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 forthe process is :
(Take : 1L-atm= 100J)`

Options

A. 4997 J
B. 4970 J
C. 9996 J
D. none of these

Answer

To solve the problem step-by-step, we will calculate the work done during the conversion of 60% of a 100 g sample of water (H₂O) into gas at constant pressure.

### Step 1: Convert the mass of water to kilograms
Given mass of water = 100 g

\[ \text{Mass in kg} = \frac{100 \text{ g}}{1000} = 0.1 \text{ kg} \]

### Step 2: Calculate the initial volume of water
Using the formula for volume:

\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]

The density of liquid water is given as 1000 kg/m³.

\[ \text{Initial Volume} (V_i) = \frac{0.1 \text{ kg}}{1000 \text{ kg/m}^3} = 0.0001 \text{ m}^3 \]

### Step 3: Determine the mass of water converted to gas
60% of the liquid is converted to gas:

\[ \text{Mass of gas} = 0.6 \times 0.1 \text{ kg} = 0.06 \text{ kg} \]

The remaining mass of water is:

\[ \text{Mass of liquid} = 0.1 \text{ kg} - 0.06 \text{ kg} = 0.04 \text{ kg} \]

### Step 4: Calculate the volume of the gas produced
Using the density of the gas, which is 0.60 kg/m³:

\[ \text{Volume of gas} (V_g) = \frac{0.06 \text{ kg}}{0.60 \text{ kg/m}^3} = 0.1 \text{ m}^3 \]

### Step 5: Calculate the volume of the remaining liquid
Using the density of the liquid water:

\[ \text{Volume of liquid} (V_l) = \frac{0.04 \text{ kg}}{1000 \text{ kg/m}^3} = 0.00004 \text{ m}^3 \]

### Step 6: Calculate the final volume
The final volume (V_f) is the sum of the volume of the gas and the volume of the remaining liquid:

\[ V_f = V_g + V_l = 0.1 \text{ m}^3 + 0.00004 \text{ m}^3 \approx 0.1 \text{ m}^3 \]

### Step 7: Calculate the change in volume (ΔV)
The change in volume is:

\[ \Delta V = V_f - V_i = 0.1 \text{ m}^3 - 0.0001 \text{ m}^3 = 0.0999 \text{ m}^3 \]

### Step 8: Calculate the work done (W)
Using the formula for work done at constant pressure:

\[ W = P \Delta V \]

Given that the pressure (P) is 1 atm, we convert this to Joules:

\[ 1 \text{ atm} = 101.325 \text{ J/L} \quad \text{(or 101325 Pa)} \]

Since we need to convert ΔV from m³ to L (1 m³ = 1000 L):

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

Now substituting the values:

\[ W = 101.325 \text{ J/L} \times 99.9 \text{ L} \approx 10125.67 \text{ J} \]

### Final Answer
The magnitude of the work done for the process is approximately **10125.67 J**.

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