# Why in an irreversible expasion, the pressure is constant

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Why in an irreversible process, is the pressure constant? I understood why I should use Wirrev = -PΔV, but I did not understand why the pressure is constant in this process.

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7 minutes ago, Scienc said:

Why in an irreversible process, is the pressure constant? I understood why I should use Wirrev = -PΔV, but I did not understand why the pressure is constant in this process.

Pressure is not constant in an irreversible expansion. The pressure that you've got there is not a state variable of the system that's expanding; it's the external pressure of the environment, which is constant. Thereby the subindex "ex" in the formula for work.

In fact, during an irreversible expansion, the pressure of the system is not even well defined. It's only the external pressure what's well-defined as a thermodynamical variable.

Edited by joigus
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15 minutes ago, joigus said:

Pressure is not constant in an irreversible expansion. The pressure that you've got there is not a state variable of the system that's expanding; it's the external pressure of the environment, which is constant. Thereby the subindex "ex" in the formula for work.

In fact, during an irreversible expansion, the pressure of the system is not even well defined. It's only the external pressure what's well-defined as a thermodynamical variable.

I think joigus has your answer but I would like to elaborate a bit further.

It's very often the external pressure that counts, especially in chemcial reactions since the reactants in the open test tube are pushing against the armousphere, which is sensible at constant pressure.

So this will be PV work done by the system on the surroundings.

As joigus notes the internal pressure of the system is not measurable.

Thermodynamics (first law)  is about the energies and masses crossing the system boundary and you will often find the calculation easier on one side or the other of this boundary.

Finally this is for irreversible work.

A reversible change would be carried out so slowly that the internal pressure always has time to equilibrate at each infinitesimal expansion or contraction.

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13 hours ago, studiot said:

A reversible change would be carried out so slowly that the internal pressure always has time to equilibrate at each infinitesimal expansion or contraction.

Good point. I forgot to mention this, which is essential, I think, to understanding why in irreversible processes it doesn't work that way.

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On 8/5/2022 at 10:31 PM, Scienc said:

Why in an irreversible process, is the pressure constant? I understood why I should use Wirrev = -PΔV, but I did not understand why the pressure is constant in this process.

What makes this an irreversible process?

As @joigus states, (if V2 = Vex) then P2*(V2-V1) can represent the work done on the environment (hence the minus sign) in making space for the expansion from V1 to V2. However, this is not in principle an irreversible process.

The PV path you indicate represents at least in part, a conversion of internal energy to bulk kinetic energy, possibly involving a frictionless piston. What happens when the system reaches the pressure of the external environment? Does it just stop in its tracks? Or does its own momentum carry it into the vacuum zone enabling the environment to reverse the piston and return in full the energy that it borrowed?

If you want to establish irreversibility, you must demonstrate a global entropy increase involving the transfer of heat across a non-trivial temperature difference. Nothing else is relevant. Certainly as regards teaching the subject. I do remember sitting through this lecture nearly 50 years ago and being confused by its weird assumptions. The damage done was fortunately reversible.

Also bear in mind that if you want to understand where the energy in this system is going, you need to integrate the VdP side of the graph in addition to the PdV side.

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Erratum

For 'if V2 = Vex' read 'If P2 = Pex'

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17 hours ago, sethoflagos said:

If you want to establish irreversibility, you must demonstrate a global entropy increase involving the transfer of heat across a non-trivial temperature difference.

But irreversibility is a premise of the OP.

Also, entropy increase doesn't only happen because of heat transfer. Certainly, entropy change will happen whenever there is heat transfer. But it will also happen in situations in which there is irreversible work involved, like eg when you stir the fluid with a blender or a rapid fan. The fact that irreversible work leads to entropy increase is betrayed by the fact that, after a while, the temperature increases, as this irreversible work is quickly converted to heat. If you wait a couple of seconds, say, and measure the temperature, you'll see that it's decreased (expansion) or increased (compression), and it's no longer possible to know whether this change in temperature has come from irreversible work or from heating with, eg, a Bunsen burner --or from cooling through a wall.

It's only that, in this particular example, it's much easier to calculate the work without involving energy at all, thus reducing the calculational work to a minimum.

Now, what you seem to be demanding from the OP is to:

1) Express the entropy as a function of V, T, and the number of molecules; or perhaps P, V, and the number of molecules, if it's a P,V,T,N system.

2) Calculate the values of S1 and S2, given that S is a state function --it sure is--, and compute it.

