Moving an adiabatic partition in an adiabatic container

[FranzDiCoccio]

Homework Statement

A horizontal cylindrical container is divided into two equal chambers by an airtight movable partition. The walls of the cylinder and the partition are adiabatic. Each chamber is filled with 1 mol of a monoatomic ideal gas. The gas in the left chamber is hotter, the one in the right chamber is colder.
The partition moves until the pressure in the two chambers is the same. Find the final temperatures in the two chambers.

Relevant Equations

Equation of state for an ideal gas. Relations between p, V and T for an adiabatic process. Internal energy for a monoatomic ideal gas. First law of thermodynamics.

The actual data for the problem and my (and my friend's) attempt at a solution are in the attached file.
In a nutshell, this is what happened.

I obtained a solution based on the fact that the system is isolated. Thus the initially hot gas moves the partition doing work onto the initially cold gas, but the total amount of internal energy stays the same. I thought I used the conservation of the total volume, but somehow this does not appear to be the case.

A friend of mine obtained a different solution explicitly requiring that the total volume of the cylindrical container should be the same for the initial and final states of the systems.

My solution has the problem that the final volume is not the same as the initial one.
My friend's solution has the problem that the total internal energy of the final state is not the same as in the initial state.

So, probably both of us are making a (different) mistake.
Possibly, both of us are introducing different assumptions on how the process comes about (although we both initially assumed that the system is isolated and that the volume does not vary).

Can someone advise, please?

 

[Chestermiller]

Homework Statement:: A horizontal cylindrical container is divided into two equal chambers by an airtight movable partition. The walls of the cylinder and the partition are adiabatic. Each chamber is filled with 1 mol of a monoatomic ideal gas. The gas in the left chamber is hotter, the one in the right chamber is colder.
The partition moves until the pressure in the two chambers is the same. Find the final temperatures in the two chambers.
Homework Equations:: Equation of state for an ideal gas. Relations between p, V and T for an adiabatic process. Internal energy for a monoatomic ideal gas. First law of thermodynamics.

The actual data for the problem and my (and my friend's) attempt at a solution are in the attached file.
In a nutshell, this is what happened.

I obtained a solution based on the fact that the system is isolated. Thus the initially hot gas moves the partition doing work onto the initially cold gas, but the total amount of internal energy stays the same. I thought I used the conservation of the total volume, but somehow this does not appear to be the case.

A friend of mine obtained a different solution explicitly requiring that the total volume of the cylindrical container should be the same for the initial and final states of the systems.

My solution has the problem that the final volume is not the same as the initial one.
My friend's solution has the problem that the total internal energy of the final state is not the same as in the initial state.

So, probably both of us are making a (different) mistake.
Possibly, both of us are introducing different assumptions on how the process comes about (although we both initially assumed that the system is isolated and that the volume does not vary).

Can someone advise, please?

Your mistake was assuming that the equations for an adiabatic reversible process (involving $\gamma$) apply to this problem. They do not. The process involved here is highly irreversible. For the details on how to solve this exact problem for the case of an insulated partition, see this thread: https://www.physicsforums.com/threads/irreversible-expansion-of-gas-against-gas.981339/#post-6270858. If the partition is not insulated, you can find the final temperature easily by requiring that the internal energy does not change. This then tells you the final pressure and volumes.

 

[FranzDiCoccio]

Hi Chestermiller,
thanks a lot for your help!

I took a look at the post you linked. Funny that someone had exactly the same problem as me, one day before... I do not think that the OP in that thread was my friend. Of course I searched the forums, but must have used the wrong keywords, because I did not find this thread. It would have saved me a lot of time.

Anyway, there is something I do not understand in your solution, and I have a lot of questions.

Your key assumption in this post seems to be that the work involved in the transformation is the same as in a isobaric process at the final pressure.

What is the reason for this assumption, other than it actually seems to provide reasonable and self-consistent results? The process is not isobaric. Pressure varies in both containers.Are there different reasonable assumptions that produce a (possibly different) reasonable result?

Also, how do I know that the process is not reversible, other than from the lack of consistency of the results I get by using the equation for reversible adiabatic processes?
Perhaps because if I move the partition back to its original position the gases can be in different states depending on how I acted on the partition?

