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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 cyclic process Heat supplied to the system = W Q in Efficiency η = 3 nRT A ln 3 - 2 nRT A 5 nRT A + 3 nRT A ln 3 = 3 ln 3 - 2 5 + 3 ln 3

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