Now, step 1) isn't elementary. It certainly can be done, and could be needed in more complicated irreversible processes. But for expansion against a fixed external pressure it's more simply done with the method that the OP proposed.

45 minutes ago, joigus said:

without involving energy at all,

I meant "without involving entropy at all".

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4 hours ago, joigus said:

But irreversibility is a premise of the OP.

Granted, but that leaves us having to guess the context in which the shaded area is indeed an irreversible quantity. Which requires an initial reasonably deep understanding of the nature of irreversibility. Which in turn makes it a very poor way to introduce the concept of irreversibilty to new students, doesn't it? And yet this is what we seem to present to them.

4 hours ago, joigus said:

Also, entropy increase doesn't only happen because of heat transfer. Certainly, entropy change will happen whenever there is heat transfer. But it will also happen in situations in which there is irreversible work involved, like eg when you stir the fluid with a blender or a rapid fan. The fact that irreversible work leads to entropy increase is betrayed by the fact that, after a while, the temperature increases, as this irreversible work is quickly converted to heat. If you wait a couple of seconds, say, and measure the temperature, you'll see that it's decreased (expansion) or increased (compression), and it's no longer possible to know whether this change in temperature has come from irreversible work or from heating with, eg, a Bunsen burner --or from cooling through a wall.

Well this paragraph begins with a statement of fact that I've never observed. The concept of irreversible work is interesting - could this be a conflation of reversible work and irreversible heat transfer occurring more or less simultaneously? I think that by the end of the paragraph you've come to realise that it might be.

5 hours ago, joigus said:

It's only that, in this particular example, it's much easier to calculate the work without involving entropy* at all, thus reducing the calculational work to a minimum.

Certainly, the OP has a nice simple equation to use to calculate a part of the work function. But why has he posted it here? And is it going to help him get to grips with the more interesting part of the integration above the shaded area? It seems that that isn't happening. Why?

5 hours ago, joigus said:

Now, what you seem to be demanding from the OP is to:

1) Express the entropy as a function of V, T, and the number of molecules; or perhaps P, V, and the number of molecules, if it's a P,V,T,N system.

2) Calculate the values of S1 and S2, given that S is a state function --it sure is--, and compute it.

Like it or not, that is the nature of the field. But here I'm simply asking why the OP has labelled an area I would normally expect to be usable work as not so. And maybe elaborate both on the practical context of the example and on his understanding of what's happening in the unshaded area below the PV curve.

6 hours ago, joigus said:

Now, step 1) isn't elementary. It certainly can be done, and could be needed in more complicated irreversible processes. But for expansion against a fixed external pressure it's more simply done with the method that the OP proposed.

I don't see fluid pressure acting on anything here but itself until it falls to P2.

It's likely that my beef is more about the way the OP is being taught the subject.

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5 hours ago, sethoflagos said:

I don't see fluid pressure acting on anything here but itself until it falls to P2.

It's likely that my beef is more about the way the OP is being taught the subject.

Not sure I understand the meaning of 'fluid pressure acting on itself'  ?

We don't actually know much about the way this student is being taught.
The one liner statement isn't sufficient to go on. There is, for instance no mention of a piston or description of what the diagram represents or where it comes from.
Also sometimes students miscopy what is written or mishear what is said, but are intelligent enough to realise later that something is amiss with their notes.

I have already noted my guessed interpretation.

If the diagram represents the expansion of a fluid into the atmosphere, from a small volume at high pressure (points 1 on the diagram) to a larger volume and lower pressure (points 2 on the diagram) and this lower pressure is the expansion into the atmosphere this process is irreversible.

The work done is the work done against the constant atmospheric pressure, represented by the blue horizontal line and the volume change, delta V is the difference between the volumes at point 1 and point 2 on the diagram.
This product PdeltaV is correctly shown by the blue hatched rectangle.

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2 hours ago, studiot said:

Not sure I understand the meaning of 'fluid pressure acting on itself'  ?

Consider a tyre burst.

Allowed near free expansion, the air in the tyre converts itself to a considerably colder, high velocity stream at a pressure suitable to interact with the external environment with balanced forces. It is this sense of internal self-readjustment that I intended. The reason neither P1 nor P2 appear explicitly in the final equation.

This process, the conversion of internal energy to kinetic energy is a work term and isentropic to a first order engineering approximation. However, in the downstream end of the process, instead of driving a turbine or piston, the kinetic energy is simply allowed to return back to internal energy once more, a dissipative heating process that is most definitely irreversible.

