Identify the process

The process is vaporization at the boiling point: $\mathrm{H_2O(l) \to H_2O(g)}$ at $T = 100^\circ \mathrm{C}$ and $P = 1\,$atm.

Sign of entropy change of the system

When liquid water changes into steam, the molecules go from a more ordered liquid state to a much more disordered gaseous state.

Therefore, $\Delta S_{\text{system}} > 0$

Sign of entropy change of the surroundings

Vaporization is an endothermic process. The system absorbs heat from the surroundings: $q_{\text{system}} > 0$ Hence the surroundings lose heat: $q_{\text{surroundings}} < 0$

Entropy change of surroundings is $\Delta S_{\text{surroundings}} = \frac{q_{\text{surroundings}}}{T}$ Since $q_{\text{surroundings}} < 0$ and $T>0$, $\Delta S_{\text{surroundings}} < 0$

Check spontaneity condition at phase equilibrium

At $100^\circ \mathrm{C}$ and 1 atm, water and steam are in equilibrium, so for a reversible phase transition: $\Delta G = 0$ and $\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} = 0$ This is consistent with equal magnitude and opposite signs of the two entropy changes.

Evaluate options

• A: $\Delta S_{\text{system}} > 0$ and $\Delta S_{\text{surroundings}} > 0$ ❌
• B: $\Delta S_{\text{system}} > 0$ and $\Delta S_{\text{surroundings}} < 0$ ✅
• C: $\Delta S_{\text{system}} < 0$ and $\Delta S_{\text{surroundings}} > 0$ ❌
• D: $\Delta S_{\text{system}} < 0$ and $\Delta S_{\text{surroundings}} < 0$ ❌

Final answer

The correct choice is: $\boxed{\text{B}}$