 

[Chestermiller]

Hi Chestermiller,
thanks a lot for your help!

I took a look at the post you linked. Funny that someone had exactly the same problem as me, one day before... I do not think that the OP in that thread was my friend. Of course I searched the forums, but must have used the wrong keywords, because I did not find this thread. It would have saved me a lot of time.

Anyway, there is something I do not understand in your solution, and I have a lot of questions.

Your key assumption in this post seems to be that the work involved in the transformation is the same as in a isobaric process at the final pressure.

What is the reason for this assumption, other than it actually seems to provide reasonable and self-consistent results? The process is not isobaric. Pressure varies in both containers.

The process is not isobaric, but this does not rule out the possibility that the force per unit area on both sides of the partition changes to the final pressure virtually immediately after the partition is released, and stays at that value on both sides of the partition throughout the expansion/compression. This is what happens in an analogous situation where a single gas is compressed in a cylinder with an external atmosphere at a constant pressure and the massless frictionless piston is released at time zero.

Are there different reasonable assumptions that produce a (possibly different) reasonable result?

Not in my judgment. This is one of those situations in which the problem cannot be solved accurately solely using thermodynamics. Without this assumption, we would have to solve the full compressible viscous fluid Navier Stokes partial differential equations as a function of time and spatial position within the chambers in conjunction with the continuity (local mass balance) equation and the thermal energy balance differential equation.

Also, how do I know that the process is not reversible, other than from the lack of consistency of the results I get by using the equation for reversible adiabatic processes?
Perhaps because if I move the partition back to its original position the gases can be in different states depending on how I acted on the partition?

The process that takes place is spontaneous and rapid. To make it reversible, you would have to manually control the motion of the partition so that the deformation takes place quasi-statically. This would involve your doing work of the combined system which is not a feature of what is happening physically.

 

[FranzDiCoccio]

Hi, thanks again for your help

The process is not isobaric, but this does not rule out the possibility that the force per unit area on both sides of the partition changes to the final pressure virtually immediately after the partition is released, and stays at that value on both sides of the partition throughout the expansion/compression.

But why should the partition move (or, for that matter, stop moving) if the pressure (and hence the force) on both its sides is the same?
I find this hard to understand.

Perhaps this has to do with the fact that during (this) irreversible process the pressure and temperature are not really well defined?

The process that takes place is spontaneous and rapid.

But how is one (especially one high school student) supposed to know this? It should be at least stated in the text of the problem, right?

I found this problem in two high-school textbooks, and in both cases the solution is the one I derived in section 3 of the document I uploaded. I looked further into that just because a more mathematically inclined friend worked out a different solution that seemed equally reasonable at first (but we found problems for both of our solutions).

To make it reversible, you would have to manually control the motion of the partition so that the deformation takes place quasi-statically. This would involve your doing work of the combined system which is not a feature of what is happening physically.

I sort of understand this, although not fully.

Without this assumption, we would have to solve the full compressible viscous fluid Navier Stokes partial differential equations as a function of time and spatial position within the chambers in conjunction with the continuity (local mass balance) equation and the thermal energy balance differential equation.

This is interesting, I wish I had the time, skill-set and computational resources to try that.
Do you know whether such an analysis has been performed?
Would it make sense to track the evolution of the pressure and temperature while the process is happening?
Would the final state be the same as the one obtained via the "reasonable assumption"?

I mean, if this is the case I would agree that the assumption is actually reasonable.
Still, I won't include this problem in a high-school textbook (or, at least, not in its present form).

 

[Andrew Mason]

Would it make sense to track the evolution of the pressure and temperature while the process is happening?
Would the final state be the same as the one obtained via the "reasonable assumption"?

I mean, if this is the case I would agree that the assumption is actually reasonable.
Still, I won't include this problem in a high-school textbook (or, at least, not in its present form).

This is just a straight forward first law problem. Since there is no work done on the surroundings, and no heat flow into or out of the gases, the total internal energy of the gases does not change. The process details are irrelevant.

