What is the change in entropy (∆S) when 1.32 g of propane (C3H8) at 0.100 atm pressure is compressed by a factor of five at a constant temperature at 20°C? Assume that propane behaves as an ideal gas. (a) ∆S = +13 J/K(b) ∆S = -13 J/K (c) ∆S = - 0.40 J/K(d) ∆S = + 0.40 J/K

Entropy is a measure of the disorder or randomness in a system. In thermodynamics, it quantifies the amount of energy in a physical system that is not available to do work. Higher entropy indicates greater disorder, while lower entropy suggests more order. Understanding how entropy changes during processes, such as compression or expansion, is crucial for predicting the spontaneity of reactions and the direction of energy flow.

The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of an ideal gas through the equation PV = nRT. This law assumes that gas particles do not interact and occupy no volume, which simplifies calculations. In this question, the behavior of propane as an ideal gas allows us to apply this law to determine changes in state variables, such as pressure and volume, when the gas is compressed.

The change in entropy (∆S) during a process can be calculated using the formula ∆S = nR ln(Vf/Vi) for an ideal gas, where Vf and Vi are the final and initial volumes, respectively. In this case, compressing the gas by a factor of five reduces its volume, leading to a negative change in entropy, as the system becomes more ordered. Understanding this relationship is essential for solving the problem and determining the correct answer.