Question

A closed system undergoing a thermodynamic cycle consisting of two reversible isothermal and two reversible adiabatic processes is shown in the following figure. If \(\delta Q\) is the infinitesimal heat transfer and T is the instantaneous temperature, then the value of the contour integral \( \oint \frac{\delta Q}{T} \) is 

• A. is positive
• B. is negative
• C. is zero
• D. cannot be determined

Hint:

Remember the Clausius theorem: \( \oint \frac{\delta Q}{T} = 0 \) for a reversible cycle and \( \oint \frac{\delta Q}{T}<0 \) for an irreversible cycle. Since the problem specifies all processes are reversible, the integral must be zero.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The question asks for the value of the cyclic integral of \( \frac{\delta Q}{T} \) for a specific thermodynamic cycle. The cycle described, consisting of two reversible isothermal processes and two reversible adiabatic processes, is the definition of a Carnot cycle.
Step 2: Key Formula or Approach:
The integral \( \oint \frac{\delta Q}{T} \) is related to the change in entropy of the system. The Clausius inequality states that for any thermodynamic cycle: \[ \oint \frac{\delta Q}{T} \le 0 \] The equality holds for a reversible cycle, and the inequality holds for an irreversible cycle. The problem states that the cycle consists of four reversible processes (reversible isothermal and reversible adiabatic). Therefore, the entire cycle is reversible.
Step 3: Detailed Analysis:
For any reversible cycle, the Clausius theorem states that the cyclic integral of \( \frac{\delta Q}{T} \) is exactly zero. \[ \oint_{\text{rev}} \frac{\delta Q}{T} = 0 \] This is because entropy (S) is a state function. For a reversible process, \( dS = \frac{\delta Q_{rev}}{T} \). The integral over a complete cycle represents the total change in entropy from the start point back to the same start point. Since entropy is a state function, its net change over a cycle must be zero. \[ \Delta S_{cycle} = \oint dS = \oint \frac{\delta Q_{rev}}{T} = 0 \] Let's also analyze the specific cycle shown (a Carnot cycle):
- Process 4 \(⇒\) 3: Reversible isothermal heat addition at \(T_H\). \( \int \frac{\delta Q}{T} = \frac{Q_H}{T_H} \).
- Process 3 \(⇒\) 2: Reversible adiabatic expansion. \(\delta Q = 0\), so \( \int \frac{\delta Q}{T} = 0 \).
- Process 2 \(⇒\) 1: Reversible isothermal heat rejection at \(T_L\). \( \int \frac{\delta Q}{T} = \frac{Q_L}{T_L} \) (note \(Q_L\) is negative).
- Process 1 \(⇒\) 4: Reversible adiabatic compression. \(\delta Q = 0\), so \( \int \frac{\delta Q}{T} = 0 \).
The total integral is \( \oint \frac{\delta Q}{T} = \frac{Q_H}{T_H} + \frac{Q_L}{T_L} \). For a Carnot cycle, it is a fundamental property that \( \frac{Q_H}{T_H} = -\frac{Q_L}{T_L} \), so their sum is zero.
Step 4: Why This is Correct:
Because the cycle is composed entirely of reversible processes, it is a reversible cycle. According to the Clausius theorem, the cyclic integral of \( \frac{\delta Q}{T} \) for any reversible cycle is zero. This reflects the fact that entropy is a property of state.