AM

 

[Chestermiller]

Hi, thanks again for your help
But why should the partition move (or, for that matter, stop moving) if the pressure (and hence the force) on both its sides is the same?
I find this hard to understand.

The reason you are struggling with this is because you don't have the full picture of what is happening. So let me fill you in:

In a rapid irreversible expansion or compression, the behavior of an ideal gas is not described by the ideal gas law (even though it is still called an ideal gas). The pressure of the gas (or, more precisely, the normal stress at the piston face) is not equal to the RT/v, where v is the molar volume. In addition to the pressure being a function of temperature and molar volume, it is also a function of the rate at which the molar volume is changing dv/dt. A better approximation to the behavior would be $$P=\frac{RT}{v}-\eta\frac{1}{v}\frac{dv}{dt}$$ where $\eta$ is proportional to the viscosity of the gas. So, on the originally higher pressure side of the partition, the volume is increasing with time, and the pressure is less than you would calculate from the ideal gas law (for that compartment), and on the originally lower pressure side of the partition, the volume is decreasing with time, and the pressure is higher than you would calculate from the ideal gas law (for that compartment). It is the rate of change of volume on the two sides of the partition that enables the pressures on the two sides to be virtually equal. Since the partition is massless and frictionless, even the slightest differential between the pressures on the two sides will cause the partition to move at a finite rate. According to the law I wrote down, the piston must be moving (and the volumes must be changing) in order for the pressures to be essentially equal.

But how is one (especially one high school student) supposed to know this? It should be at least stated in the text of the problem, right?

I found this problem in two high-school textbooks, and in both cases the solution is the one I derived in section 3 of the document I uploaded. I looked further into that just because a more mathematically inclined friend worked out a different solution that seemed equally reasonable at first (but we found problems for both of our solutions).

I can't account for why a problem like this would be presented to high school students. The only thing I can conclude is that the person to posed the problem did not himself understand the irreversible nature of this process, and what it entailed. The same goes for the person who derived the solution in the textbooks.

I sort of understand this, although not fully.

What I am saying is that the solution you derived would be correct if supplementary external work were being applied to the partition. However, in that case, the internal energy of the system would have to change.

This is interesting, I wish I had the time, skill-set and computational resources to try that.
Do you know whether such an analysis has been performed?
Would it make sense to track the evolution of the pressure and temperature while the process is happening?
Would the final state be the same as the one obtained via the "reasonable assumption"?

Calculations of this type can routinely be applied by engineers using Computational Fluid Dynamics (CFD). This usually involves commercially available computer software that automatically solves the complicated partial differential equations once the input parameters and computational grid is specified. If this particular problem were solved using CFD, the results for the final state would not be exactly the same as the "reasonable assumption," but my experience tells me it would be pretty close.

I mean, if this is the case I would agree that the assumption is actually reasonable.
Still, I won't include this problem in a high-school textbook (or, at least, not in its present form).

I would never dream of including a problem like this in a high-school textbook. However, I would consider including a version in which the partition movement is controlled quasi-statically by an external agent. Of course, in that case, part of the problem solution would be to determine the change in internal energy of the system (equal to the work done by the external agent).

 

[FranzDiCoccio]

Thanks Chestermiller,
this is really helpful.

I would never dream of including a problem like this in a high-school textbook. However, I would consider including a version in which the partition movement is controlled quasi-statically by an external agent. Of course, in that case, part of the problem solution would be to determine the change in internal energy of the system (equal to the work done by the external agent).

You actually anticipated a further question I'd like to ask, if you are available.

The wording of the textbook problem might actually leave room for that. It actually says "the partition is moved quasi-statically".
My friend and I supposed that all the job was done by the pressure differential alone. One reason for doing so was that we thought we needed to know how the partition was moved to answer the question.

In light of your clarifications I would say that the solution should be the one in section 4 of the document I uploaded (that is my friend's "second attempt").
I guess that the system loses energy because the external agent is countering the rapid expansion of the gas, and hence the latter has to perform extra work that lowers its internal energy.

My first attempt, i.e. the solution provided in the textbook, seems to be kind of hopeless. While the internal energy is conserved, the system has a final volume that is less than the initial one, which makes no sense.

