@MetalStorm If you are claiming in the isochoric case that Delta U ≠ Q, then you are in violation of the very first law of thermodynamics. If you apply zero work and only heat in an isochoric case reversible or not, then by energy conservation -- which does not care about reversibility but is a universal law in every case -- the internal energy of the system must increase by the amount of heat energy introduced or decrease by that same amount.

The issue is that $Delta U = Q$ even in very irreversible cases is obvious to show. Consider the irreversible isochoric heating of a system from $T_1$ to $T_2$ by a heat bath "already at $T_2$" to begin with. This is not a quasi-static process and the temperature is not uniform (system is at $T_1$, heat bath is always at a $T_2 \gg T_1$).

Nonetheless, we will still transfer $\Delta U = nC_v*\Delta T = nC_v*(T_2-T_1) = Q$ amount of heat from the reservoir to the system. Or, are you claiming we do not transfer this amount?

Also clearly this is irreversible because $\Delta S_{sys} = nC_v*\ln(T_2/T_1)$ but $\Delta S_{surr} = n*C_v*(T_1-T_2)/T_2$ which are not equal and hence entropy has been generated in the system. But, nonetheless, an amount of heat $Q = T_2 \Delta S_{surr} = \int_{T_1}^{T_2} TdS_{sys}dT$ has been transferred to the system.

Also, the only requirement an isobaric process must fulfill is that the external pressure is always constant. This is the only requirement in either the reversible or irreversible cases.

And according to Chet Miller himself we can always use the external pressure, if well-defined, to calculate work regardless of reversibility / irreversibility. And for all isobaric cases, this external pressure $P_{ext}$ is always constant and thus perfectly well-defined.

Here is another link where Chet Miller explains why we can always use the external pressure (whenever well-defined) to calculate the work performed in all cases reversible or not. And in isobaric scenarios this is especially the case since $P_{ext}$ is a simple complete constant (hence likewise "well-defined") in those cases anyway.

Before answering, I think you are avoiding the texts that I have sent you. It is healthy to look first at the texts. You can argue that the texts are wrong, and I don't have nothing against that. But I am sending books, that everyone knows from university, which claims the same thing I am telling you.

I am not an authorative source, books are.

*authoratitive

But I also looked at your texts too. If you read it closely, Atkins in section 2.3 does not say that we cannot use the external pressure only to compute dW. In the image you provided here he is clearly not making any statements at all about the specifically external pressure. You just have to notice very carefully between the lines that Atkins is not making any statements about the explicitly external pressure.

In the image there it says: only for MECHANICALLY REVERSIBLE, constant-pressure process can heat and work be calculated by the equation $ Q = n \Delta H$

it literally says reversible

It literally excludes EXTERNAL pressure

That image is not from the Atkins' book

Because only for reversible processes you can say that p_{ext} = p_{int}

But in isobaric scenarios you don't have to use $p_{int}$

So you can freely use $p_{ext}$ which Atkins has no problem with

Atkins says you cannot use $p_{int}$ unless the process is reversible

If the process is irreversible, you don't even know the value of p_{int}

If you know p_{out}, you can calculate w as you know for sure

But to say that \Delta H is going to be the same as that work you calculated, is wrong

You cannot establish that equality between work and enthalpy change, the same as for heat and internal change, for constant-P and constant-V processes

But is the same! The "work performed" is always p_{out}dV regardless. You can only use $p_{int}$ when it's reversible, but we don't need to worry about $p_{int}$ because $p_out$ is always well-defined and constant in isobaric situations

So in this case Atkins doesn't refute any of my claims

I said that I agree with that

you can calculate work like that

but that value is not equal to \Delta H

But then the math disproves your claim ---> physics.stackexchange.com/a/799753/168695

I have seen the math and I will repeat it again

Since PdV in the enthalpy formula is equivalent to dW from the dU

And opposite in sign

when you write dU, you have two terms, w and q

when you say that w=-pdV for irreversible processes, that is NOT TRUE. You can only say w=-p_{ext}dV

