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`

Answer

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 process.

### Step-by-Step Solution:

1. **Understanding the Process**:
- One mole of water (H₂O) is being converted to vapor at its boiling point (100°C) and at 1 atmospheric pressure.
- At this point, the water is transitioning from a liquid state to a gaseous state.

2. **Phase Change and Equilibrium**:
- At the boiling point, the vapor pressure of the liquid (water) equals the external atmospheric pressure. This creates an equilibrium between the liquid and vapor phases.
- The process of boiling involves both liquid water and water vapor existing simultaneously.

3. **Gibbs Free Energy (ΔG)**:
- At equilibrium, the change in Gibbs free energy (ΔG) is zero. This is a fundamental principle of thermodynamics, indicating that the system is at a state of balance.
- The equation for Gibbs free energy is given by:
\[ \Delta G = \Delta H - T\Delta S \]
- At equilibrium, ΔG = 0, which implies:
\[ \Delta H = T\Delta S \]

4. **Entropy Change (ΔS)**:
- The change in entropy (ΔS) for the process of boiling water is positive. This is because the transition from liquid to gas increases the randomness or disorder of the system.

5. **Enthalpy Change (ΔH)**:
- The enthalpy change (ΔH) for the process of converting water to vapor is also positive, indicating that the process is endothermic (heat is absorbed).

6. **Internal Energy Change (ΔE)**:
- The change in internal energy (ΔE) is also positive during this process since energy is required to break intermolecular hydrogen bonds in liquid water to form vapor.

### Conclusion:
- The correct statement regarding the conversion of one mole of water to vapor at its boiling point and 1 atmospheric pressure is that **ΔG = 0** at equilibrium.