Do you agree with these two comments?

Thanks again for the really precious insight you gave me.

 

[Chestermiller]

I agree with your two comments. The sum of the two volumes will not change while the partition does external work so that the internal energy of the system decreases. Nice reasoning.

 

[FranzDiCoccio]

Yes. Also, it just occurred to me that in the "first attempt" it makes little sense requiring that the internal energy be conserved.
The fact that the partition is "controlled" by an external agent (as stated in the problem) very likely means that the internal energy is going to change.
This is what confused me in the text of the problem. I checked that my (initial) solution was the same as the one provided by the book. Then I concluded that all the work should be ascribed to the pressure differential alone, and the external agent had no role.

Now I definitely agree also with your comment

The only thing I can conclude is that the person to posed the problem did not himself understand the irreversible nature of this process, and what it entailed. The same goes for the person who derived the solution in the textbooks.

A partial excuse for the latter is that he/she probably was very young and underpaid :D
Thanks again!

 

[Andrew Mason]

Yes. Also, it just occurred to me that in the "first attempt" it makes little sense requiring that the internal energy be conserved.
The fact that the partition is "controlled" by an external agent (as stated in the problem) very likely means that the internal energy is going to change.

The internal energy cannot change because there is no external work done on or by the system and no heatflow into/out of the system. It does not matter how fast or slow the partition moves.

So the decrease in internal energy of one side has to equal the increase of internal energy of the other side. Since they are monatomic gases of equal measure (1 mole), this means that the temperature decrease of the high temperature side will be equal to the temperature increase of the low temperature side. Since the total volume is fixed, the increase in volume of the high temperature side has to equal the decrease in volume of the low temperature side.

AM

 

[hutchphd]

Work was done (negatively) by an external agent to keep the partition from accelerating. I believe the solution arrived at is correct.

 

[Chestermiller]

The internal energy cannot change because there is no external work done on or by the system and no heatflow into/out of the system. It does not matter how fast or slow the partition moves.

So the decrease in internal energy of one side has to equal the increase of internal energy of the other side. Since they are monatomic gases of equal measure (1 mole), this means that the temperature decrease of the high temperature side will be equal to the temperature increase of the low temperature side. Since the total volume is fixed, the increase in volume of the high temperature side has to equal the decrease in volume of the low temperature side.

AM

Andrew,

In order to cause the change to be quasi-static, you are going to need to apply an external force to the partition, say by passing a rod through the wall of the enclosure. Otherwise, if you allow the partition to move on its own, the change will not be quasi-static.

 

[Andrew Mason]

It doesn't matter whether the change is quasi-static or not. It is still an unbalanced pressure so the work done by the left side on the right side is determined by the pressure of the gas in the right side (we will say the initial high pressure side is on the left). It doesn't matter whether the system is in equilibrium while the process occurs.

Let's suppose that there are a thousand little notches and the partition moved in 1000 steps (somebody pulls a pin out to move to the next notch). No work is done by stopping the partition (we are ignoring the kinetic energy of the partition itself as there is no mention of its mass, so assume its mass is negligible i.e. the notches do not do any work in stopping the partition). All the work is being done within the system. There is no change in total internal energy. The work done is determined entirely by the pressure of the gas in the right side whether it happens quickly or very slowly.

AM

 

[Chestermiller]

It doesn't matter whether the change is quasi-static or not. It is still an unbalanced pressure so the work done by the left side on the right side is determined by the pressure of the gas in the right side (we will say the initial high pressure side is on the left). It doesn't matter whether the system is in equilibrium while the process occurs.

Let's suppose that there are a thousand little notches and the partition moved in 1000 steps (somebody pulls a pin out to move to the next notch). No work is done by stopping the partition (we are ignoring the kinetic energy of the partition itself as there is no mention of its mass, so assume its mass is negligible i.e. the notches do not do any work in stopping the partition). All the work is being done within the system. There is no change in total internal energy. The work done is determined entirely by the pressure of the gas in the right side whether it happens quickly or very slowly.

