Question

A thermodynamic system is taken through the cycle PQRSP process. The net work done by the system is

• A. 20 J
• B. -20 J
• C. 400 J
• D. -374 J

Hint:

Negative work means work done on the system.

Answer 1

The Correct Option is B

To determine the net work done by the system in the cycle PQRSP, we need to analyze the given pressure-volume (P-V) diagram. The work done by the system in a thermodynamic cycle is the area enclosed by the cycle on a P-V diagram.

The cycle PQRSP is a rectangular loop in the P-V diagram. We will calculate the work done during each segment of the cycle:

1. The segment PQ is an isobaric expansion at 100 kPa from 100 cc to 300 cc.

The work done in an isobaric process is given by:

\(W_{PQ} = P \times \Delta V\)

Substituting the values:

\(W_{PQ} = 100 \, \text{kPa} \times (300 - 100) \, \text{cc} = 100 \times 200 \, \text{cc} = 20 \, \text{J}\) (since 1 cc = 1 mL = 0.001 L and 1 kPa = 0.001 J/cc)

1. The segment QR is an isochoric process, where volume remains constant, so no work is done: \(W_{QR} = 0\)
2. The segment RS is an isobaric compression at 200 kPa from 300 cc to 100 cc.

The work done during compression is:

\(W_{RS} = P \times \Delta V = 200 \, \text{kPa} \times (100 - 300) \, \text{cc} = 200 \times (-200) \, \text{cc} = -40 \, \text{J}\)

1. The segment SP is also an isochoric process, hence \(W_{SP} = 0\)

Summing up the work done in each segment, we get the net work done by the system:

\(W_{\text{net}} = W_{PQ} + W_{QR} + W_{RS} + W_{SP} = 20 + 0 - 40 + 0 = -20 \, \text{J}\)

Therefore, the correct answer is -20 J.