1 mole of `H_(2)O` is vaporised at `100^(@)C` and 1 atm pressure. If the latent heat of vaporisation of water is x J/g, then `Delta S` in terms of J/mole K is given by

Options

A. `(x)/(373)`
B. `(18x)/(300)`
C. `(18x)/(373)`
D. `(1)/(2)xx(18x)/(373)`

Answer

To solve the problem, we need to calculate the change in entropy (ΔS) when 1 mole of water (H₂O) is vaporized at 100°C and 1 atm pressure. The latent heat of vaporization is given as x J/g.

### Step-by-Step Solution:

1. **Identify the Mass of Water**:
- We know that 1 mole of water has a molar mass of approximately 18 g/mol.

2. **Calculate the Total Latent Heat of Vaporization**:
- The latent heat of vaporization for 1 gram of water is given as x J/g.
- Therefore, for 18 grams (1 mole) of water, the total latent heat (ΔH) can be calculated as:
\[ \Delta H = 18 \, \text{g} \times x \, \text{J/g} = 18x \, \text{J} \]

3. **Convert Temperature to Kelvin**:
- The temperature given is 100°C. To convert this to Kelvin:
\[ T = 100 + 273 = 373 \, \text{K} \]

4. **Use the Formula for Change in Entropy**:
- The change in entropy (ΔS) during a phase change at constant temperature can be calculated using the formula:
\[ \Delta S = \frac{\Delta H}{T} \]
- Substituting the values we have:
\[ \Delta S = \frac{18x \, \text{J}}{373 \, \text{K}} = \frac{18x}{373} \, \text{J/mol K} \]

5. **Final Answer**:
- Therefore, the change in entropy (ΔS) when 1 mole of water is vaporized at 100°C and 1 atm pressure is:
\[ \Delta S = \frac{18x}{373} \, \text{J/mol K} \]