The answer in Arihant's book is given as (b), but wouldn't (c) be correct as well? This is a question of IIT JAM Chemistry.

A system undergoes two cyclic processes 1 and 2. Process 1 is reversible and Process 2 is irreversible. The correct statement relating to the two processes is
(a) $\delta S=0$ for Process 1, while $\Delta S\neq0$ for Process 2
(b) $Q_\text{cyclic}=0$ for Process 1 and $Q_\text{cyclic}\neq0$ for Process 2
(c) More heat can be converted to work in Process 1 than in Process 2
(d) More work can be converted to heat in Process 1 than in Process 2

  • $\begingroup$ Yes, you are correct $\endgroup$ – Yusuf Hasan Dec 19 '18 at 7:45

Answer b is incorrect for the reversible Carnot cycle. The book has given the wrong answer. Answer c is correct if only one cycle of each process is carried out. You can obviously convert more heat to work if you run process 2 through the same cycle several times, which also counts as a cyclic process.

  • $\begingroup$ Well Sir, it's given that from Clausius inequality, for reversible process, q(cycle)=0 and for irreversible process , q(cycle) less than 0 . I agree with you on the latter. $\endgroup$ – user67074 Dec 19 '18 at 15:13
  • $\begingroup$ That is NOT what the Clausius inequality says. Are you saying that you think that Qcyclic is zero for the reversible Carnot cycle. $\endgroup$ – Chet Miller Dec 19 '18 at 18:08
  • $\begingroup$ So, what do you think? Do you still stand by what you said about the Clausius inequality indicating that, for a reversible process q(cycle) = 0? $\endgroup$ – Chet Miller Dec 20 '18 at 13:15
  • $\begingroup$ Yes sir, actually this answer given in the book got me a bit confused. I understood that q(cyclic) is not 0 for a reversible cyclic process. But, would q(cyclic) for an irreversible process be less than q(cyclic) for reversible process? $\endgroup$ – user67074 Dec 20 '18 at 14:11
  • $\begingroup$ Depending on the reversible path selected, it could be less or it could be more. But, the change in entropy for any reversible cyclic path is zero. Please understand that I am asserting that, contrary to what your book says, answer b is incorrect. $\endgroup$ – Chet Miller Dec 20 '18 at 14:36

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