Work done for converson of 0.5 mole of water of `100^(@)C` to steam at 1 atm pressure is (heat of vaporisation of water at `100^(@)C` is `4070J"mol"^(-1)`)

Options

A. `-1.54kJ`
B. `1.54kJ`
C. `1.25kJ`
D. `-1.35kJ`

Explanation

To calculate the work done for the conversion of 0.5 moles of water at 100°C to steam at 1 atm pressure, we can follow these steps:

### Step 1: Calculate the volume of steam produced
We can use the ideal gas equation \( PV = nRT \) to find the volume of the steam produced.

1. **Given:**
- Number of moles (n) = 0.5 moles
- R (ideal gas constant) = 0.0821 L·atm/(K·mol)
- Temperature (T) = 100°C = 373 K (convert Celsius to Kelvin by adding 273)
- Pressure (P) = 1 atm

2. **Calculate Volume (V):**
\[ V = \frac{nRT}{P} \]
\[ V = \frac{0.5 \, \text{mol} \times 0.0821 \, \text{L·atm/(K·mol)} \times 373 \, \text{K}}{1 \, \text{atm}} \]
\[ V = \frac{15.3 \, \text{L·atm}}{1 \, \text{atm}} = 15.3 \, \text{L} \]

### Step 2: Calculate the work done
The work done during the phase change can be calculated using the formula:
\[ W = -P_{\text{external}}(V_2 - V_1) \]
Where:
- \( V_2 \) is the final volume of steam (15.3 L)
- \( V_1 \) is the initial volume of water (which is negligible)

Since the initial volume of water is negligible, we can simplify the equation to:
\[ W = -P_{\text{external}} \times V_2 \]

3. **Substituting values:**
\[ W = -1 \, \text{atm} \times 15.3 \, \text{L} \]
\[ W = -15.3 \, \text{atm·L} \]

### Step 3: Convert work to Joules
To convert work from atm·L to Joules, we use the conversion factor:
1 atm·L = 101.325 J

4. **Convert:**
\[ W = -15.3 \, \text{atm·L} \times 101.325 \, \text{J/(atm·L)} \]
\[ W = -1550.5 \, \text{J} \]

### Final Answer:
The work done for the conversion of 0.5 moles of water at 100°C to steam at 1 atm pressure is approximately:
\[ W \approx -1550.5 \, \text{J} \]