Entropy, reversible process

[eprparadox]

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.

 

[Hurkyl]

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?

 

[vela]

The final states are different. Though they end at the same temperature, the two paths end at different pressures and volumes.

 

[eprparadox]

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.

 

[rwisz]

I would say that the answer to this question depends on the specific conditions and assumptions of the problem. If we assume that the gas is an ideal gas and that the process is reversible, then the answer is indeed choice a. This is because for a reversible process, the heat transferred to the system (Qr) is equal to the change in entropy (delta S) divided by the temperature (T). Since the temperatures are different for paths A and B, the change in entropy will also be different.

However, if we relax the assumption of reversibility and consider an irreversible process, then the answer would be choice b. This is because in an irreversible process, the heat transferred to the system is not equal to the change in entropy divided by temperature. In fact, the change in entropy for an irreversible process is always greater than or equal to the change in entropy for a reversible process between the same two states. In this case, the specific paths taken from A to B do not matter, as the change in entropy will be the same regardless.

Therefore, it is important to consider the specific conditions and assumptions of the problem when determining the relationship between entropy changes for different paths.