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: \[ \boxed{2PV}. \]

Answer 2

To determine the work done during the cycle ABCA of an ideal gas shown in the P-V diagram, we must calculate the net work done over one complete cycle. In a P-V diagram, work done is represented by the area enclosed by the cycle path.

Assuming a rectangular path ABCA where:

• A to B: Isobaric expansion at pressure \( P \), changing volume from \( V_1 \) to \( V_2 \).
• B to C: Isochoric process (constant volume).
• C to A: Isobaric compression at pressure \( P \), changing volume from \( V_2 \) to \( V_1 \).
• A to A: Isochoric process (completing the cycle).

The work done during the isochoric processes (B to C and A to A) is zero because the volume does not change. Therefore, the net work done is the area enclosed by the rectangle.

Area is calculated as:
\[\text{Area} = \text{Width} \times \text{Height} = (V_2 - V_1) \times P\]

Since the width of the rectangle (change in volume \( V_2 - V_1 \)) is from \( V_1 \) to \( V_2 \) and the rectangle height is \( 2P \) (as it spans from P and drops back to 0), the work done for one complete cycle is twice the initial estimation:

\[W = P \times (V_2 - V_1) \times 2 = 2PV\]

This is the total work done over one cycle in the P-V diagram, thus the correct answer is: 2PV.