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How does the random motion of countless molecules give rise to the pressure and temperature we measure?

The molecular kinetic theory model, the assumptions behind it, the kinetic theory equation, root mean square speed, and the link between mean kinetic energy of a molecule and absolute temperature.

A focused answer to AQA A-Level Physics 3.6.2.3 and 3.6.2.4, covering the kinetic theory equation, the assumptions of an ideal gas, root mean square speed, and the relationship between mean molecular kinetic energy and absolute temperature.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. Assumptions of the model
  3. The kinetic theory equation
  4. Root mean square speed
  5. Linking molecular energy to temperature
  6. Try this

What this dot point is asking

AQA specification points 3.6.2.3 and 3.6.2.4 want you to state the assumptions of the molecular kinetic theory model, use the kinetic theory equation, define and calculate root mean square speed, and link the mean kinetic energy of a molecule directly to absolute temperature.

Assumptions of the model

These assumptions let the gas be treated statistically: although individual molecules move chaotically, their average behaviour is well defined and gives the smooth, reproducible pressure and temperature we measure.

The kinetic theory equation

By considering the change in momentum each time a molecule collides elastically with a wall and rebounds, and summing over all molecules, the theory derives the pressure of the gas in terms of the molecular motion.

The factor of one third arises because the molecules move in three dimensions, but only the component of motion perpendicular to a given wall contributes to the pressure on it.

Root mean square speed

Molecules have a wide spread of speeds (the Maxwell-Boltzmann distribution), so we cannot use a single speed. The relevant average is the rms speed, because pressure depends on the mean of the squared speeds.

The rms speed is slightly larger than the simple mean speed, because squaring weights faster molecules more heavily.

Linking molecular energy to temperature

Comparing the kinetic theory equation with the molecular form of the ideal gas equation, pV=NkTpV = NkT, gives the central result of the model.

So the mean kinetic energy of a molecule depends only on the absolute temperature, not on the type of gas. This is why all gases at the same temperature have the same mean molecular kinetic energy, and why lighter molecules (smaller mm) move faster on average than heavier ones at the same temperature.

Try this

Q1. State two assumptions of the molecular kinetic theory model. [2 marks]

  • Cue. Random motion, negligible molecular volume, elastic collisions, or negligible intermolecular forces (any two).

Q2. Calculate the mean kinetic energy of a gas molecule at 300 K300 \text{ K} (k=1.38×10−23 J K−1k = 1.38 \times 10^{-23} \text{ J K}^{-1}). [2 marks]

  • Cue. 32kT=32×1.38×10−23×300=6.2×10−21 J\tfrac{3}{2}kT = \tfrac{3}{2} \times 1.38 \times 10^{-23} \times 300 = 6.2 \times 10^{-21} \text{ J}.

Q3. State how the mean kinetic energy of a molecule depends on the absolute temperature. [1 mark]

  • Cue. It is directly proportional to the absolute temperature.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksCalculate the root mean square speed of oxygen molecules at a temperature of 290 K290 \text{ K}. The mass of one oxygen molecule is 5.3×10−26 kg5.3 \times 10^{-26} \text{ kg} and k=1.38×10−23 J K−1k = 1.38 \times 10^{-23} \text{ J K}^{-1}.
Show worked answer →

Use the molecular energy relation 12mc2‾=32kT\tfrac{1}{2}m\overline{c^2} = \tfrac{3}{2}kT, rearranged to c2‾=3kTm\overline{c^2} = \dfrac{3kT}{m}.

c2‾=3(1.38×10−23)(290)5.3×10−26=1.20×10−205.3×10−26=2.27×105 m2 s−2\overline{c^2} = \dfrac{3(1.38 \times 10^{-23})(290)}{5.3 \times 10^{-26}} = \dfrac{1.20 \times 10^{-20}}{5.3 \times 10^{-26}} = 2.27 \times 10^{5} \text{ m}^2 \text{ s}^{-2}.

Taking the square root: crms=2.27×105=4.8×102 m s−1c_{rms} = \sqrt{2.27 \times 10^5} = 4.8 \times 10^2 \text{ m s}^{-1}.

Markers reward the energy relation, rearrangement for the mean square speed, and taking the square root at the end (a common error is to forget the final root).

AQA 20214 marksState two assumptions made in the molecular kinetic theory of an ideal gas, and explain how the model accounts for the pressure a gas exerts on the walls of its container.
Show worked answer →

Two valid assumptions: the molecules are in continuous random motion; collisions between molecules and with the walls are perfectly elastic (others acceptable: negligible molecular volume, negligible intermolecular forces except during collisions, negligible collision time).

The molecules collide with the walls and rebound, each collision producing a change in momentum. By Newton's second law, the rate of change of momentum equals a force, and many such collisions per second produce a steady average force. Pressure is this force divided by the wall area.

Markers reward two correct assumptions and explaining pressure through the rate of change of momentum during elastic collisions with the walls.

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