Entropy & second law of thermodynamics

[sskk221]

Homework Statement

Consider all possible isothermal contractions of an ideal gas. The change in entropy of the gas:
A) does not increase for any of them
B) increases for all of them
C) is zero for all of them
D) decreases for all of them
E) does not decrease for any of them

Homework Equations

$$\Delta S\geq 0$$
$$\Delta S = Sf -Si$$
$$\Delta S = Q/T$$ (isothermal process)

The Attempt at a Solution

I chose answer E since my physics book states that "in a closed system, the entropy of the system increases for irreversible processes and remains constant for reversible processes. It never decreases." However, the answer key has the correct answer as D. I think it might have something to do with the isothermal modifier, but I'm not exactly sure of the logic.

 

[Fightfish]

It's not a closed system when you perform an isothermal process on an ideal gas - you have to supply or remove heat from the system.
In an isothermal contraction, the entropy decreases, as while the average kinetic energy of the molecules remain the same, the gas now occupies a smaller volume - no of states decreases.

 

[eaglelake]

Homework Statement

Consider all possible isothermal contractions of an ideal gas. The change in entropy of the gas:
A) does not increase for any of them
B) increases for all of them
C) is zero for all of them
D) decreases for all of them
E) does not decrease for any of them

Homework Equations

$$\Delta S\geq 0$$
$$\Delta S = Sf -Si$$
$$\Delta S = Q/T$$ (isothermal process)

The Attempt at a Solution

I chose answer E since my physics book states that "in a closed system, the entropy of the system increases for irreversible processes and remains constant for reversible processes. It never decreases." However, the answer key has the correct answer as D. I think it might have something to do with the isothermal modifier, but I'm not exactly sure of the logic.

When the gas contracts during an isothermal process heat energy flows out of the gas. Thus, Q is negative and so is DeltaS=Q/T. Therefore, the entropy S decreases.