When the dust settles the leakage performs Pex*(V2-V1) Joules of work and absorbs Pex*(V2-V1) Joules of heat, restoring the original ambient temperature. (But not the original entropy).

If I've read your posts correctly, then we have essentially the same picture of the context whereby the process is indeed irreversible. And we agree on the value of W.

However, what is missing is any mention of Q, the dissipated heat that balances the 1st Law (Q=W) and creates the 2nd Law consequences of the tyre burst via dS=dQ/T. This seems to indicate a failure to appreciate that there are (at least!) two distinct processes going on here. And they really are as different as chalk and cheese.

2 hours ago, studiot said:

We don't actually know much about the way this student is being taught.

After many years of mentoring junior graduate engineers, the phrase 'irreversible work' speaks volumes.

Edited by sethoflagos
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8 hours ago, sethoflagos said:

Consider a tyre burst.

Allowed near free expansion, the air in the tyre converts itself to a considerably colder, high velocity stream at a pressure suitable to interact with the external environment with balanced forces. It is this sense of internal self-readjustment that I intended. The reason neither P1 nor P2 appear explicitly in the final equation.

This process, the conversion of internal energy to kinetic energy is a work term and isentropic to a first order engineering approximation. However, in the downstream end of the process, instead of driving a turbine or piston, the kinetic energy is simply allowed to return back to internal energy once more, a dissipative heating process that is most definitely irreversible.

When the dust settles the leakage performs Pex*(V2-V1) Joules of work and absorbs Pex*(V2-V1) Joules of heat, restoring the original ambient temperature. (But not the original entropy).

If I've read your posts correctly, then we have essentially the same picture of the context whereby the process is indeed irreversible. And we agree on the value of W.

However, what is missing is any mention of Q, the dissipated heat that balances the 1st Law (Q=W) and creates the 2nd Law consequences of the tyre burst via dS=dQ/T. This seems to indicate a failure to appreciate that there are (at least!) two distinct processes going on here. And they really are as different as chalk and cheese.

After many years of mentoring junior graduate engineers, the phrase 'irreversible work' speaks volumes.

I saw no mention of a burst tyre in the OP.

We don't know that heat is not being considered or other matters on other screens/snapshots of the lecturer's presentation.

In any case you have to calculate the heat and work terms separately.

We don't know if Ideal or other gas laws are being invoked.

In fact we don't know much about the problem at all.

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Here's my analysis of the situation. I did this many years ago to help myself understand the details of this --very well known but quite academic, and at the same time somewhat puzzling-- example.

Diagramatics:

Obviously, there must be an initial pressure in the gas (P1) that is in excess to the environment's pressure (P1>Pex.) Otherwise the gas wouldn't expand --let's assume is an expansion. One problem is this pressure is not mentioned in the statement. That's one reason why it's confusing. There is, of course, a fiducial isotherm wich goes through this point (P1,V1), and we're not going to use either.

Isotherms 1 and 2 are there just as a reference, two coordinate curves on a P,V diagram if you will, but they play no role in the problem. They would look as straight lines on a P,T diagram, and they look like hyperbolae on a P,V diagram. That's perhaps another reason why it's confusing.

Now, very important: The straight line in red does not represent the actual trajectory on the P-V plane of the system. The system is jumping outside of the P,V,T surface of possible states of the system, the reason being that it's going through a series of states that are not equilibrium states. It would look something like the  orange line that joins (P1,V1) and (P2,V2). So @sethoflagos has a point, if I understood them correctly. Something's going on of which we're given no account.

What happens in between? That's the final and most important reason why discussing the problem of the "in between non-thermodynamic states" is so confusing. During this time, the system is no longer described by equilibrium thermodynamics, and sure enough time starts playing a role.

We must appeal to a mix between thermodynamical and mechanical arguments, if only to qualitatively understand what's going on.

Here's what's going on: If the initial and final internal energies of the gas are resp. U1 and U2, we have an energy tradeoff that looks like:

$U\left(t\right)=U_{1}+\delta U\left(t\right)+\delta K_{\textrm{piston}}\left(t\right)$

But it stands to reason that the energy balance can be expressed as a function of only initial and final states, because the "mechanical boundary conditions" if you will, are constrained by the thermodynamics of the problem.

