For each of the following pairs, which substance has the greater value of S? a. \(\mathrm{N}_{2} \mathrm{O}(\text { at } 0 \mathrm{K})\) or He (at 10 \(\mathrm{K} )\) b. \(\mathrm{N}_{2} \mathrm{O}(g)\) (at $1 \mathrm{atm}, 25^{\circ} \mathrm{C} )\( or He(g) (at 1 atm, \)25^{\circ} \mathrm{C} )$ c. \(\mathrm{NH}_{3}(s)\) (at 196 $\mathrm{K} ) \longrightarrow \mathrm{NH}_{3}(l)(\text { at } 196 \mathrm{K})$

In case a, we have \(\mathrm{N}_{2} \mathrm{O}(\text { at } 0 \mathrm{K})\) and He(at \(10 \mathrm{K}\)). Since entropy (S) generally increases with temperature, He at 10 K will have higher entropy than N2O at 0K: Result for case a: Helium (He) at 10 K has greater value of S.

In case b, we have \(\mathrm{N}_{2} \mathrm{O}(g)\) (at \(1 \mathrm{atm}, 25^{\circ} \mathrm{C})\) and He(g) (at 1 atm, \(25^{\circ} \mathrm{C}\)). Both substances are in the gaseous state and at the same temperature and pressure. Thus, we can compare their entropies based on the molecular complexity. Greater molecular complexity is associated with greater entropy. Nitrous oxide (N2O) is more complex than helium (He), a monoatomic gas: Result for case b: Nitrous oxide (N2O) at 1 atm and \(25^{\circ} \mathrm{C}\) has a greater value of S.

In case c, we have the phase transition \(\mathrm{NH}_{3}(s)\) (at 196 \(\mathrm{K}\)) \(\longrightarrow \mathrm{NH}_{3}(l)(\text { at } 196 \mathrm{K})\). When comparing the same substance but in different states at the same temperature, we can determine which state has a greater entropy by considering the order within the states. Entropy increases as the order in a system decreases. The solid state has more order compared to the liquid state: Result for case c: Ammonium (NH3) in its liquid state at 196 K has a greater value of S. To summarize: a. He (at 10 K) has a greater value of S. b. N2O (at 1 atm, \(25^{\circ} \mathrm{C}\)) has a greater value of S. c. NH3 in its liquid state (at 196 K) has a greater value of S.