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Irreversible Quasistatic PV Work

  1. Mar 21, 2014 #1

    1) I am reading Pippard's Classical Thermodynamics and was confused by one of his examples in the attachment.

    What I do not understand is the concept of using P' (I am thinking P' as = P_fric + P_interface) for work calculations. If I take the system as composed of just the gas inside the cylinder and assuming this system is insulated and the process quasistatic, pressure at the gas-piston interface should be equal (P_ interface) and the system only sees P_ interface applying work onto it during compressing and applying P_ interface during expansion. I get that P' is important because it includes friction, but the system/gas doesn't see this since it is happening outside its system boundaries.

    2) For the same insulated piston cylinder system, if the compression-expansion process is quasistatic but the process contain irreversibilities like friction which is converted into heat and absorbed by the system since it is insulated, would first law become Qfric - W = E_2 - E_1 where W = P_ interfaceΔV and external applied pressure equal to P_interface + Pfriction?
    I'm having trouble imagining how this system is irreversible since all of the applied energy in one form or another is added to the system; why wouldn't it be able to return to its initial state?

    Thanks very much

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  3. Mar 22, 2014 #2
    They're assuming that all the frictional heat goes into the gas. If this happens, then the approach of treating the gas as the system doesn't work because its temperature is affected by this heat.
    No. It would become - W = E_2 - E_1 , where W=P'ΔV.
    If it's done isothermally, no problem. But, if it's done adiabatically (which apparently is the case they're looking at), the frictional work contributes increases in internal energy during both expansion and compression. So this can't be reversed.

  4. Mar 22, 2014 #3
    The way I am imaging this, P' is applied to the piston, and the piston compresses the gas at P_int < P' and the rest used to overcome friction between the piston and cylinder wall outside the system (taking system as only gas). To me, the gas only sees P_int applied onto it and doesn't see P' or friction. What are other options for defining a system in this case where it would make sense to include P'? What happens if frictional heat is assumed to go elsewhere and not the gas?

    Why is it incorrect to treat friction as a heat term and use W=P_intΔV; also does |P'ΔV| = |Qfric| + |P_intΔV|?

  5. Mar 22, 2014 #4
    You ask such good questions!!! The "system" they are using includes the piston (which is assumed to be massless, and to have zero heat capacity). Their analysis assumes that all the frictional heat goes into the gas.

    If the frictional heat is assumed to go elsewhere and not the gas, then the problem reduces to your analysis (which, of course, for that case, is correct).

    You don't have direct access to Qfric. You don't even have direct access to (P' - P), unless you put a pressure gage inside the cylinder. It is, of course, possible to calculate P incrementally, driven by the work (P'-P)dV, and to solve for P along the path as well as for the incremental amounts of frictional heat along the path. But, this is not as convenient as just measuring P' and calculating the integral of P'dV. This will give you what you want much more easily. That's what they do in your book.

  6. Mar 23, 2014 #5
    I was told to never include two separate components in one system which was why I never thought to include the piston; so the piston is basically a 'void' in the system that does nothing thermodynamically (i.e. work cannot be done by/onto it, change in its KE, PE is insignificant etc.)?

    For irreversible, non-quasistatic processes, wouldn't the above imply that W = ∫P'dV? I've read for rreversible, non-quasistatic, internal gas pressure is usually undefined, and work does not equal to ∫PdV, even if it can be computed. Instead, from this post and this post (where the system appear to be gas only), work is compared to ∫PdV as an inequality, although I'm not sure which P is being used. My confusion is now basically which pressure is used in PdV calculations for irr. quasistatic and non-quasistatic cases (on the piston, inside, or outside of the gas-piston boundary), and if W = ∫P'dV is true for non-quasistatic cases, does this contradict two posts I mentioned above.

    If friction goes elsewhere, the system would be the gas and W = ∫P_interfacedV = E_2 - E_1? Would this case still be irreversible as friction existed in the process but outside the system?

    Thanks very much
    Last edited: Mar 23, 2014
  7. Mar 23, 2014 #6
    Yes, if the system as taken as the gas plus piston (as your book does).
    I think that all this can be cleared up for you by reading my 2 page Blog at my personal PF area. But, in summary, what you said is correct: W = ∫P'dV applies both to reversible and irreversible processes (if P' is taken as the force per unit area at the interface between the system and surroundings). It's just that, in a reversible process, P = P' everywhere throughout the system.
    Yes. If you chose the gas as your system and the deformation were quasistatic, this would be correct.

