For the process H₂O(l) → H₂O(g), at T = 100°C and 1 atmosphere pressure, the correct choice is:

At 100°C and 1 atmosphere pressure H₂O(l)⇌H₂O(g) is at equilibrium. For equilibrium, ΔG = 0. Since ΔG = ΔH - TΔS, at equilibrium, 0 = ΔH - TΔS, therefore ΔH = TΔS. For the process H₂O(l) → H₂O(g), the enthalpy change (ΔH) is positive because energy is absorbed to overcome intermolecular forces in the liquid to convert it to gas (endothermic process). Thus, ΔH > 0. Since ΔH = TΔS, and ΔH > 0, then ΔS must also be positive (ΔS > 0). The system's entropy increases because the gaseous state is more disordered than the liquid state. The surroundings' entropy change depends on the heat absorbed (ΔH). Since the process is endothermic, heat is absorbed from the surroundings, therefore causing a decrease in the entropy of the surroundings (ΔSsurroundings < 0). Therefore, the correct choice is ΔSsystem > 0 and ΔSsurrounding < 0.