A mol of `Al_(3)C_(4)(s)` reacts with water in a closed vessel at `27^(@)C` against atmospheric pressure, work is doens

Options

A. `-1800` cal
B. `-600` cal
C. `+1800` cal
D. zero

Explanation

To solve the problem of calculating the work done when one mole of \( \text{Al}_3\text{C}_4(s) \) reacts with water in a closed vessel at \( 27^\circ C \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**:
- We have a solid \( \text{Al}_3\text{C}_4 \) reacting with water in a closed vessel. The term "closed vessel" indicates that no gas can escape and the volume of the system remains constant.

2. **Identify the Conditions**:
- The reaction occurs at a constant temperature of \( 27^\circ C \) and against atmospheric pressure.

3. **Determine Work Done**:
- The formula for work done (\( W \)) in a chemical reaction is given by:
\[ W = -P \Delta V \]
- Where \( P \) is the pressure and \( \Delta V \) is the change in volume.

4. **Analyze Volume Change**:
- Since the reaction occurs in a closed vessel, the volume does not change. Therefore, \( \Delta V = 0 \).

5. **Calculate Work Done**:
- Substituting \( \Delta V = 0 \) into the work done formula:
\[ W = -P \times 0 = 0 \]
- Thus, the work done in this reaction is \( 0 \).

6. **Conclusion**:
- The total work done when one mole of \( \text{Al}_3\text{C}_4(s) \) reacts with water in a closed vessel is \( 0 \).

### Final Answer:
The work done is \( 0 \).