Work Done in Cyclic Process — PV Diagram Trick 🔥

 

❓ Question

An ideal gas undergoes a cyclic process as shown in the figure.

Work done by the gas in the entire cycle is:

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🖼 Question Image

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✍️ Short Solution

This is a PV diagram cycle question.

Key Idea:

$$\textbf{Work done in one complete cycle = Area enclosed in PV graph}$$

Direction matters:

• Clockwise → Work positive

• Anticlockwise → Work negative

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🔷 Step 1 — MOST IMPORTANT RULE 💯

For cyclic process:

$$W = \oint P\, dV$$

Graphically:

$$\boxed{W = \text{Area enclosed by the cycle}}$$

No need to calculate individual paths unless asked separately.

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🔷 Step 2 — Identify Shape of Cycle

Look carefully at the diagram:

• If rectangle → Area = $\Delta P \times \Delta V$
  

• If triangle → Area = $\frac{1}{2} \times \text{base} \times \text{height}$
  

• If irregular → Break into simple shapes

👉 Base = change in Volume
👉 Height = change in Pressure

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🔷 Step 3 — Check Direction of Cycle

• Clockwise loop → Gas expands at higher pressure

  $$W > 0$$
  

• Anticlockwise loop → Gas compressed at higher pressure

  $$W < 0$$
  

JEE loves this sign trap ⚠️

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🔷 Step 4 — Final Calculation

$$W = \text{Area enclosed}$$

(Substitute values from graph)

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✅ Final Answer

$$\boxed{W = \text{Area of enclosed region (with proper sign)}}$$

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⭐ Golden JEE Insight

In cyclic process:

$$\Delta U = 0$$

So from First Law:

$$Q = W$$

🔥 That means net heat supplied equals net work done.