First law of thermodynamics — dU = dQ − dW, sign conventions: Worked examples

Before anything, recall the only three quantities and one rule we use everywhere on this page.

Recall The one rule (from the parent)

$Δ U = Q - W$

$Q$ = heat added to the gas ( $+$ in, $-$ out).
$W$ = work done by the gas ( $+$ when it expands, $-$ when compressed).
$Δ U$ = change in internal energy ( $+$ = hotter for an ideal gas). Everything below is just plugging into this, being careful with the two signs.

The scenario matrix

The First Law has exactly two inputs that carry a sign: heat $Q$ and work $W$ . Each can be positive, negative, or zero. That is a $3 \times 3$ grid — the complete universe of cases. Below, each cell names the physical situation and points to the example that nails it.

	$W > 0$ (gas expands)	$W = 0$ (fixed volume)	$W < 0$ (gas compressed)
$Q > 0$ (heat in)	Ex 1 — heat in, work out	Ex 3 — isochoric heating	Ex 2 — heat in and squeezed
$Q = 0$ (adiabatic)	Ex 5 — adiabatic expansion	Ex 6 — degenerate: free expansion	Ex 4 — adiabatic compression
$Q < 0$ (heat out)	Ex 7 — heat out yet expands	Ex 3 (variant)	Ex 8 — both signs negative

Two more cells that don't fit the 3×3 grid but must be shown:

Special-shape work — isobaric ( $W = P Δ V$ directly), isothermal ( $Δ U = 0$ ). → Ex 9
Word problem / exam twist — a cyclic process where $Δ U = 0$ over a full loop. → Ex 10

The figure below is the same grid drawn as a picture — memorise the diagonal (top-left to bottom-right) as "the boring cases where $Q$ and $W$ push $Δ U$ opposite ways."

Figure — First law of thermodynamics — dU = dQ − dW, sign conventions

The examples

Cell $(Q > 0, W > 0)$ — heat in, gas pushes out

Cell $(Q > 0, W < 0)$ — heat in AND squeezed

Cell $(Q > 0, W = 0)$ — fixed volume (isochoric)

Worked example Example 3 — Isochoric heating (and its heat-out variant)

A gas in a rigid sealed container absorbs $80$ J of heat. Find $Δ U$ . Then: if instead it loses $80$ J, find $Δ U$ .

Forecast: Rigid means the walls can't move, so the gas can't push anything. With no work, all heat becomes internal energy. Guess $Δ U = \pm 80$ J.

Why is $W = 0$ ? Volume is fixed ⇒ $d V = 0$ ⇒ $W = \int P d V = 0$ . Why this step? Work only happens through volume change; no volume change, no work — see Work done in thermodynamic processes (PV diagrams).
Heat-in case. $Δ U = Q - 0 = + 80$ J.
Heat-out variant. Now $Q = - 80$ : $Δ U = - 80 - 0 = - 80$ J (gas cools).

Verify: With $W = 0$ , $Δ U$ tracks $Q$ exactly, sign and all. $+ 80$ heats, $- 80$ cools. ✓ This is why the heat capacity at constant volume $C_{v}$ is the "cleanest" one — see Heat capacities Cp and Cv.

Cell $(Q = 0, W < 0)$ — adiabatic compression

Cell $(Q = 0, W > 0)$ — adiabatic expansion

Degenerate cell $(Q = 0, W = 0)$ — free expansion into vacuum

Worked example Example 6 — Free expansion (both terms zero)

An ideal gas rushes into an evacuated chamber (vacuum). No heat is exchanged and it pushes on nothing. Find $Δ U$ and $Δ T$ .

Forecast: If nothing pushed back, no work was done — and no heat flowed. Nothing crossed the boundary. So $Δ U$ must be exactly $0$ .

Why $W = 0$ ? The gas expands into vacuum — there is no piston, no surroundings resisting. Force it pushes against $= 0$ , so $W = 0$ even though $d V > 0$ . Why this step? This is the trap: volume does increase, but work needs something to push on. Zero opposing pressure ⇒ zero work.
Why $Q = 0$ ? Insulated / sudden ⇒ no heat crosses.
Plug in. $Δ U = 0 - 0 = 0$ J.
Temperature. For an ideal gas $U = U (T)$ only, so $Δ U = 0 \Rightarrow Δ T = 0$ — see Internal energy and degrees of freedom.

