Entropy change involve in conversation of `1`mole of liquid water at `373K`to vapour at the same temperature (latent heat of vaporisation of water=`2.257kJg^(-1))`

Options

A. `105.9 JK^(-1)`
B. `107.9 JK^(-1)`
C. `108.9 JK^(-1)`
D. `109.9 JK^(-1)`

Answer

To find the entropy change involved in the conversion of 1 mole of liquid water at 373 K to vapor at the same temperature, we can follow these steps:

### Step 1: Understand the formula for entropy change
The entropy change (ΔS) for the vaporization process can be calculated using the formula:
\[ \Delta S = \frac{\Delta H_{\text{vap}}}{T} \]
where ΔH_vap is the enthalpy change (latent heat of vaporization) and T is the temperature in Kelvin.

### Step 2: Convert latent heat of vaporization to kJ per mole
The latent heat of vaporization of water is given as 2.257 kJ/g. To convert this to kJ per mole, we need to multiply by the molar mass of water (18 g/mol):
\[ \Delta H_{\text{vap}} = 2.257 \, \text{kJ/g} \times 18 \, \text{g/mol} = 40.626 \, \text{kJ/mol} \]

### Step 3: Substitute values into the entropy change formula
Now that we have ΔH_vap in kJ/mol, we can substitute it into the entropy change formula along with the temperature (373 K):
\[ \Delta S = \frac{40.626 \, \text{kJ/mol}}{373 \, \text{K}} \]

### Step 4: Calculate the entropy change
Perform the calculation:
\[ \Delta S = \frac{40.626}{373} \approx 0.1089 \, \text{kJ/mol K} \]
To convert this to J/mol K, we multiply by 1000:
\[ \Delta S \approx 108.9 \, \text{J/mol K} \]

### Final Answer
The entropy change involved in the conversion of 1 mole of liquid water at 373 K to vapor at the same temperature is approximately:
\[ \Delta S \approx 108.9 \, \text{J/mol K} \]

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