Compare liquid-pressure changes across stated layers
Apply the uniform-liquid equation separately to supplied non-mixing layers and keep every reference explicit.
Section P1.8 · Pressure
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
- Use Δp = ρgΔh for one uniform liquid.
- Identify vertical layer thicknesses and non-mixing density order.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For a non-mixing layer model, why not use one layer's density over every depth?
What you will learn
- Add pressure increments for stated uniform-density layer segments.
- Compare complete depth/density accounts without inventing a shared boundary.
Separate segments when density changes
In an original application of the syllabus equation, a stationary upper layer has density 800 kg/m³ and thickness 0.50 m, with a denser lower layer of 1000 kg/m³ and thickness 0.30 m. For g = 10 N/kg, the upper increment is 4000 Pa and the lower increment is 3000 Pa, giving 7000 Pa from the top surface to the bottom. The layers are non-mixing, each density is uniform, and interface curvature effects are negligible in this stated model.
Pressure continues from the interface into the next layer, so add the increments rather than restarting each segment at zero total pressure. With a supplied top-surface pressure of 100 kPa, bottom total pressure is 107 kPa. Applying 1000 kg/m³ across all 0.80 m would give 8000 Pa and incorrectly replace the upper density. Layering is an application example, not an additional separately numbered syllabus requirement.
Choose just the segments between the stated points
For a move from 0.20 m below the top surface to 0.70 m in the same layered system, the route crosses 0.30 m of the upper liquid and 0.20 m of the lower liquid. The increase is 800 × 10 × 0.30 + 1000 × 10 × 0.20 = 4400 Pa. Using the full surface-to-bottom thickness would describe different endpoints. Reversing the interval gives a −4400 Pa change.
A uniform 1000 kg/m³ liquid to depth 0.80 m has an 8000 Pa liquid contribution for the same g, larger than this layered 7000 Pa contribution. That total-pressure comparison needs equal surface pressure. If the layered surface were 102 kPa and the uniform surface 100 kPa, their bottom totals would instead be 109 and 108 kPa. A smaller liquid increment can coexist with a larger total because the boundary differs.
Sources for this lesson
Pause and explain
Does a smaller liquid increment always imply smaller total pressure if surface pressures differ?
Worked example
For upper 0.50 m at 800 kg/m³ and lower 0.30 m at 1000 kg/m³, find the liquid pressure increment with g = 10 N/kg, then total from 100 kPa at the top.
Show the worked solution
- Upper increment = 800 × 10 × 0.50 = 4000 Pa.
- Lower increment = 1000 × 10 × 0.30 = 3000 Pa.
- Combined increment = 7000 Pa; total = 107 kPa.
Answer 7000 Pa increase; total 107 kPa.
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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 the bottom increment and total pressure for upper 0.50 m at 800 kg/m³, lower 0.30 m at 1000 kg/m³, g = 10 N/kg and top-surface pressure 100 kPa.
Give me a hint
Calculate each uniform segment, then add the common top reference once.
Compare my reasoning
- Upper gives 4000 Pa.
- Lower gives 3000 Pa; combined increment is 7000 Pa.
- Total is 100 000 + 7000 = 107 000 Pa.
7000 Pa increase; total 107 kPa.
Look for these in your work
- Retain each layer density.
- Add the surface pressure once.
In the same layer model, calculate the signed pressure change from depth 0.20 m to 0.70 m, then for the reverse move.
Give me a hint
Split the interval at the 0.50 m interface rather than using the whole tank.
Compare my reasoning
- Upper crossed thickness is 0.30 m and lower is 0.20 m.
- Downward increase = 2400 + 2000 = 4400 Pa.
- Reversing the endpoints gives −4400 Pa.
+4400 Pa; −4400 Pa.
Look for these in your work
- Use the stated endpoints.
- Preserve the reverse sign.
Compare the layered bottom with top pressure 102 kPa and liquid increment 7000 Pa against uniform liquid with surface pressure 100 kPa and increment 8000 Pa. Which total is greater?
Give me a hint
Add each boundary contribution before ranking.
Compare my reasoning
- Layered total is 102 + 7 = 109 kPa.
- Uniform total is 100 + 8 = 108 kPa.
- The layered total is greater despite its smaller liquid increment.
Layered 109 kPa versus uniform 108 kPa; layered total is greater.
Look for these in your work
- Convert both increments to kPa.
- Compare complete totals, not increments alone.
Common mistakes
- Use each segment's stated density.
- Add increments without resetting total at the interface.
- Total-pressure rankings also need the surface reference.
What can you explain now?
In that system, what pressure change occurs from depth 0.20 m to 0.70 m?
Compare with the explanation
+4400 Pa.
The interval includes 0.30 m upper liquid and 0.20 m lower liquid.
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Make it stick
Calculate a stated layer application by segments, reverse a point-to-point change and compare totals with different surface references.
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.