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A fixed mass of oxygen gas performs a cyclic process ABCA as shown in the figure. Efficiency of the cycle will be

Options

  1. A3 ln 3 - 2 5 + 3 ln 3
  2. B3 ln 4 - 2 4 + 5 ln 3
  3. C3 ln 4 - 3 4 + 5 ln 3
  4. D3 ln 3 - 1 6 + 3 ln 4

Correct answer

A. 3 ln 3 - 2 5 + 3 ln 3

Step-by-step solution

A  →  B isochoric as P is directly proportional to T W AB = 0 Q AB = n C v 3 T   - T   = 5 2 nR 2 T   = 5 nRT   B   →   C   isothermal and thus Q   =   W as there won't be any change in internal energy. W BC = nR 3 T   ln P B P C = 3 nRT   ln 3 = 3 nRT   ln 3 = Q BC C   → A   isobaric process W CA = P V A - V C = nR T A - 3 T   = - 2 n RT   Efficiency of a cyclic process is given by η = Work done in

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