Edexcel IGCSE Physics 5(b) Density and Pressure — Notes
Written for Pearson Edexcel International GCSE Physics (4PH1), Science Double Award (4SD0) and Science Single Award (4SS0), linear and modular. This page also covers the 5(a) units for Topic 5.
Which parts do I need?
Single Award candidates need only pressure \(p=F/A\), the idea that pressure in a fluid acts equally in all directions, and the Topic 5 units. They skip all of density (the equation, the unit kg/m³ and the density practical), the pressure difference equation \(p=h\times\rho\times g\) and the unit J/kg °C.
Double Award candidates skip only the unit J/kg °C, which belongs with specific heat capacity in 5(c).
What this page covers
| Spec | Modular no. | You need to be able to… | Route |
|---|---|---|---|
| 5.1 | 4.1 | use the Topic 5 units: °C, K, J, kg, kg/m³, m, m², m³, m/s, m/s², N and Pa (Single Award: not kg/m³) | Single · Double · Physics |
| 5.2P | 4.2P | use the unit of specific heat capacity, J/kg °C | Physics only · 2P |
| 5.3 | 4.3 | recall and use density = mass ÷ volume | Double · Physics |
| 5.4 | 4.4 | find the density of solids and liquids by measuring mass and volume (practical) | Double · Physics |
| 5.5 | 4.5 | recall and use pressure = force ÷ area | Single · Double · Physics |
| 5.6 | 4.6 | understand that at any point in a liquid or gas at rest, the pressure acts equally in every direction | Single · Double · Physics |
| 5.7 | 4.7 | recall and use pressure difference = height × density × g | Double · Physics |
Modular candidates: density and pressure are in 4XPH1 Unit 1 (4WPH1) and 4XSD1 Unit 5 (4WSD5), numbered 4.1–4.7 under “Solids, liquids and gases: Part 1”. The units in 4.1 are assessed again in 4XPH1 Unit 2 (4WPH2) and 4XSD1 Unit 6 (4WSD6), and 4.2P again in 4XPH1 Unit 2, with the ideal gas work (5(d)).
In one minute
- Density = mass ÷ volume, \(\rho=\dfrac{m}{V}\), in kg/m³ (or g/cm³). 1 g/cm³ = 1000 kg/m³. Double · Physics
- Find a density by measuring mass on a balance and volume with a ruler or by displacement in a measuring cylinder. Double · Physics
- Pressure = force ÷ area, \(p=\dfrac{F}{A}\), in pascals: 1 Pa = 1 N/m². A smaller area gives a larger pressure for the same force.
- At any point in a liquid or gas at rest, the pressure acts equally in all directions, and the force on any surface acts at right angles to it.
- The pressure difference between two depths is \(p=h\times\rho\times g\): it grows with vertical depth and with the density of the liquid. Double · Physics
Which units do I need for Topic 5? Single · Double · Physics
These are the units used across the whole of Topic 5 — density and pressure here, change of state in 5(c) and gases in 5(d). Learn each symbol and what it measures.
| Quantity | Unit | Symbol | Route |
|---|---|---|---|
| temperature | degree Celsius | °C | All |
| temperature (absolute scale) | kelvin | K | All |
| energy | joule | J | All |
| mass | kilogram | kg | All |
| length, height, depth | metre | m | All |
| area | metre² | m² | All |
| volume | metre³ | m³ | All |
| speed (of molecules) | metre/second | m/s | All |
| acceleration | metre/second² | m/s² | All |
| force | newton | N | All |
| pressure | pascal | Pa | All |
| density | kilogram/metre³ | kg/m³ | Double · Physics |
| specific heat capacity | joule/kilogram degree Celsius | J/kg °C | Physics only · 2P |
Single Award: your units list for Topic 5 is the same except that it does not include kg/m³, because density is not in Single Award. J/kg °C is Physics only and is used in 5(c).
- Areas and volumes: 1 m² = 100 × 100 = 10 000 cm², and 1 m³ = 100 × 100 × 100 = 1 000 000 cm³. Square or cube the length conversion — do not just multiply by 100.
- Pressure: 1 Pa = 1 N/m². Large pressures are often given in kPa (× 1000).
