Question

An ideal gas is taken through a cyclic process ABCA shown in a P–V diagram. The net work done by the gas during the complete cycle is:

• A. \( 4 P_0 V_0 \)
• B. \( 2 P_0 V_0 \)
• C. \( P_0 V_0 \)
• D. \( 6 P_0 V_0 \)

Hint:

For cyclic processes, never integrate blindly—directly use enclosed area. Clockwise cycle ⇒ positive work, anticlockwise ⇒ negative work.

Answer 1

The Correct Option is B

Concept: In a P–V diagram, the net work done by a gas in a cyclic process is equal to the area enclosed by the cycle: \[ W_{\text{net}} = \oint P \, dV \] For a closed loop, this equals the geometric area enclosed.

Step 1: Nature of cycle The given cycle ABCA forms a triangular loop in the P–V plane. From the diagram description:

• Volume varies from \(V_0\) to \(3V_0\)

• Pressure varies from \(P_0\) to \(3P_0\)
So: \[ \text{Base} = 3V_0 - V_0 = 2V_0 \] \[ \text{Height} = 3P_0 - P_0 = 2P_0 \]

Step 2: Area of triangle \[ W_{\text{net}} = \frac{1}{2} \times \text{Base} \times \text{Height} \] \[ W_{\text{net}} = \frac{1}{2} \times (2V_0) \times (2P_0) \]

Step 3: Final result \[ W_{\text{net}} = 2 P_0 V_0 \] Thus, the net work done is: \[ \boxed{2 P_0 V_0} \]