A kettle containing 1kg of water is heated open to atmosphere until evaporation is complete. The work done during this process is

Explanation

To solve the problem of calculating the work done during the evaporation of 1 kg of water in a kettle heated open to the atmosphere, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Process**:
- The process involves heating 1 kg of water until it completely evaporates. This is an open system where the water turns from liquid to gas.

2. **Identify the Work Done Formula**:
- The work done (W) during expansion against the atmospheric pressure can be calculated using the formula:
\[ W = -P_{\text{external}} \Delta V \]
- Here, \(P_{\text{external}}\) is the external pressure (atmospheric pressure), and \(\Delta V\) is the change in volume.

3. **Determine the Change in Volume**:
- The change in volume (\(\Delta V\)) when water evaporates can be expressed as:
\[ \Delta V = V_{\text{gas}} - V_{\text{liquid}} \]
- Since the volume of gas is significantly larger than the volume of liquid, we can approximate:
\[ \Delta V \approx V_{\text{gas}} \]

4. **Calculate the Volume of Gas**:
- Using the Ideal Gas Law, we can express the volume of gas as:
\[ PV = nRT \]
- Rearranging gives:
\[ V = \frac{nRT}{P} \]

5. **Determine the Number of Moles (n)**:
- For 1 kg of water, the number of moles \(n\) is calculated as:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ mol} \]

6. **Substitute Values**:
- Use \(R = 8.314 \text{ J/(mol K)}\), \(T = 373 \text{ K}\) (the boiling point of water), and \(P \approx 101325 \text{ Pa}\) (atmospheric pressure).
- Substitute these values into the volume equation:
\[ V_{\text{gas}} = \frac{(55.56 \text{ mol})(8.314 \text{ J/(mol K)})(373 \text{ K})}{101325 \text{ Pa}} \]

7. **Calculate the Volume**:
- Performing the calculation:
\[ V_{\text{gas}} \approx \frac{(55.56)(8.314)(373)}{101325} \approx 1.72 \text{ m}^3 \]

8. **Calculate Work Done**:
- Now, substitute \(V_{\text{gas}}\) back into the work done formula:
\[ W = -P_{\text{external}} V_{\text{gas}} \approx -101325 \text{ Pa} \times 1.72 \text{ m}^3 \approx -174,000 \text{ J} \]
- Since we are interested in the magnitude of work done:
\[ W \approx 174,000 \text{ J} \]

### Final Answer:
The work done during the evaporation of 1 kg of water is approximately **174,000 Joules**.