Boyle's Law (Edexcel IGCSE Physics)

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Katie M

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Katie M

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Boyle's law

  • For a fixed mass of a gas held at a constant temperature, the Boyle's law formula is:

pV = constant

  • Where:

    • p = pressure in pascals (Pa)

    • V = volume in metres cubed (m3)

  • This means that the pressure and volume are inversely proportional to each other

    • When the volume decreases (compression), the pressure increases

    • When the volume increases (expansion), the pressure decreases

  • This is because when the volume decreases, the same number of particles collide with the walls of a container but more frequently as there is less space

    • However, the particles still collide with the same amount of force meaning greater force per unit area (pressure)

  • The key assumption is that the temperature and the mass (and number) of the particles remains the same

Gas Laws Molecular Model (1), downloadable AS & A Level Physics revision notes

Increasing the volume of a gas decreases its pressure

  • This equation can also be rewritten for comparing the pressure and volume before and after a change in a gas:

p1V1 = p2V2

  • Where:

    • p1 = initial pressure in pascals (Pa)

    • V1 = initial volume in metres cubed (m3)

    • p2 = final pressure in pascals (Pa)

    • V2 = final volume in metres cubed (m3)

  • This equation is sometimes referred to as Boyle's Law

Pressure-vs-Volume, IGCSE & GCSE Physics revision notes

Initial pressure and volume, p1 and V1, and final pressure and volume, p2 and V2. When volume decreases, pressure increases

Worked Example

A gas occupies a volume of 0.70 m3 at a pressure of 200 Pa. Calculate the pressure exerted by the gas if it is compressed to a volume of 0.15 m3.Assume that the temperature and mass of the gas stay the same.

Answer:

Step 1: List the known quantities

  • Initial volume, V1 = 0.70 m3

  • Initial pressure, p1 = 200 Pa

  • Final volume, V2 = 0.15 m3

Step 2: Write the relevant equation

p subscript 1 V subscript 1 space equals space p subscript 2 V subscript 2

Step 3: Rearrange for the final pressure, p2

  • Divide both sides by V2 to get the p2 term on its own

fraction numerator p subscript 1 V subscript 1 over denominator V subscript 2 end fraction space equals space fraction numerator p subscript 2 up diagonal strike V subscript 2 end strike over denominator up diagonal strike V subscript 2 end strike end fraction

fraction numerator p subscript 1 V subscript 1 over denominator V subscript 2 end fraction space equals space p subscript 2

Step 4: Substitute in the values

p subscript 2 space equals space fraction numerator p subscript 1 V subscript 1 over denominator V subscript 2 end fraction space equals space fraction numerator 200 space cross times space 0.70 over denominator 0.15 end fraction

p subscript 2 space equals space 930 space Pa (2 s.f.)

Examiner Tips and Tricks

Always check whether your final answer makes sense. If the gas has been compressed, the final pressure is expected to be more than the initial pressure (like in the worked example). If this is not the case, double-check the rearranging of any formulae and the values put into your calculator.

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Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.