A) \[\Delta S\] (for process I) = 0, while \[\Delta S\] (for process II) \[\ne 0\]
B) \[{{q}_{cycle}}=0\] for process I and \[{{q}_{cycle}}\ne 0\] for process II
C) more heat can be converted to work in process I than in process II
D) more work can be converted to heat in process I than in process II
Correct Answer: B
Solution :
From clausius inequality, \[ds\ge \frac{dq}{T}\] For reversible process, \[\oint\limits_{cycle}{ds}\ge \frac{{{q}_{cycle}}}{T}\] \[0\ge \frac{{{q}_{cycle}}}{T}\] \[{{q}_{cycle}}\le 0\] So for reversible process, \[{{q}_{cycle}}=0\] and for irreversible process, \[{{q}_{cycle}}<0\]
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