The density of ice at `0^(@)C` is `0.99987g//"cc"` . The work done for melting mole of ice at `1.00` bar (assuming work is done only due to expansion) is approximately

Options

A. `0.17J`
B. `1.7xx10^(3)J`
C. `1.7xx10^(-6)J`
D. can't be determined

Explanation

To find the work done for melting one mole of ice at 1 bar, we can follow these steps:

### Step 1: Understand the Work Done Formula
The work done (W) during a process at constant pressure can be calculated using the formula:
\[ W = -P \Delta V \]
However, since we are considering work done due to expansion, we will use:
\[ W = P \Delta V \]

### Step 2: Calculate the Volume of Ice
We know the density of ice at \(0^{\circ}C\) is \(0.99987 \, \text{g/cm}^3\). The molar mass of ice (water) is approximately \(18 \, \text{g/mol}\).

To find the volume of 1 mole of ice:
\[ \text{Volume of ice} = \frac{\text{mass}}{\text{density}} = \frac{18 \, \text{g}}{0.99987 \, \text{g/cm}^3} \]

Calculating this gives:
\[ \text{Volume of ice} \approx 18.0002 \, \text{cm}^3 \]

### Step 3: Calculate the Volume of Water
Next, we need to calculate the volume of the water produced from melting the ice. The density of water at \(0^{\circ}C\) is approximately \(1.000 \, \text{g/cm}^3\).

To find the volume of 1 mole of water:
\[ \text{Volume of water} = \frac{18 \, \text{g}}{1.000 \, \text{g/cm}^3} = 18 \, \text{cm}^3 \]

### Step 4: Calculate the Change in Volume
The change in volume (\(\Delta V\)) when ice melts into water is:
\[ \Delta V = \text{Volume of water} - \text{Volume of ice} \]
\[ \Delta V = 18 \, \text{cm}^3 - 18.0002 \, \text{cm}^3 \]
\[ \Delta V \approx -0.0002 \, \text{cm}^3 \]

### Step 5: Convert Volume to Cubic Meters
Since \(1 \, \text{cm}^3 = 10^{-6} \, \text{m}^3\):
\[ \Delta V \approx -0.0002 \, \text{cm}^3 \times 10^{-6} \, \text{m}^3/\text{cm}^3 = -2.0 \times 10^{-10} \, \text{m}^3 \]

### Step 6: Calculate Work Done
Now, substituting the values into the work done formula:
- Pressure \(P = 1 \, \text{bar} = 10^5 \, \text{Pa}\)
- Change in volume \(\Delta V = -2.0 \times 10^{-10} \, \text{m}^3\)

Thus,
\[ W = P \Delta V = (10^5 \, \text{Pa}) \times (-2.0 \times 10^{-10} \, \text{m}^3) \]
\[ W = -0.02 \, \text{J} \]

### Step 7: Final Result
The work done for melting one mole of ice at 1 bar is approximately:
\[ W \approx 0.02 \, \text{J} \]