An engine cycle on a $p$–$V$ graph turns into geometry: the work is an area, and each leg's heat comes from the first law of The First Law of Thermodynamics. The efficiency of Heat Engines then uses only the heat taken in. Each problem here is one short scene and a chain of questions. Answer them in order: each later question uses what you found before, so a slip early on carries through. Keep full precision between steps and round only at the end.
Key idea
Net work $W$ = the area inside the loop (positive when it runs clockwise). For a monatomic ideal gas $U = \tfrac{3}{2}pV$, so constant volume: $Q = \tfrac{3}{2}V\,\Delta p$; constant pressure: $Q = \tfrac{5}{2}p\,\Delta V$. $$e = \frac{W}{Q_{\text{in}}}.$$
Worked example: A rectangular cycle
A monatomic gas goes A ($1 \times 10^5\ \text{Pa}$, $2 \times 10^{-3}\ \text{m}^3$) → B ($3 \times 10^5\ \text{Pa}$, same volume) → C (same pressure, $5 \times 10^{-3}\ \text{m}^3$) → D → A. Find (1) the net work, (2) the heat taken in and (3) the efficiency.
- (1) $W = (2 \times 10^5)(3 \times 10^{-3}) = 600\ \text{J}$.
- (2) A → B: $\tfrac{3}{2}(2 \times 10^{-3})(2 \times 10^5) = 600\ \text{J}$. B → C: $\tfrac{5}{2}(3 \times 10^5)(3 \times 10^{-3}) = 2250\ \text{J}$. In all, $2850\ \text{J}$.
- (3) $e = 600/2850 = 0.211$, about $21.1\%$.
Answer: $600\ \text{J}$, $2850\ \text{J}$, about $21.1\%$
Common mistake
Dividing the work by all the heat that crosses the boundary, in and out. Efficiency compares the work with the heat you pay for, $Q_{\text{in}}$. Check: $e$ must come out below 1.