    No. For the gas as the system, the process would be reversible.

  8. Mar 23, 2014 #7
    In both your above and blog post, is the interface between the system and surroundings taken at the gas-piston interface (gas only system) or at the piston-outside (gas + piston)? How come work and ∫PdV are expressed by an inequality in the two posts I referenced above?

    In your blog post, for non-quasistatic cases, is PI(t) taken on the gas side of the interface or the piston side? What if PI(t) varies spatially in all directions, how would one define a PI(t)?

    So in essence, how the system is defined will dictate whether it is reversible or irreversible even if the process as a whole (piston-cylinder plus gas) is irreversible?

    Thanks very much
  9. Mar 23, 2014 #8
    Wherever you choose to specify. You pick the system that's most convenient, and for which you can most easily determine the pressure at the interface.
    I don't know. I wasn't able to follow the details of these developments. All I know is that, to get the work, you need to evaluate it at the interface between the system and the surroundings. That's the work that's being done on your system.
    Whichever is more convenient to do the analysis. Here is a reference to a current thread that Loopint and I have been working on to apply these principles to some specific problems. I think you will find it very interesting and educational.
    Are you saying, "what if the pressure at the interface varies with spatial position along the interface?"
    In this particular case, yes. In other cases, it is often not possible to identify any subsystem in which a reversible change is occurring. See, for example, some of the problems analyzed in the thread I referenced above.

    P.S. Have you checked out that thread that I referenced in my previous post. We are currently working on a problem where a mass M is suddenly released onto a piston of mass m enclosing a gas inside a cylinder. The deformation is irreversible for the gas.

    Last edited: Mar 23, 2014
  10. Mar 23, 2014 #9
    From this post:
    which is what I was confused by, as this statement states W ≠ ∫PdV.

    From what I read (such as in this post),"external" pressure applied to the gas must be used to evaluate work in non-quasistatic cases. But what is "external" pressure? If the system is gas only, is it that applied onto the piston or just on the outside of the gas-piston interface? Which pressure is PI(t) referring to in your blog?

    Yes that's what I meant, I would guess that an additional spatial integration to account for pressure variation would be needed?

    Thanks very much
  11. Mar 23, 2014 #10
    First you specify what your system is, and then you focus of the boundary of your system. This is the interface between your system and the surroundings. This is where you have to specify PI. If you can't specify the pressure on this boundary, then you have to identify a different system to use that has a boundary on which you do know the pressure.

    I disagree with the last sentence in the quote. If you choose the system (as you did with the gas) where you know the pressure throughout the system (and, of course, at its boundary), then ∫PdV is the work done by your system on the surroundings (provided that you take into account the effect of the frictional temperature rise of the gas on the pressure, or if none of the frictional heat goes into the gas).

    Yes indeed. Very perceptive. For a possibly curved interface where the pressure varies with location at the interface, the instantaneous rate of doing work on the surroundings is equal to the component of velocity normal to the interface times the local pressure, integrated over the interface.

  12. Mar 24, 2014 #11
    What about the case of free expansion? In most examples the gas is taken as the system and at the interface of the gas during the expansion process, PI on the gas side is finite (and difficult to define) while PI on the vacuum side is 0. I believe the argument of using "external"/vacuum pressure arises from this example, as that will give W = 0 as expected. However this case is contrary to what you said as boundary pressure is undefined and discontinuous; is there another way to look at this problem?

    In rapid/non-quasistatic compression/expansion, does taking the gas as a system mean that pressure is discontinuous at the system/gas-piston boundary (pressure on gas side is higher/lower than on piston side in expansion/compression) due to the lack of pressure equilibrium? I was thinking this may be a reason why work is expressed as an inequality to ∫P_extdV since PI isn't well defined. I am under the impression that for PI to be useful pressure should be at local equilibrium at the boundary.

    Thanks very much
  13. Mar 24, 2014 #12
    I never said that the boundary pressure is undefined and discontinuous. You must have gotten that from someone else's quote. Also, if the piston is massless, the gas pressure on the bottom of the piston is the same as the external pressure at the top of the piston, i.e., zero.
    As I said, if the piston is massless, the forces per unit area (pressure) on both sides of the piston are the same. On the gas side of the piston, even though the pressure within the gas is non-uniform, PI still represents the gas pressure at the interface.
    I don't quite follow what you are saying here, but, if you are saying that there is some sort of pressure discontinuity at the base of the piston, this is not the case. If your system is the gas, then PI represents the value of the gas pressure at the base of the piston, and this is just the pressure value at that location.

