Spontaneity at Constant Volume

[@vicelabae]

Hi Mr. Steven. Thank you so much for the videos, your effort is much appreciated. That said, there is something in the video bugging me and i hoped you could help me sort it out. The clausius inequality integral has it's argument evaluated at the boundary of the system. Since entropy definition comes from the inequality, entropy is also evaluated at the boundary. For an irreversible process it'd be like dS = (đq_irr / Tb) + đσ Where Tb is the boundary's temperature, and đσ is the entropy generated by internal reversibilities. Now, in the reversible limit dS = (đq_rev / T) since in a reversible process theres no temperature gradient, Tb = T, where T is the temperature if the system were in equilibrium. In general for an irreversible process Tb ≠ T, then why are you not making the distinction between T and Tb when writing the expression in 1:37? Shouldn't it be dS >= (đq / Tb)?

[@gelomelo3426]

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[@PhysicalChemistry]

You're much too kind, thanks

[@mortezakhoshbin]

Double like!

[@PhysicalChemistry]

Haha, double thanks!

[@rinsim]

Hi Steven, In one of the previous lessons we demonstrated that dU = T*dS - p*dV for a reversible process. But because we have only exact differentials here, we said (kzbin.info/www/bejne/jJjPeJpnmtOtrpo at 4:38) that this relationship is always valid. I'm confused because that is in conflict with this video: if that was true, dU - T*dS would be always 0 at constant volume. I suspect the formula for dU is only valid for reversible processes. Because we ended up with exact differentials under some assumptions, we cannot then generalise. Am I missing something?

[@PhysicalChemistry]

This is indeed a subtle point. I think the best way to resolve the apparent contradiction is not to think about whether the process is reversible, but whether the endpoints of the process are in equilibrium or not. (These are similar, but not identical, concepts.) For example, if I have some constant-volume process A → B, where both the initial and final states ARE in equilibrium, then dU = -T dS will be true, regardless of whether the path is reversible or irreversible. U is a state function, so ΔU does not depend on the path. On the other hand, suppose A is some non-equilbrium microstate. Perhaps, in violation of Boltzmann's expectations, I have prepared the system so all of the particles are in some excited state, and none are in the ground state. Then the internal energy of this system ( Σ pᵢ Eᵢ ) will not be the same as we would predict from the state variables S, T, P, and V. I.e. the state function U(S,V) is incorrect because we are not in equilbrium. (Think of the state function U(S,V) as a "macrostate" function. It is useless if we can't summarize our microstate { pᵢ } well with a macrostate {S,V}.) Then, if I let the system relax to some final state B that IS in Boltzmann equilibrium (by letting the proper fraction of molecules fall to the ground state, for example) then it will be true that dU - T dS < 0 for this spontaneous process. So you're right that dU = T dS - P dV (and its equivalent for finite differences) is not always true. But it fails when the endpoints are non-equilibrium. (In which case the path will also be irreversible.) It holds when the endpoints are in equilibrium, even if the path between them is irreversible (i.e. passes through some non-equilibrium states).

[@rinsim]

@@PhysicalChemistry Thanks for the detailed reply! The difference between equilibrium and reversibility is indeed tricky. And I very often forget that most of the relations only hold for states of equilibrium.

[@rinsim]

One more thing: for what we've said, when we have dU - TdS != 0 then at least one of the initial and final states must be of non-equilibrium. For example liquid water at -10 C. If I understand correctly, when we go from a state of equilibrium A to a state of equilibrium B I don't think we can say that dU = TdS in any intermediate state, or at least we need to interpret this. In my understanding we can say that exists a path (a reversible path) where dU = TdS in any intermediate state, and so U - TS is constant along this path. Any intermediate state is a state of equilibrium. In other words when I say dU = TdS this expresses the infinitesimal increment of U from the state of equilibrium {S} to the state of equilibrium {S+dS}, so how U changes along the reversible path. But because we only care about start and end point, thinking that I'm going from A to B in a reversible way is easier, as I can use dU = TdS to integrate from A to B. But I'm not following the real path, just a path convenient to me. If the process is irreversible, as U - TS in A and B is the same and somewhere in the path I have dU - TdS < 0 (an intermediate state of non-equilibrium), then I must have done something non-spontaneous in the process (dU - TdS > 0). Did I understand correctly?

[@kylecatman7738]

uh, far out man, are you like a doctor or something?? :o