Irreversible Quasistatic PV Work

[Red_CCF]

Hi

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

 

[Chestermiller]

Hi

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.

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.

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?

No. It would become - W = E_2 - E_1 , where W=P'ΔV.

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

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.

Chet

 

[Red_CCF]

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.

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?

No. It would become - W = E_2 - E_1 , where W=P'ΔV.

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

Thanks

 

[Chestermiller]

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?

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).

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

Thanks

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.

Chet

 

[Red_CCF]

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.

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 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).

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

 

[Chestermiller]

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?

Yes, if the system as taken as the gas plus piston (as your book does).

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.

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.

If friction (heat) goes elsewhere, the system would be the gas and W = ∫P_interfacedV = E_2 - E_1?

Yes. If you chose the gas as your system and the deformation were quasistatic, this would be correct.

Would this case still be irreversible as friction existed in the process but outside the system?

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

Chet

 

[Red_CCF]

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.

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 P_I(t) taken on the gas side of the interface or the piston side? What if P_I(t) varies spatially in all directions, how would one define a P_I(t)?

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

Chet

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

 

[Chestermiller]

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)?

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.

How come work and ∫PdV are expressed by an inequality in the two posts I referenced above?

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.

In your blog post, for non-quasistatic cases, is P_I(t) taken on the gas side of the interface or the piston side?

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.
https://www.physicsforums.com/showthread.php?t=743675

What if P_I(t) varies spatially in all directions, how would one define a P_I(t)?

Are you saying, "what if the pressure at the interface varies with spatial position along the interface?"

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?

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.

Chet

 

[Red_CCF]

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.

From this post:

In case of non-quasistatic processes you can no longer apply the usual thermodynamic equations. Work done is then no longer given by the integral of P dV. In general you cannot even define the pressure for non-equilibrium processes in unambiguous way. But even if the pressure is well defined and the integral of P dV is well defined, it is not the same as the work done.

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 P_I(t) referring to in your blog?

Are you saying, "what if the pressure at the interface varies with spatial position along the interface?"

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

 

[Chestermiller]

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 P_I(t) referring to in your blog?

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

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.

Chet

 

[Red_CCF]

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 P_I. 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).

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, P_I on the gas side is finite (and difficult to define) while P_I 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 P_I isn't well defined. I am under the impression that for P_I to be useful pressure should be at local equilibrium at the boundary.

Thanks very much

 

[Chestermiller]

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, P_I on the gas side is finite (and difficult to define) while P_I 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?

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.

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?

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, P_I still represents the gas pressure at the interface.

I was thinking this may be a reason why work is expressed as an inequality to ∫P_extdV since P_I isn't well defined. I am under the impression that for P_I to be useful pressure should be at local equilibrium at the boundary.

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 P_I 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 P_extdV (=P_IdV). If your system does not include the piston (and is only the gas), then P_IdV ≠ P_extdV. In this case, P_IA=mg+P_extA+ma, where A is the area of the piston, and a is the acceleration of the mass.

Chet

 

[Chestermiller]

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?

Chet

 

[Red_CCF]

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.

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.

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.

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, P_I still represents the gas pressure at the interface.

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?

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 P_extdV (=P_IdV). If your system does not include the piston (and is only the gas), then P_IdV ≠ P_extdV. In this case, P_IA=mg+P_extA+ma, where A is the area of the piston, and a is the acceleration of the mass.

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

Thanks very much

 

[Chestermiller]

Hi Chet

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.

Another excellent question. Rather than a membrane, let's 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, let's 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

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?

As I said, there are going to be two pressure regions within the gas, separated by a rapid transition region that propagates backwards.

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

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.

Chet

 

[Chestermiller]

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.

Chet

 

[Red_CCF]

Another excellent question. Rather than a membrane, let's 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, let's 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

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?

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.

Chet

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

 

[Chestermiller]

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?

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.

What happens if the other side was vacuum?

I want to take some more time to formulate my answer for this case.

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?

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.

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?

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.

Chet

 

[Chestermiller]

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.

Chet

 

[Red_CCF]

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.

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?

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.

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

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

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.

Chet

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 = ∫P_IdV, is P_I = 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

 

[Chestermiller]

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?

Yes, exactly, except I would change the > sign to an = sign.

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

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

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

In that case, the value of P_I averaged over the area of the piston (assumed rigid) has to equal P_ext.

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 = ∫P_IdV, is P_I = 0 then?

Yes.

This would imply that the gas at the system boundary has 0 pressure during expansion, which doesn't make sense intuitively.

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.

Chet

 

[Red_CCF]

In that case, the value of P_I averaged over the area of the piston (assumed rigid) has to equal P_ext.

So does this mean that P_I can still vary spatially (so long as the P_I averaged over the piston is equal to P_ext) even if P_ext is perfectly uniform? If so wouldn't this mean that locally, P_I may not equal P_ext?

Yes, exactly, except I would change the > sign to an = sign.

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.

Chet

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, P_I,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

 

[Chestermiller]

So does this mean that P_I can still vary spatially (so long as the P_I averaged over the piston is equal to P_ext) even if P_ext is perfectly uniform?

Yes.

If so wouldn't this mean that locally, P_I may not equal P_ext?

Sure, but so what. The piston is rigid. So P_I averaged over the piston area is equal to P_ext 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).

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, P_I,gas = 0 because they shoot straight out and don't hit/interact with anything?

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.

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

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.

Chet

 

[Red_CCF]

Sure, but so what. The piston is rigid. So P_I averaged over the piston area is equal to P_ext 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).

If P_ext is applied uniformly across the piston but P_I,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?

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.

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

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.

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

 

[Chestermiller]

If P_ext is applied uniformly across the piston but P_I,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?

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.

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

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).

Chet

 

[Red_CCF]

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.

Would this imply that for an ideal, rigid piston, a uniform exerted pressure must equal a uniform P_I (i.e. the two equal at any point)? I just don't get why it is possible that pressure is discontinuous locally but yet on an average basis must be equal on either side of the gas-piston boundary (how come the boundary condition can be restrictive yet somewhat lenient)?

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.

So for free expansion, is P_I,gas = 0 the case locally (at any point in along the boundary between gas/vacuum) or just on an average basis (like in the above piston-cylinder case)?

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).

Chet

If the problem essentially expansion of a piston into vacuum, the cylinder itself is uninsulated (to allow heat transfer), volume change is known.

