PV diagrams — how to read work, heat, and process type from graph

Question

A gas undergoes a cyclic process shown on a PV diagram. The cycle consists of: an isobaric expansion from state A to B, an isochoric (constant volume) decrease in pressure from B to C, and an isothermal compression from C back to A. Identify each process, determine the sign of work done in each leg, and find the net work done per cycle if the enclosed area is 500 J.

(JEE Main 2023 pattern)

Solution — Step by Step

On a PV diagram, the shape of the curve tells us the process type:

Horizontal line (constant P) = isobaric process
Vertical line (constant V) = isochoric process
Hyperbola ( $PV = \text{constant}$ ) = isothermal process
Steeper curve ( $PV^\gamma = \text{constant}$ ) = adiabatic process

So A to B (horizontal) is isobaric, B to C (vertical) is isochoric, and C to A (hyperbolic curve) is isothermal.

Work done BY the gas = area under the curve on a PV diagram.

A to B (isobaric expansion): Volume increases, so $W_{AB} = P \Delta V > 0$ (positive work by gas)
B to C (isochoric): Volume is constant, so $W_{BC} = 0$
C to A (isothermal compression): Volume decreases, so $W_{CA} < 0$ (gas is compressed, work done ON gas)

For any cyclic process, the net work equals the area enclosed by the cycle on the PV diagram.

Clockwise cycle = net positive work done by gas. Anticlockwise cycle = net negative work (work done on gas).

Since the cycle goes A $\to$ B $\to$ C $\to$ A, we trace clockwise. Net work = +500 J.

flowchart TD
    A["Read the PV diagram"] --> B{"What is the curve shape?"}
    B -->|"Horizontal line"| C["Isobaric (constant P)"]
    B -->|"Vertical line"| D["Isochoric (constant V)"]
    B -->|"Hyperbola PV=k"| E["Isothermal (constant T)"]
    B -->|"Steep curve PVγ=k"| F["Adiabatic (Q=0)"]
    C --> G["W = P ΔV"]
    D --> H["W = 0"]
    E --> I["W = nRT ln(V₂/V₁)"]
    F --> J["W = (P₁V₁ - P₂V₂)/(γ-1)"]
    G --> K["Net work = enclosed area"]
    H --> K
    I --> K
    J --> K

Why This Works

A PV diagram is essentially a work diagram. Since $W = \int P \, dV$ , the area under any curve on this diagram directly gives the work. When we complete a full cycle and return to the starting state, the net internal energy change is zero ( $\Delta U = 0$ for a cycle). By the first law, $Q_{net} = W_{net}$ , so the enclosed area also equals the net heat absorbed.

The direction of traversal matters because it tells us whether the gas is a heat engine (clockwise, net work output) or a refrigerator/heat pump (anticlockwise, net work input).

Alternative Method — Calculate Each Leg Separately

If numerical values are given for each state (P, V, T), calculate the work for each leg individually:

W_{net} = W_{AB} + W_{BC} + W_{CA}

For isobaric: $W = P(V_2 - V_1)$ . For isochoric: $W = 0$ . For isothermal: $W = nRT \ln(V_2/V_1)$ .

Add them up. This must equal the enclosed area on the graph.

Quick sign check: if the cycle goes clockwise on a PV diagram, the net work must be positive. If your calculation gives a negative net work for a clockwise cycle, recheck your signs — something went wrong in one of the legs.

Common Mistake

Students often confuse “work done BY the gas” with “work done ON the gas.” In physics (especially JEE and NEET), the convention is $W = \int P \, dV$ which gives work done by the gas. If volume increases, work by gas is positive. If someone asks for “work done on the gas,” just flip the sign. Mixing up these conventions leads to sign errors in first law calculations: $\Delta U = Q - W_{by}$ or $\Delta U = Q + W_{on}$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Calculate Each Leg Separately

Common Mistake

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