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time601
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A simple example at: http://blog.sciencenet.cn/blog-629442-600007.html
I will translate the talk later.
I will translate the talk later.
To what work are you referring?...the difference of two energy states must do work to others.
As Simon has suggested, whether or not work is involved depends on the choice of the system (or subsystems). All choices must lead to the same result for the final solution to the problem, but some choices (such as selecting the constant volume box as the system) make solving the problem much easier than others.Simon Bridge said:Or you can take the system as one side or the other - in which case the volume does change since the "membrane with mass" moves.
Some notes:
The mass on the membrane only slightly modifies the statements the others have been making.
To show work in your diagram, you want to explicitly show the mass (draw a square with "m" in it where the mass is), and you want the movement to be "upwards" (against gravity) and not sideways: this is so the final state of the system is clearly different from the initial.
The small hole in the membrane is a minor technicality.
You could have started with a chamber divided into two volumes by a removable wall.
One side is very low pressure and the other has a much higher pressure.
On the low-pressure side there is a small cart with a sail attached.
Quickly remove the dividing wall - now the gas on the two sides can be considered part of a single system much better than the case where there is a small hole. As gas rushes from one side to the other - the cart gets pushed along: accelerated.
Thus the gas is doing work on the cart.
Is this how you are thinking?
Are you also thinking that this violates some rule in thermodynamics about isochoric processes doing no work?
Wikipedia is not a good place to learn thermodynamics.time601 said:
The system has to be describable by a state equation. This means it has to have the same temperature and pressure throughout the volume at any particular time. This means that it has to be in thermal equilibrium with itself.The isochoric process does not include the non-equilibrium process? Why?
[/quote]Then your thinking is mistaken ;)The isochoric process must be a quasi-static process? I don't think so.
I think that the non-equilibrium process on floor #1 and #3 belongs to the isochoric process too.
Well quite a lot of people make mistakes.time601 said:To Simon Bridge:
We both know no work can be done in a quasistatic isochoric process. However, a few (perhaps a lot of) persons generalize the result to all fixed volume processes including non-equilibrium processes.
I may be interpreting you incorrectly, but I think you are talking about fixed mass here, not fixed volume. To me, the term "fixed volume" means isochoric. If you take the entire atmosphere as your system, then this system does no work on its surroundings (outer space).It is not surprising that a gas in a fixed enclosed volume may do work - we see it everyday around us: it's called "weather". Nobody wonders at the wind moving stuff about, thermals lifting things, etc. Therefore it makes no sense to talk about every possible fixed volume doing no work does it?
The parcels of gas inside the shipping container are not undergoing constant-volume processes, but, if your system is the entire contents of the shipping container, and the container is air tight and insulated, then the overall combined contents undergo a constant volume process, there is no work done by the system on the surroundings, and the change in internal energy from the initial equilibrium state (car not yet started, and gasoline and air unreacted) to the final equilibrium state (gasoline or oxygen fully consumed and contents equilbrated) is zero. Is this your understanding?It is always possible to put a fixed volume around any thermodynamic process - that does not magically make everything an isochoric process, even if you do drill a small hole in the container. If I put a car in a sealed shipping container and turn on the engine, are the engine gasses undergoing constant-volume processes?
For this discussion I am making a distinction between things that occur within a fixed geometric volume and an isochoric process.Chestermiller said:I may be interpreting you incorrectly, but I think you are talking about fixed mass here, not fixed volume. To me, the term "fixed volume" means isochoric. If you take the entire atmosphere as your system, then this system does no work on its surroundings (outer space).
That sounds about right. This is probably a better way of putting it too.The parcels of gas inside the shipping container are not undergoing constant-volume processes, but, if your system is the entire contents of the shipping container, and the container is air tight and insulated, then the overall combined contents undergo a constant volume process, there is no work done by the system on the surroundings, and the change in internal energy from the initial equilibrium state (car not yet started, and gasoline and air unreacted) to the final equilibrium state (gasoline or oxygen fully consumed and contents equilbrated) is zero. Is this your understanding?