W = m_pistong/A_piston*(Vf-Vi)
Q = W + m_gas(uf-ui)

I end up with two unknowns and one equation and not sure where to go from here to find Tfinal and hence uf. Given that 2 state variables (final pressure and specific volumes) are known I should be able to define a Tfinal, but for ideal gas, tables only relate u with T and I haven't found any tables relating T with p and v.

Thanks very much

 

[Chestermiller]

Would this imply that for an ideal, rigid piston, a uniform exerted pressure must equal a uniform P_I (i.e. the two equal at any point)?

No. All that is required is that, for a massless piston, the forces on both sides of the piston must be equal. This is just application of Newton's second law for a rigid body. The force on either side of the massless piston can be distributed in any way we want, as long as the integrals of the force distributions are equal.

I just don't get why it is possible that pressure is discontinuous locally but yet on an average basis must be equal on either side of the gas-piston boundary (how come the boundary condition can be restrictive yet somewhat lenient)?

Again, just Newton's second law for a rigid body. I don't know how to answer the question as to how come the boundary condition can be restrictive, yet lenient. Personally, that doesn't strike me strange at all.

So for free expansion, is P_I,gas = 0 the case locally (at any point in along the boundary between gas/vacuum) or just on an average basis (like in the above piston-cylinder case)?

In the limit of free expansion, P_I,gas approaches zero everywhere at the interface, either for a cylinder or for a gas/vacuum. But, as soon as the piston reaches the far end of the cylinder (or the leading molecules begin to reach the far end of the cylinder), the pressure starts building up again. In the case of free expansion, it can't vary laterally at the piston because this would require negative pressures so that the average is zero.

If the problem essentially expansion of a piston into vacuum, the cylinder itself is uninsulated (to allow heat transfer), volume change is known.

W = m_pistong/A_piston*(Vf-Vi)
Q = W + m_gas(uf-ui)

I end up with two unknowns and one equation and not sure where to go from here to find Tfinal and hence uf. Given that 2 state variables (final pressure and specific volumes) are known I should be able to define a Tfinal, but for ideal gas, tables only relate u with T and I haven't found any tables relating T with p and v.

I was asking about the case of an insulated piston, not an uninsulated piston. Am I correct in saying that you are comfortable with the insulated case, and you now want to look at the "uninsulated" case?

Chet

 

[Red_CCF]

No. All that is required is that, for a massless piston, the forces on both sides of the piston must be equal. This is just application of Newton's second law for a rigid body. The force on either side of the massless piston can be distributed in any way we want, as long as the integrals of the force distributions are equal.

Again, just Newton's second law for a rigid body. I don't know how to answer the question as to how come the boundary condition can be restrictive, yet lenient. Personally, that doesn't strike me strange at all.

Would another valid explanation for P_ext = P_I,gas on an average basis be Newton's Third Law applied at the gas-piston boundary?

I accept that on an average basis that P_ext = P_I,gas, but what I find strange is that the distribution on either side of the gas-piston boundary may not identical. If the pressure distribution around the piston face is such that at there exists some point where the gas pressure < P_{I,gas average} and thus < P_ext (applied uniformly), at that location there is two unequal forces acting on zero mass since it's at the gas-piston boundary. I have trouble seeing how this is physically possible and how locally Newton's Third Law isn't violated.

In the limit of free expansion, P_I,gas approaches zero everywhere at the interface, either for a cylinder or for a gas/vacuum. But, as soon as the piston reaches the far end of the cylinder (or the leading molecules begin to reach the far end of the cylinder), the pressure starts building up again. In the case of free expansion, it can't vary laterally at the piston because this would require negative pressures so that the average is zero.

Thanks for clearing this up for me, I had always thought that P_gas ≠ 0. So I basically just plug P_I,gas into W = ∫PdV to justify that W = 0 in free expansion to vaccum?

I was asking about the case of an insulated piston, not an uninsulated piston. Am I correct in saying that you are comfortable with the insulated case, and you now want to look at the "uninsulated" case?

Chet

I think I didn't understand the question properly. What is the system and the process?

Thanks very much

 

[Chestermiller]

Would another valid explanation for P_ext = P_I,gas on an average basis be Newton's Third Law applied at the gas-piston boundary?

No. Newton's third law says that the upward pressure of the gas on the base of the piston P_I is equal to the downward pressure of the piston (at its base) on the gas P_I. It also says that the downward pressure exerted by the surroundings on the top face of the piston P_ext is equal to the upward pressure exerted by the top face of the piston on the surroundings P_ext.

I accept that on an average basis that P_ext = P_I,gas, but what I find strange is that the distribution on either side of the gas-piston boundary may not identical. If the pressure distribution around the piston face is such that at there exists some point where the gas pressure < P_{I,gas average} and thus < P_ext (applied uniformly), at that location there is two unequal forces acting on zero mass since it's at the gas-piston boundary. I have trouble seeing how this is physically possible and how locally Newton's Third Law isn't violated.

Really? I'm very surprised, because, in freshman physics, you did a zillion problems where exactly this same type of situation prevailed, and you had absolutely no trouble with it at all. Suppose you have a rigid massless table top supported by cinder blocks at its four corners, and you put a big heavy sheet birthday cake on top of the table. The cake covers the entire surface of the table, and exerts a downward pressure of W/A uniformly over the entire table top, but the cinder blocks exert much higher upward pressures over their respective areas A' at the corners of the table: W/(4A'), where A >>4A'. In the portion of the table away from the corners, the pressure on the top of the table is W/A, while the pressure on the bottom of the table is zero.

Thanks for clearing this up for me, I had always thought that P_gas ≠ 0. So I basically just plug P_I,gas into W = ∫PdV to justify that W = 0 in free expansion to vaccum?

Yes.

I think I didn't understand the question properly. What is the system and the process?

You have an insulated, frictionless piston with mass inside a horizontal insulated cylinder. The piston is initially constrained not to move. There are two chambers in the cylinder, one with an ideal gas in it (to the left of the piston), and one with vacuum in it (to the right of the piston). At time = 0, you remove the constraint and wait a long enough time for the system to equilibrate to a new equilibrium state. How much work does the combination of gas and piston do on their surroundings? How much work does the gas do on the piston (over the entire re-equilibration process)? How does the temperature of the gas at the new equilibrium state compare with the temperature at the original equilibrium state?

