The magnitude of heat exchanged by a system for the given cyclic process ABCA (as shown in figure) is (in SI unit) :

Solution

Physics sir · 4 steps

The cycle closes along line CA, so the heat is just the area of the top half of the loop.

1. Heat equals work. In a cycle the gas returns to its starting state, so Δ U = 0. By the first law, Q = W. The work equals the area enclosed by the loop on the P–V graph.
2. Which area. The path goes from A up through B to C along the arc. It returns from C to A along the straight line at 300 kPa. So the loop is only the upper half of the curve, and the lower part through D is never traced.
3. Semi-axes in SI. Along V, the half-width is 100 cc = 100 × 10⁻⁶ m³ = 10⁻⁴ m³. Along P, the half-height is 100 kPa = 10⁵ Pa. In SI units the curve is an ellipse, so its full area is π a b.

   |Q| = 1/2π a b = 1/2π × 10⁻⁴ × 10⁵ = 5π J

4. So. The magnitude of heat exchanged in the cycle is 5π J, option (c). The loop runs anticlockwise, so the net work is negative, but its magnitude is still 5π.

The idea to keep. For any cycle, |Q| equals the area enclosed by the path actually traced. Check which part of the curve the arrows follow, and convert both axes to SI before you multiply.

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