If gas volume remains constant, it can do work?

In summary, the conversation discusses a simple example of a system with two chambers separated by a membrane with a small hole. The gas in one chamber is at high pressure while the other is at low pressure. When the membrane is removed, the gas expands and does work on a cart with a sail attached. This process is not an isochoric process and can result in a change in temperature for real gases. However, for an ideal gas, there is no change in temperature or internal energy. The concept of work in this scenario is also discussed.
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  • #2
Let me guess - the amount of gas in the volume is not constant?
 
  • #3
My English is not good, but I think I can describe it clear.
The boxes in figure are rigid. There is a membrane with mass dividing the box into two parts. The left part is filled with high pressure gas, and the right part is filled with low pressure gas (or vacuum). The only difference between the upper fig and the down fig is: there is a small hole in the membrane in the down fig. There is no friction between membrane and the box. Because there is a hole in the membrane, the gas in left part and the right part can be seen a whole gas. And the whole gas volume is constant. The high pressure gas in left part must push the membrane with mass to move. The membrane’s energy is from the gas’s work, but the volume is constant.
 

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  • #4
The total volume may be constant but the volumes of the two regions are not and I think they need to be considered separately. If you want to find where the energy comes from, to move the membrane, you could look at any temperature changes involved.
 
  • #5
What he said :)

What you actually have there is two chambers - and it looks like gas can flow from one to the other. The volume occupied by the gas changes. Another way of looking at it is that the system is not in thermal equilibrium with itself - since it does not have the same pressure or temperature everywhere.

Whichever way you look at it - this is not an isochoric process so of course it is possible for it to do work.
http://en.wikipedia.org/wiki/Isochoric_process
 
  • #6
It looks like the OP is describing a free expansion of a gas.

It doesn't matter if the gas expands by the ( massless ) membrane moving, or through the pinhole, it ends up as being the same thing. The chosen system consists of both volumes on either side of the membrane to simplify the calculation.

There is nothing on the other side, ie the vacuum, to provide a force against expansion, and so it happens quickly, and uncontrolled.

Membranes are usually considered massless. So thus no work is being done.
It happens quickly, with no time for heat to flow in or out.

Thus, no energy crosses the system boundary, the internal energy of the gas is constant, and thus so is temperature constant for the gas.

That is for an ideal gas.

For a real gas, we see a Joule Thompson effect.
http://www.etomica.org/app/modules/sites/JouleThomson/Background2.html
a real gas has a slight change in temperature during free uncontrolled expansion.
 
  • #7
the process is not quasi-static, the equation dw=pdV is not correct.
the high energy state relaxes to low energy state, the process usually is non-equilibrium. If the volume remains constant and it is no heat transfer, the difference of two energy states must do work to others.
 
  • #8
Of course its not quasi-static. An expansion into a vacuum is not quasi-static.

The temperature of an ideal gas in this scenario does not change from state 1 to state 2, and therefor neither does the internal energy.

From the first law
ΔE = Q+W
In this case,
ΔU = Q+W, as no other forms of energy are crossing the boundary.
ΔU = 0, --> the temperatue did not change.
Q = 0
It follows that W = 0
...the difference of two energy states must do work to others.
To what work are you referring?
 
  • #9
If you take as the system the overall box, then no work is done on the system. And, if the overall box is insulated, there is no heat entering or leaving. Therefore, from the first law (as 256bits points out), there is no change in total internal energy of the system U in going from the initial equilibrium state to the final equilibrium state.

Chet
 
  • #10
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?
 
  • #11
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?
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.

Chet
 
  • #12
An isochoric process, also called a constant-volume process, an isovolumetric process, or an isometric process, is a thermodynamic process during which the volume of the closed system undergoing such a process remains constant.
http://en.wikipedia.org/wiki/Isochoric_process

The isochoric process does not include the non-equilibrium process? Why?
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.
 
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  • #13
time601 said:
Wikipedia is not a good place to learn thermodynamics.
The isochoric process does not include the non-equilibrium process? Why?
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.

You will notice that the wikipedia article you cite describe the "isochoric process" in terms of a P and T which does not vary across the volume.

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?

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?

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.
[/quote]Then your thinking is mistaken ;)

I'm not saying you are wrong to say that a gas enclosed in a fixed volume container may do work. What I am saying is that you have misunderstood the definition of an isochoric process. To be fair - most books are quite relaxed about desribing it.
 
