In a reversible process, the value of `Delta S _(sys) + Delta S_(surr) is`

Options

A. `gt 0`
B. `lt 0`
C. `= 0`
D. All of these

Answer

To solve the question regarding the value of \( \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \) in a reversible process, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Entropy**:
- Entropy (\( S \)) is a measure of the disorder or randomness in a system. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time.

2. **Identify the Reversible Process**:
- A reversible process is an idealized process that occurs infinitely slowly, allowing the system to remain in equilibrium at all times. In such processes, the system can be returned to its initial state without any net change in the surroundings.

3. **Apply the Second Law of Thermodynamics**:
- According to the second law, for any process, the change in entropy of the universe (\( \Delta S_{\text{universe}} \)) is given by:
\[ \Delta S_{\text{universe}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \]
- For irreversible processes, \( \Delta S_{\text{universe}} > 0 \) (the entropy of the universe increases).
- For reversible processes, \( \Delta S_{\text{universe}} = 0 \) (the entropy of the universe remains constant).

4. **Set Up the Equation for Reversible Process**:
- Since we are dealing with a reversible process, we can write:
\[ \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = 0 \]

5. **Conclusion**:
- Therefore, in a reversible process, the value of \( \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \) equals zero.

### Final Answer:
\[ \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = 0 \]