ManRow

So, does this mean you finally agree with me then? That an irreversible, isobaric process will have $\Delta H = Q$ if there is no non-PV work? Remember, the process is still both "irreversible" and "isobaric", even if non-PV work is zero! Because $\Delta S_{univ} > 0$ and $P_{ext}$ is constant the whole entire time.

In chat.stackexchange.com/transcript/message/65169603#65169603 I clearly emphasized and explicitly clarified that this requires that "non-PV work is zero".

So, when you have something where both $\Delta S_{univ} > 0$ and $P_{ext}$ is constant, then you have an irreversible isobaric process. Correct? My claim is that it is possible for an irreversible isobaric process to have $\Delta H = Q$.

An "isobaric" only has only one condition -- that $P_{ext}$ is constant throughout the whole process, regardless of reversibility.

An "irreversible" process has only one condition -- that $\Delta S_{sys} + \Delta S_{surr} = \Delta S_{univ} > 0$. That is, it increases the total entropy of the universe. There are many ways to do irreversible processes.

@MetalStorm But how I am being dishonest? My claim this whole entire time is that it is possible for an irreversible, isobaric process to have $\Delta H = Q$. That is what it has always been from the beginning. Does this mean you finally agree with me?

@ChetMiller (I'm asking for clarification as I believe MetalStorm has "misinterpreted" your example, what you wrote earlier.)

@ChetMiller It seems that the book linked by @MetalStorm only requires a "mechanically reversible" isobaric process for $\Delta H = Q$. But, it is possible for even an "irreversible" isobaric process to still be "mechanically reversible" (have $P_{int} = P_{ext}$ throughout the whole process) as your earlier example clearly demonstrates. Does this mean that my claim that $\Delta H_{irr} = Q$ for irreversible isobaric processes with zero non-PV work is true?

@MetalStorm So, in summary --> this is exactly what I am claiming here : chat.stackexchange.com/transcript/message/65169631#65169631 and your book is in full agreement with my statement too because "zero non-PV work" automatically means "mechanically reversible" even for an "irreversible isobaric process" where $\Delta S_{univ} > 0$ anyway...

@MetalStorm So therefore, an irreversible isobaric process "with zero non-PV work" will always have $\Delta H = Q$ because "zero non-PV work" always means "mechanically reversible" anyway just like your book requires. Which means my claim is correct. A process can be "mechanically reversible" but still "overall irreversible" because it instead can just be thermally irreversible instead.

@MetalStorm Also, "mechanically reversible" also seems to follow from the requirement that "non-PV work is zero". If all work is PV-only, then we are obviously in a "mechanically reversible" process anyway since "non-PV work equals zero" means that the piston is also massless and friction-less as well, which already implies $P_{int} = P_{ext}$ aka "mechanical reversibility".

But, it is still an "overall irreversible" process because $\Delta S_{univ} > 0$...

@MetalStorm So I guess your whole confusion was simply assuming that "mechanically reversible" means a process is "wholly reversible", when it does not. It is possible for an irreversible process where $\Delta S_{univ} > 0$ to also be "mechanically reversible" the whole entire time, just like in Chet's example.

Your book is written very carefully, so you cannot assume "mechanically reversible" means "also completely reversible in every possible way", because the latter would have to add "also thermally reversible" as well...

Looks like this exactly what Chet Miller was saying too -- you can have an overall "irreversible isobaric process" with $\Delta H = Q$ if $P_{int} = P_{ext} = $ constant throughout the whole process. But, the process is still irreversible, as Chet Miller said, due to the "very large heat flux". Mechanically reversible yes, thermally reversible no. So therefore, still an irreversible isobaric process with $\Delta H = Q$

So, your book is in perfect agreement with me. An irreversible process that is "mechanically reversible" and at constant pressure can certainly have $\Delta H = Q$. Just a matter of reading closely -- your book requires it to be "only mechanically" reversible, but the process can still be "thermally irreversible" and therefore still an irreversible process overall.

@MetalStorm So, a "mechanically reversible but thermally irreversible" isobaric process can easily have $\Delta H = Q$. Just like my example of the the piston earlier, without any contradiction from your book. This means an irreversible isobaric process can have $\Delta H = Q$, since your book requires only "mechanical" but "not thermal" reversibility, but a "fully reversible" has to have "both reversibilities"...

@MetalStorm Also, I think you have to reread your own book very closely again too. What the book says is for a "mechanically reversible, constant-pressure process", meaning that $P_{int} = P_{ext} = $ constant throughout the whole process. It is possible for an irreversible isobaric process to be "mechanically reversible" as well.

