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How does the random motion of molecules produce the pressure of a gas, and how is temperature linked to molecular kinetic energy?

Kinetic theory: the assumptions of the kinetic model, the derivation of pV = (1/3)Nm<c^2>, and the link between absolute temperature and the mean kinetic energy of a molecule.

A focused answer to the Eduqas A-Level Physics Component 1 kinetic theory content, covering the assumptions of the kinetic model of a gas, the derivation of the kinetic theory equation pV = (1/3)Nm<c^2>, and the link between absolute temperature and the mean kinetic energy of a molecule.

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What this dot point is asking

Eduqas wants you to state the assumptions of the kinetic model of an ideal gas, outline the derivation of the kinetic theory equation pV=13Nmc2‾pV = \frac{1}{3}Nm\overline{c^2}, and relate the absolute temperature of a gas to the mean kinetic energy of its molecules.

The answer

Assumptions of the kinetic model

Deriving the kinetic theory equation

Temperature and molecular kinetic energy

Examples in context

The kinetic theory links the microscopic world of molecules to the macroscopic gas laws, explaining why pressure rises with temperature and why gases diffuse and exert pressure. It underlies the operation of engines, refrigerators and the behaviour of the atmosphere. The result that mean kinetic energy depends only on temperature explains why light molecules (such as hydrogen and helium) move fastest and can escape a planet's gravity, shaping planetary atmospheres.

Try this

Q1. State two assumptions of the kinetic model of an ideal gas. [2 marks]

  • Cue. Any two: continuous random motion; negligible molecular volume; perfectly elastic collisions; negligible collision time; no intermolecular forces between collisions.

Q2. State the relationship between the mean kinetic energy of a gas molecule and the absolute temperature. [1 mark]

  • Cue. Ek‾=32kT\overline{E_k} = \frac{3}{2}kT, directly proportional to the absolute temperature.

Q3. Find the mean kinetic energy of a molecule at 400 K400\ \text{K} (k=1.38×10−23 J K−1k = 1.38 \times 10^{-23}\ \text{J K}^{-1}). [2 marks]

  • Cue. Ek‾=32kT=32(1.38×10−23)(400)=8.3×10−21 J\overline{E_k} = \frac{3}{2}kT = \frac{3}{2}(1.38 \times 10^{-23})(400) = 8.3 \times 10^{-21}\ \text{J}.

Exam-style practice questions

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

Eduqas 20194 marksCalculate the mean kinetic energy of a molecule of an ideal gas at 27 ∘C27\ ^\circ\text{C}, and hence the root-mean-square speed of a nitrogen molecule of mass 4.7×10−26 kg4.7 \times 10^{-26}\ \text{kg}. Take the Boltzmann constant k=1.38×10−23 J K−1k = 1.38 \times 10^{-23}\ \text{J K}^{-1}.
Show worked answer →

Convert the temperature to kelvin: T=27+273=300 KT = 27 + 273 = 300\ \text{K}.

Mean kinetic energy per molecule: Ek‾=32kT=32(1.38×10−23)(300)=6.21×10−21 J\overline{E_k} = \tfrac{3}{2}kT = \tfrac{3}{2}(1.38 \times 10^{-23})(300) = 6.21 \times 10^{-21}\ \text{J}.

Root-mean-square speed from Ek‾=12mc2‾\overline{E_k} = \tfrac{1}{2}m\overline{c^2}: c2‾=2Ek‾m=2×6.21×10−214.7×10−26=2.64×105\overline{c^2} = \dfrac{2\overline{E_k}}{m} = \dfrac{2 \times 6.21 \times 10^{-21}}{4.7 \times 10^{-26}} = 2.64 \times 10^{5}, so crms=2.64×105=514 m s−1c_\text{rms} = \sqrt{2.64 \times 10^{5}} = 514\ \text{m s}^{-1}.

Markers reward T=300 KT = 300\ \text{K}, Ek‾=32kT≈6.2×10−21 J\overline{E_k} = \frac{3}{2}kT \approx 6.2 \times 10^{-21}\ \text{J}, and the rms speed about 510 m s−1510\ \text{m s}^{-1}.

Eduqas 20214 marksState three assumptions of the kinetic model of an ideal gas, and explain in terms of molecular motion why a gas exerts a pressure on the walls of its container.
Show worked answer →

Three assumptions (any three): the molecules are in continuous random motion; the molecules are point particles whose total volume is negligible compared with the container; collisions with the walls and each other are perfectly elastic; there are no intermolecular forces except during collisions; the duration of a collision is negligible compared with the time between collisions.

Pressure arises because molecules continually collide with the walls. Each collision reverses the molecule's momentum component perpendicular to the wall, so by Newton's second law the wall exerts a force on the molecule and (third law) the molecule exerts an equal force on the wall. The many collisions per second average out to a steady force over the wall area, which is the pressure.

Markers reward three valid assumptions, and explaining pressure as the rate of change of momentum from many elastic collisions producing a force per unit area on the wall.

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