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 expansion is

Options

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

Answer

To solve the problem of calculating the total work done during the expansion of an ideal gas from a 5-liter cylinder into an evacuated vessel of 550 liters capacity, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**: We have an ideal gas in a 5-liter cylinder at a pressure of 15 atm. The gas is allowed to expand into an evacuated vessel (which means it is a vacuum).

2. **Identify External Pressure**: Since the gas is expanding into a vacuum, the external pressure (P_external) acting on the gas is 0 atm. This is a crucial point because work done by a gas during expansion is dependent on the external pressure.

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

4. **Calculate Change in Volume**:
- Initial volume (V_initial) = 5 L (the volume of the gas in the cylinder).
- Final volume (V_final) = 5 L + 550 L = 555 L (the total volume after expansion into the evacuated vessel).
- Therefore, the change in volume (\( \Delta V \)) is:
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 555 \, \text{L} - 5 \, \text{L} = 550 \, \text{L} \]

5. **Substitute Values into the Work Done Formula**:
- Since \( P_{\text{external}} = 0 \, \text{atm} \):
\[ W = -0 \times 550 \, \text{L} = 0 \, \text{L atm} \]

6. **Conclusion**: The total work done during the process of expansion is 0 L atm.

### Final Answer:
The total work done during the process of expansion is **0 L atm**.