Final answer:
The entropy change of the gas system when the volume is halved and the pressure is doubled under constant temperature is a finite negative value, reflecting a small decrease in entropy due to a reversible isothermal compression.
Step-by-step explanation:
The question discusses a scenario where the volume of a gas X at different pressures is described and asks about the change in entropy. The relation between pressure (P) and volume (V) of a gas can be described by Boyle's Law, which states that for a fixed amount of gas at constant temperature, the product of the pressure and volume is constant (P1V1 = P2V2). In this case, when the volume of the gas is halved while the pressure is doubled (V is K and becomes K/2 when P goes from 3.2 atm to 6.4 atm), and temperature and the number of moles of gas are held constant, it implies an isothermal process for an ideal gas.
The entropy change (ΔS) under these conditions for an ideal gas can be calculated using the formula ΔS = nRln(V2/V1), where n is the number of moles of gas and R is the gas constant. Here, since the volume is halved, V2/V1 = (K/2)/K = 1/2. However, since there is no information on the number of moles of gas, the exact value of the entropy change cannot be calculated, but the process describes a reversible isothermal compression, which leads to a decrease in entropy.
Therefore, the answer is (b) Becomes a small change, because the entropy change is finite and negative, reflecting the orderly arrangement of the gas particles upon compression.