Question

An ideal monoatomic gas is taken around the cycle ABCDA as shown in the P.V diagram. The work done during the cycle is

• A. \(-P_0V_0\)
• B. \(2P_0V_0\)
• C. \(-2P_0V_0\)
• D. \(6P_0V_0\)

Hint:

Work in a cycle equals the signed area enclosed. The direction ABCDA decides the sign.

Answer 1

The Correct Option is C

Step 1: Area of the loop:
The rectangle spans a volume width $3V_0 - V_0 = 2V_0$ and a pressure height $2P_0 - P_0 = P_0$. Its area is $2V_0\cdot P_0 = 2P_0V_0$.

Step 2: Sign from the direction:
The cycle A, B, C, D runs from bottom left to bottom right, up to top right, then to top left. That is anticlockwise on the P-V plane. In an anticlockwise cycle the compression happens at the higher pressure, so the net work done by the gas is negative.

Step 3: Result:
$W = -2P_0V_0$.

Step 4: Cross-check:
Expansion on the low pressure side gives $+2P_0V_0$, and compression on the high pressure side gives $-4P_0V_0$. The sum is $-2P_0V_0$.

Final Answer:
Option (C). \[ \boxed{-2P_0V_0 \text{ (C)}} \]