Pressure and volume

J-shaped glass tube, filled with mercury, and a trapped air bubble at the smaller end.

If a balloon is squeezed it will get smaller. If the pressure is increased, the volume will decrease.

The Irish scientist, Robert Boyle, originally investigated this relationship in the 17th century. Boyle carried out an experiment that gave one of the first pieces of experimental evidence for the particle theory.

By pouring mercury into a J-shaped tube that was sealed at one end, Boyle was able to trap a bubble of air. He then poured more mercury in slowly and watched what happened to the volume of the air bubble.

The higher the column of mercury in the left hand side of the tube, the greater the pressure the trapped air was experiencing and the smaller the bubble became.

Boyle was able to show that volume is inversely proportional to pressure.

For a fixed mass of gas at a constant temperature:

pressure × volume = constant

p~V = constant

This is when:

  • pressure (p) is measured in pascals (Pa)
  • volume (V is measured in metres cubed (m3)

If the temperature of a gas stays the same, the pressure of the gas increases as the volume of its container decreases. This is because the same number of particles collides with the walls of the container more frequently as there is less space. However, the particles still collide with the same amount of force.

The change in volume or pressure for gas at a constant temperature can be calculated using the equation:

p_{1}V_{1} = p_{2}V_{2}

p1 and V1 are the pressure and volume before either are changed, p2 and V2 are the pressure and volume after the change.


A gas occupies a volume of 0.50 m³ at a pressure of 100 Pa. Calculate the pressure exerted by the gas if it is compressed to a volume of 0.25 m³. Assume that the temperature and mass of the gas stay the same.

Rearrange p1 V1 = p2 V2 to find p2:

p_2 = \frac{p_1 \times v_1}{V_2}

p_2 = \frac{100 \times 0.50}{0.25}

New pressure: p_{2} = 200~Pa


If a gas has a pressure of 200,000 Pa when it is in a volume of 10 m3, what will its pressure be if the volume is reduced to 2.5 m3?

Rearrange p1 V1 = p2 V2 to find p2

p_2 = \frac{p1 \times V1}{V_2}

p_2 = \frac{200,000 \times 10}{2.5}

New pressure: p_2 = 800,000~Pa or 800~kPa