Heat & work — the sign convention

Every thermodynamic problem is the same one equation, ΔQ = ΔU + ΔW — the first law. Get the signs of heat and work right and the algebra tells you whether the gas warms, cools, expands or is squeezed. The rule is simple: heat in is +, work done by the gas is +. Drag the two sliders below and watch the internal energy — and the temperature — follow.

01 — the whole idea

Heat and work are inputs; internal energy is the score

You choose two things — how much heat you pour into the gas (ΔQ) and how much work the gas does on its surroundings (ΔW). The first law then fixes the change in internal energy, ΔU = ΔQ − ΔW. Since for an ideal gas U depends only on temperature, a positive ΔU means the molecules speed up — the gas gets hotter.

heat supplied ΔQ{{ qTxt }}

− out of gasinto gas +

work done by gas ΔW{{ wTxt }}

− compressedexpands +

The first law, live

{{ duTxt }} = {{ qTxt }} − {{ wTxt }}

02 — the three quantities you plug in

ΔQ, ΔW and ΔU — where each sign comes from

STATE vs PATH ΔU is a state function: it depends only on the start and end states, never on how you got there. ΔQ and ΔW are path functions — the same two end-states reached along a different route trade different amounts of heat and work, but their difference ΔQ − ΔW always lands on the same ΔU. Over a complete cycle the gas returns to its start, so ΔU = 0 and all the heat taken in becomes net work out.

03 — where the work's sign lives

Work is the area under the P–V curve

The gas does work only by pushing its boundary: ΔW = ∫ P dV. Move right on the P–V diagram (volume grows) and the gas does positive work; move left (compression) and the work is negative. The direction you travel sets the sign — the shaded area only ever gives the magnitude.

Expansion · ΔW > 0

Volume increases, the gas pushes the piston out — it spends energy on the surroundings.

Compression · ΔW < 0

The surroundings squeeze the gas in — work is done on it, so ΔW carries a minus sign.

A cycle · net area

Clockwise loop ⇒ net work out (an engine); anticlockwise ⇒ net work in (a fridge). ΔU = 0 each cycle.

04 — the payoff

One law, every process — let the signs work

Worked: heated gas that expands

120 J of heat is supplied to a gas, which then does 40 J of work pushing its piston out. Plug in the signs: ΔQ = +120 J, ΔW = +40 J.

ΔU=+120−40=+80 J

The internal energy rises by 80 J — the plus sign tells you the gas ends up hotter, even though it spent some energy on work. You never had to decide that yourself.

The sign, decoded

05 — quick check

Call the sign

Six one-tap questions. Don't compute anything — just read the situation and name the sign. {{ quizSolved }}/6 right.

Q{{ q.n }}. {{ q.q }}

Recap card — the convention

◦ First law: ΔQ = ΔU + ΔW.

◦ Heat into the system is +, out is −.

◦ Work done by the gas (expansion) is +, on it is −.

◦ ΔU is a state function; for an ideal gas it tracks temperature only.

◦ Over a full cycle ΔU = 0.

◦ Work = area under P–V; direction sets the sign.

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