Suppose we vaporize a mole of liquid water at \(25^{\circ} \mathrm{C}\) and another mole of water at \(100{ }^{\circ} \mathrm{C}\). (a) Assuming that the enthalpy of vaporization of water does not change much between \(25^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C},\) which process involves the larger change in entropy? (b) Does the entropy change in either process depend on whether we carry out the process reversibly or not? Explain.

First, let's recall the relationship between enthalpy change (∆H) and entropy change (∆S) for a phase transition at constant temperature, T: \[\Delta G = \Delta H - T \Delta S\] Where ∆G is the Gibbs free energy change. For a phase transition, such as vaporization, ∆G = 0. Therefore, we can rearrange the equation to solve for ∆S: \[\Delta S = \frac{\Delta H}{T}\] Since we are given the assumption that the enthalpy of vaporization does not change significantly between the two temperatures, we can use a constant value for ∆H. The enthalpy of vaporization for water at 100°C is approximately 40.79 kJ/mol. We can then convert the given temperatures to Kelvin by adding 273.15: T1 = 298.15 K and T2 = 373.15 K.

Now, we'll use the enthalpy of vaporization and the temperatures to calculate the entropy change for each process: Process 1: Vaporizing water at 25°C (298.15 K) \[\Delta S_1 = \frac{\Delta H}{T_1} = \frac{40.79 \, \text{kJ/mol}}{298.15 \, \text{K}} = 0.1369 \, \text{kJ/mol K}\] Process 2: Vaporizing water at 100°C (373.15 K) \[\Delta S_2 = \frac{\Delta H}{T_2} = \frac{40.79 \, \text{kJ/mol}}{373.15 \, \text{K}} = 0.1093 \, \text{kJ/mol K}\]

Now that we have calculated the entropy change for both processes, we can directly compare them: \[\Delta S_1 = 0.1369 \, \text{kJ/mol K} > \Delta S_2 = 0.1093 \, \text{kJ/mol K}\] The entropy change of vaporizing water at 25°C is greater than the entropy change of vaporizing water at 100°C.

Entropy change depends on the heat transfer between the system and its surroundings. In a reversible process, the heat exchange is carried out infinitely slowly, allowing for the system to always be in equilibrium with the surroundings. In an irreversible process, the heat exchange occurs at a finite rate, and the system may not always be in equilibrium. However, in both processes (reversible and irreversible), the initial and final states of the system are the same, which means the values of the initial and final entropies do not depend on the path (reversible or irreversible) taken between them. Thus, the entropy change in either process does not depend on whether we carry out the process reversibly or not.