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
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For calculation questions, marks are typically awarded for correct substitution of values, correct use of units, and the final answer. For explanation questions, marks are given for clarity, accuracy, and the inclusion of relevant scientific principles.
What the command words demand
- Calculate
- Perform the necessary mathematical operations to find a numerical answer.
- Explain
- Provide a clear and concise explanation, using appropriate scientific terminology.
- Derive
- Show the step-by-step process of deriving an equation from given principles or assumptions.
- Compare
- Identify similarities and differences between two or more concepts or phenomena.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate approximately 5-7 minutes per question to ensure you have enough time to carefully read the question, perform calculations, and check your work.
- Use the equation pV = ⅓Nm(c rms ) 2 to find the pressure.1 mark
- 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 .1 mark
- Calculate the right-hand side: ⅓Nm(c rms ) 2 = (1/3) × 1.0 × 10 23 × 4.0 × 10 -26 kg × (500 m/s) 2 .0 marks
- Simplify: (1/3) × 1.0 × 10 23 × 4.0 × 10 -26 × 250000 = 3.33 × 10 2 Pa·m 3 .1 mark
- Divide by the volume to find the pressure: p = (3.33 × 10 2 Pa·m 3 ) / 0.02 m 3 = 16650 Pa .1 mark
Final answer: 16650 Pa
Work through every step correctly and you earn all 4 marks.
Another worked example
A gas has a root mean square speed of 600 m/s and an average molecular mass of 3.2 × 10 -26 kg. Calculate the temperature of the gas in Kelvin.
- Use the equation for average kinetic energy: E k = (1/2) m c rms 2 and set it equal to (3/2) kT .1 mark
- Substitute the given values: m = 3.2 × 10 -26 kg , c rms = 600 m/s , and k = 1.38 × 10 -23 J/K .1 mark
- Calculate the kinetic energy: (1/2) × 3.2 × 10 -26 kg × (600 m/s) 2 .0 marks
- Simplify: (1/2) × 3.2 × 10 -26 × 360000 = 5.76 × 10 -21 J .1 mark
- Set this equal to (3/2) kT : 5.76 × 10 -21 J = (3/2) × 1.38 × 10 -23 J/K × T .0 marks
- Solve for T : T = (5.76 × 10 -21 J) / ((3/2) × 1.38 × 10 -23 J/K) = 5.76 × 10 -21 / 2.07 × 10 -23 ≈ 278 K .1 mark
Final answer: 278 K
Work through every step correctly and you earn all 4 marks.
Common mistakes
Confusing the root mean square speed with the average speed of gas molecules.
Why it happens: Students often mix up these two concepts, thinking they are the same. The root mean square speed is a specific measure that takes into account the distribution of molecular speeds.
Fix: 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.
Why it happens: Students sometimes forget to convert units, leading to incorrect calculations. For example, using pressure in kPa instead of Pa or volume in cm 3 instead of m 3 .
Fix: 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.
Why it happens: Students may not fully understand the significance of each assumption, leading to confusion in applying the theory.
Fix: 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.
Why it happens: Students might use E k = (1/2) m v 2 instead of E k = (3/2) kT or (3/2) RT / N A .
Fix: 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.
Why it happens: Students often overlook the need to convert between molar and molecular quantities, leading to incorrect calculations.
Fix: 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.
Why it happens: Students sometimes use the ideal gas law ( pV = nRT ) instead of the kinetic theory equation ( pV = ⅓Nm(c rms ) 2 ), leading to incorrect results.
Fix: 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
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Kinetic Theory Pressure | Use the kinetic theory equation to calculate the pressure of an ideal gas. | 4 |
| Calculate Gas Temperature | Use the kinetic theory equation to find a gas temperature from rms speed and molecular mass. | 4 |
| Calculate RMS Speed | Use the kinetic theory equation to find the root mean square speed of gas molecules. | 4 |
| Calculate RMS Speed | Find the root mean square speed of gas molecules from temperature and molecular mass. | 3 |
| Ideal Gas Calculation | Use the ideal gas equation with molecule number to find the gas temperature in Kelvin. | 5 |
| Total across these question types | 20 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.