In the following p-V diagram the equation of state along the curved path is given by (V-2)²=4ap where a is a constant. The total work done in the closed path is

Solution

Physics sir · 4 steps

Find the pressure at A and C from the curve, then add the work on each leg; this loop runs anticlockwise.

1. Pressures at corners. On the curve, p = (V-2)²/4a. At V = 1 and at V = 3 this gives p = 1/4a, so the top line AC sits at p = 1/4a. At B (V = 2) the pressure is zero.
2. Curved leg ABC. The gas expands from V = 1 to V = 3 along the curve, so this work is positive. Put u = V - 2, so u runs from -1 to 1.

   W_(ABC) = ∫₁³ (V-2)²/4a dV = 1/4a∫₋₁¹ u² du = 1/4a×2/3 = 1/6a

3. Straight leg CA. The gas is compressed at constant pressure 1/4a from V = 3 to V = 1. So W_(CA) = 1/4a×(1 - 3) = -1/2a.
4. So. W = 1/6a - 1/2a = -1/3a. The size is the area enclosed by the loop, and the sign is negative because the cycle runs anticlockwise. The answer is (d).

The idea to keep. Net work in a cycle equals the area enclosed on the p-V diagram. A clockwise loop gives positive work and an anticlockwise loop gives negative work.

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