Edexcel IGCSE Physics 5.7 Pressure difference equation

Practise using Δp = hρg in manometers, bottles and liquids, including rearranging, unit conversion and choosing suitable liquids.

Exam points

state or rearrange Δp = hρg using symbols for height, density and gravitational field strength
calculate height, depth or pressure difference from liquid data, converting kPa or cm where needed
explain why a denser manometer liquid gives a smaller height difference for large pressures

[Maximum number: 3]

The diagram shows a manometer, a device used for measuring differences in pressure.

Figure for Question 5.7 Pressure difference equation question 1 — Edexcel IGCSE Physics

One side of the manometer has some trapped gas. The other side is left open to the atmosphere.

The difference in pressure can be calculated using this formula.
[difference in pressure = height × density × 10 ]
The density of the liquid is $1.3 \times 10^{4} \mathrm{~kg} / \mathrm{m}^{3}$ .
The difference in the levels of the liquid is 3.8 cm.
Calculate the difference in pressure between the atmosphere and the trapped gas.
difference in pressure = Pa

Marking points:
- conversion of cm to m
- substitution into the given formula
- evaluation

Example:
$3.8\ \mathrm{cm}=0.038\ \mathrm{m}$ $\text{pressure difference}=0.038\times1.3\times10^{4}\times10$ $\text{pressure difference}=4900\ \mathrm{Pa}$

Marking guidance:
- power-of-ten error gives a 1 mark penalty
- 494000 gains 2 marks

Accept:
- $4940\ \mathrm{Pa}$
- 4800, 4840, 4846, or $4841\ \mathrm{Pa}$ for use of g=9.8 or 9.81

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5.7 Pressure difference equation question 1