The `Delta S` for the vaporisation of 1 mol of water is 88.3 J/mole K. The value of `Delta S` for the condensation of 1 mole of water vapour will be

Options

A. 88.3 J/mol K
B. `(88.3)^(2)` J/mol K
C. `-88.3` J/mol K
D. `(1)/(88.3)` J/mol K

Answer

To solve the problem, we need to understand the relationship between the entropy change (ΔS) for the vaporization and condensation processes of water.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The ΔS for the vaporization of 1 mole of water is given as 88.3 J/mole K.

2. **Understand the Processes:**
- Vaporization is the process where liquid water (H₂O in liquid state) converts to water vapor (H₂O in gaseous state).
- Condensation is the reverse process, where water vapor (H₂O in gaseous state) converts back to liquid water (H₂O in liquid state).

3. **Recognize the Relationship Between ΔS for Vaporization and Condensation:**
- ΔS is a state function, meaning it depends only on the initial and final states, not on the path taken.
- When a process is reversed, the sign of ΔS changes. Therefore, the ΔS for condensation will be the negative of the ΔS for vaporization.

4. **Calculate ΔS for Condensation:**
- Since the ΔS for vaporization is +88.3 J/mole K, the ΔS for condensation will be:
\[ \Delta S_{\text{condensation}} = -\Delta S_{\text{vaporization}} = -88.3 \, \text{J/mole K} \]

5. **Final Answer:**
- The value of ΔS for the condensation of 1 mole of water vapor is **-88.3 J/mole K**.