Question

P(atm)

V(Litre)

Answer 1

The net work done during the cycle is approximately $1 + \pi$ L atm or 419.6 J, assuming the curved path A-B-C is a semi-ellipse.

Answer 2

The problem presents a P-V diagram of a thermodynamic cycle. To find the net work done, we calculate the area enclosed by the cycle.

1. Work Done in Straight Line Segments:

   • C -> D: $W_{C \to D} = 1.5$ L atm
   • D -> A: $W_{D \to A} = 1.5$ L atm

2. Area Under the Lower Path (C-D-A):

   • Total area = $1.5 + 1.5 = 3$ L atm

3. Approximation of the Curved Path (A-B-C):

   • Assuming the curve A-B-C is a semi-ellipse with center at (3, 2), semi-major axis of 2 (along the P-axis), and semi-minor axis of 1 (along the V-axis).
   • The area enclosed by a semi-ellipse is $A = \frac{1}{2} \pi a b = \frac{1}{2} \pi (1)(2) = \pi$.

4. Net Work Calculation:

   • The area under the semi-ellipse (A-B-C) from V=2 to V=4 is then calculated as: $\int_{2}^{4} P_{upper} dV = 4 + \pi$.
   • $W_{cycle} = \int_{2}^{4} P_{upper} dV - 3$.
   • $W_{cycle} = (4 + \pi) - 3 = 1 + \pi \approx 4.14$ L atm.

5. Conversion to Joules:

   • $1 \text{ L atm} = 101.325 \text{ J}$
   • $W_{cycle} = (1 + \pi) \times 101.325 \approx 419.6 \text{ J}$

Thus, the net work done during the cycle is approximately $1 + \pi$ L atm or 419.6 J, assuming the curved path A-B-C is a semi-ellipse. The positive value indicates that the cycle is clockwise, and the system performs work on the surroundings.