AM

In my judgment, for whatever it is worth, the process you describe with 1000 pins being pulled would be every bit as dissipative (during each pin interval) as the process of simply releasing the partition at time zero and allowing nature to take its course. So again, even for this process, in my judgment, the $pv^{\gamma}$ equations for a reversible process would not apply.

 

[Andrew Mason]

The adiabatic condition would apply to the right side because the work done in each step is against the internal pressure of the compressing gas. Since the work done on the right side is by the gas in the left side and there is no external work or heatflow, the $\Delta Us$ are equal and opposite.

A similar effect would occur with an adiabatic free expansion. If there was no gas in the right half, it would not matter if the partition was allowed to move all at once or in small increments. No work would be done so there would be no temperature change. If you add some gas to the right side the internal energy, hence temperature of the left side would decrease by exactly the amount that the internal energy of the right side increased. It is just first law: conservation of energy.

If the partition is allowed to move freely, it would oscillate back and forth until everything settled down in reaching equilibrium. A similar thing happens with a sound wave in air. Each compression or rarefaction can be closely approximated by reversible adiabatic processes.

AM

 

[Chestermiller]

The adiabatic condition would apply to the right side because the work done in each step is against the internal pressure of the compressing gas. Since the work done on the right side is by the gas in the left side and there is no external work or heatflow, the $\Delta Us$ are equal and opposite.

A similar effect would occur with an adiabatic free expansion. If there was no gas in the right half, it would not matter if the partition was allowed to move all at once or in small increments. No work would be done so there would be no temperature change. If you add some gas to the right side the internal energy, hence temperature of the left side would decrease by exactly the amount that the internal energy of the right side increased. It is just first law: conservation of energy.

If the partition is allowed to move freely, it would oscillate back and forth until everything settled down in reaching equilibrium. A similar thing happens with a sound wave in air. Each compression or rarefaction can be closely approximated by reversible adiabatic processes.

AM

I'm not arguing that if the partition is released and the system is allowed to re- equilibrate, the change in internal energy will be zero. What I am saying is that this process will not be even close to reversible, to the extent that the equations for a reversible process can not be used. And I am also arguing that, if the pins-and-notches version that you suggested will also not be even close to reversible.

 

[Andrew Mason]

I'm not arguing that if the partition is released and the system is allowed tore- equilibrate, the change in internal energy will be zero. What I am saying is that this process will not be even close to reversible, to the extent that the equations for a reversible process can not be used. And I am also arguing that, if the pins-and-notches version that you suggested will also not be even close to reversible.

The left side expansion is definitely not reversible. But the right side compression would be very close to reversible because the right side would be close to equilibrium during the compression if it occurred where the speed of the partition was much less than the average speed of the gas molecules (so that the right side gas responds more rapidly than the applied change initiated by the moving partition.

AM

 

[Chestermiller]

The left side expansion is definitely not reversible. But the right side compression would be very close to reversible because the right side would be close to equilibrium during the compression if it occurred where the speed of the partition was much less than the average speed of the gas molecules.

AM

I guess we are just going to have to agree to disagree.

 

[Andrew Mason]

I guess we are just going to have to agree to disagree.

All I am saying is that the work done on the right side is $-\int P_{int}dV = \int C_vdT$ . What do you say it would be?

AM

 

[Chestermiller]

All I am saying is that the work done on the right side is $-\int P_{int}dV = \int C_vdT$ . What do you say it would be?

AM

This comes down to just a judgment call. The approximation that I thought would probably be pretty accurate would be the one I suggested using in post #9 of this thread: https://www.physicsforums.com/threads/irreversible-expansion-of-gas-against-gas.981339/#post-6270858. Based on this approximation, there would be entropy increase in both chambers. As I said, it really comes down to a judgment call. At least this approach would be guaranteed to be consistent with the exact final pressure.

 

[Andrew Mason]

All one has to do is assume that the right side reacts very quickly to the moving adiabatic partition so that $P_{int}V=RT$ and $\Delta (P_{int}V) = R\Delta T$ applies, to a reasonable approximation, on the right side during compression. This means that the work done on the right side will be $W_R = \Delta U_R = -\int P_{R_{int}}dV = C_v \Delta T$. This relation does not apply on the left side because the work done BY the left side is $W_L = -\Delta U_L \ne \int P_{L_{int}}dV$

As a result, the change in entropy of the left side is positive (there is no reversible adiabatic process by which it can be returned to its original state) but the change in entropy of the right side is very close to 0.