Chet

 

[Red_CCF]

Really? I'm very surprised, because, in freshman physics, you did a zillion problems where exactly this same type of situation prevailed, and you had absolutely no trouble with it at all. Suppose you have a rigid massless table top supported by cinder blocks at its four corners, and you put a big heavy sheet birthday cake on top of the table. The cake covers the entire surface of the table, and exerts a downward pressure of W/A uniformly over the entire table top, but the cinder blocks exert much higher upward pressures over their respective areas A' at the corners of the table: W/(4A'), where A >>4A'. In the portion of the table away from the corners, the pressure on the top of the table is W/A, while the pressure on the bottom of the table is zero.

My apologies, I have no idea where my head has been the last few days, this explanation clears everythings up.

One last thing on this, you mentioned that Newton's Second Law is the reason why the total forces on either side of a massless piston must be equal; is this because if the forces are not equal then we an undefined solution (Fnet is finite, but m*a is 0 or undefined)?

You have an insulated, frictionless piston with mass inside a horizontal insulated cylinder. The piston is initially constrained not to move. There are two chambers in the cylinder, one with an ideal gas in it (to the left of the piston), and one with vacuum in it (to the right of the piston). At time = 0, you remove the constraint and wait a long enough time for the system to equilibrate to a new equilibrium state. How much work does the combination of gas and piston do on their surroundings? How much work does the gas do on the piston (over the entire re-equilibration process)? How does the temperature of the gas at the new equilibrium state compare with the temperature at the original equilibrium state?

Chet

a) For the gas+piston combined system, W = 0 since P_ext = 0

b) For the work the gas does on the piston, I am tempted to say that W = 0 and P_I = 0, but am not confident in this. I think at the beginning of the process an infinitesimal pressure difference will occur between the two sides of the piston just to get it moving, and I'm not sure if the inertia of the piston will hinder the expansion process if it is moving slower than required for P_I to be 0.

c) The final temperature of the gas should be the same as the initial if no work is done (internal energy constant) as an ideal gas's internal energy is a function of temperature only

Thanks very much

 

[Red_CCF]

No. Newton's third law says that the upward pressure of the gas on the base of the piston P_I is equal to the downward pressure of the piston (at its base) on the gas P_I. It also says that the downward pressure exerted by the surroundings on the top face of the piston P_ext is equal to the upward pressure exerted by the top face of the piston on the surroundings P_ext.

I was thinking that the piston would be part of the surrounding if the gas is the system, so I visualized this as the surrounding exerting P_ext onto the gas, which by third law must exert a force/pressure P_I = P_ext back onto the piston; is this a valid reason for P_I = P_ext?

Thanks very much

 

[Chestermiller]

I was thinking that the piston would be part of the surrounding if the gas is the system, so I visualized this as the surrounding exerting P_ext onto the gas, which by third law must exert a force/pressure P_I = P_ext back onto the piston; is this a valid reason for P_I = P_ext?

Thanks very much

Sure.

Chet

 

[Chestermiller]

One last thing on this, you mentioned that Newton's Second Law is the reason why the total forces on either side of a massless piston must be equal; is this because if the forces are not equal then we an undefined solution (Fnet is finite, but m*a is 0 or undefined)?

Let's come back to this after we do the case of finite piston mass.

a) For the gas+piston combined system, W = 0 since P_ext = 0

Correct.

b) For the work the gas does on the piston, I am tempted to say that W = 0 and P_I = 0, but am not confident in this. I think at the beginning of the process an infinitesimal pressure difference will occur between the two sides of the piston just to get it moving, and I'm not sure if the inertia of the piston will hinder the expansion process if it is moving slower than required for P_I to be 0.

All these uncertainties can be cleared up by applying Newton's second law to the piston. Let P_I be the (unknown) gas pressure on the left side of the piston and let A be the area of the piston. Let m be the mass of the piston, and let x be the displacement of the piston to the right relative to its initial location when the original constraint on the piston is removed. Write out Newton's second law, and we will be able to continue from there.

Chet

 

[Red_CCF]

All these uncertainties can be cleared up by applying Newton's second law to the piston. Let P_I be the (unknown) gas pressure on the left side of the piston and let A be the area of the piston. Let m be the mass of the piston, and let x be the displacement of the piston to the right relative to its initial location when the original constraint on the piston is removed. Write out Newton's second law, and we will be able to continue from there.

Chet

F(t) = m_pistona = P_I(t)A_piston
dW(t) = P_I(t)A*dx

Is the above correct?

 

[Chestermiller]

F(t) = m_pistona = P_I(t)A_piston
dW(t) = P_I(t)A*dx

Is the above correct?

Yes, so far. The acceleration can be expressed in terms of x as d²x/dt². So:
m\frac{d^2x}{dt^2}=P_IA
If we multiply both sides of this equation by dx/dt, and integrate with respect to time, we get:
W(t)=\frac{1}{2}mv^2(t)
where W(t) is the work done by the gas on the piston up to time t, v(t) = dx/dt, and the right hand side is recognized as the kinetic energy of the piston.

Here are some questions to consider:
What happens when the piston collides with the far end of the cylinder (if the collision is elastic)? Does the piston keep moving forever (even if the piston is frictionless)? If the frictionless piston eventually stops, what causes it to slow down and stop?

Chet

 

[Red_CCF]

Yes, so far. The acceleration can be expressed in terms of x as d²x/dt². So:
m\frac{d^2x}{dt^2}=P_IA
If we multiply both sides of this equation by dx/dt, and integrate with respect to time, we get:
W(t)=\frac{1}{2}mv^2(t)
where W(t) is the work done by the gas on the piston up to time t, v(t) = dx/dt, and the right hand side is recognized as the kinetic energy of the piston.

I got a little mixed up, how did you go from the first to the second equation? Also, for this integral to work one has to assume that a and P_I are non-zero for at least some time during the time interval; how is this known?

Here are some questions to consider:
What happens when the piston collides with the far end of the cylinder (if the collision is elastic)? Does the piston keep moving forever (even if the piston is frictionless)? If the frictionless piston eventually stops, what causes it to slow down and stop?

Chet

If the piston collides at at the end of the cylinder, I would assume that some energy is transferred to the cylinder even if the collision is elastic (i.e. if the cylinder was on a frictionless surface it would gain kinetic energy). If there are no irreversibilities within the gas itself, I think the piston would initially move back until the gas pressure slows it down and pushes it towards the far wall again, and this behaviour should oscillate with decreasing range of motion for the piston. I think this behaviour would continue as t → ∞, so maybe the piston never stops?

