A thermodynamic system is taken through cyclic process. The total work done in the process is :

Solution

Physics sir · 3 steps

In a cycle, the work done is the area enclosed by the loop, and here that loop is a triangle.

1. Read the corners. From the graph: C = (2 m³, 100 Pa), D = (2 m³, 400 Pa), E = (4 m³, 100 Pa). The path goes C → D → E → C, which is clockwise, so the net work done by the gas is positive.

   Segment	Process	Work by gas
   C → D	volume fixed	0
   D → E	expansion	1/2(400 + 100) × 2 = +500 J
   E → C	compression at 100 Pa	100 × (2 - 4) = -200 J

2. Check with area. The enclosed triangle has base 4 - 2 = 2 m³ and height 400 - 100 = 300 Pa.

   W = 1/2 × 2 m³ × 300 Pa = 300 J

3. So. Adding the segments gives 0 + 500 - 200 = 300 J, the same as the enclosed area. The total work done is 300 J, option (c).

The idea to keep. For any cyclic process on a P–V graph, net work equals the area enclosed by the loop. Clockwise means positive work by the gas, anticlockwise means negative.

Answer from the NTA final answer key. Solution by PrepWiser.