Only when you write -pdV, it cancels out with the other pdV that is on the right

So you are assuming that the $P$ in the enthalpy definition $H = U + PV$ refers to the internal pressure of the system?

the p in that formula refers to the system. But the p that appears in U is only the system's pressure for reversible process

The SAME happens with T

When it is irreversible, you have T_{ext}S

Only when the heat transfer is reversible, you have that T_{ext} = T , and something similar takes place

If it is irreversible, you get an inequality, at most

But you cannot identify those terms as HEAT or WORK

I know this seems very subtle, but it is what it is

Do you have a source for that assignment of $P$? from $H$? Perhaps something in Atkins?

I will send you two pages from a book that clarifies this

are you a chemist?

No, but even if it refers to the "system" pressure, it is still irrelevant because the start and end points of even an irreversible isobaric process are still equal to $P_{ext}$

Hence the change in enthalpy, as a state function, cannot distinguish between reversible or irreversible path taken between them

If you have the expression "(x^2) dy - (x)dy", do you think that is zero?

You are saying this "p_{ext}dV - pdV = 0"

Yes because p and p_ext are identical and have the same power

Unlike your (x^2) and (x) formula

And (x)*dy - (x)*dy = 0 always

They are identical for mechanically reversible processes

because those pressures are equal

Only in that case

No they are identical always because p_ext = p_ext

And they refer to the same thing when p = p_ext at the endpoints

If p = p_ext, like for an irreversible isobaric process only at the endpoints only, then still, at the endpoints, pdV - p_extdV = 0

And you only care about endpoints because enthalpy is a state function regardless of path always

*anyway

If H = U + PV then ΔH = ΔU + Δ(PV), and Δ(PV) = P_fV_f - P_iV_i = P_ext*(V_f-V_i) = P_ext * ΔV

Because P_f = P_i = P_ext

regardless of anything that happens "during" the process the start and end points -- endpoints only and nothing else -- are the same as P_ext

But you only care about endpoints because H is a state function anyway, so we are good here

So, we have ΔH = ΔU + P_ext * ΔV for isobaric processes. So, there is no difference then in ΔH between reversible and irreversible processes if both occur under the same P_ext and ΔV, unless you are claiming ΔU_{irr} ≠ ΔU_{rev} ??

(After all, mathematically that is the only way your claim that ΔH_{irr} ≠ ΔH_{rev} can work!)

Well, I think that there is not much I can do, I think you are not reading what I am saying. I can send you you two pages from the book that explain this. The book is "Molecular Thermodynamics from Fluid Phase Equilibria" by Prausnitz , Lichtenhaler, and de Azevedo.

I don't know if you have an email account

Sure, can you send them to [email protected] ?

I have sent them

Well, I am going to rest, have a nice day!

@MetalStorm Well, regarding "what you are saying" -- it doesn't seem to line up with the math. Hence I am contradicting your claims (which Atkins doesn't seem to 100% support either upon "close" inspection)

Just looked at the sections you sent -- none of them contradict what I am saying upon the same inspection

The books are correct but they don't contradict what I'm saying

Just like how Atkins says you can't use internal pressure for irreversible scenarios, it's irrevelant since I don't need "internal" (only external) pressure anywhere at all. Just like how the chapter you just sent says "dU < T_b * dS" for irreversible processes, the "dS" is for system not the environment and also irrelevant because "dQ = T*dS" always when both T and S refer to the same thing

Regardless of irreversibility -- that is simply per the definition of entropy

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

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

Maybe we can arrange a meeting through zoom google meet, you can you me how you go with the formulas and try to work it out

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

This seems quite mathematically trivial actually

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

Yes, state functions will yield the same. But they are not longer equal to work and heat if they are not reversible

Well, then that's fine

I think we can just disagree, I guess

You say that those equalities hold for irreversible processes, I say they do not (although, you can calculate \Delta U and \Delta H without a problem)

Those are the stances... Hope you have a good day

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

@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!