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

Step 1: Understand work in a cycle.
An ideal gas goes around a closed loop ABCA on a pressure volume diagram. The net work done in one full cycle equals the area enclosed by the loop.

Step 2: Write the work rule.
The work done is the area under the path, and for a closed loop it is the area inside: \[ W_{net} = \oint P\,dV = \text{area of the loop} \]

Step 3: Find the shape of the loop.
The cycle ABCA forms a triangle on the diagram. The volume goes from $V_0$ to $3V_0$ and the pressure goes from $P_0$ to $3P_0$.

Step 4: Find the base of the triangle.
The base lies along the volume axis: \[ \text{base} = 3V_0 - V_0 = 2V_0 \]

Step 5: Find the height of the triangle.
The height lies along the pressure axis: \[ \text{height} = 3P_0 - P_0 = 2P_0 \]

Step 6: Compute the area.
The area of a triangle is half the base times the height: \[ W_{net} = \frac{1}{2}\times 2V_0 \times 2P_0 = 2 P_0 V_0 \] So the net work done by the gas in one cycle is two times $P_0 V_0$. \[ \boxed{2 P_0 V_0} \]