Density and Pressure

Edexcel 4PH1 awards marks for the correct unit on every numerical answer. For Topic 5:

• Temperature in °C or K (kelvin); kelvin scale used for ideal-gas calculations (spec 5.17).
• Energy in joules (J).
• Mass in kg.
• Density in kg/m³ (NOT g/cm³ for SI answers — but 1 g/cm³ = 1000 kg/m³).
• Length in m; area in m²; volume in m³.
• Speed in m/s; acceleration in m/s².
• Force in N; pressure in pascals (Pa = N/m²).

Paper 2 only (spec 5.2P). Specific heat capacity has unit J/(kg °C) — joules per kilogram per degree Celsius. Used in spec 5.13P calculations.

Definition. The MASS of a substance contained in unit VOLUME.

$\rho = \dfrac{m}{V}$

• $\rho$ in kg/m³ (or g/cm³)
• $m$ in kg (or g)
• $V$ in m³ (or cm³)

Key values to know.

• Water: 1000 kg/m³ (= 1.0 g/cm³). Define by historical convention.
• Air at room temperature: ≈ 1.2 kg/m³ (much less than water — that's why air rises in water).
• Ice: 920 kg/m³ — less than water, so ice floats.
• Aluminium: 2700 kg/m³.
• Steel: 7800 kg/m³.
• Lead: 11 300 kg/m³.

Why density matters. Floats and sinks: an object less dense than the fluid it's in will FLOAT; one denser will SINK. Hot air rises in cooler air (lower density when hot, spec 4.6 — convection).

Volume of regular solids.

• Cuboid: $V = lwh$.
• Cylinder: $V = \pi r^2 h$.
• Sphere: $V = \tfrac{4}{3}\pi r^3$.

Volume of liquid. Read from a measuring cylinder — at eye level, at the BOTTOM of the meniscus.

Volume of irregular solid. Water displacement: $V = V_{2} - V_{1}$ (spec 5.4 required practical).

Spec 5.4 lists THREE situations for measuring density. Memorise all three methods.

(1) Regular solid (e.g. a metal cuboid).

1. Use a balance to measure mass $m$.
2. Use a ruler (or vernier callipers for accuracy) to measure dimensions.
3. Calculate volume from the appropriate formula ($lwh$ for cuboid, $\pi r^2 h$ for cylinder).
4. $\rho = m/V$.

(2) Irregular solid (e.g. a stone).

1. Use a balance to measure mass $m$.
2. Half-fill a measuring cylinder with water. Read initial volume $V_1$ at eye level.
3. Gently submerge the object using a thread (no splashing!).
4. Read final volume $V_2$.
5. Volume of object = $V_2 - V_1$.
6. $\rho = m/(V_2 - V_1)$.

(3) Liquid.

1. Find the mass $m_1$ of an EMPTY measuring cylinder.
2. Pour in a known volume of liquid $V$.
3. Find the new mass $m_2$.
4. Mass of liquid = $m_2 - m_1$.
5. $\rho = (m_2 - m_1) / V$.

Sources of error.

• Parallax when reading the meniscus → eye level matters.
• Air bubbles clinging to an irregular solid → over-estimate volume → under-estimate density.
• For a regular solid, ruler resolution (0.5 mm) can be significant on small objects.

Improvement. Repeat each measurement at least 3 times and take the mean. For higher accuracy, use vernier callipers for solid dimensions and an electronic balance reading to 0.01 g.

Definition. Pressure = force applied PERPENDICULAR to a surface, divided by the area of that surface.

$p = \dfrac{F}{A}$

• $p$ in pascals (Pa = N/m²)
• $F$ in newtons (N) — force component normal to the surface
• $A$ in square metres (m²)

Intuitive picture. The same force spread over a LARGER area gives LOWER pressure.

• Walking on snow with snowshoes: large area → low pressure → don't sink in.
• High-heel shoes: tiny area → enormous pressure (~MPa) → can dent floors.
• Drawing pin: blunt end vs sharp end concentrates force on a tiny area at the tip.
• Tractor tyres: large area → low pressure → don't compact soil.
• Knives: sharp edge has small contact area → high pressure → cuts easily.

1 Pa = 1 N/m². Small unit — atmospheric pressure is ~10⁵ Pa, and tyre pressure ~200 kPa.

Pressure also applies to fluids. A gas in a sealed container pushes on the walls. The force per unit area on each wall is the gas pressure.

Spec 5.6 statement. The pressure at a point in a gas or liquid which is at rest acts EQUALLY in all directions.

What this means. Imagine a tiny spherical balloon submerged in water. Water molecules hit every part of the balloon's surface. At the same DEPTH, the force per unit area (pressure) is the SAME no matter which direction you choose to measure it.

Practical examples.

• A diving bell with windows on all sides: water pressure pushes inwards on every window equally (at the same depth).
• A balloon underwater: squeezed uniformly from all sides.
• A submerged object experiences a NET UPWARD force (buoyancy = upthrust) because pressure increases with depth — the bottom is at higher pressure than the top — but at any one depth the pressure is direction-independent.

Why dams are thicker at the base. Pressure increases with depth (spec 5.7), so the bottom of the wall holds back more pressure. Pressure pushes HORIZONTALLY on the wall at depth — so the wall must be strong horizontally, not just vertically.

Formula. The pressure DIFFERENCE between a point at depth $h$ and the surface of a static fluid of density $\rho$:

$p = h \rho g$

• $p$ in pascals (Pa)
• $h$ = VERTICAL depth (m)
• $\rho$ = density of the fluid (kg/m³)
• $g$ = 9.8 N/kg (often 10 in 4PH1 questions)

Derivation (not required but illustrative). A column of fluid of cross-sectional area $A$, height $h$, has volume $V = A h$, mass $m = \rho V = \rho A h$, weight $W = mg = \rho A h g$. The force is spread over area $A$, so pressure = $W/A = h \rho g$.

Worked examples.

• 10 m of water: $p = 10 \times 1000 \times 10 = 10^5$ Pa. So every 10 m of depth adds ~1 atmosphere of pressure. A diver at 30 m experiences ~3 atmospheres of water pressure (+ 1 atmosphere of air → total 4 atmospheres).
• The Mariana Trench is ~11 km deep — pressure ~1100 atmospheres (over 100 MPa).

Key features of $p = h\rho g$.

1. Pressure depends ONLY on depth $h$, density $\rho$ and $g$ — NOT on the shape of the container.
2. Two different-shaped vessels filled with water to the same height have the same pressure at the bottom (the 'hydrostatic paradox').
3. Pressure due to atmosphere (~$10^{5}$ Pa) adds on top if you want TOTAL pressure.

Application: U-tube manometer. A U-shaped tube with liquid in it. Connect one side to a gas → the height difference $h$ between the two columns gives the gas pressure relative to atmospheric: $p_{\text{gas}} - p_{\text{atm}} = h\rho g$.