An ideal gas is kept in a 5 litre cylinder at a pressure of 15 atm. In 30 minutes the gas is allowed to enter slowly into an evacuated vessel of 550 L capacity. Total work done during the process of energy is

Options

A. 2500 L atm
B. 7500 L atm
C. 7425 L atm
D. 0

Answer

To solve the problem of calculating the work done by the ideal gas when it expands into an evacuated vessel, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**: We have an ideal gas in a cylinder with a volume of 5 liters and a pressure of 15 atm. The gas is allowed to expand into an evacuated vessel with a capacity of 550 liters.

2. **Identify the Process**: The gas is expanding into a vacuum (evacuated vessel). In this scenario, the external pressure is zero because there is no opposing force to the expansion of the gas.

3. **Work Done Formula**: The work done (W) by a gas during expansion can be calculated using the formula:
\[ W = -P_{\text{ext}} \Delta V \]
where \( P_{\text{ext}} \) is the external pressure and \( \Delta V \) is the change in volume.

4. **Determine External Pressure**: Since the gas is expanding into a vacuum, the external pressure \( P_{\text{ext}} \) is 0 atm.

5. **Calculate Change in Volume**: The change in volume \( \Delta V \) can be calculated as:
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 550 \, \text{L} - 5 \, \text{L} = 545 \, \text{L} \]

6. **Substituting Values into the Work Formula**: Now substituting the values into the work done formula:
\[ W = -0 \, \text{atm} \times 545 \, \text{L} = 0 \]

7. **Conclusion**: The total work done during the process is 0.

### Final Answer:
The total work done during the process is **0**.

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