    If the piston has mass, then you have to decide whether you want to take the boundary of your system so as to include the piston or to exclude the piston. In the case where your system includes the piston, then the work is PextdV (=PIdV). If your system does not include the piston (and is only the gas), then PIdV ≠ PextdV. In this case, PIA=mg+PextA+ma, where A is the area of the piston, and a is the acceleration of the mass.

  14. Mar 25, 2014 #13
    Hi Red_CCF,

    The questions you are asking are great, but it might be easier to explain how this plays out if we focus on a specific example. A few posts ago, I mentioned another thread where we are working on the solution to an irreversible adiabatic compression problem. The problem consists of a cylinder of gas oriented vertically within an evacuated room. Initially, there is a frictionless piston (with mass m) sitting on top of the gas, and the piston and gas are in equilibrium. We initially hold another mass M just barely touching the top of the piston, and at time zero, we release the mass M. We wish to determine the amount of work done by the gas on the base of the piston over the course of time until the system has re-equilibrated. We also wish to determine the temperature of the gas in the new equilibrium state. We can rework on this problem here in this thread, or you can look over what was done so far in the other thread and ask questions. We can also work this problem with your choice of the "system," or with several different choices of the system. Or, we can continue proceeding the way we have been. Either way, I'm glad to answer your questions. What are your thoughts?

  15. Mar 25, 2014 #14
    Hi Chet

    Thank you very much for your help thus far. For now I would prefer getting a more general understanding before going into specific examples, as I usually confuse myself even more when I try to draw a general understanding from specific examples. However I am trying to think of examples that will illustrate my confusion more clearly.

    So in the case of free expansion (adiabatic cylinder with gas and vacuum separated by a membrane), once the membrane breaks and as the gas expands, if I take the gas as the system, isn't there a jump from finite gas pressure to vacuum or P=0 at the system boundary at any time before final equilibrium, unless the pressure drop at the gas boundary to vacuum is infinitesimal? This is where I get the idea of pressure discontinuity from for this and the piston-cylinder case.

    This is essentially what I was confused about. If the pressure on both sides of the piston are same and uniform, and that of the gas is not (in rapid compression/expansion), how does one say that the pressure at the gas side equal to the pressure on the piston side at the gas-piston boundary?

    Is the bolded equation valid for any process (reversible or non-quasistatic)? Also for quasistatic process I am imagining that Pext+mg/A exceeds PI for only dt (and the piston accelerates) and the gas immediately comes into equilibrium again at an infinitesimally higher PI; is this correct?

    Thanks very much
  16. Mar 26, 2014 #15
    Another excellent question. Rather than a membrane, lets just imagine a massless piston that has been held in place between the gas and the vacuum, and released at time t = 0. (It's really pretty much the same thing, but gives us something a little more substantial to work with). Also, lets assume that, rather than a vacuum, the pressure on the other side of the piston is a tiny value ε (so we can continue to treat the gas as a continuum). We are going to use Gas Dynamics to help us out. So, at the instant that the piston is released, the gas pressure at the boundary becomes ε (mathematically), and the gas immediately adjacent to the wall starts expanding. But, because all the gas has inertia (mass), not all the gas expands instantly. In the idealized situation, the expansion region of the gas propagates backwards along the cylinder traveling at a velocity comparable to the speed of sound. So, in the ideal case, there is going to be a discontinuity in the pressure within the cylinder, from the initial pressure far from the piston, to the pressure ε close to the piston. Again, there will be two pressure regions within the cylinder, one where the pressure has not yet changed from the initial pressure, and the other where the pressure had dropped to ε. The movement of the piston that is observed experimentally is the result of the lower pressure region growing in extent and the higher pressure region becoming smaller as time progresses. In real life, the gas within the cylinder is going to have viscosity, so that boundary between the two regions is not going to be sharp like in the ideal case, but there will be a transition region where the pressure is varying very rapidly with location. On either side of this transition region, the pressure is equal to either the original pressure or ε. I don't want to go into further detail on the gas dynamics here except to reiterate that:

    At the very instant that the piston is released, a "discontinuity" in pressure develops within the gas on the gas side of the interface

    There will be two regions of gas pressure within the cylinder, so the gas pressure is non-uniform during the expansion

    The discontinuity will propagate backwards along the cylinder
    As I said, there are going to be two pressure regions within the gas, separated by a rapid transition region that propagates backwards.
    The bolded equation is valid only for non-quasistatic. It is not valid for quasistatic. In quasistatic, the ma term is zero, and has to be replaced by an actual force F that you apply (say, manually) to move the piston gradually.