Thanks very much

 

[Chestermiller]

I got a little mixed up, how did you go from the first to the second equation?

m\frac{d^2x}{dt^2}=P_IA
Multiply both sides by dx/dt:
m\frac{dx}{dt}\frac{d^2x}{dt^2}=P_IA\frac{dx}{dt}
Substitute v for dx/dt on the left side of the equation and dV/dt = Adx/dt on the right side of the equation:
mv\frac{dv}{dt}=P_I\frac{dV}{dt}
Now, integrate with respect to t:
\frac{m}{2}v^2(t)=\int{P_IdV}

Also, for this integral to work one has to assume that a and P_I are non-zero for at least some time during the time interval; how is this known?

For the case of a piston with mass, P_I is never going to be zero. The gas is always going to be accelerating the piston toward the far end of the cylinder.

If the piston collides at at the end of the cylinder, I would assume that some energy is transferred to the cylinder even if the collision is elastic (i.e. if the cylinder was on a frictionless surface it would gain kinetic energy).

No. The cylinder is assumed to be anchored to the lab, so, in an elastic collision, the piston retains all its kinetic energy, and its velocity simply reverses sign.

If there are no irreversibilities within the gas itself, I think the piston would initially move back until the gas pressure slows it down and pushes it towards the far wall again, and this behaviour should oscillate with decreasing range of motion for the piston. I think this behaviour would continue as t → ∞, so maybe the piston never stops?
Thanks very much

By "irreversibilities within the gas," I assume you mean viscous dissipation. Without viscous dissipation, the cylinder and the gas would continue moving back and forth forever, interchanging compression with kinetic energy. In our case, the air within the cylinder does have viscosity, and eventually, the piston velocity will slow down to zero, and the piston will come to rest at the far end of the cylinder. This will be the condition of our system at final steady state. We already said earlier this when we analyzed the problem taking the combination of the piston and the gas as our system.

So, in our adiabatic irreversible problem, we now know that the work done by the gas on the piston is zero, and the heat transferred is zero, so the change in internal energy is zero, and the change in temperature is therefore also zero. So, even in the case of a finite mass for the piston, we still get the same answer.

Chet

 

[Red_CCF]

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.
Chet

I imagine that the molecules shooting out once the membrane is gone aren't all parallel to each other so some should collide with each other and create pressure; is the number density in the first layer so low that this can be ignored?

For the case of a piston with mass, P_I is never going to be zero. The gas is always going to be accelerating the piston toward the far end of the cylinder.

Out of curiosity, is there a theoretical velocity (order of magnitude wise) in which the piston can move such that P_I = 0; I am guessing it is comparable to gas expanding against a massless piston? Is this affected by the initial pressure?

So, in our adiabatic irreversible problem, we now know that the work done by the gas on the piston is zero, and the heat transferred is zero, so the change in internal energy is zero, and the change in temperature is therefore also zero. So, even in the case of a finite mass for the piston, we still get the same answer.

Why is W = 0 when the piston has mass, as I thought the equation derived for this case was W(t)=\frac{1}{2}mv^2(t)?

A couple of aside questions:

How was Newon's Second law used to justify P_ext = P_I?

In general, is it valid to treat friction heat addition as Q in first law or for ΔS = ∫δQ/T + σ, or does Q have to be heat transfer via finite temp gradient between the system and surrounding? In my OP you mentioned that first law was simply W = ∫P_extdV= E_2 - E_1 if piston was included in the system; I took this to mean that the work that turned into frictional heat that went into the system still counted as work and thus ∫δQ/T = 0, is this correct?

Thanks very much

 

[Chestermiller]

I imagine that the molecules shooting out once the membrane is gone aren't all parallel to each other so some should collide with each other and create pressure; is the number density in the first layer so low that this can be ignored?

Yes. That's the idea. It's just a limiting case.

Out of curiosity, is there a theoretical velocity (order of magnitude wise) in which the piston can move such that P_I = 0; I am guessing it is comparable to gas expanding against a massless piston?

Sure. That's what it is. We would have to solve the gas dynamics problem in detail to pin all this down accurately.

Is this affected by the initial pressure?

Probably.

Why is W = 0 when the piston has mass, as I thought the equation derived for this case was W(t)=\frac{1}{2}mv^2(t)?

This is a correct equation for time t. However, in our analysis, we are looking for W at infinite time W(∞). By that time, as we said earlier, because of viscous dissipation in the gas, the piston velocity v is going to be equal to zero, and the piston is going to stay at the dead end of the cylinder.

A couple of aside questions:

How was Newton's Second law used to justify P_ext = P_I?

For a massless frictionless piston, the forces on both sides of the piston are equal.

In general, is it valid to treat friction heat addition as Q in first law or for ΔS = ∫δQ/T + σ, or does Q have to be heat transfer via finite temp gradient between the system and surrounding?

I don't quite understand this question. What are you including in your "system"? Gas or gas plus piston?

In my OP you mentioned that first law was simply W = ∫P_extdV= E_2 - E_1 if piston was included in the system; I took this to mean that the work that turned into frictional heat that went into the system still counted as work and thus ∫δQ/T = 0, is this correct?

I think there should a minus sign in front of the W and the integral. What you said is correct regarding the frictional heat.

Chet

 

[Red_CCF]

This is a correct equation for time t. However, in our analysis, we are looking for W at infinite time W(∞). By that time, as we said earlier, because of viscous dissipation in the gas, the piston velocity v is going to be equal to zero, and the piston is going to stay at the dead end of the cylinder.

As the piston oscillates and the gas does work on the piston and vice versa back and forth, the end result is that at the end no net work is done by either one? I noticed that W in that equation is always positive, but isn't there a period where the piston does work on the gas (after colliding at the wall) such that there should be a sign change?

I understand that viscous dissipation is a loss, but like the example in my OP, if some of the applied work is turned into friction heat and still goes into the system, the system's temperature and pressure will go up; how does one show that this gain in pressure and hence ∫PdV over a cycle is less than ∫PdV if the process was reversible without using second law?

For a massless frictionless piston, the forces on both sides of the piston are equal.

Is this because a massless object cannot have a finite net force acted up on it?

If the piston does have mass but frictionless, how does P_ext relate to P_I,gas in quasistatic and non-quasistatic compression/expansion?

