A heating coil is immersed in a 100 g sample of `H_(2)O` (l) at a 1 atm and `100^(@)` C in a closed vessel. In this heating process , `60%` of the liquid is converted to the gaseous form at constant pressure of 1 atm . The densities of liquid and gas under these conditions are 1000 `kg//m^(3)` and 0.60 `kg//m^(3)` respectively . Magnitude of the work done for the process is :
(Take : 1L-atm= 100 J)`

Options

A. 4997 J
B. 4970 J
C. 9994 J
D. None of these

Explanation

To solve the problem, we need to calculate the work done during the conversion of 60% of a 100 g sample of water from liquid to gas at constant pressure. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the initial volume of the liquid water
We know the mass of the water and its density. The formula for volume (V) is given by:

\[ V = \frac{m}{\rho} \]

Where:
- \( m = 100 \, \text{g} = 0.1 \, \text{kg} \) (convert grams to kilograms)
- \( \rho = 1000 \, \text{kg/m}^3 \)

Substituting the values:

\[ V_{\text{initial}} = \frac{0.1 \, \text{kg}}{1000 \, \text{kg/m}^3} = 0.0001 \, \text{m}^3 \]

### Step 2: Determine the mass of water converted to gas
60% of the liquid water is converted to gas:

\[ \text{Mass of gas} = 60\% \times 100 \, \text{g} = 60 \, \text{g} = 0.06 \, \text{kg} \]

### Step 3: Calculate the volume of the gas produced
Using the density of the gas (0.60 kg/m³), we can find the volume of the gas:

\[ V_{\text{gas}} = \frac{m_{\text{gas}}}{\rho_{\text{gas}}} = \frac{0.06 \, \text{kg}}{0.6 \, \text{kg/m}^3} = 0.1 \, \text{m}^3 \]

### Step 4: Calculate the final volume
The final volume (V_final) is the sum of the volume of the remaining liquid and the volume of the gas:

- Remaining mass of liquid = 40 g = 0.04 kg
- Volume of remaining liquid:

\[ V_{\text{liquid}} = \frac{0.04 \, \text{kg}}{1000 \, \text{kg/m}^3} = 0.00004 \, \text{m}^3 \]

Now, adding the volumes:

\[ V_{\text{final}} = V_{\text{gas}} + V_{\text{liquid}} = 0.1 \, \text{m}^3 + 0.00004 \, \text{m}^3 \approx 0.1 \, \text{m}^3 \]

### Step 5: Calculate the change in volume (ΔV)
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 0.1 \, \text{m}^3 - 0.0001 \, \text{m}^3 = 0.0999 \, \text{m}^3 \]

### Step 6: Calculate the work done (W)
The work done at constant pressure is given by:

\[ W = P \Delta V \]

Where:
- \( P = 1 \, \text{atm} = 101.325 \, \text{kPa} \) (but we will use the conversion factor directly)
- \( \Delta V = 0.0999 \, \text{m}^3 \)

To convert the volume change into liters, we multiply by 1000:

\[ \Delta V = 0.0999 \, \text{m}^3 \times 1000 = 99.9 \, \text{L} \]

Now, using the conversion \( 1 \, \text{L-atm} = 100 \, \text{J} \):

\[ W = 99.9 \, \text{L} \times 1 \, \text{atm} \times 100 \, \text{J/L-atm} = 9990 \, \text{J} \]

### Final Answer
The magnitude of the work done for the process is approximately:

\[ W \approx 9990 \, \text{J} \]