OK. Good. So it looks like we are dealing with terminology here. When you refer to isochoric, you are referring to no volume change for a gas. When I think isochoric, I am thinking of a fixed geometric volume.Simon Bridge said:For this discussion I am making a distinction between things that occur within a fixed geometric volume and an isochoric process.
Some additional care being made to be sure that the volume of gas inside this fixed geometric volume is also fixed - although the density of the gas through the volume need not be. The geometric volume can also contain anything else. My example was a sailcar.
That sounds about right. This is probably a better way of putting it too.
I think there is a fundamental and common misunderstanding here and I'm trying to express it in terms the OP already is familiar with.
The concept of mechanical work in thermodynamics is something a lot of students wrestle with.
I suspect OP has become confused by sources saying that the work "done by the gas" (implicitly: "on anything") is zero.
A classical closed system cannot do work on anything external to the system: it's closed! Thermodynamic "systems" tend not to be like that. We may be able to add heat to them and extract mechanical work, for instance. In order to extract mechanical work, the volume of the system must change.
In OPs examples, one part of the system loses energy and another part gains it - since the part that loses energy is the gas (goes to KE in the mass), the assertion is that the gas is doing work. But, then, the gas is not undergoing an isochoric process. (The entire system may be though.)
Is that better?
Executive version - it is not surprising that OP finds the ability to get a mass in a fixed enclosed volume to move a mass about inside the same volume.
You are talking about the difference between the closed system version of the first law of thermodynamics and the open system version of the first law. In the development of the open system version of the first law (see Smith and van Ness, Introduction to Chemical Engineering Thermodynamics), the work is divided into two parts: work to push material into and out of the system (control volume) and all other work. The all other work is called the "shaft work." The work to push material into and out of the system is lumped together with the internal energy of the material entering and leaving to give the enthalpy of the material entering and leaving. You need to study some additional descriptions of the development of the open system version until you get comfortable with it and can reconcile it with the closed system version.EM_Guy said:In a control volume, Pv represents flow energy (or flow work); that is, it represents the energy required to push fluid into the control volume. But in a closed system, Pv is not doing any kind of work. Can we call it a flow potential energy (not to be confused with gravitational potential energy or other kinds of potential energy)? Clearly it is some kind of energy, but where did the increase of Pv come from? All we added was qin to the system, and all of it went into u. u2 - u1 = qin. h2 - h1 > qin. P2v2 - P1v1 > 0. What does this mean?
Chestermiller said:You are talking about the difference between the closed system version of the first law of thermodynamics and the open system version of the first law. In the development of the open system version of the first law (see Smith and van Ness, Introduction to Chemical Engineering Thermodynamics), the work is divided into two parts: work to push material into and out of the system (control volume) and all other work. The all other work is called the "shaft work." The work to push material into and out of the system is lumped together with the internal energy of the material entering and leaving to give the enthalpy of the material entering and leaving. You need to study some additional descriptions of the development of the open system version until you get comfortable with it and can reconcile it with the closed system version.
Chet
EM_Guy said:Chet,
I understand that, and I have studied open systems (control volumes). However, whether the system is a closed system or an open system, specific enthalpy is an intensive property of the system. By the state postulate, the state of a simple compressible system is completely specified by two independent, intensive properties. So, in any particular state, a closed system does have a certain value for the enthalpy property, which is to simply say that it has an internal energy, a pressure, and a specific volume. h = u + Pv
And during a constant volume heat addition of a closed system, the change of internal energy is equal to the heat transfer to the system. But in addition to this, the pressure also increases. Therefore, delta(h) > delta(u). And this is my conundrum.
I don't think so. A and G are just defined potentials that are more convenient to work with in certain kinds of problems than U, H, and S.I think the conundrum would be resolved if I understood Helmholtz free energy and/or Gibb's free energy. As of right now, I'm having trouble wrapping my head around those concepts. But apparently there as many as five thermodynamic potentials. http://en.wikipedia.org/wiki/Thermodynamic_potential I think the answer to my question lies in understanding the nature of these various potentials.
