@MetalStorm And I have left comments on those two answers as well. It seems you're just overestimating the role of internal pressure and underestimating that of external pressure, which Chet Miller's answer here chemistry.stackexchange.com/a/69834/43057 covers quite nicely actually.

External pressure $P_{ext}$ is always a well-defined constant in isobaric situations regardless of reversibility. And similarly, the work is always $P_{ext}dV$ regardless of any reversibility, no matter what any "poorly defined" $P_{internal}$ is. So, in any isobaric case, we are free to completely ignore anything related to $P_{internal}$ and just use $P_{external}$ everywhere in all calculations instead.

And all your links are consistent with this claim as well, since isobaric processes always have constant well-defined $P_{external}$ anyway which allows us to compute $\Delta H$ using only $P_{external}$ and completely ignore any issues of $P_{internal}$ and anything related to $P_{internal}$ as well.

So, any issues related to $P_{internal}$ mentioned in your books, other answers, and links can simply be ignored for isobaric processes as they have no bearing in any isobaric $\Delta H$ calculations anyway (between equilibrium states) regardless of whether the process is reversible or not.

How come you are able to cancel "-p{ext}dV + pdV" in the derivation? What does the "p" refer to, in your opinion?

@MetalStorm These "new" messages merely claim the same things as my earlier messages. Perhaps you found this re-phrasing easier to understand?

Regarding your last question here -- in the absence of non-PV work, the internal and external pressures are always the same regardless of reversibility simply by Newton's 2nd law as well. You can also refer to the same Chet Miller's explanation for this over here as well too ---> physicsforums.com/threads/…

I assume your whole claim that $\Delta H_{irr} \ne Q$ does not rely on the presence of "some non-PV work", right? Because that would be really silly -- it is very trivial, after all, to just come up with an "irreversible isobaric process" that simply does absolutely zero "non-PV" work whatsoever!

Just place an ideal gas initially at $T_i$ into a heat bath of constant $T_f > T_i$ and watch the piston expand against $P_{ext}$. And this is an isobaric process because $P_{ext}$ is constant throughout the whole process, from start to finish.

But you must note that this is also an irreversible process as well! Because the "opposite direction" will not occur -- the gas will not magically "become cold again" (and contract the piston back to the original position). And there is also "zero non-PV work" as well. And in this process, both the internal and external pressure are always the same the whole time -- the piston is massless and friction-less as well.

So, this is an irreversible isobaric process that does zero non-PV work and where the internal and external pressures are the same. So $p = p_{ext} = $ "constant" throughout the whole process the whole entire time as well. And hence we also have "dH = dq" since dP = 0 by the isobaric criterion.

Yes, now it is much clearer to me. I understand your point. I pinged Chet Miller but he did not respond. May be you can try that yourself? In addition, did you find any book which explicity says that the proof holds for reversible and irreversible processes?

I could find one book, which is the one I pointed in this photo postimg.cc/bSYg5C1t, that specifically says that only for a mechanically reversible process that equation is true.

I tried searching The Atkins' book in insight and there is no comment about the reversibility of the process. I checked other books as well.

@ChetMiller Can you read these messages?

@ManRow May be you can find a book that has that proof and says that it is also true for irreversible processes, apart from your thoughts.

I am 100% sure that Chet Miller has read Smith, Van Ness, and Abbott's book a lot, because we are both chemical engineers. Thus, he might also disagree with what they say

My head is spinning from reading these discussions. H = U +PV is independent of any process or whether a process is reversible or irreversible. The parameters in this equation refer to a thermodynamic equilibrium state. You speak of $P_{ext}$ and $P_{int}$, but you are imprecise as to what these represent. Does $P_{int}$ represent the force per unit area of the gas on the insidede face of the piston?

Sowa P_{wzt}$ represent the force per unit area of the inside face of the piston on the gas. Or does it represent the force per unit area exerted on the outside face of the piston, or something else? If you do a force balance on the piston using Newton's 2nd law of motion for an irreversible expansion or compression, what do you get?

@ChetMiller I summarise. I think that the equation $Q = \Delta H$ is valid for an isobaric, and mechanically reversible process. If the process is irreversible, that equation is NOT true. I leave you an image from Smith, Van Ness, and Abbott's book that state this in Chapter 2: postimg.cc/bSYg5C1t

On the other hand, @ManRow thinks that the equation $Q = \Delta H$ is valid for processes that are isobaric, but reversibility doesn't matter. Thus, the equation is also valid for irreversible processes.

To sum it up, is $Q = \Delta H$ valid for constant-pressure and irreversible processes?

I agree with what you say.