  • #14
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.
 
  • #15
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.
Well quite a lot of people make mistakes.
Can you provide a reference to a reputable source that makes that generalization?
Then we can see better what you are trying to talk about.
 
  • #16
Hi Simon,

I think you need to clarify what you're saying a little.
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?
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 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?
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?

Chet
 
  • #17
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).
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.

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?
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.
 
  • #18
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.
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.

Also, when you refer to a closed system, you are referring to what I call an isolated system. When I refer to a closed system, I am referring to a system for which no mass enters or leaves, but one which can do work and exchange heat with the surroundings.

Engineers and Physicists on the Tower of Babel.

Chet
 
  • #19
Say you have a closed system, and you have an constant-volume heat addition. The increase of the internal energy (delta u) = qin. But enthalpy (a thermodynamic property) increases by an amount that is greater than the internal energy. delta (h) > delta (u).

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?
 
  • #20
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?
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
 
  • #21
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

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 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.
 
  • #22
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 quite understand why this is a problem. In a closed system at constant volume, ΔU=CvΔT=Q, so ΔH=ΔU+VΔP. So what? What am I missing? Are you asking what the significance of H increasing is?

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.
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.

Chet
 
  • #23
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.
 
  • #24
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.
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
 
  • #25
I'm playing around with some differentials here, and I think I'm onto something.

dw = Pdv (dw is an inexact differential)
dq = Tds (dq is an inexact differential)

By the chain rule:
d(Pv) = Pdv + vdP

And
d(Ts) = Tds + sdT

Equivalently,

d(Pv) = dw + vdP

d(Ts) = dq + sdT

Now, du = dq - dw (depending on the sign convention)

Energy is conserved.

But entropy is not conserved.

delta(s) = int(dq/T) (internally reversible)

For all processes, entropy is either generated (irreversible) or remains the same (reversible).

So, for any heat transfer process involving no work, delta(Ts) >= q. Entropy is generated.

For closed or open systems, and for reversible or irreversible processes,

Tds >= du + Pdv = dh - vdP

Or

ds >= du/T + Pdv/T = dh/T - vdP/T

If specific volume is held constant,

ds >= du/T = dh/T - v*dP/T

Thus,

vdP/T >= dh/T - ds

vdP >= dh - Tds

vdP >= dh - dq

vdP = dh - dq + Tdsgen

I'm not sure that all of this differential math is right, but it appears that the increase of pressure arises due to the generation of entropy in the system.
 
  • #26
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

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.
 
  • #27
EM_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.
If you want to get a better understanding of all this, see my recent Physics Forums Insight article at
https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

Incidentally, the effect of pressure on the entropy of a material is given by the general equation

$$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$

Chet
 
  • #28
Chet,

Agreed. And, of course, the integration would have to be performed along a reversible path.
 
  • #29
Stirling engine. One side hot, one side cold, sliding piston. Total volume of system is geometrically constant. Work is performed.
 
  • #30
Nuke said:
Stirling engine. One side hot, one side cold, sliding piston. Total volume of system is geometrically constant. Work is performed.
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
 
  • #31
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
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?
 
Last edited:
  • #32
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?
Nuke,

I think you understood the context of the OP's original question, and that he was referring to a closed system in which the volume of the system remains constant throughout the process. I also think you understand that in a stirling engine, the movement of the piston causes the volume of the closed system and the volume of the gas to change with time over each cycle. So the only thing I can conclude is either that you were trying to mislead and confuse the OP, or you were trying to show everyone how smart you think you are by bringing up this example that does not even match the context of the original question. But, it is not very smart to get into a game of words with a Mentor. I am issuing you some Warning points and am closing this thread to further responses. In the future, please use better judgement.
 
  • Like
Likes Nugatory

1. What is the definition of gas volume remaining constant?

Gas volume remaining constant means that the amount of gas in a closed system does not change, even if the pressure or temperature changes.

2. How does gas volume remaining constant relate to work?

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.

3. Can gas volume remaining constant be used to do work in everyday situations?

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.

4. What happens if gas volume is not constant?

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.

5. How is gas volume remaining constant related to the laws of thermodynamics?

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.

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