# Pressure

> CIE IGCSE Physics · 2026-2028 CIE IGCSE Physics 0625
> Source: https://www.owlsprep.com/study/cie-0625-u1-pressure/

This guide covers all Core and Extended pressure content for CIE IGCSE Physics 0625 Unit 1, including formula application, liquid pressure rules, atmospheric pressure, and exam-standard worked examples.

**Prerequisites:** [Knowledge of SI units for force, mass and length](https://www.owlsprep.com/study/cie-0625-u1-si-units/); [Understanding of gravitational field strength $g = 9.8$ N/kg for IGCSE calculations](https://www.owlsprep.com/study/cie-0625-u1-mass-weight/)

## Learning objectives

- Define pressure and recall the $p = F/A$ formula for pressure in solids
- Perform pressure calculations using correct SI units and unit conversions
- Explain qualitative properties of pressure in liquids and atmospheric pressure
- (Extended Only) Apply $p = h\rho g$ and Boyle's Law ($p_1V_1 = p_2V_2$) to solve numerical problems
- Identify common applications of pressure in real-world and exam contexts

## Pressure in Solids (Core)

**Pressure** — Pressure is the force applied per unit area perpendicular to a surface.

*Notation:* $p$

*Example:* A sharp knife cuts better than a blunt knife because force acts over a smaller contact area, creating higher pressure.

The core formula for calculating pressure in solid contexts is:

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

Where $p$ = pressure in pascals (Pa), $F$ = perpendicular force in newtons (N), and $A$ = cross-sectional contact area in square metres ($m^2$). 1 Pa = 1 N/m².

**Worked example:** A student of mass 50 kg stands on one foot, which has a contact area of 0.015 m² with the floor. Calculate the pressure exerted on the floor (use $g = 9.8$ N/kg).

1. First calculate the student's weight, which is the force exerted on the floor:

   $$F = m \times g = 50 \times 9.8 = 490 N$$
2. Substitute values into the pressure formula:

   $$p = \frac{F}{A} = \frac{490}{0.015} \approx 32700 Pa (or 3.3 \times 10^4 Pa, 2 s.f.)$$

> **Exam tip:** Always convert area to $m^2$ before calculating pressure: 1 cm² = $1 \times 10^{-4} m^2$, this is one of the most common unit conversion errors in pressure questions.

## Pressure in Liquids (Core + Extended)

Liquids are incompressible, so they exert pressure equally in all directions at the same depth. Three core rules for liquid pressure you must know:

- Pressure acts equally in all directions at the same depth in a uniform liquid
- Pressure increases linearly with depth below the liquid surface
- Higher density liquids exert higher pressure at the same depth

> **Extended Only**
>
> For Extended students, the formula for pressure difference at depth in a liquid is:

$$p = h \rho g$$

Where $h$ = depth below surface (m), $\rho$ = liquid density (kg/m³), $g$ = gravitational field strength (N/kg).

**Worked example:** Calculate the pressure difference between the surface of the sea and a point 20 m below the surface. Density of seawater = 1025 kg/m³, $g = 9.8$ N/kg.

1. Substitute values directly into the liquid pressure formula:

   $$p = 20 \times 1025 \times 9.8 = 200900 Pa = 2.0 \times 10^5 Pa$$

## Atmospheric Pressure (Core)

**Atmospheric Pressure** — Pressure exerted by the weight of the Earth's atmosphere on all surfaces it contacts.

*Example:* A drinking straw works because you reduce pressure inside the straw, so higher atmospheric pressure pushes liquid up the straw into your mouth.

Atmospheric pressure decreases as altitude increases, because there is less air above a surface at higher heights. Standard atmospheric pressure at sea level is approximately $1 \times 10^5$ Pa, which you must memorize for exams.

**Worked example:** A mercury barometer measures atmospheric pressure as 760 mm of mercury. If mercury density = 13600 kg/m³, $g = 9.8$ N/kg, calculate atmospheric pressure in Pa (Extended Only question).

