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NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice

Heat absorbed by a system in going through a cyclic process shown in figure is

Options

  1. A1 0 7 π J
  2. B1 0 6 π J
  3. C1 0 2 π J
  4. D1 0 4 π J

Correct answer

C. 1 0 2 π J

Step-by-step solution

We know that, Δ u = 0 in cyclic process So Δ Q = Δ u + Δ w Δ Q = Δ w We know, work done = Area under P-V curve = Area of circle = π r 2 4 = π 4 × ( 30 - 10 ) × 1 0 3 ( 30 - 10 ) × 1 0 - 3 = π 4 × 20 × 20 = π 4 × 400 = 100 π = 1 0 2 π J So Heat energy absorbed Δ Q = Δ w = 1 0 2 π J

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