The first law of thermodynamics
Energy bookkeeping for a gas
- Heat a gas, or squeeze it, and its internal energy changes.
- The first law of thermodynamics is just energy conservation for a gas.
- It tracks two ways energy gets in: heating and work.
Work done by a gas
- A gas changing volume against a pressure does work: $W = p\,\Delta V$.
- Expanding → it does work on the surroundings; compressing → work is done on it.
The work done by a gas expanding at constant pressure is:
Force = $pA$, distance = $\Delta x$, so work $= pA\Delta x = p\,\Delta V$.
The sign convention
- We write $W$ as the work done on the gas: positive when compressed, negative when it expands.
- $q$ is the heat added: positive in, negative out.
When a gas expands, it does work on its surroundings.
Yes — it pushes the piston/atmosphere outward. So the work done on the gas is negative.
The first law
- $\Delta U = q + W$ — the rise in internal energy equals heat in plus work done on the gas.
- It is conservation of energy: the same $\Delta U$ can come from different mixes of $q$ and $W$.
The first law of thermodynamics: $\Delta U = q +$ ____.
$\Delta U = q + W$, where $W$ is the work done on the gas and $q$ is the heat added.
$100\ \text{J}$ of heat is added to a gas while $30\ \text{J}$ of work is done on it. What is the rise in internal energy?
$\Delta U = q + W = 100 + 30 = 130\ \text{J}$.
Standard processes
- Isothermal (constant $T$): $\Delta U = 0$, so $q = -W$.
- Constant volume: $W = 0$, so $q = \Delta U$. Adiabatic (no heat): $q = 0$, so $\Delta U = W$.
Match each process to what it makes zero.
Constant $T$ → no change in internal energy; constant $V$ → no work; adiabatic → no heat flow.
In an isothermal change of an ideal gas, the internal energy does not change.
For an ideal gas $U \propto T$, so constant $T$ means $\Delta U = 0$ (and then $q = -W$).
You've got it
- work done by/on a gas: $W = p\,\Delta V$
- first law: $\Delta U = q + W$ ($W$ = work on the gas) — conservation of energy
- isothermal $\Delta U = 0$; constant volume $W = 0$; adiabatic $q = 0$