Is Entropy Change Different for Reversible Paths in Ideal Gas?

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The discussion centers on the entropy change of an ideal gas transitioning from an initial temperature Ti to a final temperature Tf along two reversible paths: constant pressure (Path A) and constant volume (Path B). The consensus is that the entropy change for Path A is greater than that for Path B, as indicated by the answer choice a (delta S(A) > delta S(B)). The confusion arises from the understanding that entropy is a state variable, which typically depends only on the endpoints, leading to the assumption that the entropy changes should be equal regardless of the path taken. However, the final states differ in pressure and volume, which affects the entropy calculation. Ultimately, the conclusion is that while entropy is a state function, the specific paths taken can lead to different entropy changes due to the varying conditions of pressure and volume.
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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?
 
The final states are different. Though they end at the same temperature, the two paths end at different pressures and volumes.
 
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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