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

Answer

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 \( 27^\circ C \) against atmospheric pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a closed vessel where a solid \( \text{Al}_3\text{C}_4 \) is reacting with water. Since the vessel is closed, no gas can escape or enter.

2. **Identifying Work Done**:
- The work done in a thermodynamic process is given by the formula:
\[ W = -P \Delta V \]
where \( W \) is the work done, \( P \) is the pressure, and \( \Delta V \) is the change in volume.

3. **Analyzing the Change in Volume**:
- In a closed vessel, the volume cannot change because the system is not allowed to expand or contract. Therefore, \( \Delta V = 0 \).

4. **Substituting Values into the Work Formula**:
- Since \( \Delta V = 0 \), we can substitute this into the work formula:
\[ W = -P \times 0 = 0 \]

5. **Conclusion**:
- The work done during the reaction in the closed vessel is zero. Thus, the answer is:
\[ \text{Work done} = 0 \]

### Final Answer:
The work done in the closed vessel is \( 0 \).

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