A cycle on the P–V diagram
"A cycle on the P–V diagram" is a free, interactive Class 11 Physics simulation for NEET and JEE, built on the NCERT section "Isochoric, isobaric and cyclic processes" in Thermodynamics. Drag the corners of a cycle on a P-V diagram; net work = enclosed area, positive clockwise; ΔU over the cycle is zero. NEET has asked about this idea 2 times in 10 years of papers. JEE Main asked 48 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Isochoric, isobaric and cyclic processes in Thermodynamics. Change the values and watch what happens.
What it shows
NCERT §11.8.6
Drag the corners of a cycle on a P-V diagram; net work = enclosed area, positive clockwise; ΔU over the cycle is zero.
Controls
3 to try
- Drag3–4 state points
- Toggleclockwise / anticlockwise
- ReadoutW per leg, net Q
Try this
Draw a cycle that does 500 J of net work, then reverse it and read the sign.
Common mistake
Work in a cycle is zero because the gas returns to its start.
In NEET
10 years of papers
Asked 2 times in 10 years of NEET.
See the past questions: Laws of Thermodynamics
In JEE Main
42 shifts analysed
48 questions from Thermodynamics across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §11.8.4, §11.8.5, §11.8.6
- Isochoric (V constant): the gas neither does work nor has work done on it. All the heat absorbed goes into internal energy and raises the temperature; the rise for a given heat is set by the specific heat at constant volume.
- Isobaric (P constant): the gas does work W = P(V₂ − V₁) = μR(T₂ − T₁).
- In an isobaric process the temperature changes, so U changes too. The heat absorbed goes partly into internal energy and partly into work; the temperature rise for a given heat is set by the specific heat at constant pressure.
- That is why C_p > C_v: at constant pressure part of the heat leaves as work, so more heat is needed for the same temperature rise.
- Cyclic process: the system returns to its initial state. U is a state variable, so ΔU = 0 over a cycle, and the net heat absorbed equals the net work done by the system.
- On a P–V diagram the work in a process is the area under its curve; for a cycle, the net work is the area enclosed by the loop.
- Summary of the first law in the four special processes: isothermal ΔU = 0, Q = W; adiabatic Q = 0, W = −ΔU; isochoric W = 0, Q = ΔU; isobaric Q = ΔU + PΔV.
Questions
5 answers
- What does the "A cycle on the P–V diagram" simulation show?
- Drag the corners of a cycle on a P-V diagram; net work = enclosed area, positive clockwise; ΔU over the cycle is zero.
- What mistake does the "A cycle on the P–V diagram" simulation fix?
- A common mistake it clears up: work in a cycle is zero because the gas returns to its start.
- Has NEET asked about this idea?
- Yes. NEET has asked about this idea 2 times in 10 years of papers.
- Does JEE Main ask questions from Thermodynamics?
- Yes. JEE Main asked 48 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Isochoric, isobaric and cyclic processes" in Thermodynamics (Class 11 Physics).
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