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Molecular kinetic theory modelAQA A-Level Physics: Revision notes

Section 1

Brownian motion and evidence for molecules

Brownian motion is the random, zigzag motion of small visible particles (such as smoke or pollen) suspended in a fluid. Smoke in air viewed under a microscope shows bright specks changing direction at random.

The explanation is that the fluid is made of tiny, invisible molecules moving rapidly and randomly, which collide with the visible particle from all sides. At any instant the collisions are unbalanced, giving a resultant force in a random direction. It is therefore evidence for the existence of atoms and molecules and for their random motion.

Heating the fluid makes the molecules faster, so the motion becomes more vigorous; larger particles move less because the collisions average out more.

Key termsBrownian motion
Common mistake

Saying the particles move because they are 'attracted' or 'pushed by currents'. The cause is random molecular bombardment.

Section 2

The kinetic theory assumptions and gas behaviour

The kinetic theory model of an ideal gas assumes:

  • a very large number of identical molecules moving randomly
  • molecules have negligible volume compared with the container
  • collisions are perfectly elastic, and last a negligible time
  • no intermolecular forces except in collisions
  • Newton's laws apply

Explaining the gas laws: pressure is the average force per unit area from molecular collisions. Halving the volume at constant temperature doubles the collision rate with the walls, so pp doubles (Boyle). Raising the temperature increases the mean speed, so collisions are harder and more frequent, and pp rises at constant volume (pressure law).

The gas laws are empirical (found from experiments), whereas the kinetic theory is a theoretical model from which the gas laws are derived.

Key termsideal gas modelelastic collisionempirical
Exam tip

In an answer explaining pressure, say harder AND more frequent collisions, with the molecular speed or number density named.

Section 3

Derivation of pV=13Nmcrms2pV = \frac{1}{3}Nmc_{rms}^2

Take a cube of side LL with NN molecules each of mass mm. For one molecule with velocity component cxc_x:

  1. Momentum change at one wall: 2mcx2mc_x.
  2. Time between collisions with the same wall: 2L/cx2L/c_x.
  3. Force on the wall =2mcx/(2L/cx)=mcx2/L= 2mc_x/(2L/c_x) = mc_x^2/L.

For NN molecules the total force is Nmcx2‾L\frac{Nm\overline{c_x^2}}{L}, and the pressure is force divided by L2L^2: p=Nmcx2‾L3=Nmcx2‾Vp = \frac{Nm\overline{c_x^2}}{L^3} = \frac{Nm\overline{c_x^2}}{V}.

Motion is random, so c2‾=cx2‾+cy2‾+cz2‾=3cx2‾\overline{c^2} = \overline{c_x^2} + \overline{c_y^2} + \overline{c_z^2} = 3\overline{c_x^2}, so cx2‾=13c2‾\overline{c_x^2} = \frac{1}{3}\overline{c^2} and

pV=13Nmc2‾=13Nmcrms2pV = \frac{1}{3}Nm\overline{c^2} = \frac{1}{3}Nmc_{rms}^2

Worked example: N=2.0×1022N = 2.0 \times 10^{22}, m=4.65×10−26m = 4.65 \times 10^{-26} kg, V=1.0×10−3V = 1.0 \times 10^{-3} m³, p=1.00×105p = 1.00 \times 10^5 Pa gives crms=5.7×102c_{rms} = 5.7 \times 10^2 m s⁻¹.

Key termsroot mean square speed
Common mistake

Writing crmsc_{rms} as the average speed. It is the square root of the mean of the squared speeds, which is slightly larger.

Section 4

Temperature and molecular kinetic energy

Combining pV=13Nmcrms2pV = \frac{1}{3}Nmc_{rms}^2 with pV=NkTpV = NkT gives

12mcrms2=32kT=3RT2NA\frac{1}{2}mc_{rms}^2 = \frac{3}{2}kT = \frac{3RT}{2N_A}

So the mean kinetic energy of a molecule is proportional to the kelvin temperature and does not depend on the gas.

In an ideal gas there are no intermolecular forces, so there is no potential energy, and the internal energy is the total kinetic energy of the molecules: U=32NkT=32nRTU = \frac{3}{2}NkT = \frac{3}{2}nRT.

Worked example: helium at 300 K: mean kinetic energy =6.21×10−21= 6.21 \times 10^{-21} J, crms=1.37×103c_{rms} = 1.37 \times 10^3 m s⁻¹, and UU for 1.00 mol =3.74×103= 3.74 \times 10^3 J.

Key termsmean kinetic energyinternal energy of an ideal gas
Exam tip

Lighter gas molecules have the same mean kinetic energy at a given temperature, so they move faster.

Section 5

How understanding of gases has changed

In the 17th to 19th centuries scientists such as Boyle and Charles found the gas laws by experiment. Kinetic theory, developed in the 19th century, offered a theoretical model that explained them from the motion of molecules.

Brown observed pollen grains moving randomly in 1827, which had no explanation. In 1905 Einstein showed that it was caused by molecular bombardment, and Perrin confirmed this experimentally, giving convincing evidence for atoms and molecules, which some scientists had still doubted.

The ideal gas model has limits: real gases deviate at high pressure and low temperature, when molecular volume and forces matter. A theory is accepted when its predictions are confirmed by experiment and checked by other scientists, and refined when it fails.

Key termstheoretical model

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Molecular kinetic theory model

  1. A student places smoke particles in a small glass cell filled with air and observes them through a microscope, lit from the side. Each smoke particle is seen as a tiny bright speck that moves along an irregular, zigzag path.
    Explain how this observation provides evidence for the existence of atoms and molecules.2 marks
  2. A fixed mass of air is sealed inside a syringe. The temperature of the air can be kept constant, and the syringe can also be heated while its plunger is held fixed in position.
    The experiments on the syringe lead to the gas laws. Explain the difference between the gas laws and the kinetic theory model of a gas.2 marks
  3. An ideal gas is modelled as NN identical molecules, each of mass mm, moving randomly in a cubic container of side LL and volume VV. Collisions with the walls are perfectly elastic. In part (b), the container holds 2.0 × 10²² molecules of nitrogen, each of mass 4.65 × 10⁻²⁶ kg, in a volume of 1.0 × 10⁻³ m³ at a pressure of 1.00 × 10⁵ Pa.
    A molecule of velocity component cxc_x in the xx direction collides with the wall perpendicular to xx and rebounds along the same line. Show that the average force exerted on this wall by this molecule is mcx2/Lmc_x^2/L.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).