Edexcel IGCSE Physics 5.22 Pressure-volume relationship

Practise applying the inverse pressure-volume relationship for a fixed gas at constant temperature in bubbles, syringes and containers.

Exam points

use p1V1 = p2V2 to calculate pressure or volume for trapped gas
calculate a changed container volume from diagram dimensions before substituting
explain a halved gas volume as doubled pressure with more frequent wall collisions

[Maximum number: 4]

This question is about gas pressure.

The width of the container is slowly decreased so that the volume of the container is smaller than before.

Diagram 2 shows the width of the container before and after this change. All other dimensions of the container remain the same.

The initial volume of the gas is 130 cm^3.
The initial pressure of the gas is 100 kPa.
Calculate the pressure of the gas after the width of the container is decreased.
Assume the temperature of the gas remains constant.

pressure = ............ kPa

Marking points:
- evaluation of new volume
- substitution into $p_{1}V_{1}=p_{2}V_{2}$
- rearrangement
- evaluation of new pressure

Example:
$V_{2}=130\times\frac{5.0}{8.4}=77\ \mathrm{cm^{3}}$ $100\times130=p_{2}\times77$ $p_{2}=\frac{100\times130}{77}$ $p_{2}=170\ \mathrm{kPa}$

Accept:
- $V_{2}=77.38\ldots\ \mathrm{cm^{3}}$
- any value that rounds to $170\ \mathrm{kPa}$
- ECF different yet incorrect volume

Marking guidance:
- Deduct 1 mark for a power-of-ten error.

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5.22 Pressure-volume relationship question 1