1. Convert depth from mm to m:

   $$h = 760 mm = 0.76 m$$
2. Apply liquid pressure formula:

   $$p = 0.76 \times 13600 \times 9.8 = 101293 Pa ≈ 1 \times 10^5 Pa$$

## Gas Pressure & Boyle's Law (Core + Extended)

For Core students, you only need to know that for a fixed mass of gas at constant temperature, pressure increases when volume decreases (inverse proportionality).

> **Extended Only**
>
> For Extended students, this relationship is called Boyle's Law, with the formula:

$$p_1 V_1 = p_2 V_2$$

Where $p_1, V_1$ = initial pressure and volume, $p_2, V_2$ = final pressure and volume, for a fixed mass of gas at constant temperature. Units for pressure and volume can be any consistent units, no conversion to SI required if both values use the same units.

**Worked example:** A gas syringe holds 40 cm³ of gas at pressure $1.0 \times 10^5$ Pa. The plunger is pushed in to reduce volume to 10 cm³ at constant temperature. Calculate the new gas pressure.

1. Rearrange Boyle's Law to solve for final pressure:

   $$p_2 = \frac{p_1 V_1}{V_2}$$
2. Substitute values (volume units are consistent in cm³, no conversion needed):

   $$p_2 = \frac{1.0 \times 10^5 \times 40}{10} = 4.0 \times 10^5 Pa$$

## Common pitfalls

- **Wrong:** Using area in cm² directly in $p=F/A$ calculations
  - Why it fails: Pascals are defined as N/m², so using cm² will give a value 10,000 times larger than the correct answer
  - Correct: Convert cm² to m² by multiplying by $10^{-4}$ before substituting into the pressure formula
- **Wrong:** Assuming hydraulic systems transmit force equally, not pressure
  - Why it fails: Pressure is transmitted equally through incompressible liquid, not force, so output force depends on piston area ratio
  - Correct: Calculate pressure from the input piston first, then apply that same pressure to the output piston to find output force
- **Wrong:** Using Boyle's Law when gas temperature or mass changes
  - Why it fails: Boyle's Law only applies for fixed mass of gas at constant temperature, changing these breaks the inverse proportionality
  - Correct: Only use $p_1V_1 = p_2V_2$ if the question explicitly states temperature and mass of gas are constant
- **Wrong:** Forgetting to convert depth to metres for $p = h\rho g$ calculations
  - Why it fails: The formula requires depth in metres to match units of density (kg/m³) and g (N/kg) to produce pressure in Pa
  - Correct: Convert all depth values to metres before substituting into the liquid pressure formula
- **Wrong:** Ignoring atmospheric pressure when asked for total pressure under a liquid
  - Why it fails: Pressure difference is just the liquid pressure, but total pressure adds atmospheric pressure at the surface
  - Correct: Read questions carefully to check if you are being asked for pressure difference or total absolute pressure

## Cheatsheet

| Concept | Formula | Units | Tier |
| --- | --- | --- | --- |
| Pressure in solids | $p = F/A$ | p: Pa, F: N, A: m² | Core |
| Liquid pressure difference | $p = h\rho g$ | h: m, ρ: kg/m³, g: N/kg | Extended |
| Boyle's Law | $p_1V_1 = p_2V_2$ | p: same units, V: same units | Extended |
| Sea level atmospheric pressure | ≈ $1 \times 10^5$ Pa | Pa | Core |

## What's next

Now that you have mastered pressure concepts for CIE IGCSE Physics 0625, you can move on to related topics in Unit 1: Motion, Forces and Energy. Next, explore energy transfer, work done and power, which build on your understanding of forces and force calculations. If you are sitting the Extended paper, make sure you practice extra numerical problems for liquid pressure and Boyle's Law, as these topics often appear in 4-6 mark long answer calculation questions. You can also practice past paper questions focused on pressure to test your understanding and identify gaps before your exam.

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