`2kg` of water is converted into steam by bolling at atmosperic pressure. The volume changes from `2xx10^(-3) m^(-3)` to `3.34 m^(3)` . The work done by the system is about

Options

A. `-340 kJ`
B. `-170 kJ`
C. `170 kJ`
D. `340 kJ`

Explanation

To find the work done by the system when converting 2 kg of water to steam at atmospheric pressure, we can follow these steps:

### Step 1: Identify the given values
- Mass of water, \( m = 2 \, \text{kg} \)
- Initial volume of water, \( V_i = 2 \times 10^{-3} \, \text{m}^3 \)
- Final volume of steam, \( V_f = 3.34 \, \text{m}^3 \)
- Atmospheric pressure, \( P = 1.013 \times 10^5 \, \text{Pa} \)

### Step 2: Calculate the change in volume
The change in volume (\( \Delta V \)) can be calculated as:
\[ \Delta V = V_f - V_i \]
Substituting the values:
\[ \Delta V = 3.34 \, \text{m}^3 - 2 \times 10^{-3} \, \text{m}^3 \]
Calculating this gives:
\[ \Delta V = 3.34 \, \text{m}^3 - 0.002 \, \text{m}^3 = 3.338 \, \text{m}^3 \]

### Step 3: Use the formula for work done
The work done (\( W \)) by the system at constant pressure can be calculated using the formula:
\[ W = P \Delta V \]
Substituting the values:
\[ W = (1.013 \times 10^5 \, \text{Pa}) \times (3.338 \, \text{m}^3) \]

### Step 4: Calculate the work done
Now, we perform the multiplication:
\[ W = 1.013 \times 10^5 \times 3.338 \]
Calculating this gives:
\[ W \approx 338,000 \, \text{J} = 338 \, \text{kJ} \]

### Step 5: Round the answer
Since the question asks for the work done, we can round it to:
\[ W \approx 340 \, \text{kJ} \]

### Final Answer
The work done by the system is approximately **340 kJ**.

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