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 `ΔS_(sys) + ΔS_(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 Components**:
- In this question, we have two components:
- ΔS_(sys): Change in entropy of the system.
- ΔS_(surr): Change in entropy of the surroundings.

3. **Apply the Second Law of Thermodynamics**:
- The second law states that for any process, the change in entropy of the universe (which includes both the system and the surroundings) must be greater than or equal to zero for spontaneous processes.
- Mathematically, this can be expressed as:
\[ ΔS_(universe) = ΔS_(sys) + ΔS_(surr) \geq 0 \]

4. **Consider Reversible Processes**:
- For reversible processes, the entropy change of the universe is exactly zero:
\[ ΔS_(universe) = ΔS_(sys) + ΔS_(surr) = 0 \]
- This implies that the change in entropy of the system and the surroundings must balance each other out.

5. **Conclude the Result**:
- Therefore, for a reversible process:
\[ ΔS_(sys) + ΔS_(surr) = 0 \]

### Final Answer:
In a reversible process, the value of `ΔS_(sys) + ΔS_(surr)` is **0**.

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