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 \(100^\circ C\) given that the molar heat of vaporization is \(8710 \, \text{cal mol}^{-1}\), we can follow these steps:

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin**:
\[ T = 100^\circ C + 273 = 373 \, K \]

2. **Use the Formula for Entropy Change**:
The change in entropy (\(\Delta S\)) during a phase transition can be calculated using the formula:
\[ \Delta S = \frac{Q}{T} \]
where \(Q\) is the heat absorbed (which is equal to the molar heat of vaporization, \(\Delta H_{vaporization}\)).

3. **Substitute the Values**:
Here, \(Q = \Delta H_{vaporization} = 8710 \, \text{cal mol}^{-1}\) and \(T = 373 \, K\).
\[ \Delta S = \frac{8710 \, \text{cal mol}^{-1}}{373 \, K} \]

4. **Perform the Calculation**:
\[ \Delta S = \frac{8710}{373} \approx 23.35 \, \text{cal mol}^{-1} K^{-1} \]

5. **Final Result**:
The entropy of vaporization of water at \(100^\circ C\) is approximately:
\[ \Delta S \approx 23.35 \, \text{cal mol}^{-1} K^{-1} \]