I don't quite understand this question. What are you including in your "system"? Gas or gas plus piston?
Chet

I was thinking in a general case where friction heat is generated at the system boundary (perhaps the piston-cylinder boundary if the piston is included in the system), or internal irreversibilities that causes heat generation (ie viscous dissipation). In these cases does ∫δQ/T = 0 even if the system is insulated?

Thanks very much

 

[Chestermiller]

As the piston oscillates and the gas does work on the piston and vice versa back and forth, the end result is that at the end no net work is done by either one?

Yes.

I noticed that W in that equation is always positive, but isn't there a period where the piston does work on the gas (after colliding at the wall) such that there should be a sign change?

There would be a decrease in W(t) during this period. I'm not sure if W(t) ever actually changes sign, but, fortunately, we don't have to resolve this by solving the gas dynamics equations. All we need to know is that, in the end, W = 0.

I understand that viscous dissipation is a loss, but like the example in my OP, if some of the applied work is turned into friction heat and still goes into the system, the system's temperature and pressure will go up; how does one show that this gain in pressure and hence ∫PdV over a cycle is less than ∫PdV if the process was reversible without using second law?

You need to solve the gas dynamics equations to flesh out all the details. But, for our purposes we don't need to do this. During the transient phase of the process, the gas expands as it does work on the piston, and, as a result it cools. But, after the kinetic energy of the piston (and gas) is dissipated as heat, the gas gets reheated back to its original temperature.

Is this because a massless object cannot have a finite net force acted up on it?

Yes. Otherwise it would have to have an infinite acceleration, which it doesn't.

If the piston does have mass but frictionless, how does P_ext relate to P_I,gas in quasistatic and non-quasistatic compression/expansion?

In quasistatic, P_ext and P_I,gas differ by an insignificant amount (for the horizontal cylinder case). In non-quasistatic, (P_I,gas-P_ext)A = ma if the cylinder is horizontal, and (P_I,gas-P_ext)A = m(g+a) if the cylinder is vertical.

I was thinking in a general case where friction heat is generated at the system boundary (perhaps the piston-cylinder boundary if the piston is included in the system), or internal irreversibilities that causes heat generation (ie viscous dissipation). In these cases does ∫δQ/T = 0 even if the system is insulated?

Sure. If both the gas and the piston are included in the system, then Q = 0 if the system is insulated (by definition). But, there is extra work done to move the piston against the friction.

Chet

 

[Red_CCF]

There would be a decrease in W(t) during this period. I'm not sure if W(t) ever actually changes sign, but, fortunately, we don't have to resolve this by solving the gas dynamics equations. All we need to know is that, in the end, W = 0.

I was thinking that, when the piston first reaches the far end of the cylinder by the gas, the gas is doing work on it and when it hits the wall and moves back, it would be doing work onto the gas and hence there should be a sign reversal?

With the W(t) equation, v = 0 when the piston is slowed to a stop by the gas as it pushes back onto the gas, how does W(t) = 0 in those scenarios since the piston doesn't push the gas all the way back to its original state due to dissipation?

You need to solve the gas dynamics equations to flesh out all the details. But, for our purposes we don't need to do this. During the transient phase of the process, the gas expands as it does work on the piston, and, as a result it cools. But, after the kinetic energy of the piston (and gas) is dissipated as heat, the gas gets reheated back to its original temperature.

What I'm confused about is how does internal irreversibility like viscous dissipation during compression/expansion cause a work loss. Does it require a higher average applied pressure over the same volume change and if so why? Also, if some of the work applied turned into frictional heat and still goes into the system, how is that a loss?

In quasistatic, P_ext and P_I,gas differ by an insignificant amount (for the horizontal cylinder case). In non-quasistatic, (P_I,gas-P_ext)A = ma if the cylinder is horizontal, and (P_I,gas-P_ext)A = m(g+a) if the cylinder is vertical.

And for the vertical case it would be P_I,gas = mg/A_piston+ P_ext?

How does friction between the gas and cylinder during compression/expansion change the relationship between P_I,gas and P_ext; is it F_friction + P_I,gasA = P_extA?

Thanks very much

 

[Chestermiller]

I was thinking that, when the piston first reaches the far end of the cylinder by the gas, the gas is doing work on it and when it hits the wall and moves back, it would be doing work onto the gas and hence there should be a sign reversal?

It would cause a sign reversal on the rate at which work is being done on the piston, not on the cumulative amount of work done on the piston.

With the W(t) equation, v = 0 when the piston is slowed to a stop by the gas as it pushes back onto the gas, how does W(t) = 0 in those scenarios since the piston doesn't push the gas all the way back to its original state due to dissipation?

This is not the proper venue for discussing the details of the gas dynamics. But, maybe an analogy will help. Think of a mass and spring connected in series on a flat frictionless table, with the other end of the spring attached to the wall (which is right next to the table). Initially, the spring is compressed, and at time t = 0, the mass is released, and the system is allowed to oscillate. It will oscillate forever in simple harmonic motion, and the kinetic energy of the mass will continually vary with time as the spring alternately stores and releases elastic energy. The spring is analogous to our gas, and the mass is analogous to our piston. Now introduce a dissipative element into the picture. Connect a viscous damper between the wall and the mass, in parallel with the spring. The characteristic of the viscous damper is that the force it exerts on the mass is proportional to the velocity of the mass (including sign), but in the opposite direction. The combination of spring and viscous damper is now analogous to our gas, rather than just the spring. Now, if the mass is released at time zero, instead of having simple harmonic motion, the mass will undergo damped oscillation, and the amplitude of the oscillation will decrease with time, so that, in the end, the mass will stop moving. This is not an outrageously bad analogy of what happens with the gas and piston. Even though, in the case of a gas, it is incapable of exerting tension on the piston, the elastic collisions of the piston with the far cylinder wall do the job of reversing the piston direction.

What I'm confused about is how does internal irreversibility like viscous dissipation during compression/expansion cause a work loss.

Does it require a higher average applied pressure over the same volume change and if so why?

Solve the spring-mass-damper oscillation problem. That should help. See how the decrease in kinetic energy of the mass comes about.

Also, if some of the work applied turned into frictional heat and still goes into the system, how is that a loss?

It's not a loss. It's just a conversion of work into internal energy. During the initial expansion, the gas is doing work on the piston, and parts of the gas cool. When, in the end, the cumulative work done on the piston drops to zero, the viscous heating has returned the gas to its original temperature.

