Work Done in Cyclic Process from PV Diagram

 

❓ Question

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

Work done by the gas in the entire cycle is:

---

🖼 Question Image

---

✍️ 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

---

🔷 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.

---

🔷 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

---

🔷 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 ⚠️

---

🔷 Step 4 — Final Calculation

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

(Substitute values from graph)

---

✅ Final Answer

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

---

⭐ Golden JEE Insight

In cyclic process:

$$\Delta U = 0$$

So from First Law:

$$Q = W$$

🔥 That means net heat supplied equals net work done.