AM

 

[Chestermiller]

All one has to do is assume that the right side reacts very quickly to the moving adiabatic partition so that $P_{int}V=RT$ and $\Delta (P_{int}V) = C_v\Delta T$ applies, to a reasonable approximation, on the right side during compression. This means that the work done on the right side will be $W_R = \Delta U_R = -\int P_{R_{int}}dV = C_v \Delta T$. This relation does not apply on the left side because the work done BY the left side is $W_L = -\Delta U_L \ne \int P_{L_{int}}dV$

As a result, the change in entropy of the left side is positive (there is no reversible adiabatic process by which it can be returned to its original state) but the change in entropy of the right side is very close to 0.

AM

Are you saying that, if you have a massless frictionless partition, the work done by the gas on the right side is not equal to minus the work done by the gas on the left side and that the magnitude of the force exerted by the gas on the right side of the partition is not equal to the magnitude of the force exerted by the gas on the left side of the partition.

If the two works are equal in magnitude, as your equations seem to imply, then the condition of constancy of the total internal energy requires that, for an ideal gas, the final pressure is the volume weighted average of the two initial pressures. Does your analysis automatically predict this?

Let's compare results. What does your analysis predict for the example in the link I provided for:

1. The final temperatures
2. The final pressure
3. The final volumes
4. The final changes in entropy

 

[Andrew Mason]

Are you saying that, if you have a massless frictionless partition, the work done by the gas on the right side is not equal to minus the work done by the gas on the left side and that the magnitude of the force exerted by the gas on the right side of the partition is not equal to the magnitude of the force exerted by the gas on the left side of the partition.

The work done by the gas by the left is exactly equal to the (negative) work done by the right side.

As far as the forces are concerned, the forces on the partition are not equal if the partition has mass, which of course it would have. If the gases on each side were in equilibrium during the process, the difference in forces (the respective pressures x area of the partition) would be equal to the mass x acceleration of the partition. The kinetic energy acquired by the partition would be imparted back to the gases when it settled down.

If the partition mass is negligible, the partition will move with a speed that results in the pressures on both immediate sides of the partition being equal during the process, but with neither side in equilibrium. In other words, there is a pressure gradient in the gases both sides of the partition. [* In the extreme case of a free expansion, the pressure on both sides would be zero.] But, at the end of the process when both sides return to equilibrium, the right side will be compressed and the left side will have expanded and the loss of internal energy by the left side will be equal to the gain in internal energy of the right side.

The actual pressure in the immediate vicinity of the partition is really not material if the reaction time of the gases is rapid (which for a small cyclinder at 300K it would be: the average speed of helium atoms at 300K is 1370 m/sec after all) because the rest of the gas - i.e. all but the atoms in the immediate vicinity of the partition - will be at pressure P = RT/V, the internal pressure of the gas ($P_{int}$). So the work on the right side is going to be very close to $\int P_{int}dV$

If the two works are equal in magnitude, as your equations seem to imply, then the condition of constancy of the total internal energy requires that, for an ideal gas, the final pressure is the volume weighted average of the two initial pressures. Does your analysis automatically predict this? Let's compare results. What does your analysis predict for the example in the link I provided for:

1. The final temperatures
2. The final pressure
3. The final volumes
4. The final changes in entropy

I'll have to work it out.

AM

 

[Chestermiller]

The work done by the gas by the left is exactly equal to the (negative) work done by the right side.

As far as the forces are concerned, the forces on the partition are not equal if the partition has mass, which of course it would have. If the gases on each side were in equilibrium during the process, the difference in forces (the respective pressures x area of the partition) would be equal to the mass x acceleration of the partition. The kinetic energy acquired by the partition would be imparted back to the gases when it settled down.

