An ideal gas is taken through the cycle \( A\rightarrow B\rightarrow C\rightarrow A \) as shown in figure. If the net heat supplied to the gas in the cycle is 10 J, the work done in the process \( C\rightarrow A \) is:
Show Hint
The net work done during a cyclic process on a P-V diagram is equal to the area enclosed by the loop. Since the cycle goes counter-clockwise here, the net work must be negative, which serves as a quick sanity check for your signs.
Step 1: Use the first law for a full cycle. In a complete cycle the gas returns to its starting state, so its internal energy does not change overall. The first law then says the total heat supplied equals the total work done: \[ Q_{net} = W_{net} = W_{AB} + W_{BC} + W_{CA} \] Step 2: Find the work in the first step A to B. Here the pressure is constant at $20\ \text{Nm}^{-2}$ and the volume grows from 1 to 2. Work at constant pressure is pressure times change in volume: \[ W_{AB} = P\,\Delta V = 20(2 - 1) = 20\ \text{J} \] Step 3: Find the work in the step B to C. Here the volume stays fixed while the pressure drops. No volume change means no work: \[ W_{BC} = 0 \] Step 4: Bring in the given net heat. We are told the net heat for the cycle is $10$ J: \[ 10 = W_{AB} + W_{BC} + W_{CA} \] Step 5: Put in the known works. \[ 10 = 20 + 0 + W_{CA} \] Step 6: Solve for the last work. \[ W_{CA} = 10 - 20 = -10\ \text{J} \] \[ \boxed{W_{CA} = -10\ \text{J}} \]