  17. Mar 26, 2014 #16
    It might be helpful to think of the gas within the cylinder as a compressed not-very-stiff spring, with non-zero mass, that is attached to the base of the cylinder, or as a sequence of masses connected by compressed massless springs. This is a pretty good analog of how the gas would behave mechanistically during an irreversible compression of expansion.

  18. Mar 26, 2014 #17
    The only thing I don't get is, why does the gas pressure (side with higher pressure) instantly drop to ε at t = 0? I get that the rest of the system require finite time to react to the expansion, but how come the boundary molecules can react instantaneously? What happens if the other side was vacuum?

    If I apply a pressure of P_ext >> P_gas to a massless frictionless piston, I should expect the gas system boundary to have a uniform P_ext but have pressure in-equilibrium within the system itself? In a more realistic case, what if this compression turbulence that reaches the boundary; does the uniform P_ext applied to the piston form a boundary condition for the gas regardless of these conditions?

    So in the quasistatic case whether the piston has mass or not doesn't affect how the process is carried out, the only change is an mg/A term added to the applied pressure?

    Thanks very much
  19. Mar 27, 2014 #18
    Because only an infinitecimal amount of mass is accelerated initially. Another way of looking at is that the molecules are already moving in all directions, so the slightest movement of the piston will reduce the momentum transfer from the ones that were moving toward the piston. This is equivalent to the pressure dropping.

    I want to take some more time to formulate my answer for this case.
    I don't quite understand this question. If I apply a pressure P_ext to a massless frictionless piston, and P_ext is higher than the initial pressure of the gas, the pressure at the interface on the gas side of the piston will jump to P_ext when the deformation begins. P_ext applied to the piston forms a boundary condition for the gas regardless of the conditions inside the cylinder. We are controlling P_ext to be anything we want it to be. This forces the gas pressure at the boundary to be P_ext.

    I definitely don't understand this question. In the quasistatic case, we are controlling the force on top of the piston such that the piston virtually does not accelerate.

  20. Mar 27, 2014 #19
    OK. Let's do your free expansion problem. You have an adiabatic cylinder with gas and vacuum separated by a membrane, and the membrane breaks, allowing the gas to fill the entire cylinder. Let's say that initially, the volumes of the two chambers are equal. We would like to determine the final equilibrium state of the system. Before we try to do the problem using the gas as the system, let's first do the problem taking the combination of the two chambers as the system. If you are already familiar with this analysis, we can go right to the case where the gas is taken as the system. Your choice.

  21. Mar 28, 2014 #20
    Is this the same as saying, as the piston moves a little, the extra volume is filled by enough gas such that both sides equilibriates to ε>0, and this takes infinitessimal time because it requires infinitessimal amount of gas. This would create a sharp pressure gradient between it and the next layer (that haven't "reacted" to the pressure drop) that propagates upstream during expansion?

    So regardless of whether I compress the piston quasistatic or non-quasistatically, pressure at the gas-piston boundary has to be Pext and we always get PI,gas = Pext (if no piston-cylinder friction)?

    I was thinking that in like a car engine, gas swirls during rapid compression; how would this affect PI? Also, if Pext is uniform, is PI also always uniform (assuming flat piston)?

    If the two chambers are the same, Q = 0, W = 0 (V is constant), and we end up with E_2 = E_1 as expected.

    The problem I am encountering is if just the gas is the system, I know the answer is the same as above (Q = 0, W = 0), but what I am not getting is why W = 0. Since W = ∫PIdV, is PI = 0 then? This would imply that the gas at the system boundary has 0 pressure during expansion, which doesn't make sense intuitively.

    Thanks very much
  22. Mar 29, 2014 #21
    Yes, exactly, except I would change the > sign to an = sign.

    Yes, if Pext represents the force per unit area applied on the other side of the piston. Also, I would add the qualifier of "massless piston."

