P5.34SD0 · Section P5: Solids, liquids and gases

Density and pressure

Key learnings

Density = mass per unit volume, ρ = m/V, in kg/m³
Density is the mass per unit volume of a material: $\rho = m/V$ , SI unit kg/m $^3$ (the symbol $\rho$ is "rho") — not "how heavy" something is. Rearrange as $m = \rho V$ and $V = m/\rho$ . To measure it, find the mass on a balance, then the volume: calculate it for a regular solid, use displacement in a measuring cylinder for an irregular solid, or weigh a known volume (subtracting the empty container) for a liquid. Water is about $1000$ kg/m $^3$ , i.e. $1$ g/cm $^3$ .
Pressure = normal force per unit area, p = F/A, in pascals
Pressure is the force acting normally (at right angles) per unit area: $p = F/A$ , SI unit pascal (Pa $=$ N/m $^2$ ). A large force over a small area gives a high pressure (a drawing pin), the same force over a large area a low pressure (a snowshoe). Rearrange as $F = pA$ and $A = F/p$ . In a fluid (liquid or gas) the pressure acts equally in all directions at a given point.
Pressure with depth: p = hρg, larger deeper down
The pressure difference across a vertical height $h$ in a fluid is $p = h\rho g$ , where $\rho$ is the fluid density and $g = 10$ N/kg (the 4PH1/4SD0 data-sheet value). Pressure increases with depth because a deeper point has a taller column of fluid above it, whose weight presses over each unit area — so a submarine or dam wall must withstand more pressure lower down. Always put a value of $g$ into the calculation, and convert Pa to kPa if asked ( $1$ kPa $= 1000$ Pa).

Common mistakes

Drawn from real examiner reports.

Definitions need "per unit"

Definition items score zero for everyday language. Density is mass per unit volume ( $\rho = m/V$ , unit kg/m $^3$ ) — not "how heavy" something is. Pressure is the force acting normally per unit area ( $p = F/A$ , unit pascal) — not just "force on an area", which drops the "per unit". A pascal is $1$ N/m $^2$ ; quoting a pressure in newtons loses the unit mark.

Convert to SI before ρ = m/V

Density calculations are marked in SI units. Convert mass grams $\rightarrow$ kg and volume cm $^3$ $\rightarrow$ m $^3$ before using $\rho = m/V$ (shortcut: $1$ g/cm $^3 = 1000$ kg/m $^3$ ). Weaker candidates who skip this lose a mark even with correct arithmetic. Sanity-check the size: a metal coming out at $2.7$ kg/m $^3$ instead of $2700$ signals a forgotten conversion.

Jun 2024 1P Q10a

In p = hρg, always substitute g

In $p = h\rho g$ you must substitute a value for $g$ ( $10$ N/kg) — leaving it out scores zero on the whole calculation. Then watch the final unit: convert Pa $\rightarrow$ kPa when asked ( $1$ kPa $= 1000$ Pa). The pressure here is the difference between two points separated by a vertical height $h$ ; deeper means a taller fluid column above, so a larger pressure.

Nov 2024 1P Q7b

In p = F/A the force is the weight

In $p = F/A$ the force is the object's weight, not its mass. Convert first with $W = mg$ (using $g = 10$ N/kg), then divide by the contact area. Substituting the mass in kilograms straight into $p = F/A$ is the common error — it gives the wrong size and the wrong unit. Where the question asks, describe measuring the weight (with a force meter) and the area.

Jun 2023 1P Q5b

Pressure change → resultant force

When a change in pressure acts on an object — e.g. the outside pressure rising on a sealed bottle as it descends — explain it as a resultant force on the surface (inward when the outside pressure exceeds the inside). Writing only that the bottle is "crushed" or "collapses" repeats the stem and misses the mark; name the resultant force that causes it.

Jun 2024 1P Q9b

Gas pressure: collisions → force → pressure

A gas exerts pressure because its particles are in constant random motion and collide with the container walls. Stating that particles "collide with the walls" is not enough on its own — complete the chain: each collision exerts a force, and the total force spread over the wall area is the pressure ( $\text{collisions} \rightarrow \text{force} \rightarrow \text{pressure}$ ).

Nov 2024 1P Q7c(i)

Identify a material from its density

To identify a material from density measurements, calculate the density of every sample, then give a definite conclusion that refers to the data — e.g. "sample B matches aluminium because its density is $2700$ kg/m $^3$ ". Candidates lose marks by computing only one density, or by not stating a clear data-based conclusion and assuming the marker will infer it.

Jun 2024 1PR Q5b

Exam tips

Calculation scaffold
Write the equation ( $\rho = m/V$ , $p = F/A$ or $p = h\rho g$ ), convert to SI first (g $\rightarrow$ kg, cm $^3$ $\rightarrow$ m $^3$ ), substitute the numbers — with $g = 10$ N/kg for depth-pressure — then convert to the unit asked (Pa $\rightarrow$ kPa if needed) and state it.
Pick the right pressure formula
Use $p = F/A$ for a solid pushing on a surface — force spread over the contact area. Use $p = h\rho g$ for the pressure difference with depth in a fluid. They answer different questions, so do not mix them: a block uses $F/A$ , a diver's depth uses $h\rho g$ .
Density practical scores the method
Marks target how the measurements are taken. Find the mass on a balance; for an irregular solid find the volume by displacement in a measuring cylinder, fully submerged, reading the meniscus bottom to avoid parallax. Reciting only $\rho = m/V$ earns little.

Revision notes