JEE Mains Physics · Thermodynamics

Cyclic Processes

A cycle returns the gas to its starting state, so ΔU = 0 and the net heat absorbed equals the net work done, which is the area enclosed on the P–V diagram.

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Why this matters

Twelve PYQs, eleven of them multiple choice and two from 2026, and eleven come with a diagram of the cycle. Six read the net work off the enclosed area: a triangle, a circle or ellipse, or a region bounded by a curve. Six work leg by leg, adding isothermal, isobaric and adiabatic works, or using the heat in each leg to find the one that is missing.

Concept 1 of 2: Net work of a cycle as the enclosed area

A cycle ends where it began, so ΔU = 0 and the net heat absorbed equals the net work done. On a P–V diagram that work is the area inside the loop. Going clockwise, the gas expands at high pressure and is compressed at low pressure, so it does net positive work and absorbs net heat. Going anticlockwise, both are negative.

Definition

$\Delta U_{\text{cycle}} = 0$ , so $Q_{\text{net}} = W_{\text{net}} =$ the area enclosed on the P–V diagram.
With P up and V across: clockwise means W > 0 and heat absorbed; anticlockwise means W < 0 and heat rejected.
Triangle: $\tfrac{1}{2}\,\Delta V\,\Delta P$ . Rectangle: $\Delta V\,\Delta P$ .
Circle or ellipse: $\pi ab$ , with a the semi-axis along V (in m³) and b the semi-axis along P (in Pa): half of each full width.
A curved side: integrate $\int P\,dV$ along it, then add the straight legs.
If a diagram puts V up and P across, the direction that gives positive work is reversed.

Net work of a cycle

W_{\text{net}} = Q_{\text{net}} = \oint P\,dV \qquad W_{\text{ellipse}} = \pi ab

Worked example

A gas is taken anticlockwise round a rectangle on a P–V diagram whose sides run from 2 L to 5 L and from 100 kPa to 300 kPa. Find the net work done by the gas and the net heat.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q22Moderate

Example 1 · Thermodynamics · Cyclic Processes

A thermodynamic system is taken through the cyclic process ABC as shown in the figure. The total work done by the system during the cycle ABC is

\_\_\_\_

Half the width, not the width

πab uses semi-axes. Using the full width on one axis doubles the area, and on both axes makes it four times too big.

The direction sets the sign

The area gives only the size of the work. Clockwise on a P–V diagram is positive work by the gas; anticlockwise is negative.

Drill 5 more on net work of a cycle as the enclosed area

Concept 2 of 2: A cycle worked leg by leg

When a cycle is built from standard processes, find the work on each leg with that process's formula and add them. Pressures and volumes at the corners come from the process rules: PV is fixed on an isotherm, P on an isobar, V on an isochor. If heats are given instead, use net heat = net work, or ΔU summing to zero, to find the leg that is missing.

Definition

Isothermal: $nRT\ln\dfrac{V_2}{V_1}$ . Isobaric: $P\Delta V$ . Isochoric: 0. Adiabatic: $\dfrac{P_1V_1 - P_2V_2}{\gamma - 1}$ .
Find the corner pressures first: after an isotherm, $P_2 = \dfrac{P_1V_1}{V_2}$ .
Over a cycle $\sum W = \sum Q$ and $\sum \Delta U = 0$ .
On each leg $Q = \Delta U + W$ ; on an adiabatic leg $\Delta U = -W$ .
In a cycle of two isotherms and two adiabats, the two adiabatic works cancel, since each depends only on the same temperature change.

Net work leg by leg

W_{\text{net}} = \sum_{\text{legs}} W_i = \sum_{\text{legs}} Q_i

Worked example

A monoatomic ideal gas goes round a cycle ABCA. AB is at constant volume and absorbs 600 J. BC is an adiabatic expansion. CA, at constant pressure, brings it back to A and rejects 500 J. Find the work on each leg and the net work.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q80Moderate

Example 2 · Thermodynamics · Cyclic Processes

An ideal gas exists in a state with pressure

P_{0}

, volume

V_{0}

. It is isothermally expanded to 4 times of its initial volume

\left( V_{0} \right)

, then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is

Find the corner pressure before using PΔV

After an isothermal expansion to k times the volume, the pressure is P/k. An isobaric leg that follows runs at that new pressure, not at the starting one.

The volume ratio in the right order

Isothermal work by the gas is nRT ln(V_final/V_initial), which is negative for a compression. Writing the ratio upside down flips the sign.

Drill 5 more on a cycle worked leg by leg

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Net work of a cycle as the enclosed area
Net work of a cycle
$W_{\text{net}} = Q_{\text{net}} = \oint P\,dV \qquad W_{\text{ellipse}} = \pi ab$
A cycle worked leg by leg
Net work leg by leg
$W_{\text{net}} = \sum_{\text{legs}} W_i = \sum_{\text{legs}} Q_i$

Watch out for (4)

Half the width, not the width
→ Net work of a cycle as the enclosed area
The direction sets the sign
→ Net work of a cycle as the enclosed area
Find the corner pressure before using PΔV
→ A cycle worked leg by leg
The volume ratio in the right order
→ A cycle worked leg by leg

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