B.4.1 (HL)—First law of thermodynamics

First law for a closed system

Using the convention that W is work done by the system,

$$Q=\Delta U+W$$

So energy supplied as thermal transfer is split between internal-energy change and work done by the gas.

Rearrange for the unknown

$$\Delta U=Q-W$$

If heat is removed, Q is negative. If the gas expands and does work on its surroundings, W is positive. If work is done on the gas, W is negative in this convention.

Use the process information

For an adiabatic process Q=0, so $$\Delta U=-W$$. An expanding adiabatic gas does positive work and its internal energy decreases.

Worked example from the mapped local textbook

A gas receives $Q=+120\,\mathrm{J}$ and does $W=+80\,\mathrm{J}$ of work. With work done by the gas defined as positive,

$$\Delta U=Q-W=120-80=+40\,\mathrm{J}$$

The positive result means the gas's internal energy increases; the remaining input energy left the system as work.

Sign boundary

Always state whether W means work done by the gas or work done on the gas before using a memorized first-law equation.