Entropy of isothermal process reversible\irreversible

  • Thread starter oferon
  • Start date
  • #1
30
0
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?
 

Answers and Replies

Related Threads on Entropy of isothermal process reversible\irreversible

Replies
9
Views
65K
Replies
1
Views
1K
  • Last Post
Replies
1
Views
12K
Replies
2
Views
9K
  • Last Post
Replies
12
Views
21K
Replies
5
Views
2K
  • Last Post
Replies
3
Views
4K
Replies
5
Views
13K
Replies
2
Views
7K
Top