Given an idealgas is expanded adiabatically and irreversibley form volume `V_(1)` to `V_(2))` , then which one of the following is correct ?

Options

A. `DeltaS` ( system ) `= 0 & DeltaS` ( surroundings ) `= + ve`
B. `DeltaS` ( system ) `= +ve & DeltaS` ( surroundings ) `= 0`
C. `DeltaS` ( system ) `= 0 & DeltaS` ( surroundings ) `= 0`
D. `DeltaS` ( system ) `= +ve & DeltaS` ( surroundings ) `= -ve`

Explanation

To solve the problem of an ideal gas expanding adiabatically and irreversibly from volume \( V_1 \) to \( V_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Adiabatic Process**:
- In an adiabatic process, there is no heat exchange with the surroundings. Thus, \( Q = 0 \).

2. **Change in Entropy of the System**:
- The change in entropy (\( \Delta S \)) of the system can be calculated using the formula:
\[ \Delta S_{\text{system}} = \frac{Q}{T} \]
- Since \( Q = 0 \) for an adiabatic process, we have:
\[ \Delta S_{\text{system}} = \frac{0}{T} = 0 \]

3. **Total Change in Entropy**:
- The total change in entropy (\( \Delta S_{\text{total}} \)) is given by the sum of the change in entropy of the system and the surroundings:
\[ \Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} \]
- Since we found \( \Delta S_{\text{system}} = 0 \), we can rewrite this as:
\[ \Delta S_{\text{total}} = 0 + \Delta S_{\text{surroundings}} = \Delta S_{\text{surroundings}} \]

4. **Entropy of the Surroundings**:
- For an irreversible process, the change in entropy of the surroundings is positive. Therefore:
\[ \Delta S_{\text{surroundings}} > 0 \]

5. **Conclusion**:
- Since \( \Delta S_{\text{system}} = 0 \) and \( \Delta S_{\text{surroundings}} > 0 \), we conclude that:
\[ \Delta S_{\text{total}} > 0 \]
- This is consistent with the second law of thermodynamics, which states that the total entropy of an isolated system always increases.

### Final Answer:
- The correct statement is that the change in entropy of the system is zero (\( \Delta S_{\text{system}} = 0 \)), while the change in entropy of the surroundings is positive (\( \Delta S_{\text{surroundings}} > 0 \)).