If the partition mass is negligible, the partition will move with a speed that results in the pressures on both immediate sides of the partition being equal during the process, but with neither side in equilibrium. In other words, there is a pressure gradient in the gases both sides of the partition. [* In the extreme case of a free expansion, the pressure on both sides would be zero.] But, at the end of the process when both sides return to equilibrium, the right side will be compressed and the left side will have expanded and the loss of internal energy by the left side will be equal to the gain in internal energy of the right side.

I agree with most this if we replace in your wording the "pressures on both sides" with the "normal stresses on both sides" which include the viscous contribution to the stresses. This all is consistent with the approximation I described as well.

The actual pressure in the immediate vicinity of the partition is really not material if the reaction time of the gases is rapid (which for a small cyclinder at 300K it would be: the average speed of helium atoms at 300K is 1370 m/sec after all) because the rest of the gas - i.e. all but the atoms in the immediate vicinity of the partition - will be at pressure P = RT/V, the internal pressure of the gas ($P_{int}$). So the work on the right side is going to be very close to $\int P_{int}dV$

This is the part that I'm not convinced about. The argument would be more convincing to me if it addressed quantitatively the reason that the viscous dissipation on the right side of the partition would be negligible compared to the viscous dissipation on the left side. In any event, I don't think it is going to matter much when we look at the actual final temperatures and volumes predicted by the two approximations. I still feel that, at this point, it is a judgment call.

I'll have to work it out.

AM

I've worked it out already, and the final predicted temperatures and volumes from the two approaches are relatively close to one another. I'm anxious to see the numbers you come up with. I think it will also be interesting to look at the predicted entropy changes for the two compartments.

 

[FranzDiCoccio]

By the way, I found this problem in the Italian version of "Physics - 10th edition" by Cutnell et al.
I just got a look at the original version of the 9th edition of the same textbook, and I see that this is problem 33 of chapter 15. The solution they provide is the wrong one, 477 K and 323K, where the total volume is not conserved.

 

[Andrew Mason]

I agree with most this if we replace in your wording the "pressures on both sides" with the "normal stresses on both sides" which include the viscous contribution to the stresses. This all is consistent with the approximation I described as well.

This is the part that I'm not convinced about. The argument would be more convincing to me if it addressed quantitatively the reason that the viscous dissipation on the right side of the partition would be negligible compared to the viscous dissipation on the left side. In any event, I don't think it is going to matter much when we look at the actual final temperatures and volumes predicted by the two approximations. I still feel that, at this point, it is a judgment call.I've worked it out already, and the final predicted temperatures and volumes from the two approaches are relatively close to one another. I'm anxious to see the numbers you come up with. I think it will also be interesting to look at the predicted entropy changes for the two compartments.

Since the final pressure on both sides is the same and there is no change in total internal energy, the increase in temperature of the right side (call it B) is the equal and opposite to the decrease in temperature of the left side (A). As you have shown, this means that:

$P_f = (P_A + P_B)/2$ so:

(1) $\frac{P_f}{P_B} = \frac{1}{2}\left(\frac{P_A}{P_B} + 1\right)$

Since side B is compressed adiabatically:

$\frac{P_f}{P_B} = \left(\frac{V}{V - \Delta V}\right)^{\gamma}$ so from (1):

(2) $\frac{1}{2}\left(\frac{P_A}{P_B} + 1\right) = \left(\frac{V}{V - \Delta V}\right)^{\gamma}$

and, therefore:

(3) $\Delta V = V(1 - [\frac{1}{2}(\frac{P_A}{P_B} +1)]^{\frac{-1}{\gamma}})$

To find $\Delta T$:

$R\Delta T = P_f(V-\Delta V) - P_BV = -P_f(V+\Delta V) - P_AV$ so:

(4) $\Delta T = \frac{1}{R}(P_f(V - \Delta V) - P_BV) = -\frac{1}{R}(P_f(V + \Delta V) - P_AV)$

Example:

So if $P_A = 2P_B$ (and, therefore, $T_A = 2T_B$) with a monatomic gas ($\gamma = 5/3$):

$\Delta V = V(1 - [\frac{1}{2}(2 + 1)]^{\frac{-3}{5}}) = V(1-.785) = .215V$

$R\Delta T = P_f(V - \Delta V) - P_BV = P_BV[\frac{3}{2}(1 - .215) -1] = .1775P_BV$

The change in entropy is almost all on the left side (side A). This is because side B can be returned to its original state by a reversible/quasi-static adiabatic expansion, but side A cannot be returned to its original state by a reversible process without heat flow (which means that the reversible process from its initial state to expanded state would require positive heat flow).

