The First Law of Thermodynamics

Energy is conserved, and for a gas there are two ways for it to cross the boundary: heat $Q$, which flows because of a temperature difference, and work $W$, a force acting through a distance, which you met in Work Done on a Gas. What arrives is stored as internal energy $U$, the energy of the molecules' motion, which for a monatomic gas is $\tfrac{3}{2}nRT$ from Temperature and Molecular Energy.

Why it works

Treat the gas like a bank account. Deposits are heat flowing in and work done on the gas; withdrawals are heat flowing out and work the gas does. The balance can change only by the deposits minus the withdrawals. Written with signs, that is the first law. It is conservation of energy, nothing more, but it lets heat and work trade places: a gas can take in heat and spend it all as work, so its temperature need not rise.

Key idea

$$\Delta U = Q + W$$ $Q > 0$ when heat flows into the gas; $W > 0$ when work is done on the gas ($W = -P\Delta V$). For an ideal gas, $U$ depends only on $T$: $\Delta U = 0$ whenever the temperature ends where it started.

Two ways of writing the same law. Since $W = -W_{\text{by}}$, they always give the same $\Delta U$.
	This course (and AP Physics 2)	Science Lab and OpenStax College Physics
First law	$\Delta U = Q + W$	$\Delta U = Q - W_{\text{by}}$
The work in it	$W$ = work done on the gas $= -P\Delta V$	$W_{\text{by}}$ = work done by the gas $= +P\Delta V$
Gas expands	$W < 0$	$W_{\text{by}} > 0$
Heat $Q$	positive into the gas	positive into the gas

Try the Energy and the First Law chapter of the Science Lab, which keeps its ledger as $\Delta U = Q - W$ with $W$ done by the gas. Same physics, opposite sign for the work.

Along an isotherm, every joule of heat that flows in leaves again as work.

Worked examples

Worked example: Heat in, work out

A gas absorbs $500\,\text{J}$ of heat and does $200\,\text{J}$ of work pushing a piston. Find $\Delta U$.

Signs: $Q = +500\,\text{J}$. The gas does work, so the work done on it is $W = -200\,\text{J}$.
$\Delta U = 500 + (-200) = +300\,\text{J}$. The gas warms.

Answer: $+300\,\text{J}$

Worked example: Squeezed and cooled

A gas is compressed with $300\,\text{J}$ of work while it gives off $100\,\text{J}$ of heat. Find $\Delta U$.

$W = +300\,\text{J}$ (done on the gas), $Q = -100\,\text{J}$ (heat out).
$\Delta U = -100 + 300 = +200\,\text{J}$. It got warmer even though heat left.

Answer: $+200\,\text{J}$

Worked example: Heat without warming

An ideal gas expands at constant temperature while absorbing $400\,\text{J}$ of heat. How much work does it do?

Constant temperature means $\Delta U = 0$ for an ideal gas.
$0 = Q + W$, so $W = -400\,\text{J}$: the gas does $400\,\text{J}$ of work.

Answer: $400\,\text{J}$ of work by the gas

Common mistake

Thinking adding heat must raise the temperature. Heat raises $U$ only if the gas does not spend it as work. In the third example all $400\,\text{J}$ left as work and the temperature never changed; and in the second, the gas warmed while losing heat. Check: write down $Q$ and $W$ with their signs before deciding what happens to $T$.

The First Law of Thermodynamics

Worked examples

Practice

Builds on

Leads to

Sources and further reading