AQA GCSE Physics Physics Paper 1 (Higher), November 2021: Question 9

9 marks · Standard Demand difficulty · Short Answer

Analyze an experiment investigating the relationship between gas pressure and volume, including range, control variables, a Boyle's law calculation, and particle model explanation.

Practise this question

Question

Question 9 shows Figure 12, a diagram of apparatus used to demonstrate the relationship between gas pressure and volume. A syringe with a plunger has weights hanging from the bottom, and its tip is connected via tubing to a dial pressure gauge. Four question parts follow: 09.1 asks for the range of force used from 0/2 N to 12.0 N; 09.2 asks for one control variable; 09.3 asks candidates to calculate the gas pressure at 40 cm³ given that the pressure was 60 kPa at 45 cm³; 09.4 asks candidates to explain why pressure on the inside walls decreased as volume increased.
Question text

09 A teacher demonstrated the relationship between the pressure in a gas and the

volume of the gas.

Figure 12 shows the equipment used.

Figure 12

This is the method used.

1. Record the initial volume of gas in the syringe and the pressure reading before

any weights are attached.

2. Attach a 2.0 N weight to the syringe.

3. Record the volume of the gas and the reading on the pressure gauge.

4. Repeat steps 2 and 3 until a weight of 12.0 N is attached to the syringe.

09.1 What was the range of force used?

[1 mark]

From N to N

09.2 Give one control variable in the investigation.

[1 mark]

09.3 When the volume of gas in the syringe was 45 cm3, the pressure gauge showed a

value of 60 kPa.

*32* Calculate the pressure in the gas when the volume of gas in the syringe was 40 cm3.

[4 marks]

Pressure = kPa

09.4 When the volume of gas in the syringe increased, the pressure on the inside walls of

the syringe decreased.

Explain why.

[3 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 9: 09.1 awards 1 mark for 0(.0) to 12(.0) (allow 2(.0) to 12(.0)). 09.2 awards 1 mark for mass of gas (in the syringe) or temperature (of the gas). 09.3 awards 4 marks: constant = 60 × 45 or 2700 (1 mark), 2700 = p × 40 (1 mark), p = 2700 / 40 (1 mark), p = 67.5 (kPa) (1 mark, allow 68). 09.4 awards 3 marks: more time between collisions or less frequent collisions with walls (1 mark), causing a lower average force on walls (1 mark), and pressure is total force per unit area (1 mark).

Question 9

AO /

Question Answers Extra information Mark

Spec. Ref.

09.1 0(.0) to 12(.0) allow 2(.0) to 12(.0) (N) 1 AO1

4.3.3.2

09.2 mass of gas (in the syringe) 1 AO3

or 4.3.3.2

temperature (of the gas)

09.3 constant = 60 × 45 1 AO2

4.3.3.2

or

constant = 2700

2700 = p × 40 1

2700 1

p =

p = 67.5 (kPa)

allow 68 (kPa) 1

09.4 there is more time between 1 AO1

collisions of particles and the 4.3.3.2

walls of the syringe

or

there are less frequent collisions

between the particles and the

walls of the syringe

(causing) a lower (average)

force on the walls of the syringe

(and) pressure is the total force

per unit area 1

Total 9

How to answer it

Gas Pressure, Volume & Boyle's Law

📋 What this question tests

This question assesses knowledge from Topic 4.3 (Particle Model of Matter - Gas Pressure and Temperature):

  • Identifying independent variables, range of values, and control variables in a practical investigation.
  • Applying Boyle’s Law: p × V = constant (or p₁V₁ = p₂V₂ ) for a fixed mass of gas at constant temperature.
  • Particle-level explanation of how gas volume affects pressure (collision rate, force, and surface area).
Question 09.1 [1 Mark]

Determining the Range of Force

Identifying the minimum and maximum force values used

✅ Correct Answer

From 0 N to 12 N (or 0.0 N to 12.0 N)

Also allowed: 2 N to 12 N (or 2.0 to 12.0 N)

🧠 Exam Technique: Finding the Range

  • "Range" means stating both the lowest and highest values: from minimum to maximum.
  • Step 1 states: "before any weights are attached" → that represents 0 N.
  • Step 4 states: "until a weight of 12.0 N is attached" → that represents 12.0 N.

❌ Common Errors

  • Writing just the difference (e.g. stating "12 N" instead of the range "0 to 12 N").
  • Writing "2.0 N to 10.0 N" by misreading step 4.
Question 09.2 [1 Mark]

Identifying a Control Variable

Maintaining fair testing conditions for gas laws

✅ Correct Answer

Either of the following:

  • Temperature (of the gas / room)
  • Mass of gas (or number/amount of gas particles / gas trapped in the syringe)

💡 Key Knowledge

Boyle's Law only holds true under two strict conditions:

  • Fixed mass of gas (no leaks in the tubing or syringe seal).
  • Constant temperature (weights must be added slowly so gas does not change temperature).

❌ Common Errors

  • Naming "volume" or "pressure" — these are the independent and dependent variables.
  • Saying "the syringe" or "equipment" without stating the specific physical property (e.g. temperature or mass).
Question 09.3 [4 Marks]

Boyle's Law Calculation

Calculating pressure after a change in volume

📐 Step-by-Step Calculation

  1. Recall the relationship:
    p × V = constant  or  p₁ × V₁ = p₂ × V₂
  2. Calculate the constant using initial conditions (Mark 1):
    Given: V₁ = 45 cm³, p₁ = 60 kPa
    constant = 60 × 45 = 2700
  3. Substitute new volume (V₂ = 40 cm³) into the equation (Mark 2):
    2700 = p × 40
  4. Rearrange to solve for p (Mark 3):
    p = 2700 / 40
  5. Calculate final value (Mark 4):
    p = 67.5 kPa (or rounded to 68 kPa )

🧠 Exam Technique: Unit Checks

  • The question already provides units in kPa and asks for the answer in kPa.
  • Volumes are both given in cm³. Because the units match on both sides, there is no need to convert cm³ to m³ or kPa to Pa!
  • Sanity Check: Volume decreased (45 → 40 cm³), so pressure must increase (60 → 67.5 kPa).

❌ Common Errors

  • Direct proportion error: calculating (60 / 45) × 40 = 53.3 kPa. Gas volume and pressure are inversely proportional!
  • Rounding 67.5 incorrectly or making simple arithmetic errors when dividing 2700 by 40.
Question 09.4 [3 Marks]

Explaining Gas Pressure Using Particle Theory

Why pressure decreases when volume increases

✅ Model Answer (3 Marks Breakdown)

  • Mark 1: There are less frequent collisions between the gas particles and the inside walls of the syringe (or more time between collisions).
  • Mark 2: This causes a lower total / average force exerted on the inside walls of the syringe.
  • Mark 3: Since pressure = force / area, the smaller force (over a larger area) results in decreased pressure.

💡 Key Knowledge: The Chain of Logic

Whenever an exam asks you to explain pressure changes using particles, always use this 3-step chain:

  1. Collisions: How often do particles hit the walls? (Frequency / rate of collisions).
  2. Force: Each collision exerts a tiny force; fewer collisions mean less overall force.
  3. Definition: Link directly to Pressure = Force / Area .

❌ Common Errors & Examiner Pitfalls

  • Saying "fewer collisions" without stating per second or less frequent collisions.
  • Claiming that the particles "slow down" or have "less kinetic energy" — temperature is constant, so particle speed remains the same!
  • Describing collisions between particles rather than collisions between particles and the walls of the syringe.

Topics

Physics · P3: Particle Model of Matter

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Higher), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.