And for the vertical case it would be P_I,gas = mg/A_piston+ P_ext?

Yes, for quasistatic.

How does friction between the gas and cylinder during compression/expansion change the relationship between P_I,gas and P_ext; is it F_friction + P_I,gasA = P_extA?

Yes, for quasistatic horizontal.

Chet

 

[Red_CCF]

It would cause a sign reversal on the rate at which work is being done on the piston, not on the cumulative amount of work done on the piston.

Solve the spring-mass-damper oscillation problem. That should help. See how the decrease in kinetic energy of the mass comes about.

I am currently imagining this as, when the piston moves back to compress the gas, once v = 0 all of the work energy the gas gave to it during expansion has been returned, but some are added as heat (viscous dissipation/damper), while some as PdV work (spring), which the gas applies back to the piston. Each cycle the amount of PdV work exchanged decreases until it is 0 when the piston stops. Is this correct?

It's not a loss. It's just a conversion of work into internal energy. During the initial expansion, the gas is doing work on the piston, and parts of the gas cool. When, in the end, the cumulative work done on the piston drops to zero, the viscous heating has returned the gas to its original temperature.

In the P-V graph in my OP, compared to the reversible case, the irreversible case over the same volume change requires a higher exerted pressure during compression, and extracts a lower pressure during expansion. If viscosity is the only irreversibility, I kind of get how a higher pressure is needed during the compression stage (viscous dissipation generates heat -> higher T -> higher P during the process). But this doesn't explain why pressure is lower during expansion?

Yes, for quasistatic horizontal.

In this case, unlike our previous examples, P_ext ≠ P_I due to friction between the gas and wall. Should I use P_I,gas (if the system gas only and insulated) to calculate ∫PdV or is P_ext more appropriate since the system is insulated hence the frictional heat is going into the gas?

From a Newton's third law perspective, what are the action-reaction pairs? I see F_fric between gas-cylinder, but the piston is exerting P_ext on the gas, yet P_I,gas < P_ext is applied to the piston by the gas, where does the difference come in?

Thanks very much

 

[Chestermiller]

Hi Red_CCF,

I want to cover a couple of things in this post. First of all, I would like to temporarily table our discussion of the frictional heating associated with the piston. I feel we need to focus on the other issue we have been discussing. We can come back to the piston friction soon.

Secondly, I wanted to make you aware of the fact that, when discussing the force per unit area at the interface (what we have been calling P_I), we have been employing the elementary thermodynamics approximation of the situation. This description assumes that P_I is the thermodynamic pressure at the interface, determined by the static PVT behavior of the gas:

P_I=ρ_IRT_I/M

where T_I is the temperature of the gas at the interface, M is the molecular weight of the gas, and ρ_I is the gas density at the interface. This description is OK for a quasistatic process where the gas is not deforming rapidly, but for an irreversible process, it is not.

The correct way to describe the boundary condition at the piston interface is to use the compressive stress σ_I at the interface (i.e., to replace P_I with σ_I). For a non-quasistatic deformation, the compressive stress at the interface is the linear sum of two contributions: the thermodynamic pressure ρ_IRT_I/M plus a viscous stress contribution -2μ∂v/∂x, where ∂v/∂x is the rate of change of gas velocity with respect to position in the gas (in the immediate vicinity of the interface). So, mechanistically, the force per unit area at the interface is the result both of the compression/expansion deformation of the gas, plus the rate of deformation of the gas. The viscous part describes the effect of the rate of deformation of the gas. This part is insignificant for a quasistatic deformation.

In an irreversible expansion or compression, the viscous portion of the stress is present everywhere within the gas, and not just at the interface. It results in a dissipation of mechanical energy, and does not only give rise to an increase in the local temperature of the gas. For our purposes, its most important effect is the dissipation of mechanical energy, and the temperature rise is somewhat secondary.

Now, let's get back to the spring-mass-damper analogy that I was talking about. This analogy should give you an idea of what is happening mechanicsically in our gas/piston system. The combination of the spring and damper is supposed to be analogous to our gas. It exhibits a linear combination of elastic and dissipative behavior. If F is the force exerted by the combination of spring and damper on the mass, then:

F = k(x₀-x) - Cdx/dt

where x₀ is the location of the end of the spring when there is no tension in it, and C is the dissipative damper constant. F is analogous to σ_IA, k(x₀-x) is analogous to the thermodynamic PVT stress contribution, and - Cdx/dt is analogous to the viscous stress contribution. From the force balance on the mass:

md²x/dt²= F

or,

m\frac{d^2x}{dt^2}=k(x_0-x)-C\frac{dx}{dt}

The work done by the "gas" on the "piston" is given by dW = Fdx. If we multiply both sides of the above equation by dx/dt, we obtain:

\frac{dW}{dt}+\frac{k}{2}\frac{d(x_0-x)^2}{dt}=-Cv^2

Note from this equation that the combination of kinetic energy of the mass plus compressional energy of the spring is decreasing with time, irrespective of the direction that the mass is moving. Only when the velocity of the mass reaches zero does the compressional energy of the spring reach its final value (assuming, as in our problem, that there is an elastic wall causing the mass to rebound periodically, and located such that the spring never goes into tension...the initial location of the mass in our situation requires the spring to be in compression).

At infinite time, W = 0, so, according to the above equation:

\frac{k}{2}[(x_0-x_∞)^2-(x_0-x_{init})^2]=-\int_0^∞{Cv^2dt}
That is, the energy stored in the spring at infinite time is less than the energy stored in the spring initially.

I think I'll stop here and let you digest what I've written.

Chet

 

[Red_CCF]

Thank you for such a detailed explanation.

For a non-quasistatic deformation, the compressive stress at the interface is the linear sum of two contributions: the thermodynamic pressure ρ_IRT_I/M plus a viscous stress contribution -2μ∂v/∂x, where ∂v/∂x is the rate of change of gas velocity with respect to position in the gas (in the immediate vicinity of the interface). So, mechanistically, the force per unit area at the interface is the result both of the compression/expansion deformation of the gas, plus the rate of deformation of the gas. The viscous part describes the effect of the rate of deformation of the gas. This part is insignificant for a quasistatic deformation.

What is the origin of -2μ∂v/∂x and what is the physical mechanism by which the rate of compression induces stress such that they can be superpositioned to P_I instead of changing P_I? Is this effect negligible in irreversible but quasistatic cases?

