Thermodynamic Cycles and Enclosed Area
For any closed thermodynamic cycle, ΔU_cycle = 0 and Q_net = W_net under the heat-in/work-by convention; on a P-V plot W_net = ∮P dV, positive for a clockwise loop.
Why this shows up in the exam
Circular, rectangular, and triangular P-V cycles · Summing heat and work over several legs · Determining engine-cycle efficiency
Learn the idea
Over a complete cycle, internal energy returns to its initial value, so net heat equals net work. A cycle comes back to the same state; the enclosed P-V area measures useful net work and its direction sets the sign.
🧠 Memory hook: Closed loop: no net U, so net Q becomes net W.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- ΔU_cycle = 0 — state-function change over a complete cycle
- Q_net = W_net = ∮ P dV — net heat and signed enclosed P-V area
- η = W_net/Q_in = 1 - Q_out/Q_in — engine efficiency with Q_in and Q_out as positive magnitudes
How to approach it
- 1Mark the traversal direction
- 2Compute signed work leg by leg or by enclosed area
- 3Use ΔU_cycle = 0 and separate heat input from heat rejection
Common slip-ups that cost marks
- •Adding unsigned leg areas
- •Setting ΔU to zero on each leg instead of only the cycle
- •Using net heat instead of total heat input in efficiency
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A gas absorbs 500 J of heat and does 200 J of work. What is the change in its internal energy?
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