The heat absorbed by a system in going through the given cyclic process is :

Solution

Physics sir · 4 steps

In a full cycle Δ U = 0, so heat absorbed equals the area of the loop: just find the circle's area.

1. Cycle means ΔU zero. The gas returns to its starting state, so Δ U = 0. The first law then gives Q = W, and the work in a cycle is the area enclosed on the P-V diagram.
2. Sign of work. The arrow runs clockwise with volume on the horizontal axis, so the net work done by the gas is positive. The system absorbs heat. The axis labels in the figure have their units swapped, but the area is still pressure in kPa times volume in cc.
3. Find the radii. Pressure goes from 60 to 340 kPa, so the half-width along P is 140 kPa = 1.4 × 10⁵ Pa. Volume goes from 60 to 340 cc, so the half-width along V is 140 cc = 1.4 × 10⁻⁴ m³.

   Q = W = π (Δ P/2)(Δ V/2) = π × (1.4 × 10⁵) × (1.4 × 10⁻⁴) = 19.6π J

4. So. Q = 19.6 × 3.14 ≈ 61.6 J, which is option (a).

The idea to keep. For any cyclic process, heat absorbed equals net work, the area inside the loop. Clockwise on a P-V graph means work by the gas is positive. Convert kPa and cc to SI before multiplying.

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