Question

An ideal gas is taken around ABCA as shown in the P-V diagram. The work done during a cycle is:

• A. \( 2PV \)
• B. \( PV \)
• C. \( \frac{1}{2} PV \)
• D. Zero

Hint:

For a cyclic process in a P-V diagram, the work done is given by the area enclosed. For a triangular cycle, use the area formula \( W = \frac{1}{2} \times \text{Base} \times \text{Height} \).

Answer 1

The Correct Option is A

Step 1: Understanding Work Done in a Cycle - The work done in a closed cycle in a P-V diagram is equal to the area enclosed by the cycle.
- The given process forms a right-angled triangle in the P-V diagram.
Step 2: Calculating the Enclosed Area - The base of the triangle along the V-axis extends from \( V \) to \( 3V \), so the length is: \[ \Delta V = (3V - V) = 2V. \] - The height of the triangle along the P-axis extends from \( P \) to \( 3P \), so the height is: \[ \Delta P = (3P - P) = 2P. \]
Step 3: Using the Area Formula \[ \text{Area of Triangle} = \frac{1}{2} \times \text{Base} \times \text{Height}. \] \[ \text{Work Done} = \frac{1}{2} \times 2V \times 2P. \] \[ W = \frac{4PV}{2} = 2PV. \] Thus, the correct answer is: \[ \boxed{2PV}. \]