A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Brownian Motion and Kinetic Theory of Gases — mark scheme explained
The short answer
Brownian motion is a phenomenon observed in the random movement of particles suspended in a fluid (a liquid or gas). This motion provides strong evidence for the existence of atoms and molecules, as it can be explained by the continuous bombardment of these particles by the much smaller and faster-moving molecules of the fluid.
The question
A container holds 1.0 × 10 23 molecules of an ideal gas with a mass of 4.0 × 10 -26 kg each. The root mean square speed of the molecules is 500 m/s. Calculate the pressure exerted by the gas if the volume of the container is 0.02 m 3 . [Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
What each mark is really for — in plain English — and the wording trap that loses it.
- S1
Use the equation pV = ⅓Nm(c rms ) 2 to find the pressure.
- S2
Substitute the given values: N = 1.0 × 10 23 , m = 4.0 × 10 -26 kg , c rms = 500 m/s , and V = 0.02 m 3 .
- S3
Calculate the right-hand side: ⅓Nm(c rms ) 2 = (1/3) × 1.0 × 10 23 × 4.0 × 10 -26 kg × (500 m/s) 2 .
- S4
Simplify: (1/3) × 1.0 × 10 23 × 4.0 × 10 -26 × 250000 = 3.33 × 10 2 Pa·m 3 .
- S5
Divide by the volume to find the pressure: p = (3.33 × 10 2 Pa·m 3 ) / 0.02 m 3 = 16650 Pa .
Model answer
Worked through, with each step tagged to the mark it earns.
- S1
Use the equation pV = ⅓Nm(c rms ) 2 to find the pressure.
- S2
Substitute the given values: N = 1.0 × 10 23 , m = 4.0 × 10 -26 kg , c rms = 500 m/s , and V = 0.02 m 3 .
- S3
Calculate the right-hand side: ⅓Nm(c rms ) 2 = (1/3) × 1.0 × 10 23 × 4.0 × 10 -26 kg × (500 m/s) 2 .
- S4
Simplify: (1/3) × 1.0 × 10 23 × 4.0 × 10 -26 × 250000 = 3.33 × 10 2 Pa·m 3 .
- S5
Divide by the volume to find the pressure: p = (3.33 × 10 2 Pa·m 3 ) / 0.02 m 3 = 16650 Pa .
Final answer: 16650 Pa
Common mistakes
- Confusing the root mean square speed with the average speed of gas molecules. — Remember that the root mean square speed ( c rms ) is given by √(3kT/m) , where k is the Boltzmann constant, T is the temperature, and m is the mass of a molecule.
- Forgetting to use the correct units when substituting values into equations. — Always check the units before substituting values into equations and ensure they are consistent. Convert units as necessary (e.g., 1 kPa = 1000 Pa, 1 cm 3 = 1 × 10 -6 m 3 ).
- Misinterpreting the assumptions of the kinetic theory of gases. — Review and memorize the key assumptions: molecules are in constant random motion, volume occupied by molecules is negligible, collisions are perfectly elastic, no intermolecular forces except during collisions, and average kinetic energy is proportional to temperature.
- Using the wrong formula for average molecular kinetic energy. — Remember that the average molecular kinetic energy is given by E k = (1/2) m c rms 2 = (3/2) kT = (3/2) RT / N A . Use the appropriate form based on the information given in the problem.
- Forgetting to use Avogadro's number when converting between molar and molecular quantities. — When dealing with molar quantities (e.g., using R instead of k ), remember to use Avogadro's number ( N A ) to convert between the two. For example, E k = (3/2) RT / N A .
- Misapplying the ideal gas law in problems involving kinetic theory. — Understand when to use each equation. The ideal gas law is empirical and relates pressure, volume, and temperature. The kinetic theory equation provides a theoretical basis for these relationships in terms of molecular motion.
Where the marks go
- Full worked solution (all marking points)4 marks