Question:

A thermodynamic cycle is impossible if

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The Clausius inequality \( \oint \frac{dQ}{T} \leq 0 \) ensures that a cycle cannot produce a net positive entropy integral, as it would violate the second law.
Updated On: May 6, 2025
  • \( \oint \frac{dQ}{T} = 0 \)
  • \( \oint \frac{dQ}{T}<0 \)
  • \( \oint \frac{dQ}{T}>0 \)
  • \( \oint dS>0 \)
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The Correct Option is C

Solution and Explanation

Step 1: Understand the Clausius inequality.
The Clausius inequality is a fundamental principle in thermodynamics, derived from the second law. It states that for any thermodynamic cycle: \[ \oint \frac{dQ}{T} \leq 0, \] where \( dQ \) is the differential heat transfer, and \( T \) is the absolute temperature at the boundary where the heat transfer occurs. The equality holds for a reversible cycle, and the inequality holds for an irreversible cycle. Step 2: Analyze the implications for a cycle.
For a reversible cycle: \[ \oint \frac{dQ}{T} = 0. \] This is because entropy is a state function, and in a reversible cycle, the net change in entropy over a cycle is zero. For an irreversible cycle: \[ \oint \frac{dQ}{T}<0. \] This reflects the increase in entropy due to irreversibilities, as the second law requires that the total entropy of the system and surroundings must increase or remain the same. Step 3: Determine the impossible condition.
If \( \oint \frac{dQ}{T}>0 \), this violates the Clausius inequality and the second law of thermodynamics. Such a cycle would imply a decrease in the entropy of the universe, which is not possible for any real or theoretical cycle. This condition corresponds to a perpetual motion machine of the second kind, which is impossible. Step 4: Evaluate the options.
(1) \( \oint \frac{dQ}{T} = 0 \): This is possible for a reversible cycle. Incorrect.
(2) \( \oint \frac{dQ}{T}<0 \): This is possible for an irreversible cycle. Incorrect.
(3) \( \oint \frac{dQ}{T}>0 \): This violates the Clausius inequality and is impossible. Correct.
(4) \( \oint dS>0 \): Entropy change over a cycle for the system is zero (\( \oint dS = 0 \)), but this option refers to the entropy integral, which is related to \( \frac{dQ}{T} \). This is not the best way to phrase the condition; the Clausius form is more direct. Incorrect.
Step 5: Select the correct answer.
A thermodynamic cycle is impossible if \( \oint \frac{dQ}{T}>0 \), matching option (3).
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