16.2 The First Law of Thermodynamics

2026 Syllabus Objectives

By the end of these notes, you should be able to:

  1. Recall and use W = pΔV for the work done when the volume of a gas changes at constant pressure, and understand the difference between work done by the gas and work done on the gas.
  2. Recall and use the first law of thermodynamics ΔU = q + W, expressed in terms of the increase in internal energy, the heating of the system, and the work done on the system.

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1. Work Done by a Gas (and on a Gas)

What does it mean for a gas to "do work"?

Think about a gas trapped inside a cylinder with a moveable piston (like the inside of a bicycle pump). When the gas expands, it pushes the piston outward. Pushing something over a distance requires energy — so the gas is doing work on its surroundings.

The opposite is also possible: if you push the piston inward, you are doing work on the gas, squeezing it into a smaller space.

The Formula: W = pΔV

When a gas changes its volume at constant pressure (meaning the pressure stays the same throughout), the work done is calculated using:

W=pΔV\boxed{W = p\Delta V}

Where:

  • W = work done (measured in Joules, J)
  • p = pressure (measured in Pascals, Pa) — this is the pressure acting on or from the gas
  • ΔV = change in volume (measured in cubic metres, m³)

ΔV simply means "change in volume." It is calculated as: ΔV = final volume − initial volume

Where does this formula come from?

Recall that work = force × distance. If a gas with pressure p pushes against a piston with cross-sectional area A, the force exerted is:

F=p×AF = p \times A

If the piston moves outward by a distance s, the volume change is:

ΔV=A×s\Delta V = A \times s

So the work done is:

W=F×s=(p×A)×s=p×(A×s)=pΔVW = F \times s = (p \times A) \times s = p \times (A \times s) = p\Delta V

This confirms the formula neatly.

Worked Example 1

A balloon is inflated from a volume of 0.015 m³ to 0.030 m³. The atmospheric pressure is 1.0 × 10⁵ Pa. Calculate the work done in inflating the balloon.

Step 1: Write down the formula.

W=pΔVW = p\Delta V

Step 2: Calculate ΔV.

ΔV=0.030−0.015=0.015 m3\Delta V = 0.030 - 0.015 = 0.015 \text{ m}^3

Step 3: Substitute the values.

W=(1.0×105)×0.015=1500 JW = (1.0 \times 10^5) \times 0.015 = 1500 \text{ J}

The balloon (gas) does 1500 J of work on its surroundings as it expands.

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