Pressure is one of the friendliest topics in the Combined Science (Physics) syllabus. The formulas are short, the numbers are usually clean, and the marks are very predictable. Yet every year students drop marks here for the same three reasons: wrong units, forgetting to convert areas, and explaining Boyle's Law without mentioning the kinetic model properly.
Here is how we teach it at Dojo, in the order the exam actually asks it.
The two formulas you must know cold
Pressure from a solid
$$p = \frac{F}{A}$$
where $F$ is the force in newtons (usually weight, so $F = mg$) and $A$ is the contact area in square metres. The unit of pressure is the pascal (Pa), where 1 Pa = 1 N/m².
Pressure in a liquid
$$p = h\rho g$$
where $h$ is the vertical depth below the surface, $\rho$ is the density in kg/m³, and $g$ = 10 N/kg (unless your paper says 9.81).
Two things examiners love to test here:
- $h$ is the vertical depth, not the length of a slanted tube.
- Liquid pressure does not depend on the shape or the cross-sectional area of the container. Only depth, density and $g$ matter.
If a question asks for total pressure at a depth, add atmospheric pressure: $p_{\text{total}} = p_{\text{atm}} + h\rho g$. If it asks for the pressure due to the liquid, do not add it. Read the wording carefully.
Worked example: solid and liquid pressure
A block of mass 12 kg rests on a table. Its base measures 20 cm by 30 cm. Find the pressure it exerts.
- Force = weight = $12 \times 10 = 120$ N
- Area = $0.20 \times 0.30 = 0.06$ m²
- $p = 120 \div 0.06 = 2000$ Pa
Notice step 2. Converting centimetres to metres before multiplying is where most marks are lost. If you convert after, you need to divide by 10 000, not 100, and that slip costs you the final answer mark.
Manometers: the one-line method
A U-tube manometer measures the pressure of a gas supply. The rule is simple:
$$p_{\text{gas}} = p_{\text{atm}} + h\rho g$$
where $h$ is the difference in levels between the two liquid surfaces.
- If the liquid is higher on the open side, the gas pressure is greater than atmospheric, so you add.
- If the liquid is higher on the gas side, the gas pressure is less than atmospheric, so you subtract.
Say the mercury difference is 5.0 cm, mercury density is 13 600 kg/m³ and atmospheric pressure is 100 000 Pa. Then $h\rho g = 0.050 \times 13600 \times 10 = 6800$ Pa, giving a gas pressure of 106 800 Pa.
Boyle's Law: what it says and when it applies
For a fixed mass of gas at constant temperature:
$$p_1V_1 = p_2V_2$$
Those two conditions in bold are worth a mark on their own in a definition question. Write them.
The kinetic model explanation
When a question asks you to explain why pressure increases when volume decreases, examiners want this chain of reasoning:
- Gas molecules are in constant random motion and collide with the walls of the container.
- Pressure is caused by the force of these collisions per unit area.
- When the volume is reduced, the molecules travel a shorter distance between collisions with the walls.
- So the rate of collisions with the walls increases.
- Since the temperature is unchanged, the average speed and average force per collision stay the same.
- Therefore pressure increases.
The phrase that earns the mark is "rate of collisions with the walls", not "the molecules hit harder". They only hit harder when the temperature rises.
Worked example: Boyle's Law
A gas occupies 250 cm³ at a pressure of 1.2 × 10⁵ Pa. The gas is compressed at constant temperature until its pressure is 3.0 × 10⁵ Pa. Find the new volume.
$$V_2 = \frac{p_1V_1}{p_2} = \frac{(1.2 \times 10^5)(250)}{3.0 \times 10^5} = 100 \text{ cm}^3$$
Because Boyle's Law is a ratio, the units of volume cancel. You may leave volume in cm³ as long as both volumes use the same unit. The same applies to pressure. This saves time in the exam.
Sanity check: pressure went up by a factor of 2.5, so volume should go down by a factor of 2.5. It did.
Common mistakes to check before you move on
- Using diameter instead of radius when a piston area is given.
- Forgetting that cm² to m² is a division by 10 000.
- Mixing units in a Boyle's Law calculation, for example kPa on one side and Pa on the other.
- Applying $p_1V_1 = p_2V_2$ when the temperature changes. That is not Boyle's Law.
- Saying "molecules collide more" without saying "per unit time" or "rate".
- Reading a graph of $p$ against $\frac{1}{V}$ and expecting a curve. It is a straight line through the origin. The $p$ against $V$ graph is the curve.
How to practise this topic efficiently
Do five past-paper structured questions in one sitting, all on pressure. Doing them together trains your eye to spot which formula the question wants within seconds. Then mark yourself against the official mark scheme and highlight every phrase you missed. Rewrite your explanation using their words.
If pressure and Boyle's Law still feel shaky, or you want a tutor to check your working line by line, come and try a lesson with us. Start a free trial class on WhatsApp and see how a Dojo session runs.