The air from the environment acts as an inexhaustible reservoir of pressure, for lack of a better word. Both external air and internal pressure act on the piston, increasing its kinetic energy as they go. They counteract each other, but don't exactly cancel:

$\delta W_{\textrm{gas -> piston}}=-\int P_{\textrm{gas}}\left(x,t\right)dV_{\textrm{gas}}\left(t\right)$

$\delta W_{\textrm{ex ->piston}}=-\int P_{\textrm{ex}}\left(x,t\right)dV_{\textrm{ex}}\left(t\right)=+\int P_{\textrm{ex}}\left(x,t\right)dV_{\textrm{gas}}\left(t\right)$

$\delta U_{\textrm{gas}}\left(t\right)=+\int P_{\textrm{ex}}\left(x,t\right)dV\left(t\right)-\int P_{\textrm{gas}}\left(x,t\right)dV_{\textrm{gas}}\left(t\right)=-\int\left[P_{\textrm{gas}}\left(x,t\right)-P_{\textrm{ex}}\left(x,t\right)\right]dV_{\textrm{gas}}\left(t\right)$

Finally, the piston stops. How does it do that? The gas, no doubt, overshoots a little bit --see diagram--, so even though the reservoir to a good approximation stays where it is thermodynamically, locally it exerts an excess pressure to take the piston to its final position. So, initially, the piston works against the air with its local field P(x,t) averaged on the piston's surface; but finally, the air works against the piston to compensate for the overshooting and leaving it at its final rest position. We must assume also that the walls leave heat go in and out. So,

$K_{\textrm{piston}}\left(2\right)=0$

So no matter how complicated the details are, the overall effect of the air is to compensate for the unbalance in the work terms so as to deliver the unbalance back to the gas, so to speak, in such a way that, energetically, it's as if all the time the pressure Pex had been the only pressure that's been doing all the work.

Final note: The proposed curve is, of course, a simplification. Really, the gas doesn't have a thermodynamic pressure, as @studiot and myself have pointed out. But even so we can talk, I think, about an average pressure that the gas exerts on the piston near the surface. That's the pressure I'm talking about when I wrote things like Pgas(t)

Edited by joigus
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3 hours ago, joigus said:

Here's my analysis of the situation. I did this many years ago to help myself understand the details of this --very well known but quite academic, and at the same time somewhat puzzling-- example.

Diagramatics:

Obviously, there must be an initial pressure in the gas (P1) that is in excess to the environment's pressure (P1>Pex.) Otherwise the gas wouldn't expand --let's assume is an expansion. One problem is this pressure is not mentioned in the statement. That's one reason why it's confusing. There is, of course, a fiducial isotherm wich goes through this point (P1,V1), and we're not going to use either.

Isotherms 1 and 2 are there just as a reference, two coordinate curves on a P,V diagram if you will, but they play no role in the problem. They would look as straight lines on a P,T diagram, and they look like hyperbolae on a P,V diagram. That's perhaps another reason why it's confusing.

Now, very important: The straight line in red does not represent the actual trajectory on the P-V plane of the system. The system is jumping outside of the P,V,T surface of possible states of the system, the reason being that it's going through a series of states that are not equilibrium states. It would look something like the  orange line that joins (P1,V1) and (P2,V2). So @sethoflagos has a point, if I understood them correctly. Something's going on of which we're given no account.

What happens in between? That's the final and most important reason why discussing the problem of the "in between non-thermodynamic states" is so confusing. During this time, the system is no longer described by equilibrium thermodynamics, and sure enough time starts playing a role.

We must appeal to a mix between thermodynamical and mechanical arguments, if only to qualitatively understand what's going on.

Here's what's going on: If the initial and final internal energies of the gas are resp. U1 and U2, we have an energy tradeoff that looks like:

U(t)=U1+δU(t)+δKpiston(t)

But it stands to reason that the energy balance can be expressed as a function of only initial and final states, because the "mechanical boundary conditions" if you will, are constrained by the thermodynamics of the problem.

