Question

John and Jane independently performed a thermodynamic experiment, in which \(\mathbf{X}\) and \(\mathbf{Y}\) represent the initial and final thermodynamic states of the system, respectively. John performed the experiment under reversible conditions, for which the change in entropy of the system was \(\Delta S_{\text{rev}}\). Jane performed the experiment under irreversible conditions, for which the change in entropy of the system was \(\Delta S_{\text{irr}}\). Which one of the following relationships is CORRECT?

• A. \(\Delta S_{\text{rev}} = \Delta S_{\text{irr}}\)
• B. \(\Delta S_{\text{rev}}>\Delta S_{\text{irr}}\)
• C. \(\Delta S_{\text{rev}}<\Delta S_{\text{irr}}\)
• D. \(\Delta S_{\text{rev}} = 2\Delta S_{\text{irr}}\)

Hint:

State functions (e.g., \(S, U, H, G\)) have path-independent changes.
Clausius inequality \(\oint \frac{\delta q}{T} \le 0\) distinguishes reversibility via entropy \emph{generation} of the universe, not the system’s \(\Delta S\) between the same states.

Answer 1

The Correct Option is A

Step 1: Entropy \(S\) is a state function. Therefore, for fixed initial and final states \(\mathbf{X} \rightarrow \mathbf{Y}\), the change in entropy of the system depends only on the states, not on the path (reversible or irreversible).

Step 2: Hence, \[ \Delta S_{\text{system}} = S(\mathbf{Y}) - S(\mathbf{X}) \] is unique, giving \[ \Delta S_{\text{rev}} = \Delta S_{\text{irr}}. \] 
Note: The heat transfer term and the inequality \(\Delta S_{\text{univ}} \geq 0\) concern the universe, not the system’s state change.

Final Answer: \[ \boxed{\;\; \Delta S_{\text{rev}} = \Delta S_{\text{irr}} \;\;} \]