    In that case, the value of PI averaged over the area of the piston (assumed rigid) has to equal Pext.
    Maybe doping it out at the molecular level will help. I'm not too good at doing this kind of thing, so no guarantees. Before the membrane is removed, the gas has a Boltzmann distribution, and the molecules are flying around in all directions. Molecules are bouncing off the membrane and reversing direction, so that this change in momentum translates into pressure on the membrane. Further back, molecules are bouncing off each other, and this also translates into pressure. Now the membrane is removed, and no molecules are bounding back the other way in the region that was previously adjacent to the interface. Whatever molecules would have hit the membrane now continue on. This causes the molecules that were next to the membrane to get further apart and to no longer interact with each other. But, slightly further back, the molecules do not yet know that the interface has been removed. So they continue exhibiting the same molecular interactions until the dilatation region reaches them.

    I know that this is probably not exactly how it happens, but this is the gist of it.

  23. Mar 29, 2014 #22
    So does this mean that PI can still vary spatially (so long as the PI averaged over the piston is equal to Pext) even if Pext is perfectly uniform? If so wouldn't this mean that locally, PI may not equal Pext?

    So is pressure due to the momentum of gas particles when they hit each other/container wall, and in this case once the molecules are released into the vacuum, PI,gas = 0 because they shoot straight out and don't hit/interact with anything?

    Is there any difference in the expansion between having a membrane or massless/frictionless piston?

    Thanks very much
  24. Mar 29, 2014 #23
    Sure, but so what. The piston is rigid. So PI averaged over the piston area is equal to Pext averaged over the piston area. The amount of work only depends on the average pressure (i.e., the total force) and the displacement (which is the same for all areas of the rigid piston).

    Pretty much so, until they hit the far wall, and then they can't go any further, and they start bouncing off that wall, and they start getting closer together, and then they start bouncing off each other again.
    No significant difference as far as we are concerned, in terms of the amount of work done and determining the final equilibrium conditions.

    I'm really getting the feeling that, for the benefit of perceptive students like yourself, it would be helpful for someone to publish a paper presenting the detailed results of gas dynamics calculations for some problems along these lines. However, I am not aware of such publications, although that doesn't mean that they do not exist.

  25. Mar 30, 2014 #24
    If Pext is applied uniformly across the piston but PI,gas may not be uniform, doesn't this mean that it is possible for there to be pressure discontinuity locally since the pressure on either side of the gas-piston boundary is different?

    So it is physically possible for gas with finite mass to have pressure PI=0? I just have trouble visualizing how something with mass can have 0 pressure at temperature above absolute zero.

    Unfortunately I don't think there's enough of an audience for such publications to warrant professors to spend their times writing them, especially since many care more about research than teaching.

    Thanks very much
  26. Mar 30, 2014 #25
    Yes. But, so what, as long as the average pressures match. That's all we're really saying.

    If you want to get into this in greater detail, make the piston elastically deformable instead of rigid. And include the effects of the gas viscosity, so that viscous stresses occur in the gas while it is deforming. Then, the boundary condition at the gas interface with the piston will be that the local stress vector (i.e., the stress tensor contracted with the unit normal to the surface) exerted by the gas at the interface must match the local stress vector that the piston metal exerts on the gas. An analogous boundary condition must be satisfied at the exterior interface of the piston with whatever is out there. The stresses that develop within the piston as a result of its deformation must be consistent with the boundary conditions on either side. Incidentally, the gas pressure is just minus the isotropic part of the total stress tensor.

    This is getting into more detail than we really need to solve many of our problems. All we are really interested in now is the change in our system from the initial to the final equilibrium state, the total amount of work done, and the total amount of heat transferred. The beauty of the first law is that, for many problems, we can determine these things without getting into the gory details of what is happening in between at the micro-scale.

    Why does zero pressure (i.e., the molecules are so far apart that they are not colliding with one another) mean that the temperature has to be absolute zero. The gas molecules still have kinetic energy. They just don't have a high enough number density to cause a significant force.

    One thing we learned from this is that, even for irreversible path's between two specified equilibrium states, there can be multiple irreversible paths that give the same amount of work and the same heat transferred. An example of this is the difference between having no piston at all (the membrane case) and having a massless piston. Suppose that the piston did have mass (but were insulated). Would that change things if the piston were horizontal? See if you can work this problem out (to find W, Q, and Tfinal).

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