- Temperature: to convert from °C to K, add 273 (this is explained in 5(d)).
What is density and how do you calculate it? Double · Physics
Not required for Single Award (4SS0).
- \(\rho\) (the Greek letter rho) = density in kg/m³, \(m\) = mass in kg, \(V\) = volume in m³.
- In the laboratory, g/cm³ is often more convenient. 1 g/cm³ = 1000 kg/m³, so water (1.0 g/cm³) has a density of 1000 kg/m³.
- Density depends on the material, not on the size of the sample: cutting a block in half halves both its mass and its volume, so the density is unchanged.
- Rearranged: \(m=\rho\times V\) and \(V=\dfrac{m}{\rho}\).
A rectangular metal block measures 4.0 cm × 5.0 cm × 10 cm. Its mass is 1.56 kg. Calculate the density of the metal in kg/m³.
Convert the lengths to metres first: \(V=0.040\times0.050\times0.10=2.0\times10^{-4}\ \text{m}^3\)
\(\rho=\dfrac{m}{V}=\dfrac{1.56}{2.0\times10^{-4}}=\) 7800 kg/m³
Alternative: V = 200 cm³ and m = 1560 g, so ρ = 7.8 g/cm³ = 7800 kg/m³.
Practical: how do you measure density? Double · Physics
You need two direct measurements — the mass (from a balance) and the volume — then you calculate \(\rho=m/V\). The method for the volume depends on the sample.
Investigating the density of solids and liquids
- Aim: determine the density of a regular solid, an irregular solid and a liquid from measurements of mass and volume.
- Apparatus: top-pan balance, ruler (or vernier calipers), measuring cylinder, water, thread, the samples.
- Measured quantities: mass and volume of each sample. There is no independent variable to change; for a fair result keep the sample dry and the balance zeroed, and read every volume at eye level.
Method
- Regular solid (e.g. a cuboid): measure its mass on the balance. Measure length, width and height with a ruler at several places and take the means. Volume = length × width × height.
- Irregular solid (e.g. a stone): measure its mass. Part-fill a measuring cylinder with water and read the volume \(V_1\). Lower the stone on a thread until it is fully under the water and read \(V_2\). Volume of stone = \(V_2-V_1\).
- If the solid is too big for a measuring cylinder, use a displacement (eureka) can filled to the spout: lower the object in and collect the water that overflows in a measuring cylinder.
- Liquid: measure the mass of an empty measuring cylinder, pour in a measured volume of the liquid and measure the mass again. Mass of liquid = difference.
- Calculate \(\rho=m/V\) for each. Repeat each measurement and use the mean.
Errors and improvements
- Read the bottom of the meniscus with your eye level with the water surface to avoid a parallax error.
- Use the narrowest measuring cylinder the sample fits into: its scale has smaller divisions, so the volume reading has a smaller uncertainty.
- Air bubbles stuck to the stone make the measured volume too large (and the density too small) — tap the stone to free them.
- Lowering the stone on a thread avoids splashing water out, which would make the volume too small. Dry the stone before measuring its mass, or measure the mass first.
- Safety: lower heavy objects gently so the glass cylinder does not crack; wipe up spills.
A student uses the method in the diagram. The stone has a mass of 81.0 g. The water level rises from 42 cm³ to 72 cm³. Calculate the density of the stone in g/cm³ and in kg/m³.
Volume of stone = 72 − 42 = 30 cm³
\(\rho=\dfrac{m}{V}=\dfrac{81.0}{30}=\) 2.7 g/cm³ = 2.7 × 1000 = 2700 kg/m³
Common error: dividing by 72 cm³ (the final reading) instead of the change in volume.
What is pressure and how do you calculate it? Single · Double · Physics
- \(p\) = pressure in pascals (Pa), \(F\) = force in N, \(A\) = area in m². 1 Pa = 1 N/m².
- For the same force, a smaller area gives a larger pressure: a sharp knife, a drawing pin and a nail all have tiny contact areas so that they cut or push in easily.
- A larger area gives a smaller pressure: snowshoes, wide tractor tyres and a wide strap on a heavy bag spread the force.
