When 1g of water at `0^@C` and `1xx10^5(N)/(m^2)` pressure is converted into ice of volume `1.091cm^3`. The external work done will e

Options

A. 0.0091 Joule
B. 0.0182 Joule
C. `-0.0091` Joule
D. `-0.0` Joule

Explanation

To solve the problem of calculating the external work done when 1g of water at 0°C is converted into ice, we can follow these steps:

### Step 1: Determine the volume of water
Given that the mass of water is 1 gram and the density of water is 1 gram/cm³, we can calculate the volume of water.

\[ \text{Volume of water} = \frac{\text{Mass of water}}{\text{Density of water}} = \frac{1 \text{ g}}{1 \text{ g/cm}^3} = 1 \text{ cm}^3 \]

### Step 2: Identify the volume of ice
The problem states that the volume of ice formed is 1.091 cm³.

### Step 3: Calculate the change in volume (ΔV)
The change in volume (ΔV) when water converts to ice can be calculated as:

\[ \Delta V = \text{Volume of ice} - \text{Volume of water} = 1.091 \text{ cm}^3 - 1 \text{ cm}^3 = 0.091 \text{ cm}^3 \]

### Step 4: Convert the change in volume to cubic meters
Since the pressure is given in N/m², we need to convert the volume from cm³ to m³:

\[ \Delta V = 0.091 \text{ cm}^3 = 0.091 \times 10^{-6} \text{ m}^3 = 9.1 \times 10^{-8} \text{ m}^3 \]

### Step 5: Calculate the external work done (W)
The external work done (W) is given by the formula:

\[ W = P \Delta V \]

Where:
- \(P = 1 \times 10^5 \text{ N/m}^2\)
- \(\Delta V = 9.1 \times 10^{-8} \text{ m}^3\)

Substituting the values:

\[ W = 1 \times 10^5 \text{ N/m}^2 \times 9.1 \times 10^{-8} \text{ m}^3 \]

Calculating this gives:

\[ W = 1 \times 10^5 \times 9.1 \times 10^{-8} = 0.0091 \text{ Joules} \]

### Final Answer
The external work done is \(0.0091 \text{ Joules}\).

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