Calculate liquid-pressure changes from vertical depth
Use density, gravitational field strength and a signed vertical depth change with consistent units.
Extended only (Supplement); Physics 0625, 2026–2028, section 1.8.4. Recall and use Δp = ρgΔh for a stationary liquid with stated density and gravitational field. A pressure change is not automatically total pressure; use vertical depth and a consistent surface or point reference. Written work is self-checked.
Before you begin
- Explain liquid depth and density effects qualitatively.
- Convert lengths to metres and density to kg/m³.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is 75 cm in metres?
What you will learn
- Recall and use Δp = ρgΔh.
- Distinguish the liquid pressure increase from total pressure at a point.
The equation describes a pressure difference
For a stationary liquid with uniform density ρ and constant g, the pressure change between two vertical levels is Δp = ρgΔh. Take downward depth change as positive: going deeper gives a positive pressure change; going upward gives a negative change. Density in kg/m³, g in N/kg and depth in m give pressure in Pa. With ρ = 1000 kg/m³, g = 10 N/kg and downward Δh = 0.60 m, Δp = 6000 Pa.
From the liquid surface to a point at depth h, the liquid contribution is ρgh. If the supplied surface pressure is 101 000 Pa, the point's total pressure is 101 000 + 6000 = 107 000 Pa. The equation for the change does not require guessing a standard atmospheric value, and an unchanged surface reference cancels when comparing two depths.
Convert and choose the vertical change
For density 800 kg/m³, g = 9.8 N/kg and depth increase 75 cm, convert 75 cm to 0.75 m before calculating 800 × 9.8 × 0.75 = 5880 Pa. If a density is supplied as 0.80 g/cm³, it equals 800 kg/m³, because both mass and volume units must be converted. The denominator's cubic conversion is not just a centimetre-to-metre length factor.
Moving from 1.0 to 3.0 m depth in the same uniform water model with ρ = 1000 and g = 10 gives Δh = +2.0 m and Δp = +20 000 Pa. Reversing that journey gives −20 000 Pa. Neither route length nor the final depth alone replaces the difference between the two stated levels. A changing density or flowing liquid needs a suitably revised account rather than automatic use of one fixed ρ.
Pause and explain
Does Δp = 6000 Pa from the surface establish total pressure 6000 Pa when surface pressure is 101 000 Pa?
Worked example
Water has ρ = 1000 kg/m³ and g = 10 N/kg. Find the increase at 0.60 m below its surface, then total pressure if the surface is at 101 000 Pa.
Show the worked solution
- Δp = 1000 × 10 × 0.60 = 6000 Pa.
- The 6000 Pa is the liquid contribution above surface pressure.
- Total = 101 000 + 6000 = 107 000 Pa.
Answer Increase 6000 Pa; total 107 000 Pa.
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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.
Calculate the pressure increment at 0.60 m in water of density 1000 kg/m³ with g = 10 N/kg, then total pressure from surface pressure 101 000 Pa.
Give me a hint
Calculate the increment first and keep the supplied reference separate.
Compare my reasoning
- Increment = 1000 × 10 × 0.60 = 6000 Pa.
- Surface pressure = 101 000 Pa.
- Total = 107 000 Pa.
6000 Pa increase; total 107 000 Pa.
Look for these in your work
- Use the vertical depth.
- Include the supplied surface contribution.
Find the change for a downward 75 cm move in density 0.80 g/cm³ with g = 9.8 N/kg; then find the change from 1.0 to 3.0 m in density 1000 kg/m³ with g = 10 N/kg.
Give me a hint
Convert the first density and length, then use the second level difference.
Compare my reasoning
- 0.80 g/cm³ = 800 kg/m³ and 75 cm = 0.75 m.
- First change = 800 × 9.8 × 0.75 = 5880 Pa.
- Second change = 1000 × 10 × 2.0 = 20 000 Pa.
+5880 Pa; +20 000 Pa.
Look for these in your work
- Convert both quantities correctly.
- Use a difference between stated depths.
A swimmer rises from 3.0 to 1.0 m depth in stationary uniform water with ρ = 1000 kg/m³ and g = 10 N/kg. A calculation uses a 5 m swimming route. Correct the signed pressure change.
Give me a hint
Use final minus initial vertical depth, with downward positive.
Compare my reasoning
- Δh = 1.0 − 3.0 = −2.0 m.
- Δp = 1000 × 10 × (−2.0).
- Pressure decreases by 20 000 Pa; route length is irrelevant here.
−20 000 Pa; a decrease of 20 kPa.
Look for these in your work
- Retain the sign and vertical reference.
- Reject the sloping route length.
Common mistakes
- Convert density and depth units.
- Use vertical depth change, not route length.
- Do not call the liquid increment total pressure without its reference.
What can you explain now?
What is the pressure change on rising 2.0 m in the same 1000 kg/m³ liquid with g = 10 N/kg?
Compare with the explanation
−20 000 Pa.
The signed downward depth change is −2.0 m, so pressure decreases.
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Make it stick
Recall the pressure-change equation, convert depth and density, and distinguish a signed change from total pressure.
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.