# 6.5.4 Work Done by a Gas

### Work Done by a Gas

• When a gas expands, it does work on its surroundings by exerting pressure on the walls of the container it’s in
• This is important, for example, in a steam engine where expanding steam pushes a piston to turn the engine
• The work done when a volume of gas changes at constant pressure is defined as:

W = pΔV

• Where:
• W = work done (J)
• p = external pressure (Pa)
• V = volume of gas (m3)

• For a gas inside a cylinder enclosed by a moveable piston, the force exerted by the gas pushes the piston outwards
• Therefore, the gas does work on the piston The gas expansion pushes the piston a distance s

#### Derivation

• The volume of gas is at constant pressure. This means the force F exerted by the gas on the piston is equal to :

F = p × A

• Where:
• p = pressure of the gas (Pa)
• A = cross-sectional area of the cylinder (m2)
• The definition of work done is:

W = F × s

• Where:
• F = force (N)
• s = displacement in the direction of force (m)
• The displacement of the gas d multiplied by the cross-sectional area A is the increase in volume ΔV of the gas:

W = p × A × s

• This gives the equation for the work done when the volume of a gas changes at constant pressure:

W = pΔV

• Where:
• ΔV = increase in the volume of the gas in the piston when expanding (m3)

• This is assuming that the surrounding pressure p does not change as the gas expands
• This will be true if the gas is expanding against the pressure of the atmosphere, which changes very slowly
• When the gas expands (V increases), work is done by the gas
• When the gas is compressed (V decreases), work is done on the gas

#### Worked Example

When a balloon is inflated, its rubber walls push against the air around it.

Calculate the work done when the balloon is blown up from 0.015 m3 to 0.030 m3.

Atmospheric pressure = 1.0 × 105 Pa.

Step 1: Write down the equation for the work done by a gas

W = pΔV

Step 2: Substitute in values

ΔV = final volume − initial volume = 0.030 − 0.015 = 0.015 m3

W = (1.0 × 105) × 0.015 = 1500 J

#### Exam Tip

The pressure p in the work done by a gas equation is not the pressure of the gas but the pressure of the surroundings. This is because when a gas expands, it does work on the surroundings.

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