[Kyriee]

Hi!
I'm studying PChem1 and I'm doing some exercises about change in Entropy.
I cannot understand one basic and fundamental point about entropy: it is a state function and, like all state functions, what matter are the initial and the final states. If this is correct, how is it possible that the variation in entropy is different if I'm talking about, for example, a reversible or an irreversible (against an external pressure different from zero) gas expansion? The fact that the trasformation is reversible or not doesn't change my initial or final states, it only changes the way I arrive to the final state.

Sorry for my bad english and sorry if this is a stupid question, but I'm desperate.

[Enthalpy]

Hi Kyriee!

Are the final states really identical? For instance a gas could attain the same final pressure, but with a different temperature and volume, if an expansion is lossy.

[Corribus]

Are the final states really identical?

No, which resolves the apparent contradiction.

[Kyriee]

Why aren't they?
Let's suppose an isothermic expansion of an ideal gas. The states (Pi,Vi), (Pf, Vf) are the same if the expansion is reversible or irreversible; what does change is the work done. Does the variation in entropy have something to do with this work?

I repeat, sorry if these are basic questions, but I really can't understand.

[Corribus]

For example. In an irreversible process there may be heat loss from the system. Therefore the final state of the system that undergoes an irreversible process is different from the final state of the system that undergoes a reversible process, in which no heat is lost. The environment is different in the two processes. Heat, in other words, is not a state function, and the "system" is everything, including the environment. The entropy change during the two processes is not the same, because the final states (which includes the environment) is not the same.

[Kyriee]

Let's take an isothermic expansion of an ideal gas as example.
In a reversible process, the work done by the system is w=nrT*ln(Vf/Vi). Being the process isothermic, q(absorbed)=w(done), since the difference in energy is 0.
In a reversible process, the work is w=p(Vf-Vi), which is equal to q, since it is also isothermic.
In both cases, the heat absorbed by the enviroment is equal to the heat released by the system. So, in the case which T(enviroment)=T(system), the variation in entropy of the universe is zero? Regardless of the reversible/irreversible nature of the transformation?

If not, can you suggest me a book where to study or understand this basic concept? I've tried Atkins, McQuarrie, youtube videos, but I'm still stuck.

[Enthalpy]

Many situations imagined in thermodynamics books and courses aren't completely described. They often forget some elements that provide or absorb work, so for instance "working against the external pressure" isn't all. Missing energy amounts, and resulting logical contradictions, often come from these omitted elements.

And for reversibility, one also must consider other bodies than the gas, vapour... whose behaviour is described. If other components provide heat from a higher temperature or regain it at a lower temperature, a transformation isn't reversible, the entropy increases, but you may not see it at the mentioned body.

In the last example you describe, the pressure of the ideal gas varies (or nothing would happen) so it's not always equal to the outside pressure. A constant outside pressure absorbs work in this process, but more elements are needed to absorb the difference of pressure between the expanded gas and the outer fluid. It can be a piston for instance. This piston absorbs work. Maybe the work converts immediately to kinetic energy, but it's work.

Sidenotes:
"Energy" isn't accurate enough for thermodynamics. You probably meant the internal energy. There are a dozen more.
If the gas expands at constant temperature, it absorbs heat that is lost by something else.

[Kyriee]

I'm really sorry but I don't get it, even if your explanations are good and rigorous.
Thank you very much anyways.

Can you please suggest me some sources that you find useful to resolve this issue?

[Enthalpy]

I know no clear book for thermodynamics, and I regret it. Yves Rocard maybe, but it's in French.

About the last case you proposed: the hypotheses you make are not consistent, so the reasoning can't bring good conclusions.

The expansion at constant temperature needs a variable pressure. But this expansion shall work against a constant outer pressure. This needs an extra element between the inner and outer fluids, like a piston, which takes or gives work but is absent from your description.

[Kyriee]

Thank you!

[sgojja]

The simplest answer is: entropy is a quantity that measures change in system as well as surrounding that certainly makes it different from other quantities like internal energy. In case of reversible or irreversible processes, the change in system is same as it is a state function. what makes these processes different is change in surroundings. we measure both add them up and say that quantity delta S (universe). we are interested in universe's entropy which always increases in an irreversible or spontaneous process. for more explanation on the topic, i would suggest you to read Horia Metiu's book on thermodynamics. for advanced explanations, see, Klotz and Rosenberg. They took a more numerical approach.
i understand, btw, the issue with entropy is that common sense doesn't come to rescue. the first law is intuitive to all but second isn't. it takes effort.
keep grinding!

[pm133]

My understanding of this (and I'm no expert) is that in an irreversible process you can't actually say anything about the path as regards entropy because it depends solely on heat.There are no known equations to solve. Irreversible processes can cause eddy currents, fluctuating pressures and all sorts of chaos which directly affect heat in unknowable ways.

For a reversible reaction you can say something about the heat and that is why you must find a reversible path to calculate entropy changes. Because entropy is a state function you are allowed to do this.

So in summary, irreversible processes mean you don't have a stable, known equation to solve in terms of heat to get entropy changes.
Entropy is a state variable so other paths can be used.
A reversible path is the path to do that.

You'll notice that delta S is described in terms of q_rev.

Hope this helps.
Have to be proven wrong if someone knows better.

[Enthalpy]

What should be a state variable then? And how come is entropy tabulated?

Things like heat or work are not state variable. They depend on the path. That's the reason for the "reversible" dQ/T, by the way.

For H, S and some others, uniform conditions (same T everywhere, no vortices, and so on) suffice to define them, whatever the path.

[pm133]

Enthalpy, I'm afraid I am struggling to understand your post.

What do you mean by "what should a state variable then" mean?
Why is entropy having tabulated values relevant?

I'm not disputing your second paragraph where you talk about the reason for using a "reversible" path. That's essentially what I said, or at least tried to say.

Your 3rd paragraph is confusing as well. If you have uniform conditions like constant temperature, are you saying you must also have reversible conditions? If not, then you will have an irreversible process and no way of calculating those state variables as far as I understand it. That, again to my understanding, is why we find and use reversible paths whenever possible. As an example, for S, we must find a reversible path. For an irreversible path, the most we can say is that S > 0.

Again, happy to be corrected if someone knows more about this than me.

[Enthalpy]

Quite possibly I misunderstood your post.

My point was that, if we know the final state well enough, we can evaluate its entropy.

But if we know a transformation imperfectly, like initial P, T and final P, but some lossy transformation meanwhile, then we can't predict the final T nor S. Were you saying that?

In machines like turbines, we define an "isentropic efficiency" to improve the description over "some lossy transformation", and then computations become possible again.