The different in the work done when one mole of `Al_(4)C_(3)(s)` reacts with water in a closed vesselat `27^(@)C` against atmospheric pressure and that in an open vessel under the same conditions is

Options

A. greater in the open vessel by 600 cal
B. greater in the closed vessel by 600 cal
C. greater in the open vessel by 1800 cal
D. greater in the closed vessel by 1800cal

Explanation

To solve the problem, we need to determine the difference in work done when one mole of \( \text{Al}_4\text{C}_3(s) \) reacts with water in a closed vessel versus an open vessel at \( 27^\circ C \).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction of aluminum carbide with water can be represented as:
\[ \text{Al}_4\text{C}_3(s) + 12 \text{H}_2\text{O}(l) \rightarrow 4 \text{Al(OH)}_3(s) + 3 \text{CH}_4(g) \]
From this reaction, we see that 1 mole of \( \text{Al}_4\text{C}_3 \) produces 3 moles of \( \text{CH}_4 \) gas.

2. **Work Done in a Closed Vessel**:
In a closed vessel, the volume does not change because the gas cannot escape. Therefore, the change in volume \( \Delta V \) is zero:
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 0 - 0 = 0 \]
The work done \( W \) in a closed vessel is given by:
\[ W = -P_{\text{external}} \Delta V = -P_{\text{external}} \times 0 = 0 \]

3. **Work Done in an Open Vessel**:
In an open vessel, the gas produced can expand freely. The final volume can be calculated using the ideal gas equation:
\[ PV = nRT \]
Here, \( n \) is the number of moles of gas produced (3 moles of \( \text{CH}_4 \)), \( R \) is the ideal gas constant (0.0821 L·atm/(K·mol)), and \( T \) is the temperature in Kelvin (300 K).

The volume \( V \) can be calculated as:
\[ V = \frac{nRT}{P} \]
Assuming atmospheric pressure \( P = 1 \text{ atm} \):
\[ V = \frac{3 \times 0.0821 \times 300}{1} = 73.89 \text{ L} \]

4. **Calculating Work Done in Open Vessel**:
The work done in the open vessel is:
\[ W = -P_{\text{external}} \Delta V \]
Here, \( \Delta V = V_{\text{final}} - V_{\text{initial}} = 73.89 - 0 = 73.89 \text{ L} \).
Therefore,
\[ W = -1 \times 73.89 = -73.89 \text{ L·atm} \]
To convert this to calories, we use the conversion factor \( 1 \text{ L·atm} = 101.3 \text{ J} \) and \( 1 \text{ cal} = 4.184 \text{ J} \):
\[ W = -73.89 \times 101.3 \text{ J} \times \frac{1 \text{ cal}}{4.184 \text{ J}} \approx -1800 \text{ cal} \]

5. **Difference in Work Done**:
The difference in work done between the closed and open vessel is:
\[ \Delta W = W_{\text{open}} - W_{\text{closed}} = -1800 - 0 = -1800 \text{ cal} \]
Since we are interested in the magnitude, we can state:
\[ \text{Difference in work done} = 1800 \text{ cal} \]

### Final Answer:
The difference in the work done when one mole of \( \text{Al}_4\text{C}_3(s) \) reacts with water in a closed vessel and in an open vessel is **1800 calories**.