IMHO, you are trying to ascribe too much physical significance to H. It is just a function that is convenient to work with in many kinds of problems. Heating of a material at constant volume is just not one of those kinds of problems.EM_Guy said:It feels like a conservation of energy crisis - sort of.
delta(u) = q. This is the conservation of energy. You can't get something out of nothing.
delta(h) = delta(u) + delta(P)*v (for constant volume).
And there is a delta(P).
So, you add heat, and u goes up that much. But your delta(Pv) also goes up (because of the heat transfer). But it is not as if the heat addition is shared between the delta(u) and the delta(Pv). Nevertheless, it is the delta(u) that causes the delta(P). That is, the rise of internal energy causes (or is effectively the same thing as) a rise of temperature which causes (or is the same thing as) a rise of pressure (for a constant volume situation). (Incidentally, the walls of the container have to exert a greater force on the fluid in the container in order to maintain a constant volume). But if the rise of internal energy causes a rise in the (Pv) product, then I would think that would be accounted for in the COE equation. But of course, it is not, because while the pressure rises, the volume does not change, thus no work is done on the system or by the system. So, it is right that the rise in (Pv) does not show up in the COE equation. Nevertheless, delta(Pv) is positive. It is due to the heat transfer to the system. But it is not part of the COE equation. And delta(h) > delta(u).
So, to sum up, you put in q by heat transfer. delta(u) = q. But you also get an increase in the Pv product. The units of Pv are obviously specific energy units. So there is some kind of increased potential that exceeds the increase of u. This seems like you are getting something from nothing, since delta(h) > q. Energy units are energy units. We are putting in q, and delta(h) exceeds q.
Conundrum.
I think it is fair to say that this is somewhat like pressing really hard against a rigid wall. You accomplish no work when you do this, yet the harder you press, the more pressure there is, and there is some kind of potential there, because if the wall suddenly gave way, that force would cause mass to accelerate.
Chestermiller said:IMHO, you are trying to ascribe too much physical significance to H. It is just a function that is convenient to work with in many kinds of problems. Heating of a material at constant volume is just not one of those kinds of problems.
Chet
If you want to get a better understanding of all this, see my recent Physics Forums Insight article atEM_Guy said:That doesn't work. The function describes a physical phenomenon. Specific enthalpy is a physical thermodynamic property of the system. Surely, it has physical significance. And in fact, the pressure does increase. As I just mentioned, I believe that the answer to the conundrum is explained by the generation of entropy. Entropy is not a conserved property. It all comes down to the Clausius inequality.
Did you really think that, if the initial and final volumes of a system are the same, that is equivalent to a constant volume process in a rigid container?Nuke said:Stirling engine. One side hot, one side cold, sliding piston. Total volume of system is geometrically constant. Work is performed.
The Stirling engine is a rigid container. Geometry (volume) of the container remains constant for the process, initial, during, and final, even though the piston is reciprocating. Gas density is moved from end to end.Chestermiller said:Did you really think that, if the initial and final volumes of a system are the same, that is equivalent to a constant volume process in a rigid container?
Chet
Nuke,Nuke said:The Stirling engine is a rigid container. Geometry (volume) of the container remains constant for the process, initial, during, and final, even though the piston is reciprocating. Gas density is moved from end to end.
Hence, Yes. But not on a closed system.
If gas volume remains constant, it can do work?
Gas volume remaining constant means that the amount of gas in a closed system does not change, even if the pressure or temperature changes.
When gas volume remains constant, the gas molecules are confined to a fixed space and cannot expand. This means that any work done by the gas must be done within that fixed space, such as pushing a piston or turning a turbine.
Yes, gas volume remaining constant is commonly used in everyday situations, such as in car engines or in refrigeration systems. In these cases, the gas is confined to a fixed volume and is used to do work, such as moving a car or cooling food.
If gas volume is not constant, it can still do work, but the amount of work will depend on the change in volume. For example, if the gas expands, it can do more work than if it is confined to a smaller volume.
Gas volume remaining constant is related to the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred or converted. When gas volume remains constant, any work done by the gas is a result of energy being transferred or converted.