Use absolute gas pressure and consistent units
Add a supplied pressure reference and convert scales before forming a product.
Section P2.1.3 · Gases and the absolute scale of temperature
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Recall p₁V₁ = p₂V₂ for fixed mass and temperature.
- Convert pressure prefixes and cubic volume units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many cubic centimetres are in 0.20 L?
What you will learn
- Distinguish gauge and absolute pressure in a supplied gas calculation.
- Use consistent pressure and volume units across two states.
A gauge reading has a reference
A pressure gauge can report pressure relative to surrounding air. If a positive gauge reading is 50 kPa and the supplied surrounding absolute pressure is 100 kPa, the gas absolute pressure is 150 kPa. The pressure in the constant-product account must use that absolute reference. A zero gauge reading can still mean 100 kPa absolute; it does not indicate absence of particles or zero gas pressure.
For a signed gauge reading below ambient, add it algebraically to the supplied ambient absolute pressure. For example −20 kPa gauge with 100 kPa ambient gives 80 kPa absolute. Do not subtract ambient twice or use an assumed atmospheric value when the problem supplies another reference. A proposed non-positive absolute pressure does not fit the positive-pressure gas states used here.
A ratio can use any common scale
Both pressures may be in kPa and both volumes in cm³ when used consistently in p₁V₁ = p₂V₂. The product then has units kPa cm³; it is not silently a product in Pa m³. If one state is in Pa and the other in kPa, convert a common pressure scale first. Likewise, 1 L = 1000 cm³ and 1 cm³ = 10⁻⁶ m³.
With 100 kPa absolute and 120 cm³, the product is 12 000 kPa cm³. In SI units it is 100 000 Pa × 0.000120 m³ = 12 Pa m³. The different numerical products express the same account in different units. An equation using a consistent ratio can avoid unnecessary conversions, but its unit labels must remain truthful.
Sources for this lesson
Pause and explain
A gas gauge reads zero with ambient pressure 100 kPa absolute. What is the gas absolute pressure?
Worked example
A gas has 50 kPa gauge pressure with 100 kPa absolute ambient pressure and volume 80 cm³. At constant mass and temperature it expands to 120 cm³. Find final absolute and gauge pressure with the same ambient reference.
Show the worked solution
- Initial absolute pressure is 50 + 100 = 150 kPa.
- Final absolute pressure is 150 × 80/120 = 100 kPa.
- Final gauge pressure is 100 − 100 = 0 kPa, not zero absolute pressure.
Answer 100 kPa absolute; 0 kPa gauge.
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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.
Use 50 kPa gauge, 100 kPa absolute ambient and 80 cm³ to find pressure after constant-temperature expansion to 120 cm³; ambient stays the same.
Give me a hint
Convert the initial pressure to absolute before applying the product.
Compare my reasoning
- Initial absolute pressure = 150 kPa.
- Final absolute pressure = 150 × 80/120 = 100 kPa.
- Subtract ambient to obtain final gauge pressure 0 kPa.
100 kPa absolute; 0 kPa gauge.
Look for these in your work
- Use the reference correctly.
- Distinguish both final pressure scales.
A fixed gas mass at 120 000 Pa absolute occupies 0.20 L. At the same temperature its volume is 100 cm³. Find final pressure in kPa and the initial product in Pa m³.
Give me a hint
Convert to common scales, then form the requested SI product separately.
Compare my reasoning
- 120 000 Pa = 120 kPa and 0.20 L = 200 cm³.
- Final pressure = 120 × 200/100 = 240 kPa.
- Initial volume = 0.00020 m³, giving product 120 000 × 0.00020 = 24 Pa m³.
240 kPa absolute; initial product 24 Pa m³.
Look for these in your work
- Convert both types of units.
- Label the SI product correctly.
A gas at 100 kPa absolute and 120 cm³ is said to have product 12 000 Pa m³. Correct the number and explain how the original number can still be used with honest units.
Give me a hint
Track the scale of both factors before labelling their product.
Compare my reasoning
- 100 kPa × 120 cm³ = 12 000 kPa cm³.
- In SI, 100 000 Pa × 0.000120 m³ = 12 Pa m³.
- The 12 000 number is usable with its actual kPa cm³ unit, not with a falsely assigned SI unit.
12 Pa m³, equivalent to 12 000 kPa cm³.
Look for these in your work
- Correct the unit scaling.
- Preserve equivalent representations.
Common mistakes
- Gauge zero does not mean absolute zero pressure.
- Add the supplied ambient reference once.
- Label a non-SI product with its actual units.
What can you explain now?
What is −20 kPa gauge with supplied ambient 100 kPa absolute?
Compare with the explanation
80 kPa absolute.
Add the signed gauge value to the reference: 100 + (−20) = 80.
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Make it stick
Use supplied pressure references, handle signed gauge readings and keep product units consistent.
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.