(a) In a chemical reaction, two gases combine to form a solid. What do you expect for the sign of \(\Delta S ?\) (b) How does the entropy of the system change in the processes described in Exercise \(19.12 ?\)

(Begin by thinking about the degrees of disorder in both the initial and final states of the system. Gases generally have a high degree of disorder, whereas solids have a more ordered arrangement of particles. When two gases combine to form a solid, the system becomes more ordered. With this knowledge, we can determine that the entropy change \(\Delta S\) should be negative, as entropy decreases when going from a more disordered state (gases) to a more ordered state (solid). So the answer for part (a) is: \(\Delta S < 0\).)

(Exercise 19.12 states that a gas is isothermally expanded to twice its initial volume and then compressed to its original volume. First, let's analyze the entropy change for each step of the process. In the isothermal expansion, the gas expands, leading to an increase in disorder, so \(\Delta S_1 > 0\). In the compression, the gas is compressed back to its original volume, which means the system has become more ordered, and thus \(\Delta S_2 < 0\). To determine the total change in the system's entropy, we have to add both changes: \(\Delta S_{total} = \Delta S_1 + \Delta S_2\). We know that the expansion and compression processes are opposite, so the net entropy change is expected to be zero since any increase in entropy during expansion should be compensated by an equal decrease during compression: \(\Delta S_{total} = 0\).)