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A cyclic process brings a gas back to where it started, so many students assume nothing has happened overall. Internal energy is back to its old value, but the work and heat are not zero, and an engine depends on exactly that. This post shows why the net work equals the area of the loop, how to read its sign, and works a full cycle leg by leg.
Where is this in the syllabus?
The CBSE 2026-27 Class XI Physics curriculum lists "isothermal, adiabatic, reversible, irreversible, and cyclic processes". The JEE (Main) 2026 syllabus lists the first law of thermodynamics with isothermal and adiabatic processes, and the JEE (Advanced) 2026 brochure lists the first law and its applications for ideal gases. Cyclic processes are not named separately in those two, but loop questions are first-law applications. The text is section 11.8.6 of the NCERT Class 11 Physics chapter, Thermodynamics (2026-27 reprint).
Why is net work equal to the area of the loop?
NCERT writes the work done by a gas as W = ∫P dV. OpenStax University Physics Volume 2 reads this as the area under the curve on a P-V diagram, positive for expansion and negative for compression.
In a cycle, the gas expands along one path and comes back along another. The expansion adds the area under the upper path; the compression subtracts the area under the lower path. What remains is the area enclosed by the loop.
NCERT then adds the key step: internal energy is a state variable, so ΔU = 0 for a cyclic process, and the total heat absorbed equals the work done. Net Q = net W = loop area.
Which way round is the loop?
With volume on the horizontal axis and pressure on the vertical axis:
Clockwise: the gas expands at high pressure and is compressed at low pressure, so net work by the gas is positive. The gas takes in net heat and does work, as in an engine.
Anticlockwise: net work by the gas is negative. Work is done on the gas and it gives out net heat, as in a refrigerator.
If a question puts P on the horizontal axis and V on the vertical axis, the rule flips. Check the axes before you read the direction.
Worked example: a rectangular cycle
A monatomic ideal gas runs the cycle A → B → C → D → A with A (1.0 L, 3.0 × 10⁵ Pa), B (4.0 L, 3.0 × 10⁵ Pa), C (4.0 L, 1.0 × 10⁵ Pa), D (1.0 L, 1.0 × 10⁵ Pa). That is clockwise on a P-V diagram. Use 1 L = 10⁻³ m³.
Net work from the area. Width 3.0 × 10⁻³ m³, height 2.0 × 10⁵ Pa. W = 2.0 × 10⁵ × 3.0 × 10⁻³ = 600 J.
Leg by leg.
A → B, constant pressure expansion: W = 3.0 × 10⁵ × 3.0 × 10⁻³ = +900 J.
B → C, constant volume: W = 0.
C → D, constant pressure compression: W = 1.0 × 10⁵ × (−3.0 × 10⁻³) = −300 J.
D → A, constant volume: W = 0.
Sum: 900 − 300 = 600 J, matching the area.
Heat on each leg. For a monatomic ideal gas, ΔU = (3/2)Δ(PV), so at constant volume Q = (3/2)VΔP, and at constant pressure Q = (5/2)PΔV.
D → A: (3/2)(1.0 × 10⁻³)(2.0 × 10⁵) = +300 J
A → B: (5/2)(3.0 × 10⁵)(3.0 × 10⁻³) = +2250 J
B → C: (3/2)(4.0 × 10⁻³)(−2.0 × 10⁵) = −1200 J
C → D: (5/2)(1.0 × 10⁵)(−3.0 × 10⁻³) = −750 J
Net Q = 300 + 2250 − 1200 − 750 = 600 J, equal to W, as the first law demands. Heat taken in is 2550 J, so this cycle's efficiency is 600 / 2550 ≈ 23.5%.
A second shape: the elliptical loop
If a cycle is drawn as a circle on graph paper, it is an ellipse in physical units. Its area is π × (half the pressure range) × (half the volume range). For pressure from 1.0 to 3.0 × 10⁵ Pa and volume from 1.0 to 3.0 L: W = π × 1.0 × 10⁵ × 1.0 × 10⁻³ ≈ 314 J.
Why do students get this wrong?
In Meltzer's 2004 study of an introductory calculus-based physics course, a large majority of the students interviewed thought the net work done or net heat absorbed in a cyclic process must be zero. The study traces this to over-generalising the state-function idea, which is true of U but not of W or Q.
Which mistakes cost marks?
Setting net work to zero because the gas returns to its start.
Reading the sign without checking which quantity is on which axis.
Forgetting litres to cubic metres, which makes answers 1000 times too large.
Treating heat taken in as net heat when calculating efficiency.
A practice routine
Draw the rectangle above, mark the direction, and shade the area.
Reverse the cycle and redo the leg table. Every sign should flip.
Redo the heat table for a diatomic gas, using (5/2) and (7/2).
Read our posts on thermodynamics sign conventions and heat engines and the Carnot limit next.
When your leg table and the loop area disagree, the Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can find which leg went wrong.
Tomorrow, draw one triangular cycle and find its net work from the area and from the legs.