- The force on a floor from an object resting on it is its weight.
(a) A crate of weight 1200 N rests on a floor. Its base measures 0.80 m × 0.50 m. Calculate the pressure on the floor.
(b) A person of weight 600 N stands with all her weight on the heel of one shoe. The heel has an area of 1.5 cm². Calculate the pressure under the heel, and explain why this shoe can damage a wooden floor while a flat shoe does not.
(a) \(A=0.80\times0.50=0.40\ \text{m}^2\); \(p=\dfrac{F}{A}=\dfrac{1200}{0.40}=\) 3000 Pa
(b) 1.5 cm² = 1.5 ÷ 10 000 = 1.5 × 10⁻⁴ m²
\(p=\dfrac{600}{1.5\times10^{-4}}=\) 4.0 × 10⁶ Pa
The force is the same (her weight), but the area is very small, so the pressure is very large — large enough to dent the wood. Standing on two flat soles of 150 cm² each gives 600 ÷ 0.030 = 20 000 Pa, two hundred times smaller.
How does pressure act in a liquid or gas at rest? Single · Double · Physics
Liquids and gases are fluids: their particles can move past each other. In a fluid that is not flowing, the particles at any point push on anything there in every direction.
- A small object placed at X is pushed equally from above, below and the sides — so a diver feels the water pressing on every side of the body, not only from above.
- Because the pressure acts in all directions, the force on a surface is always perpendicular to the surface, whichever way the surface faces: on the side walls, the force is horizontal; on the base, it is vertical.
- At the same depth in the same fluid, the pressure is the same everywhere — on the side walls and at X.
- The same idea applies to the air: atmospheric pressure acts on every surface of your body in every direction, which is why you are not pushed over by it.
How does pressure change with depth? Double · Physics
Not required for Single Award (4SS0).
The deeper you go in a liquid, the greater the weight of liquid above you, so the greater the pressure.
\[ \text{pressure difference}=\text{height}\times\text{density}\times\text{gravitational field strength}\qquad p=h\times\rho\times g \]- \(p\) = pressure difference in Pa, \(h\) = vertical height difference in m, \(\rho\) = density of the fluid in kg/m³, \(g\) = gravitational field strength in N/kg.
- Pearson papers say “assume \(g\) = 10 N/kg”; use 10 N/kg unless told otherwise (9.8 N/kg is also accepted).
- The pressure difference depends only on the vertical depth and the density — not on the shape or width of the container.
- To find the pressure at a depth below an open surface, add the pressure difference to the atmospheric pressure acting on the surface (about 1.0 × 10⁵ Pa).
A diver swims 12 m below the surface of the sea. The density of seawater is 1030 kg/m³. Calculate the pressure difference between the surface and the diver. Assume g = 10 N/kg.
\(p=h\times\rho\times g=12\times1030\times10=123\,600\ \text{Pa}=\) 1.2 × 10⁵ Pa
So the total pressure on the diver at 12 m is about 2.2 × 10⁵ Pa — more than double atmospheric pressure: 1.0 × 10⁵ Pa from the air plus 1.2 × 10⁵ Pa from the water.
(a) A small submarine has a window of area 0.030 m². The submarine is at a depth of 45 m in seawater of density 1030 kg/m³. The air inside is at atmospheric pressure. Calculate the extra force on the window due to the water.
(b) In a tank of oil, the pressure difference between two points 0.40 m apart vertically is 3600 Pa. Calculate the density of the oil.
(a) The pressure difference across the window is due to 45 m of water:
\(p=h\rho g=45\times1030\times10=463\,500\ \text{Pa}\)
\(F=p\times A=463\,500\times0.030=13\,905\ \text{N}=\) 1.4 × 10⁴ N
(b) \(\rho=\dfrac{p}{h\times g}=\dfrac{3600}{0.40\times10}=\) 900 kg/m³
Part (a) links two equations: find the pressure from the depth, then the force from \(F=p\times A\). Atmospheric pressure acts on both sides of the window, so it cancels.
Exam tips for 5(b)
1 cm² = 1 × 10⁻⁴ m² and 1 cm³ = 1 × 10⁻⁶ m³. Dividing an area in cm² by 100 instead of 10 000 is the most common power-of-ten error in pressure questions.
