The entropy change for vapourisation of liquid water to steam `100^@C` is …. `JK^(-1) mol^(-1)`. Given that heat of vapourisation is `40.8 kJ mol^(-1)`.

Options

A. 109.38
B. 100.38
C. 110.38
D. 120.38

Explanation

To calculate the entropy change for the vaporization of liquid water to steam at 100°C, we can follow these steps:

### Step 1: Convert the heat of vaporization from kJ/mol to J/mol
The heat of vaporization (ΔH_vap) is given as 40.8 kJ/mol. To convert this to joules, we use the conversion factor \(1 \text{ kJ} = 1000 \text{ J}\).

\[ \Delta H_{vap} = 40.8 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 40800 \, \text{J/mol} \]

### Step 2: Convert the temperature from Celsius to Kelvin
The temperature (T) is given as 100°C. To convert this to Kelvin, we add 273.15.

\[ T = 100 \, \text{°C} + 273.15 = 373.15 \, \text{K} \]

### Step 3: Use the formula for entropy change
The change in entropy (ΔS) for a process at constant temperature can be calculated using the formula:

\[ \Delta S = \frac{\Delta H_{vap}}{T} \]

### Step 4: Substitute the values into the formula
Now we can substitute the values we have calculated into the formula:

\[ \Delta S = \frac{40800 \, \text{J/mol}}{373.15 \, \text{K}} \]

### Step 5: Perform the calculation
Now we perform the calculation:

\[ \Delta S \approx \frac{40800}{373.15} \approx 109.3 \, \text{J/K/mol} \]

### Conclusion
The entropy change for the vaporization of liquid water to steam at 100°C is approximately:

\[ \Delta S \approx 109.3 \, \text{J/K/mol} \]