Use the constant pressure-volume product
Calculate one changed gas state while keeping mass and temperature fixed.
Section 2.1.3 · Gases and the absolute scale of temperature
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028, section 2.1.3. Use pV = constant for a fixed gas mass at constant temperature, including the graphical relationship. Use absolute pressure and consistent units; a gauge reading alone is not the required absolute pressure. Core collision and temperature lessons should be studied first. Written practice is self-checked. Human subject-teacher review, supervised practical experience and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Explain constant-temperature compression qualitatively.
- Rearrange a product equation and keep units consistent.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which rearrangement finds p₂ from p₁V₁ = p₂V₂?
What you will learn
- Recall pV = constant and use p₁V₁ = p₂V₂.
- Explain the inverse relationship and check a calculated state against it.
The product is fixed under stated conditions
For a fixed mass of gas at constant temperature in this model, pV = constant. Two states therefore satisfy p₁V₁ = p₂V₂. Pressure is inversely proportional to volume: doubling volume halves pressure, while halving volume doubles pressure. This is a product relationship, not a constant pressure/volume ratio.
Use absolute gas pressure, measured relative to vacuum, and positive gas volumes. A gauge reading relative to surrounding air needs its reference added before it can serve as absolute pressure. The gas remains in the stated model and temperature is controlled; the equation is not an unrestricted rule for every compression, leakage or state change.
Rearrange and check the direction
A fixed sample at 100 kPa absolute pressure occupies 120 cm³. At constant temperature its volume becomes 40 cm³, so p₂ = 100 × 120/40 = 300 kPa. The product is 12 000 kPa cm³ in both states. The smaller volume must correspond to larger pressure under these controls, giving a useful direction check.
For a missing volume use V₂ = p₁V₁/p₂. If pressure rises by a factor of 1.5, volume falls to 1/1.5 of its original value, not by an equal percentage. State each input scale and the fixed conditions rather than memorising a swap of numbers with no physical interpretation.
Pause and explain
Pressure rises from 100 to 150 kPa absolute at fixed mass and temperature. What happens to a 120 cm³ volume?
Worked example
A fixed gas mass has p₁ = 100 kPa absolute and V₁ = 120 cm³. At constant temperature V₂ = 40 cm³. Find p₂.
Show the worked solution
- The fixed-mass, constant-temperature controls allow p₁V₁ = p₂V₂.
- p₂ = 100 × 120/40 = 300 kPa.
- Both states have product 12 000 kPa cm³, and pressure rises as volume falls.
Answer 300 kPa absolute.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find final absolute pressure when 100 kPa and 120 cm³ change to 40 cm³ at fixed mass and temperature.
Give me a hint
Divide the initial product by the final volume.
Compare my reasoning
- Initial product is 100 × 120 = 12 000 kPa cm³.
- p₂ = 12 000/40.
- Final pressure is 300 kPa, consistent with compression.
300 kPa absolute.
Look for these in your work
- Use the correct product.
- Check the pressure direction.
A fixed gas sample at 150 kPa absolute occupies 80 cm³. At the same temperature pressure becomes 100 kPa, then 200 kPa. Find each volume.
Give me a hint
The initial product is common to both later states.
Compare my reasoning
- The product is 150 × 80 = 12 000 kPa cm³.
- At 100 kPa, volume is 12 000/100 = 120 cm³.
- At 200 kPa, volume is 12 000/200 = 60 cm³.
120 cm³; 60 cm³.
Look for these in your work
- Keep the same product.
- Use absolute pressures.
Pressure increases by 50% under the same controls from a 120 cm³ state. A student subtracts 50% from volume. Correct the volume and explain.
Give me a hint
Use the reciprocal of the pressure factor, not an equal percentage subtraction.
Compare my reasoning
- A 50% rise multiplies pressure by 1.5.
- Volume must be divided by 1.5.
- 120/1.5 = 80 cm³, a decrease of one third rather than one half.
80 cm³; inverse factors do not mean equal opposite percentages.
Look for these in your work
- Use the multiplicative factor.
- Explain the difference from the shortcut.
Common mistakes
- Use a product rather than p/V.
- Retain fixed mass and temperature.
- Inverse factors do not give equal opposite percentage changes.
What can you explain now?
What happens to pressure when volume triples under the stated controls?
Compare with the explanation
It becomes one third of its original value.
The unchanged product requires pressure to scale by the reciprocal factor.
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Make it stick
Use and rearrange the constant product, check inverse factors and retain the required gas conditions.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.