The first law $\Delta U = Q + W$ holds for every process, but it has three unknowns. Real processes are often simple enough to fix one of them: the volume, the pressure, the temperature, or the heat. Each choice decides how the energy splits.
Why it works
For a monatomic ideal gas, $U = \tfrac{3}{2}nRT = \tfrac{3}{2}PV$. Hold $V$ fixed and no work is done, so all heat becomes internal energy. Hold $T$ fixed and $U$ cannot change, so $Q = -W$: heat must flow in to pay for the work. Allow no heat (insulate the gas, or act too fast for heat to flow) and $\Delta U = W$: work done on the gas heats it, and a gas that does work cools. Hold $P$ fixed and both happen: heat raises $U$ and pays for the work.
| Process | Held fixed | $W$ (on the gas) | $Q$ | $\Delta U$ |
|---|---|---|---|---|
| Isochoric | Volume | $0$ | $\Delta U$ | $\tfrac{3}{2}nR\Delta T$ |
| Isobaric | Pressure | $-P\Delta V$ | $\tfrac{5}{2}P\Delta V$ | $\tfrac{3}{2}P\Delta V$ |
| Isothermal | Temperature | minus the area | $-W$ | $0$ |
| Adiabatic | No heat flow | $\Delta U$ | $0$ | $W$ |
Key idea
Isochoric: $W = 0$. Isobaric: $W = -P\Delta V$. Isothermal: $\Delta U = 0$, so $Q = -W$. Adiabatic: $Q = 0$, so $\Delta U = W$. Then use $\Delta U = Q + W$ for whatever is left.
Both are in the Energy and the First Law chapter of the Science Lab, which writes $\Delta U = Q - W$ with $W$ done by the gas.
Worked examples
Worked example: Heating in a rigid tank
How much heat warms $0.50\,\text{mol}$ of argon by $40\,\text{K}$ at constant volume?
- Isochoric: $W = 0$, so $Q = \Delta U = \tfrac{3}{2}nR\Delta T$.
- $Q = \tfrac{3}{2}(0.50)(8.31)(40) = 249\,\text{J}$.
Answer: $249\,\text{J}$
Worked example: Expanding at constant pressure
A monatomic ideal gas expands at $100\,\text{kPa}$ from $2.0\,\text{L}$ to $5.0\,\text{L}$. Find $W$, $\Delta U$ and $Q$.
- $W = -P\Delta V = -(100\,\text{kPa})(3.0\,\text{L}) = -300\,\text{J}$.
- $\Delta U = \tfrac{3}{2}P\Delta V = +450\,\text{J}$ (from $U = \tfrac{3}{2}PV$).
- $Q = \Delta U - W = 450 + 300 = 750\,\text{J}$.
Answer: $W = -300\,\text{J}$, $\Delta U = 450\,\text{J}$, $Q = 750\,\text{J}$
Worked example: A bicycle pump
Air in a pump is compressed so quickly that no heat escapes, with $200\,\text{J}$ of work done on it. Find $\Delta U$.
- Adiabatic: $Q = 0$, so $\Delta U = W = +200\,\text{J}$.
- The internal energy rises, so the air gets hotter. That is why a pump's barrel warms up.
Answer: $+200\,\text{J}$
Common mistake
Mixing up isothermal and adiabatic. Isothermal means the temperature stays fixed, and that takes heat flow: heat in during expansion, out during compression. Adiabatic means no heat flows, so the temperature must change. Check: if $T$ is constant, ask where the heat went; if $Q = 0$, ask why $T$ changed.