The air from the environment acts as an inexhaustible reservoir of pressure, for lack of a better word. Both external air and internal pressure act on the piston, increasing its kinetic energy as they go. They counteract each other, but don't exactly cancel:

δWgas -> piston=Pgas(x,t)dVgas(t)

δWex ->piston=Pex(x,t)dVex(t)=+Pex(x,t)dVgas(t)

δUgas(t)=+Pex(x,t)dV(t)Pgas(x,t)dVgas(t)=[Pgas(x,t)Pex(x,t)]dVgas(t)

Finally, the piston stops. How does it do that? The gas, no doubt, overshoots a little bit --see diagram--, so even though the reservoir to a good approximation stays where it is thermodynamically, locally it exerts an excess pressure to take the piston to its final position. So, initially, the piston works against the air with its local field P(x,t) averaged on the piston's surface; but finally, the air works against the piston to compensate for the overshooting and leaving it at its final rest position. We must assume also that the walls leave heat go in and out. So,

Kpiston(2)=0

So no matter how complicated the details are, the overall effect of the air is to compensate for the unbalance in the work terms so as to deliver the unbalance back to the gas, so to speak, in such a way that, energetically, it's as if all the time the pressure Pex had been the only pressure that's been doing all the work.

Final note: The proposed curve is, of course, a simplification. Really, the gas doesn't have a thermodynamic pressure, as @studiot and myself have pointed out. But even so we can talk, I think, about an average pressure that the gas exerts on the piston near the surface. That's the pressure I'm talking about when I wrote things like Pgas(t)

+1

Personally, I found it very difficult to develop an intuitive feel for PV diagrams. In power system design, an engineer is far more likely to use HS diagrams as they indicate changes in energy so much more clearly. So the path for the OP case would be a near vertical plunge south for the expansion phase followed by a steady drift north-east for the reheating phase. No need to worry about any mysterious goings on in the P,V,T planes.

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On 8/14/2022 at 2:15 PM, joigus said:

Final note: The proposed curve is, of course, a simplification. Really, the gas doesn't have a thermodynamic pressure, as @studiot and myself have pointed out. But even so we can talk, I think, about an average pressure that the gas exerts on the piston near the surface. That's the pressure I'm talking about when I wrote things like Pgas(t)

Just to be clear, I have a different viewpoint on this. Modelling of expansion/compression processes assuming constant PV^k is just too good a predictor of real machinery performance to be ignored. Both P and V must be quite mappable through these highly dynamic changes of state irrespective of equilibrium considerations.

It is unfortunate that thermodynamics and fluid flow are treated in most universities as separate topics since scenarios such as the OP are under the control of not just the thermodynamic equations of state, but also the appropriate form of Navier-Stokes equations. And the particular example of the OP is dominated by the most challenging form of the Navier-Stokes: that for flow of compressible fluids where both inertial and viscous terms are significant in all three spatial dimensions.

Little wonder the OP is confused. And if he's doing one of the pure sciences, it's highly unlikely that he'll ever be given the analytic tools that might help him make sense of it.

He would learn far more from looking at gas expansion through a porous plug. Same process of turning U to W but so little of that nasty kinetic stuff that it can be cheerfully ignored.

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On 8/14/2022 at 6:51 PM, sethoflagos said:

+1

Personally, I found it very difficult to develop an intuitive feel for PV diagrams. In power system design, an engineer is far more likely to use HS diagrams as they indicate changes in energy so much more clearly. So the path for the OP case would be a near vertical plunge south for the expansion phase followed by a steady drift north-east for the reheating phase. No need to worry about any mysterious goings on in the P,V,T planes.

On 8/15/2022 at 6:42 PM, sethoflagos said:

Just to be clear, I have a different viewpoint on this. Modelling of expansion/compression processes assuming constant PV^k is just too good a predictor of real machinery performance to be ignored. Both P and V must be quite mappable through these highly dynamic changes of state irrespective of equilibrium considerations.

It is unfortunate that thermodynamics and fluid flow are treated in most universities as separate topics since scenarios such as the OP are under the control of not just the thermodynamic equations of state, but also the appropriate form of Navier-Stokes equations. And the particular example of the OP is dominated by the most challenging form of the Navier-Stokes: that for flow of compressible fluids where both inertial and viscous terms are significant in all three spatial dimensions.

Little wonder the OP is confused. And if he's doing one of the pure sciences, it's highly unlikely that he'll ever be given the analytic tools that might help him make sense of it.

He would learn far more from looking at gas expansion through a porous plug. Same process of turning U to W but so little of that nasty kinetic stuff that it can be cheerfully ignored.

You make some good points that the problem has other possible alternative treatments, probably more realistic, and that the real problem is more involved. But I'm not sure that using the Navier-Stokes equation would be the best approach for, eg, freshman or sophomore students. We don't know the student's level, so... It seems he's been exposed to the --probably simplistic, granted-- formula of Pex(V2-V1). Because that's the recipe that he's going to be responsible for in his assignments and exams, I suppose, trying to draw a simple intuitive reasoning of how it works and why, to a first approximation is, IMO, the way to go here.