In an irreversible expansion or compression, the viscous portion of the stress is present everywhere within the gas, and not just at the interface. It results in a dissipation of mechanical energy, and does not only give rise to an increase in the local temperature of the gas. For our purposes, its most important effect is the dissipation of mechanical energy, and the temperature rise is somewhat secondary.

If all other irreversibilities are nonexistent and only viscous dissipation exists, the viscous stress term would be the reason why a higher pressure (in compression) is needed for work compared to reversible cases? Is this why textbooks say that P_ext should always (reversible or irreversible) be used for calculating work?

Now, let's get back to the spring-mass-damper analogy that I was talking about. This analogy should give you an idea of what is happening mechanicsically in our gas/piston system. The combination of the spring and damper is supposed to be analogous to our gas. It exhibits a linear combination of elastic and dissipative behavior. If F is the force exerted by the combination of spring and damper on the mass, then:

F = k(x₀-x) - Cdx/dt

where x₀ is the location of the end of the spring when there is no tension in it, and C is the dissipative damper constant. F is analogous to σ_IA, k(x₀-x) is analogous to the thermodynamic PVT stress contribution, and - Cdx/dt is analogous to the viscous stress contribution. From the force balance on the mass:

md²x/dt²= F

or,

m\frac{d^2x}{dt^2}=k(x_0-x)-C\frac{dx}{dt}

The work done by the "gas" on the "piston" is given by dW = Fdx. If we multiply both sides of the above equation by dx/dt, we obtain:

\frac{dW}{dt}+\frac{k}{2}\frac{d(x_0-x)^2}{dt}=-Cv^2

Note from this equation that the combination of kinetic energy of the mass plus compressional energy of the spring is decreasing with time, irrespective of the direction that the mass is moving. Only when the velocity of the mass reaches zero does the compressional energy of the spring reach its final value (assuming, as in our problem, that there is an elastic wall causing the mass to rebound periodically, and located such that the spring never goes into tension...the initial location of the mass in our situation requires the spring to be in compression).

At infinite time, W = 0, so, according to the above equation:

\frac{k}{2}[(x_0-x_∞)^2-(x_0-x_{init})^2]=-\int_0^∞{Cv^2dt}
That is, the energy stored in the spring at infinite time is less than the energy stored in the spring initially.

I think I'll stop here and let you digest what I've written.

Chet

A couple of questions on this:

Is dW/dt supposed to represent the (rate of) kinetic energy (change) of the piston and how?

For this spring system, is there a simple way of showing ∫dW = 0 for t = 0 to ∞ or is the v = 0? If -Cv^2 = 0 (no damper), would the last equation be not integrable?

When you say that the spring energy is lower at t = ∞ than t = 0, is this equivalent of saying that the final gas pressure is lower than the initial and thus has less potential to push the piston if the cylinder wall was removed?

Thanks very much

 

[Chestermiller]

What is the origin of -2μ∂v/∂x and what is the physical mechanism by which the rate of compression induces stress such that they can be superpositioned to P_I instead of changing P_I?

It originates with Newton's Law of viscosity. It is part of the 3D tensorial version of the law. See Bird, Stewart, and Lightfoot, Transport Phenomena (a wonderful book that should be in everyone's library) for more details.

Is this effect negligible in irreversible but quasistatic cases?

Yes. But, in such cases, there can also be irreversibility due to conductive heat transport and chemical reaction.

If all other irreversibilities are nonexistent and only viscous dissipation exists, the viscous stress term would be the reason why a higher pressure (in compression) is needed for work compared to reversible cases?

No. During the irreversible deformation, the inertia of the gas also contributes to the force on the piston (the gas itself is experiencing spatially inhomogeneous acceleration, and is exchanging kinetic energy, in the same kind of way that the kinetic energy of the piston is contributing). However, at final steady state, of course, the gas is again stationary, and all its kinetic energy has been dissipated.

Is this why textbooks say that P_ext should always (reversible or irreversible) be used for calculating work?

Not exactly. In the irreversible case, you basically have very little control over what is happening in the gas (as you say), but you do have control over what you impose on its boundary. That's why you use P_ext, since you have control over that. In the reversible case, P_ext is the same as P of the gas (if, by P_ext, you mean the total force per unit area imposed on the boundary from outside, and the gas is your system).

Is dW/dt supposed to represent the (rate of) kinetic energy (change) of the piston and how?

Yes. We derived this a few posts ago, when we multiplied md²x/dt² by dx/dt to obtain mdv²/dt.

For this spring system, is there a simple way of showing ∫dW = 0 for t = 0 to ∞ or is the v = 0?

By solving the differential equation, you find that v = 0 at infinite time. This automatically means that ∫dW is equal to zero, since the change in kinetic energy of the piston is then zero.

If -Cv^2 = 0 (no damper), would the last equation be not integrable?

If there were no damping, then the system would oscillate forever. The equation would be integrable, and it would tell you that the sum of the kinetic energy plus the compressional energy is constant.

When you say that the spring energy is lower at t = ∞ than t = 0, is this equivalent of saying that the final gas pressure is lower than the initial and thus has less potential to push the piston if the cylinder wall was removed?

Yes.

I'll wait to see if you have any additional questions or clarifications. I also want to add one more detail to the spring damper analog model, so you can see how the mass of the gas comes into play. Tell me when you are ready to hear about this.

Chet

 

[Red_CCF]

It originates with Newton's Law of viscosity. It is part of the 3D tensorial version of the law. See Bird, Stewart, and Lightfoot, Transport Phenomena (a wonderful book that should be in everyone's library) for more details.

I always think of gas applying a force on something via pressure; how does this mechanism work such that an additional force is applied without increasing gas pressure?

No. During the irreversible deformation, the inertia of the gas also contributes to the force on the piston (the gas itself is experiencing spatially inhomogeneous acceleration, and is exchanging kinetic energy, in the same kind of way that the kinetic energy of the piston is contributing). However, at final steady state, of course, the gas is again stationary, and all its kinetic energy has been dissipated.

Assuming the process is irreversible but quasistatic, can gas inertia be neglected? I'm interested in looking at viscous dissipation by itself, and why it would required pressure is higher during compression yet lower during expansion compared to a reversible case.

