Describe how pressure varies with force, area and depth, and how it is related to the density of a gas, using everyday examples, simple calculations and experimental techniques required by the Cambridge IGCSE 0625 syllabus.
Definition & symbol
Pressure (\(\mathbf{p}\)) is the normal force applied per unit area of a surface.
\[
p = \frac{F}{A}
\]
where \(p\) = pressure (Pa), \(F\) = normal force (N) and \(A\) = contact area (m²).
Units and symbols
SI unit: pascal (Pa) = N · m⁻²
Common multiples: 1 kPa = 10³ Pa, 1 MPa = 10⁶ Pa
Other useful units: bar (1 bar = 100 kPa), atmosphere (1 atm ≈ 101.3 kPa), mm Hg (used for blood‑pressure readings)
1. Relationship between force, area and pressure
For a given area, pressure is directly proportional to force: double the force → double the pressure.
For a given force, pressure is inversely proportional to area: double the area → halve the pressure.
Consequences:
To obtain a required pressure a larger force can be applied over a larger area.
To obtain the same force with less effort the contact area must be reduced (e.g. a knife edge).
Worked numeric example (forward calculation)
A 200 N weight is placed on a flat board of area 0.020 m². The pressure is
\[
p = \frac{200\;\text{N}}{0.020\;\text{m}^2}=1.0\times10^{4}\;\text{Pa}=10\;\text{kPa}
\]
Reverse‑calculation examples (re‑arranging the formula)
Required force for a given pressure and area:
\[
F = pA
\]
If a gardener wants a pressure of 5 kPa under a foot‑plate of area 0.015 m², the needed force is
\(F = 5\times10^{3}\,\text{Pa}\times0.015\,\text{m}^2 = 75\;\text{N}\).
Required area for a given force and pressure:
\[
A = \frac{F}{p}
\]
A 300 N load must not exceed 2 kPa on a delicate glass surface. The minimum safe area is
\(A = 300\;\text{N}/(2\times10^{3}\,\text{Pa}) = 0.15\;\text{m}^2\).
2. Pressure variation with depth in a fluid (hydrostatic pressure)
In a fluid at rest the pressure increases with depth because of the weight of the fluid above.
\[
\Delta p = \rho g h
\]
\(\rho\) = density of the fluid (kg m⁻³)
g = acceleration due to gravity (≈ 9.8 m s⁻²)
h = vertical depth below the free surface (m)
Total pressure at depth: \(\displaystyle p_{\text{total}} = p_{\text{atm}} + \rho g h\)
For a fixed temperature, the pressure of a gas is directly proportional to its density (or mass per unit volume):
\[
p \propto \rho \qquad\text{or}\qquad p \propto \frac{m}{V}
\]
Quick calculation – Compare two containers of equal volume at the same temperature. Container A contains air of mass 1.2 kg (ρ ≈ 1.2 kg m⁻³); container B contains carbon‑dioxide of mass 1.8 kg (ρ ≈ 1.8 kg m⁻³). The pressure in B is \(\frac{1.8}{1.2}=1.5\) times the pressure in A.
4. Practical skills – measuring pressure with a U‑tube manometer
Equipment – U‑tube, coloured water or oil (known density), ruler, stand.
Step‑by‑step checklist
Secure the U‑tube vertically on the stand.
Zero the instrument: ensure both arms are at the same level when both are open to atmospheric pressure; record this as 0 m.
Connect the arm that will sense the unknown pressure to the system (e.g., a gas container). Keep the other arm open to the atmosphere.
Allow the liquid to settle; read the height difference \(h\) between the two arms to the nearest 0.01 m.
Calculate the pressure difference using \(\Delta p = \rho_{\text{liquid}} g h\). Add atmospheric pressure if an absolute pressure value is required.
Record the result, the liquid used, temperature (optional) and any observations.
5. Safety considerations
Never point the open end of a pressurised container at yourself or others.
Wear appropriate personal protective equipment (PPE): safety goggles, lab coat and gloves.
When working with hydraulic or pneumatic systems, release pressure slowly via the supplied bleed valve.
Check for leaks before applying force to a hydraulic piston or before connecting a manometer.
Dispose of fluids (especially coloured oils) according to the school’s waste‑disposal policy.
6. Everyday examples (mapped to syllabus sub‑points)
Example
Force (N)
Contact area (m²)
Resulting pressure (Pa)
What the example illustrates
High‑heeled shoe on sand
≈ 500
≈ 5 × 10⁻⁴
≈ 1.0 × 10⁶ Pa (1 MPa)
Force + very small area → high pressure → sinking (force & area)
Snowshoe on snow
≈ 500
≈ 0.25
≈ 2.0 × 10³ Pa (2 kPa)
Large area reduces pressure → prevents sinking (force & area)
Sharp kitchen knife cutting bread
≈ 20
≈ 1 × 10⁻⁴
≈ 2.0 × 10⁵ Pa (0.2 MPa)
Small edge area gives enough pressure to break material (force & area)
Hydraulic car jack – small piston
100
0.001
1.0 × 10⁵ Pa (100 kPa)
Pascal’s principle – same pressure acts on larger piston (hydraulic press)
Dam resisting 5 m water depth
—
—
≈ 5.0 × 10⁴ Pa (50 kPa) additional to atmospheric
Hydrostatic pressure \(\Delta p = \rho g h\) (depth‑pressure)
Human blood pressure (clinical measurement)
—
—
Systolic ≈ 120 mm Hg ≈ 1.6 × 10⁴ Pa (16 kPa)
Use of mm Hg unit, conversion to pascals; real‑world health context
In fluids at rest, total pressure at depth is \(p_{\text{total}} = p_{\text{atm}} + \rho g h\).
At constant temperature, gas pressure is directly proportional to density (\(p \propto \rho\)).
Pascal’s principle: pressure applied to a confined incompressible fluid is transmitted unchanged in all directions.
Pressure can be measured with a U‑tube manometer using the relation \(\Delta p = \rho g h\); follow the procedural checklist and observe safety rules.
Suggested diagrams (to be drawn by the teacher or included in the textbook): (a) high‑heeled shoe sinking into sand, (b) snowshoe distributing weight, (c) U‑tube manometer with height‑difference reading, (d) hydraulic press showing two pistons of different areas.