Is Entropy Change Different for Reversible Paths in Ideal Gas?

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eprparadox
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Homework Statement


An ideal gas is taken from an initial temperature Ti to a higher final temperature Tf along two different reversible paths: Path A is at constant pressure; Path B is at constant volume. The relation between the entropy changes of the gas for these paths is
a) delta S(A) > delta S(B)
b) delta S(A) = delta S(B)
c) delta S(A) < delta S(B)


Homework Equations



delta S = delta Qr / T
Qr = heat transferred to system while the system is going along a reversible path

The Attempt at a Solution


This is one of those checkpoint questions in the chapter and the answer is given as choice a (delta S(A) > delta S(B)).

I'm confused though because in this book, it says that entropy is a state variable and as such, it only depends on the endpoints and is therefore independent of the actual path taken from A to B. But here, we're taking two different paths and yet we're getting that the change in entropy going from one path is different than when we take the other path.

I think the answer should be choice b (delta S(A) = delta S(B)).

It would seem that if you're only dependent on the endpoints, then regardless of the path taken, if you're going from A to B in multiple ways, that the entropy should be the same for all cases.

Where am I going wrong in my thought process? Thanks a lot ahead of time.
 
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eprparadox said:
it says that entropy is a state variable and as such, it only depends on the endpoints
Remind me -- does state include pressure and volume too? Or does the state involve temperature alone?
 
Hurkyl said:
Remind me -- does state include pressure and volume too? Or does the state involve temperature alone?

Ah, I think I see. So if we a system taking two paths to some final state, then that final state is the same for that system only if the pressure, volume, and temperature are all the same?

And if this is true, then the entropy should be the same as well?

Thanks a lot for your quick response.