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.
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).
A 200 N weight is placed on a flat board of area 0.020 m². The pressure is
\[
p = \frac{200\;\text{N}}{0.020\;\text{m}^2}=1.0\times10^{4}\;\text{Pa}=10\;\text{kPa}
\]
\[
F = pA
\]
If a gardener wants a pressure of 5 kPa under a foot‑plate of area 0.015 m², the needed force is
\(F = 5\times10^{3}\,\text{Pa}\times0.015\,\text{m}^2 = 75\;\text{N}\).
\[
A = \frac{F}{p}
\]
A 300 N load must not exceed 2 kPa on a delicate glass surface. The minimum safe area is
\(A = 300\;\text{N}/(2\times10^{3}\,\text{Pa}) = 0.15\;\text{m}^2\).
In a fluid at rest the pressure increases with depth because of the weight of the fluid above.
\[
\Delta p = \rho g h
\]
\[
\Delta p = (1000\;\text{kg m}^{-3})(9.8\;\text{m s}^{-2})(5\;\text{m}) = 4.9\times10^{4}\;\text{Pa}=49\;\text{kPa}
\]
\[
p_{\text{total}} = 101\;\text{kPa} + 49\;\text{kPa} \approx 150\;\text{kPa}
\]
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.
Equipment – U‑tube, coloured water or oil (known density), ruler, stand.
| 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 a closed, incompressible‑fluid system the pressure is transmitted equally in all directions.
Numerical illustration – \(F{1}=100\;\text{N}\), \(A{1}=0.001\;\text{m}^2\), \(A_{2}=0.05\;\text{m}^2\):
\(p = \dfrac{100}{0.001}=1.0\times10^{5}\;\text{Pa}\)
\(F{2}=pA{2}=1.0\times10^{5}\times0.05=5.0\times10^{3}\;\text{N}\)
Assumptions: the fluid is incompressible and there are no losses due to friction.
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