Final answer:
The correct answer to the student's question is (A) I only, as for a spontaneous vaporization of a liquid, Gibbs free energy change (∆G) must be negative.
Step-by-step explanation:
Considering the spontaneous vaporization of a liquid under constant temperature and pressure, Gibbs free energy (∆G) and enthalpy (∆H) of the system are essential thermodynamic quantities for analyzing this process. The spontaneity of vaporization implies that the Gibbs free energy change must be negative (∆G < 0), which aligns with the first inequality I. ∆G < 0. According to the Gibbs free energy equation G = H - TS, where H is enthalpy, T is temperature, and S is entropy, a negative ∆G can result from either an increase in entropy (∆S), a decrease in enthalpy (∆H), or both. Since vaporization typically involves the absorption of heat, the enthalpy change is positive (∆H > 0). The internal energy change ∆E is not directly comparable to ∆H because some of the heat added goes into doing work against the atmospheric pressure during vaporization, and is not a direct indicator of spontaneity.