@MetalStorm In short, what your book states is only true if we use $P_{int}$. But am I using $P_{ext}$ and ignoring $P_{int}$, so there is no contradiction between my claim and what the book says, since $P_{int}$ is irrelevant.

@MetalStorm Also does that mean you disagree with Chet Miller? Because his last message perfectly agrees with my claim.

@MetalStorm But that book only talks about $P_{int}$, and $P_{int}$ is irrelevant. So what book says is irrelevant to my claim. It seems you have forgotten to distinguish between $P_{int}$ and $P_{ext}$ again. So therefore, that book does not refute my claim and in perfect agreement with $\Delta H = Q$ for an irreversible isobaric process because you can ignore $P_{int}$.

Nonetheless, Chet Miller describes an irreversible, isobaric process that has $\Delta H = Q$. Which is essentially what I am claiming is possible, and what you are claiming is not possible.

@MetalStorm Also regarding your book request, you cannot seem to find a book that refutes my claim either. So, it seems none of the books have anything to say on this..

@MetalStorm In his example, there is an irreversible heat flux imposed -- hence making the process irreversible. We also have $P_{int} = P_{ext} = $ constant in his answer too. So, therefore, an irreversible isobaric process with $\Delta H = Q$, exactly what you were trying to refute.

@MetalStorm Actually he agrees with me -- read his answer closely. He just said that it is possible (even if difficult) to make an irreversible isobaric process have $\Delta H = Q$

@MetalStorm I think the "@" symbol is required to notify users, so I'm not sure if you've seen Chet Miller's latest message then...

@ChetMiller I think MetalStorm posted a question in the last comment you replied to, so I'm not sure which claim you are agreeing with. Are you saying $\Delta H = Q$ is only true for reversible isobaric processes? (this is what MetalStorm is claiming) My claim is that $\Delta H = Q$ can also be true for irreversible isobaric processes as well provided that "non-PV work is zero" (i.e, that all "work" is PV-only!)

Chet Miller also clarifies that in the comments for this question too over at ---> chemistry.stackexchange.com/questions/173109/…

@MetalStorm I should emphasize the zero non-PV work requirement . If non-PV work is nonzero, i.e, $\delta w_{non-PV} \ne 0$, then $dU = \delta q - p_{ext}dV + \delta w_{non-PV}$ and isobaric $dH = \delta q + \delta w_{non-PV} \ne \delta q$. So, my claim that isobaric $\Delta H_{irr} = Q$ requires that non-PV work is zero, just like as in the "irreversible heat-bath" example.

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.

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.

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.

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!

So, to answer your question -- for isobaric processes, the "p" in your enthalpy equation dH = dq - p_{ext} + pdV + VdP is the same as p_{ext} in the absence of non-PV work.

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/…

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

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.

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.

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.

@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.

@MetalStorm Btw as a late addition -- you mentioned here it would be interesting to get Chet Miller's take on this. I think in this chat system you can even "@"-include users outside of the chat/question itself to notify them here as well!

@MetalStorm Sure, no problem -- you have a good day too**

But, what I am saying, is that this can lead to oddities if we consider, say, just for example, isochoric situations that are reversible or irreversible. If we both agree that work is zero, then we have to agree on the same "heat" regardless of reversibility / irreversibility if the state function, for example, ΔU gives us the same result for both.

@MetalStorm Well then, in that case -- then it seems like a matter of semantics perhaps. That is, only if we are not referring to the "same" work or the "same" heat. Then sure, that is fine.

@MetalStorm Don't have webcam, so can't really use those services

This seems quite mathematically trivial actually

And hence ΔH_{irr} = ΔH_{rev} because Δ(PV)_rev = Δ(PV)_irr anyway

But your text even explicitly says "Because U is a state function, this result is independent of the path of integration, and it is independent of whether the system is maintained in a state of internal equilibrium or not during the actual process; it requires only that the initial and final states be equilibrium states". So, as long the start and end points are the same, then ΔU_{irr} = ΔU_{rev} exactly directly by what your text says

And that is why $dU = TdS$ and $dU = dq$ always in every isochoric situation when your references are identical to the same thing, which is the proper definition anyway

Same as in the isochoric formula dU = T*dS = dQ --> this is always true regardless of reversibility / irreversibility when T and S both only refer to the system or both only the environment because of the definition of entropy in the first place