Five moles of an ideal gas expanded spontaneously in to vacuum.
Then which is correct ?

Options

A. `omega=0,DeltaG=0`
B. `omega=0,DeltaGlt0`
C. `omega=0,DeltaGgt0`
D. `omega!=0,DeltaG=0`

Answer

To solve the question regarding the spontaneous expansion of 5 moles of an ideal gas into a vacuum, we will analyze the concepts of Gibbs free energy (ΔG) and work done (W) during the process.

### Step-by-Step Solution:

1. **Understanding the Process**:
- The gas expands spontaneously into a vacuum. This means there is no opposing pressure acting against the gas during its expansion.

2. **Gibbs Free Energy Change (ΔG)**:
- The Gibbs free energy change (ΔG) is a thermodynamic potential that can be used to predict the direction of chemical reactions and phase changes.
- For spontaneous processes, ΔG is negative (ΔG < 0). This indicates that the process can occur without any external input of energy.

3. **Work Done (W) During Expansion**:
- The work done by the gas during expansion can be described by the equation:
\[ W = -P_{\text{external}} \Delta V \]
- Here, \(P_{\text{external}}\) is the external pressure, and \(\Delta V\) is the change in volume.

4. **Condition of Free Expansion**:
- In the case of free expansion into a vacuum, the external pressure (\(P_{\text{external}}\)) is zero because there is no resistance to the expansion of the gas.
- Therefore, substituting \(P_{\text{external}} = 0\) into the work done equation gives:
\[ W = -0 \cdot \Delta V = 0 \]
- Thus, the work done by the gas during this spontaneous expansion is zero.

5. **Conclusion**:
- Since the gas expands spontaneously into a vacuum, we conclude that:
- ΔG < 0 (indicating spontaneity)
- Work done (W) = 0

### Final Answer:
- The correct statements regarding the expansion of the gas into a vacuum are:
- ΔG < 0 (spontaneous process)
- Work done = 0