One mole of water is converted to vapour at its boiling point `100^@C` and '1' atmospheric pressure. For this process, which one of following statement is correct ?

Options

A. `DeltaS=0`
B. `DeltaG=0`
C. `DeltaH=0`
D. `DeltaE=0`

Explanation

To solve the question regarding the conversion of one mole of water to vapor at its boiling point (100°C) and 1 atmospheric pressure, we need to analyze the thermodynamic properties involved in this phase change.

### Step-by-Step Solution:

1. **Understanding the Process**:
- The process described is the phase transition of water from liquid to vapor (steam) at its boiling point. This occurs at 100°C under 1 atm pressure.

2. **Phase Change at Equilibrium**:
- At the boiling point, water is in equilibrium between its liquid and vapor phases. This means that the rate of evaporation equals the rate of condensation.

3. **Thermodynamic Properties**:
- For any phase transition that occurs at constant temperature and pressure, the change in Gibbs free energy (ΔG) is zero at equilibrium. This is because the system is at a stable state where the forward and reverse processes occur at the same rate.

4. **Identifying the Correct Statement**:
- The options provided are:
- A) ΔS = 0
- B) ΔG = 0
- C) ΔH = 0
- D) ΔE = 0
- Since we established that the Gibbs free energy change (ΔG) is zero for the phase transition at equilibrium, the correct statement is:
- **B) ΔG = 0**

### Final Answer:
The correct statement is **B) ΔG = 0**.