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The Delta S for the vaporisation of 1 mo...

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

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

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To solve the problem, we need to understand the relationship between the entropy change (ΔS) for vaporization and condensation processes. ### Step-by-Step Solution: 1. **Identify the Given Information**: - The entropy change (ΔS) for the vaporization of 1 mole of water is given as 88.3 J/mol·K. 2. **Understand the Processes**: - Vaporization is the process where liquid water (H2O) converts into water vapor (H2O gas). - Condensation is the reverse process, where water vapor (H2O gas) converts back into liquid water (H2O). 3. **Relation Between ΔS for Vaporization and Condensation**: - The entropy change for the condensation process is the negative of the entropy change for the vaporization process. This is because condensation is the reverse of vaporization. - Mathematically, this can be expressed as: \[ \Delta S_{condensation} = -\Delta S_{vaporization} \] 4. **Calculate ΔS for Condensation**: - Using the value provided for ΔS of vaporization: \[ \Delta S_{condensation} = -88.3 \, \text{J/mol·K} \] 5. **Final Answer**: - The value of ΔS for the condensation of 1 mole of water vapor is **-88.3 J/mol·K**.
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