Entropy of isothermal process reversible\irreversible

[oferon]

We were shown in class how to get those entropys.

For reversible isothermal - ΔT=0 thus ΔE=0 thus Q = -W.
ΔS(sys) = Qrev/T = nR(V1/V2)
And ΔS(surr) = -nR(V1/V2) because surroundings made opposite work.

For irreversible isothermal in vacuum - ΔT=0 thus ΔE=0.
No work is done by surroundings (vacuum) so W=0 then Q=0.
ΔS(sys) is a function state, so it remains ΔS(sys)=nR(V1/V2).
ΔS(surr) = Q/T = 0

So I have 2 questions:
1 - How come they decide that ΔS(sys) is a state function, but ΔS(surr) is not?? Can't we get ΔS(surr) = -nR(V1/V2) for exactly the same reason??

2 - I thought the defenition for entropy was integral(Qrev/T), so how come on the irreversible we use Q=0, and not the Q for the reversible process?

 

[jbsteele]

1 - The entropy of the system is a state function because it depends only on the current state of the system and does not depend on the history of the system. The entropy of the surroundings, however, does depend on the history and therefore is not a state function. In the reversible process, the entropy of the surroundings can be calculated from the entropy of the system and the work done by the surroundings, but in the irreversible process, the entropy of the surroundings cannot be calculated from the entropy of the system because no work is done by the surroundings. 2 - The definition of entropy is indeed the integral of Qrev/T, but for an irreversible process we need to consider the entropy change of the system and surroundings separately. For the system, the entropy change is given by the integral of Q/T, which in this case is zero since there is no heat transfer (Q=0). For the surroundings, the entropy change is zero since there is no work done by the surroundings (W=0).