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

Explanation

To find the work done during the evaporation of 1 kg of water in an open kettle, we can follow these steps:

### Step 1: Understand the Process
When water is heated in an open kettle, it undergoes a phase change from liquid to gas (steam). The work done during this process is primarily due to the expansion of the water vapor against the atmospheric pressure.

### Step 2: Identify the Work Done Formula
The work done (W) during the expansion can be expressed as:
\[ W = -P_{\text{external}} \Delta V \]
where \( \Delta V \) is the change in volume, and \( P_{\text{external}} \) is the external pressure (atmospheric pressure).

### Step 3: Determine the Change in Volume
The change in volume (\( \Delta V \)) during the evaporation can be approximated as:
\[ \Delta V = V_{\text{gas}} - V_{\text{liquid}} \]
Since the volume of the liquid (water) is much smaller than the volume of the gas (steam), we can simplify this to:
\[ \Delta V \approx V_{\text{gas}} \]

### Step 4: Use the Ideal Gas Law
For the gas (steam), we can use the ideal gas equation:
\[ PV = nRT \]
Rearranging gives:
\[ V = \frac{nRT}{P} \]
Thus, the work done can be expressed as:
\[ W = -P_{\text{external}} \cdot \left(\frac{nRT}{P_{\text{external}}}\right) \]
This simplifies to:
\[ W = -nRT \]

### Step 5: Calculate the Number of Moles (n)
To find \( n \) (the number of moles of water), we use the formula:
\[ n = \frac{m}{M} \]
where \( m \) is the mass of water (1 kg = 1000 g) and \( M \) is the molar mass of water (approximately 18 g/mol):
\[ n = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

### Step 6: Use the Ideal Gas Constant and Temperature
The ideal gas constant \( R \) is approximately \( 8.314 \, \text{J/(mol K)} \) and the temperature \( T \) at which water boils is \( 373 \, \text{K} \).

### Step 7: Calculate the Work Done
Now we can calculate the work done:
\[ W = -nRT = - (55.56 \, \text{mol}) \cdot (8.314 \, \text{J/(mol K)}) \cdot (373 \, \text{K}) \]

Calculating this gives:
\[ W \approx -173.2 \, \text{kJ} \]

### Final Answer
The work done during the evaporation of 1 kg of water is approximately:
\[ W \approx -173.2 \, \text{kJ} \]

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