Entropy Change Signs for Water Vaporization at Boiling Point

To determine the signs of $\Delta S_{\text{system}}$ and $\Delta S_{\text{surroundings}}$ for the vaporization of water:

$\text{H}_2\text{O} (l) \rightarrow \text{H}_2\text{O} (g)$

1. Entropy Change of the System ($\Delta S_{\text{system}}$): In transitioning from a liquid to a gaseous state, the molecules acquire significantly higher degrees of translational, rotational, and vibrational freedom. Since the gaseous state possesses greater randomness and molecular disorder than the liquid state: $S_{\text{gas}} > S_{\text{liquid}}$ Therefore, the change in entropy of the system is positive: $\Delta S_{\text{system}} = S_{\text{gas}} - S_{\text{liquid}} > 0$

2. Entropy Change of the Surroundings ($\Delta S_{\text{surroundings}}$): Vaporization is an endothermic process ($\Delta H_{\text{vap}} > 0$), meaning heat is absorbed by the system from the surroundings: $q_{\text{surroundings}} = -q_{\text{system}} = -\Delta H_{\text{vap}} < 0$ The change in entropy of the surroundings at constant temperature is given by: $\Delta S_{\text{surroundings}} = \frac{q_{\text{surroundings}}}{T} = -\frac{\Delta H_{\text{vap}}}{T}$ Since $\Delta H_{\text{vap}} > 0$ and $T > 0$, we have: $\Delta S_{\text{surroundings}} < 0$

   (Note: At the normal boiling point of $T = 100^\circ\text{C}$ and $P = 1\text{ atm}$, the liquid and vapor phases are at equilibrium, meaning $\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} = 0$, which also confirms that $\Delta S_{\text{surroundings}} = -\Delta S_{\text{system}} < 0$.)

Thus, the correct relation is: $\Delta S_{\text{system}} > 0 \quad \text{and} \quad \Delta S_{\text{surroundings}} < 0$

Therefore, the correct choice is (B).