Find the entropy change for vaporisation of water to steam at `100^@C` in `JK^(-1)mol^(-1)` if heat of vaporisation is `40.8 kJ mol^(-1)`.

Options

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

Answer

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

### Step 1: Convert Temperature to Kelvin
The temperature given is 100°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
So,
\[ T = 100 + 273.15 = 373.15 \, K \]
For simplicity, we can round it to 373 K.

### Step 2: Convert Heat of Vaporization to Joules
The heat of vaporization is given as 40.8 kJ/mol. We need to convert this to Joules:
\[ \Delta H_{vaporization} = 40.8 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 40800 \, \text{J/mol} \]

### Step 3: Use the Entropy Change Formula
The formula for calculating the change in entropy (\( \Delta S \)) during a phase change is given by:
\[ \Delta S = \frac{\Delta H}{T} \]
Where:
- \( \Delta H \) is the heat of vaporization in Joules per mole,
- \( T \) is the temperature in Kelvin.

### Step 4: Substitute the Values
Now we can substitute the values we have into the formula:
\[ \Delta S = \frac{40800 \, \text{J/mol}}{373 \, K} \]

### Step 5: Calculate the Entropy Change
Now we perform the calculation:
\[ \Delta S = \frac{40800}{373} \approx 109.38 \, \text{J/K/mol} \]

### Final Answer
The entropy change for the vaporization of water to steam at 100°C is approximately:
\[ \Delta S \approx 109.38 \, \text{J/K/mol} \]

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