There is a million-dollar prize for just proving the existence and uniqueness of the Navier-Stokes. Generations of mathematicians have failed at solving it in general.

On the other hand, H, S are quite abstract in comparison to P, V, which are a lot more intuitive.

Should we start teaching gas behaviour by the Van Der Waals equation? It's certainly closer to how real gases behave. There's the key to phase transitions, the triple point of water, and so on. But generations of students have come to understand gases first with the concept of an ideal gas for a reason.

As to using PVk, I'm all in favour of it for general quasi-static processes, when the system has a P. When the system is out of equilibrium and adjusting to a new equilibrium, I'm not sure it's the right approach from the conceptual point of view.

Unless you know of a way to derive an effective P and an effective k from the Navier-Stokes equation in out-of-equilibrium situations. If such method exists, I'm not aware of it and I would love to learn about it to the extent of my abilities, which are quite limited, I must say.

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2 hours ago, joigus said:

I'm not sure that using the Navier-Stokes equation would be the best approach for, eg, freshman or sophomore students.

It most definitely isn't. Far better to pick example thermodynamic processes whose paths are simple, well defined and relevant.

2 hours ago, joigus said:

We don't know the student's level, so... It seems he's been exposed to the --probably simplistic, granted-- formula of Pex(V2-V1). Because that's the recipe that he's going to be responsible for in his assignments and exams, I suppose, trying to draw a simple intuitive reasoning of how it works and why, to a first approximation is, IMO, the way to go here.

Excellent point. But this result could have been obtained by considering the gradual inflation of a balloon for example. I pick this because in good teaching examples of PV work, it is usually arranged that the pressure exerted by the surroundings on the surface of the system is well defined and equal to the pressure exerted by the system on the surroundings.

In the OP's example, this is not the case: P1, P2, and any possible thermodynamic path between them are complete and utter red herrings.

3 hours ago, joigus said:

Should we start teaching gas behaviour by the Van Der Waals equation? It's certainly closer to how real gases behave. There's the key to phase transitions, the triple point of water, and so on. But generations of students have come to understand gases first with the concept of an ideal gas for a reason.

I don't think I mentioned Z, did I? Bit strawmannish there, J 😉. Besides which, all the process simulators seem to be running Peng-Robinson these days. But that's for much later in the course, as you say.

3 hours ago, joigus said:

As to using PVk, I'm all in favour of it for general quasi-static processes, when the system has a P. When the system is out of equilibrium and adjusting to a new equilibrium, I'm not sure it's the right approach from the conceptual point of view.

Unless you know of a way to derive an effective P and an effective k from the Navier-Stokes equation in out-of-equilibrium situations. If such method exists, I'm not aware of it and I would love to learn about it to the extent of my abilities, which are quite limited, I must say.

By 'out of equilibrium' I'm assuming you mean that the system has significant P, T gradients, ie parts of the system are not in thermodynamic equilibrium with other parts never mind the surroundings.

Whether or not it's necessary to invoke N-S (thankfully not in most cases), a reasonable approach may be to shrink the system under consideration to a differential volume moving with the local mean fluid velocity. This unit cell can be assumed to be at some localised P, T equilibrium state with no mass interchange with the surrounding cells, and therefore the standard thermodynamic analytic techniques should be applicable to the interactions of the unit cell with its 'external environment' of neighbouring cells.

And then the maths starts, using the equation of state to relate the pressure and density terms in the applicable flow and mass continuity equations. Pointless to go into the methodologies as these are highly case dependent. Sometimes you're lucky and end up with a simple ODE. Sometimes not.

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8 hours ago, sethoflagos said:

In the OP's example, this is not the case: P1, P2, and any possible thermodynamic path between them are complete and utter red herrings.

I don't agree with this statement, in thecontext of this question.

Have you ever wondered why the OP has not come back to discuss it further, although he has been here since he posted it ?

I do also wonder if there was not a mistake on the whiteboard (or copying from it) and that the Pex under the integral sign should be just P or that there was a carry over from a previous whiteboard we haven't seen ?

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38 minutes ago, studiot said:

I don't agree with this statement, in thecontext of this question.

Perhaps @joigus put it more succintly:

On 8/14/2022 at 2:15 PM, joigus said:

I draw your attention to the phrase 'We don't know and we don't care!'

Only delta V is relevant in context.

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