Not exactly. In the irreversible case, you basically have very little control over what is happening in the gas (as you say), but you do have control over what you impose on its boundary. That's why you use P_ext, since you have control over that. In the reversible case, P_ext is the same as P of the gas (if, by P_ext, you mean the total force per unit area imposed on the boundary from outside, and the gas is your system).

In a general case (quasistatic or not), if the gas is the system, should ∫P_extdV always be used, but P_ext may or may not equal to P_I due to friction or viscous stresses etc.? In your blog, what was the assumption behind the system type such that P_I may be used for work calculations?

By solving the differential equation, you find that v = 0 at infinite time. This automatically means that ∫dW is equal to zero, since the change in kinetic energy of the piston is then zero.

Does the path dependence of work affect the way it's integrated? Can we just subtract the piston's KE at the beginning and end (0 both ways) and say that the net work done on it is 0?

I'll wait to see if you have any additional questions or clarifications. I also want to add one more detail to the spring damper analog model, so you can see how the mass of the gas comes into play. Tell me when you are ready to hear about this.

Please do.

Thanks very much

 

[Chestermiller]

I always think of gas applying a force on something via pressure; how does this mechanism work such that an additional force is applied without increasing gas pressure?

Newton's law of viscosity shows that the stress within a fluid (a gas is an incompressible fluid) is a function not only of the amount that the fluid has deformed, but also the rate at which it is deforming. For a so-called Newtonian fluid, the stress is a linear function of both the volumetric strain and the rate of strain. Almost all gases and most liquids are described extremely accurately by the Newtonian fluid model (properly expressed mathematically in 3D tensorial form). The thermodynamic pressure corresponds to the portion of the stress resulting from volumetric strain, and the viscous stresses describe the portion of the stress resulting from rate of strain. So, in summary, the pressure does not tell the whole story.

Assuming the process is irreversible but quasistatic, can gas inertia be neglected?

Yes.

I'm interested in looking at viscous dissipation by itself, and why it would required pressure is higher during compression yet lower during expansion compared to a reversible case.

One answer to this question is to say "see BSL for the molecular mechanism associated with viscosity." Another answer would be to consider a very viscous fluid bar that is being pulled on both ends. The rate of deformation of the bar results in actual tension in the bar, which is the same as negative compressive stress. Once you stop the deformation (i.e., zero rate of deformation), the tension goes away. From a fluid mechanics standpoint, liquids and gases are both described very accurately by the Newtonian fluid model.

I very strongly encourage you to read the first chapter or two in BSL, since you now seem highly motivated to learn this material. You won't be disappointed.

In a general case (quasistatic or not), if the gas is the system, should ∫P_extdV always be used, but P_ext may or may not equal to P_I due to friction or viscous stresses etc.?

As I said, if the gas is the system, P_I needs to be replaced by σ_I. This takes care of dealing with the viscous stresses that are present during irreversible. What we are dealing with here is the fact that, in most thermodynamics text developments, the authors play it fast and loose with the mathematics. They are trying to keep it simple (or are even unaware of the contribution of viscous stresses), but, as a result, some confusion ensues. If P_ext represents the force per unit area on the top of the piston, then ∫P_extdV is the work done on the system only if the massless piston is included as part of the system. Then, P_ext = F/A + σ_I, where F is the frictional force of the cylinder on the massless piston.

In your blog, what was the assumption behind the system type such that P_I may be used for work calculations?

In my Blog, I took a bit of literary license when I called P_I the pressure at the interface. I was trying to keep it simple for new students of thermodynamics, and was also trying to be somewhat consistent with what they would encounter in other elementary thermo developments. So, in a way, I was doing what I criticized above (playing a little fast and loose with the mathematics). However, I knew that it was going to be mostly novices that read my Blog, and I had some important points I wanted to make, particularly with regard to the Second Law. I didn't want to lose everyone who was reading the blog by starting to introduce concepts such as the stress tensor, deformational mechanics, and Newtonian fluids. These concepts are not typically included as part of the elementary thermo curriculum, and are not introduced until later when students start studying Fluid Mechanics, Solid Mechanics, and Continuum Mechanics.

Now that you personally have progressed beyond the novice level, you have begun to ask more perceptive questions that can be answered by starting to study these more advanced and precise disciplines.

Does the path dependence of work affect the way it's integrated?

No. The mathematics does not lie.

Can we just subtract the piston's KE at the beginning and end (0 both ways) and say that the net work done on it is 0?

Yes.

I'm going to stop here and try to field any further questions and clarifications you might have before proceeding to the more detailed mechanical analog that includes the inertia of the gas.

Chet

 

[Red_CCF]

Newton's law of viscosity shows that the stress within a fluid (a gas is an incompressible fluid) is a function not only of the amount that the fluid has deformed, but also the rate at which it is deforming. For a so-called Newtonian fluid, the stress is a linear function of both the volumetric strain and the rate of strain. Almost all gases and most liquids are described extremely accurately by the Newtonian fluid model (properly expressed mathematically in 3D tensorial form). The thermodynamic pressure corresponds to the portion of the stress resulting from volumetric strain, and the viscous stresses describe the portion of the stress resulting from rate of strain. So, in summary, the pressure does not tell the whole story.

Is gas considered incompressible in the sense that the Ma << 0.3? Is it possible to measure the force due to rate of strain alone like one would do with P_I?

As I said, if the gas is the system, P_I needs to be replaced by σ_I. This takes care of dealing with the viscous stresses that are present during irreversible. What we are dealing with here is the fact that, in most thermodynamics text developments, the authors play it fast and loose with the mathematics. They are trying to keep it simple (or are even unaware of the contribution of viscous stresses), but, as a result, some confusion ensues. If P_ext represents the force per unit area on the top of the piston, then ∫P_extdV is the work done on the system only if the massless piston is included as part of the system. Then, P_ext = F/A + σ_I, where F is the frictional force of the cylinder on the massless piston.

Is ∫σ_IdV easily solvable, as it requires the knowledge of ∂v/∂t as a function of volume? Is ∂v/∂t negative during compression?

With regards to my question about friction between the gas-cylinder (but not piston-cylinder), the pressure exerted onto the gas at the gas-piston interface is higher than P_I (or σ_I) if the piston wants to compress the gas. What are the Newton's Third Law action/reaction pairs as at the gas-piston interface, the forces applied by the two on each other in this case are not equal?

For a gas that undergoes a reversible or irreversible cycle, net work depends on the process and is not necessarily zero even though it begins and ends in the same state. How come for the piston this isn't the case?

Thanks very much