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.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.
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.
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.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.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.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.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?