Entropy of vaporisation of water at `100^(@)C`, if molar heat of vaporisation is `8710 cal mol^(-1)` will be

Options

A. `20 cal mol^(-1) K^(-1)`
B. `23.36 cal mol^(-1) K^(-1)`
C. `24 cal mol^(-1) K^(-1)`
D. `28.0 cal mol^(-1) K^(-1)`

Answer

To find the entropy of vaporization of water at 100°C, given that the molar heat of vaporization is 8710 cal/mol, we can follow these steps:

### Step 1: Identify the given values
- Molar heat of vaporization (ΔH_vaporization) = 8710 cal/mol
- Temperature (T) = 100°C

### Step 2: Convert the temperature from Celsius to Kelvin
To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
Substituting the given temperature:
\[ T = 100 + 273 = 373 \, K \]

### Step 3: Use the formula for entropy of vaporization
The relationship between the change in entropy (ΔS_vaporization) and the heat of vaporization (ΔH_vaporization) at a given temperature (T) is given by:
\[ \Delta S_{vaporization} = \frac{\Delta H_{vaporization}}{T} \]

### Step 4: Substitute the values into the formula
Substituting the values we have:
\[ \Delta S_{vaporization} = \frac{8710 \, \text{cal/mol}}{373 \, K} \]

### Step 5: Perform the calculation
Calculating the above expression:
\[ \Delta S_{vaporization} = \frac{8710}{373} \approx 23.36 \, \text{cal/mol·K} \]

### Conclusion
The entropy of vaporization of water at 100°C is approximately:
\[ \Delta S_{vaporization} \approx 23.36 \, \text{cal/mol·K} \]

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