, and relate pressure and temperature to the mean square molecular speed

What this dot point is asking

SEAB wants you to state the assumptions of the kinetic model of an ideal gas, apply the ideal gas equation in its molar and molecular forms, and use the kinetic-theory result that links pressure to the mean square molecular speed and temperature to the mean molecular kinetic energy. This is where the microscopic and macroscopic descriptions of a gas meet.

The answer

Assumptions of the kinetic model

An ideal gas is modelled as a large number of identical molecules for which:

• the molecules are in continuous 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 (with each other and the walls) are perfectly elastic,
• the duration of a collision is negligible compared with the time between collisions.

These assumptions let us treat the gas as point particles bouncing elastically, which is enough to derive the gas laws.

The ideal gas equation

The macroscopic behaviour is captured by:

where $n$ is the number of moles and $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ is the molar gas constant. Equivalently, in terms of the number of molecules $N$:

where $k = \dfrac{R}{N_A} = 1.38 \times 10^{-23}\ \text{J K}^{-1}$ is the Boltzmann constant. Temperature must always be in kelvin.

Pressure from molecular motion

Deriving the pressure from molecular collisions with the walls gives:

where $m$ is the mass of one molecule and $\langle c^2 \rangle$ is the mean square speed. This shows that pressure arises from the rate of change of momentum of molecules striking the walls.

Temperature and molecular kinetic energy

Comparing $pV = NkT$ with $pV = \tfrac{1}{3}Nm\langle c^2 \rangle$ gives:

So the mean translational kinetic energy of a molecule is $\tfrac{3}{2}kT$, depending only on the absolute temperature, not on the type of gas. The root-mean-square speed is $c_{\text{rms}} = \sqrt{\langle c^2 \rangle}$.

Examples in context

Example 1. Why hydrogen leaks faster than oxygen. At the same temperature, hydrogen and oxygen molecules have the same mean kinetic energy $\tfrac{3}{2}kT$. Because hydrogen is far lighter, its molecules move much faster ($c_{\text{rms}} \propto 1/\sqrt{m}$), so they escape through small gaps and diffuse more quickly, an everyday consequence of the kinetic theory.

Example 2. A tyre on a hot day. Heating the air in a tyre at roughly constant volume raises the pressure in proportion to absolute temperature, since $p \propto T$ at constant $V$ and $n$. A tyre inflated on a cold morning reads higher pressure after a fast highway drive, because both ambient heating and friction warm the gas.

Try this

Q1. State four assumptions of the kinetic theory of an ideal gas. [2 marks]

• Cue. Random motion; negligible molecular volume; no intermolecular forces except in collisions; perfectly elastic collisions; negligible collision time.

Q2. A gas has pressure $2.0 \times 10^5\ \text{Pa}$ and volume $1.5 \times 10^{-3}\ \text{m}^3$ at $310\ \text{K}$. Find the number of moles ($R = 8.31$). [2 marks]

• Cue. $n = \dfrac{pV}{RT} = \dfrac{(2.0 \times 10^5)(1.5 \times 10^{-3})}{8.31 \times 310} = 0.117\ \text{mol}$.

Q3. Explain why, at the same temperature, helium atoms move faster on average than nitrogen molecules. [2 marks]

• Cue. Mean kinetic energy $\tfrac{3}{2}kT$ is equal, so $\tfrac{1}{2}m\langle c^2\rangle$ is equal; the lighter helium has a larger $\langle c^2 \rangle$ and hence a higher rms speed.