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What you will learn
- The enclosed area of a p–V cycle represents net work.
- Clockwise and anticlockwise cycles have opposite work signs.
- No heat engine can exceed the Carnot efficiency for its reservoirs.
Thermodynamics · reward round
Drag the two corners of the cycle on the p–V diagram. The shaded area is the work you sell, Qₕ is the heat drawn from the hot plate along the expansion, and η = W/Qₕ chases — but never catches — the Carnot ceiling η_C = 1 − T_c/Tₕ.
Score 0 · level 1/5 · work W ≥ 100 J
Every stroke obeys the first law, Q = ΔU + W, with ΔU = (3/2)Δ(pV) for the monatomic gas. The hot stroke A→B is a straight line in the p–V plane, and because dQ along a straight segment changes direction at most once, the game splits it there and books the heat gross: Qₕ is only what flows in. The two curved flanks are exact adiabats (pV^γ constant, Q = 0), and the return stroke rides the cold plate’s isotherm, rejecting Q_c = nRT_c·ln(V_D/V_C).
The work is the enclosed area: going round the loop, ∮p dV is exactly Qₕ − |Q_c|, which is why the shaded patch and the W read-out can never disagree. Efficiency is η = W/Qₕ — work sold per joule of heat bought.
The drag limits keep the whole loop between the reservoir temperatures, and that is what makes Carnot unbeatable here: heat absorbed at T ≤ Tₕ brings entropy of at least Qₕ/Tₕ, entropy the gas can only shed by paying at least T_c·(Qₕ/Tₕ) to the cold plate. So η ≤ 1 − T_c/Tₕ with equality only for a reversible cycle hugging both isotherms — level 5’s “target” sits above that ceiling, which is why no shape of loop, in this game or in any laboratory, can reach it.
Game 22 · Thermodynamics learning guide
Learning objectiveShape a cycle on a pressure–volume diagram to produce useful work while respecting the Carnot efficiency limit.
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What geometric feature of a p–V cycle gives the net work done per cycle?
Suitable forUpper-secondary physics · thermodynamics and heat engines
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