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: Understanding the Concept:
In a P-V diagram, the work done by the gas in a cycle equals the area enclosed. It is positive for a clockwise cycle and negative for an anticlockwise cycle.

Step 2: Key Formula or Approach:
\[ W = \int P\,dV \]
On a path of constant pressure, \(W = P\,\Delta V\). On a path of constant volume, \(W = 0\).

Step 3: Detailed Explanation:
The figure shows a rectangle with A at \((V_0, P_0)\), B at \((3V_0, P_0)\), C at \((3V_0, 2P_0)\) and D at \((V_0, 2P_0)\). The cycle is taken in the order A to B to C to D to A.
A to B: pressure \(P_0\), volume rises from \(V_0\) to \(3V_0\):
\[ W_{AB} = P_0(3V_0 - V_0) = 2P_0V_0 \]
B to C: constant volume, so \(W_{BC} = 0\).
C to D: pressure \(2P_0\), volume falls from \(3V_0\) to \(V_0\):
\[ W_{CD} = 2P_0(V_0 - 3V_0) = -4P_0V_0 \]
D to A: constant volume, so \(W_{DA} = 0\).
Total:
\[ W = 2P_0V_0 - 4P_0V_0 = -2P_0V_0 \]
This equals minus the area of the rectangle, \(2V_0\times P_0\). Option (A) uses half the area, option (B) has the wrong sign, and option (D) is far too big.

Final Answer:
The work done during the cycle is \(-2P_0V_0\), option (C). \[ \boxed{-2P_0V_0 \text{ (C)}} \]