Question

In thermodynamics, if a system undergoes a reversible process from state A to B, then (ΔS)universe=0. However, in the case of an irreversible process, even if the system's entropy change can be calculated using the reversible route from A to B, the surroundings' entropy change is not the negative of the system's entropy change. This suggests that (ΔS)universe may not be a state function but depends on the path (reversible vs irreversible). Considering entropy as a state function, is this reasoning correct?

a.The reasoning is correct; (ΔS)universe is not a state function.
b.The reasoning is incorrect; (ΔS)universe is always zero regardless of the process path.
c.The reasoning is incorrect; (ΔS)universe depends on the nature of the system, not the process path.
d.The reasoning is correct; (ΔS)universe is a state function.

Answer 1

The reasoning provided in the question is incorrect; the change in entropy of the universe is a state function regardless of the process being reversible or irreversible.

Step-by-step explanation:

The reasoning in the question is incorrect because the entropy change of the universe, ΔSuniv, is indeed a state function.

While it is true that for a reversible process, the change in entropy of the universe is zero (ΔSuniv=0), this is not because entropy is not a state function, but because the entropy gained by the system is exactly balanced by the entropy lost by the surroundings.

In the case of an irreversible process, the entropy of the universe increases; however, this does not mean that ΔSuniv is not a state function.

State functions depend only on the initial and final states, not on the path taken between them. Even for irreversible processes, entropy changes in the system and surroundings can be calculated using hypothetical reversible paths, allowing for the determination of ΔSuniv based on state functions.

Therefore, the correct answer to the inquiry is: d. The reasoning is incorrect; ΔSuniv is a state function.