Verify: Both channels shut ⇒ internal energy untouched ⇒ temperature unchanged. This is the classic result: ideal free expansion does not cool the gas. ✓

Cell $(Q < 0, W > 0)$ — the surprising one: heat OUT yet gas EXPANDS

Cell $(Q < 0, W < 0)$ — both negative

Special-shape work — isobaric then isothermal

Worked example Example 9 — Isobaric expansion, then an isothermal check

(a) A gas expands at constant pressure $P = 2 \times 1 0^{5}$ Pa from $V_{1} = 1 \times 1 0^{- 3}$ m³ to $V_{2} = 3 \times 1 0^{- 3}$ m³, absorbing $Q = 700$ J. Find $W$ and $Δ U$ . (b) Separately, an ideal gas expands isothermally absorbing $Q = 400$ J. Find $W$ .

Forecast (a): Constant pressure means work is just $P$ times the volume swept — an easy rectangle on the PV diagram. Expect a few hundred Joules of work.

(a) Work at constant $P$ . $W = P Δ V = P (V_{2} - V_{1}) = 2 \times 1 0^{5} \times (3 - 1) \times 1 0^{- 3}$ . Why this step? When $P$ is constant it comes out of the integral: $W = \int P d V = P Δ V$ . The area under the PV curve is a rectangle.
Compute. $W = 2 \times 1 0^{5} \times 2 \times 1 0^{- 3} = 400$ J.
Internal energy. $Δ U = Q - W = 700 - 400 = + 300$ J.
(b) Isothermal. For an ideal gas $U = U (T)$ ; $T$ constant ⇒ $Δ U = 0$ ⇒ $0 = Q - W \Rightarrow W = Q = 400$ J. Why this step? Isothermal is the special case where the First Law forces all absorbed heat to become work.

Verify (a): $Pa \times m^{3} = \frac{N}{m ^{2}} \cdot m^{3} = N\cdotm = J$ , so $W = 400$ J is dimensionally right; $700 - 400 = 300$ J. ✓ Verify (b): $Δ U = Q - W = 400 - 400 = 0$ , consistent with constant $T$ . ✓

The figure shows the isobaric case as a rectangle whose area is the work.

Word problem / exam twist — a full cycle

Worked example Example 10 — Cyclic process (the exam favourite)

A gas is taken around a closed cycle on a PV diagram: it absorbs $500$ J of heat on the way up, expels $300$ J on the way back, and does a net $200$ J of work over the loop. (a) What is $Δ U$ for the whole cycle? (b) Check consistency with the numbers.

Forecast: A cycle returns to the same state $(P, V, T)$ . Since $U$ is a state function, going in a loop must give $Δ U = 0$ — no matter the path.

Why $Δ U = 0$ ? $U$ depends only on state; start state = end state ⇒ $Δ U_{cycle} = 0$ . Why this step? This is the whole point of "state function" from the parent's lake analogy — return to the same water level.
Net heat. $Q_{net} = + 500 - 300 = + 200$ J.
Apply the law to the cycle. $Δ U = Q_{net} - W_{net} \Rightarrow 0 = 200 - W_{net} \Rightarrow W_{net} = 200$ J.

Verify: The stated net work is $200$ J and our law predicts $200$ J — consistent. Over any cycle, net heat in = net work out ( $Q_{net} = W_{net}$ ), which is exactly the seed of the Second law of thermodynamics (you can't get more work out than heat in). ✓

Recall check

Recall Match the cell to the outcome

Adiabatic expansion — does the gas heat or cool? ::: Cools; $Q = 0$ so $Δ U = - W < 0$ . Free expansion of an ideal gas — what is $Δ T$ ? ::: Zero; $Q = 0$ , $W = 0$ ⇒ $Δ U = 0$ ⇒ $Δ T = 0$ . Isochoric process — which term vanishes? ::: $W = 0$ , so $Δ U = Q$ . Full cycle — what is $Δ U$ ? ::: Zero, because $U$ is a state function. Compression with heat loss (Ex 8) — sign of $Δ U$ ? ::: $+ 90$ J; compression added more than heat removed.

Connections

Parent: First Law & sign conventions
Work done in thermodynamic processes (PV diagrams)
Isothermal, adiabatic, isobaric, isochoric processes
Internal energy and degrees of freedom
Heat capacities Cp and Cv
Conservation of energy (mechanics)
Second law of thermodynamics