In “describe how to determine the density” questions, most marks are for the measurements: name the instrument for each (balance, ruler, measuring cylinder), say what you read (\(V_1\), \(V_2\)) and how the volume is found (\(V_2-V_1\)). Add one practical detail — reading at eye level, zeroing the balance, repeating and taking a mean.
In \(p=h\rho g\), \(h\) is measured vertically from the surface (or between the two points). In a sloping pipe or a container of odd shape, ignore horizontal distances completely.
When explaining fluid pressure, use the precise phrases: the pressure at a point acts equally in all directions; the force on a surface acts at right angles to it. Vague answers such as “the water pushes everywhere” do not score.
If the question asks for a density in kg/m³ and you worked in g/cm³, multiply by 1000 before you write the final answer. A correct number in the wrong unit loses the final mark.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A bigger piece of iron has a bigger density. | Mass and volume increase in the same proportion. | Density is a property of the material, not of the sample size. |
| Heavy objects always exert a large pressure. | Pressure depends on area as well as force. | A heavy object on a large area can give a small pressure, \(p=F/A\). |
| Water pressure only pushes downwards. | At a point in a fluid, pressure acts in all directions. | The force on any surface acts at right angles to it — sideways on walls, upwards on the underside of an object. |
| A wider tank has a greater pressure at the bottom. | More water spreads over a larger area. | The pressure difference depends only on vertical depth and density: \(p=h\rho g\). |
| 1 m² = 100 cm². | Both sides of the square are converted. | 1 m² = 100 × 100 = 10 000 cm². |
Key equations
| Equation | Use it to find | Route | On the Equation List? |
|---|---|---|---|
| density = mass ÷ volume, \(\rho=\dfrac{m}{V}\) | density, mass or volume | Double · Physics | yes — all papers |
| pressure = force ÷ area, \(p=\dfrac{F}{A}\) | pressure on a surface, or the force from a pressure | All | yes — all papers |
| pressure difference = height × density × g, \(p=h\times\rho\times g\) | pressure difference between two depths in a fluid | Double · Physics | yes — all papers |
Pearson includes the Equation List with the question papers, but the first mark in many questions is for writing the relationship yourself — learn them anyway.
Check your understanding
1. Explain why a drawing pin has a sharp point and a wide, flat head. All
The same force acts at both ends. The point has a tiny area, so the pressure is large and it goes into the board. The head has a large area, so the pressure on your thumb is small.
2. A force of 450 N acts on an area of 0.015 m². Calculate the pressure. All
\(p=\dfrac{450}{0.015}=\) 30 000 Pa (3.0 × 10⁴ Pa). (1 mark substitution, 1 mark answer with unit.)
3. A diver says, “At the bottom of the pool the water only pushes down on me.” Explain why she is wrong. All
At any point in a liquid at rest the pressure acts equally in all directions, so the water pushes on every part of her body — from the sides and from below as well as from above — with the force at right angles to each surface.
4. 250 cm³ of a liquid has a mass of 200 g. Calculate its density in kg/m³. Double · Physics
\(\rho=\dfrac{200}{250}=0.80\ \text{g/cm}^3=\) 800 kg/m³
5. At what depth in fresh water (density 1000 kg/m³) is the pressure difference from the surface 50 kPa? Use g = 10 N/kg. Double · Physics
\(h=\dfrac{p}{\rho g}=\dfrac{50\,000}{1000\times10}=\) 5.0 m
6. A tube of mercury (density 13 600 kg/m³) holds a column 0.76 m high. (a) Calculate the pressure difference the column produces. (b) A 13.5 g piece of copper has a volume of 1.5 cm³. Calculate the density of copper in g/cm³. Double · Physics
(a) \(p=0.76\times13\,600\times10=103\,360\ \text{Pa}\approx\) 1.0 × 10⁵ Pa. (b) \(\rho=13.5\div1.5=\) 9.0 g/cm³.
7. Which quantity is measured in J/kg °C? Physics only · 2P
Specific heat capacity — the energy needed to raise the temperature of 1 kg of a material by 1 °C (see 5(c)).
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