AM

 

[Chestermiller]

Since the final pressure on both sides is the same and there is no change in total internal energy, the increase in temperature of the right side (call it B) is the equal and opposite to the decrease in temperature of the left side (A). As you have shown, this means that:

$P_f = (P_A + P_B)/2$ so:

(1) $\frac{P_f}{P_B} = \frac{1}{2}\left(\frac{P_A}{P_B} + 1\right)$

Since side B is compressed adiabatically:

$\frac{P_f}{P_B} = \left(\frac{V}{V - \Delta V}\right)^{\gamma}$ so from (1):

(2) $\frac{1}{2}\left(\frac{P_A}{P_B} + 1\right) = \left(\frac{V}{V - \Delta V}\right)^{\gamma}$

and, therefore:

(3) $\Delta V = V(1 - [\frac{1}{2}(\frac{P_A}{P_B} +1)]^{\frac{-1}{\gamma}})$

To find $\Delta T$:

$R\Delta T = P_f(V-\Delta V) - P_BV = -P_f(V+\Delta V) - P_AV$ so:

(4) $\Delta T = \frac{1}{R}(P_f(V - \Delta V) - P_BV) = -\frac{1}{R}(P_f(V + \Delta V) - P_AV)$

Example:

So if $P_A = 2P_B$ (and, therefore, $T_A = 2T_B$) with a monatomic gas ($\gamma = 5/3$):

$\Delta V = V(1 - [\frac{1}{2}(2 + 1)]^{\frac{-3}{5}}) = V(1-.785) = .215V$

$R\Delta T = P_f(V - \Delta V) - P_BV = P_BV[\frac{3}{2}(1 - .215) -1] = .1775P_BV$

The change in entropy is almost all on the left side (side A). This is because side B can be returned to its original state by a reversible/quasi-static adiabatic expansion, but side A cannot be returned to its original state by a reversible process without heat flow (which means that the reversible process from its initial state to expanded state would require positive heat flow).

AM

Hi Andrew. I appreciate your running this calculation. But I had asked you to do the calculation for the problem statement in the link I provided so we could compare results. Any chance you could run your approximation for that case?

In that problem, PA=4 atm, PB=1 atm, TA=900 K, TB=300 K, VA=VB=1 liter, diatomic gas $\gamma = 1.4$

 

[Andrew Mason]

Hi Andrew. I appreciate your running this calculation. But I had asked you to do the calculation for the problem statement in the link I provided so we could compare results. Any chance you could run your approximation for that case?

In that problem, PA=4 atm, PB=1 atm, TA=900 K, TB=300 K, VA=VB=1 liter, diatomic gas $\gamma = 1.4$

The problem as stated in this thread is based on the same number of moles of monatomic ideal gas in each side. That is what I worked out. The $\Delta T$ of the two sides will not be equal and opposite if there are different quantities of gas in the two compartments, which is what those values indicate.

Using the values provided in the OP attachment ($T_A = 525K$ and $T_B = 275K$):

$P_A/P_B = 525/275 = 1.91$

$P_A = RT_A/V = 8.3145 \times 525 \div 10^{-3} = 4.365 \times 10^6 Pa$
$P_B = P_A/1.91 = 2.285 \times 10^6 Pa$
$P_f = 3.325 \times 10^6 Pa$

So: $\Delta V = V(1 - [\frac{1}{2}(\frac{P_A}{P_B} +1)]^{\frac{-1}{\gamma}}) = 1-(2.91/2)^{-.6} = .201$ L.

$R\Delta T = P_f(V - \Delta V) - P_BV$
$R\Delta T = (3.325 \times 10^6)(.799 \times 10^{-3}) - (2.285 \times 10^6)(10^{-3}) = 371.675$J.

So $\Delta T = 371.675/R = 44.7$K

AM

 

[Chestermiller]

The problem as stated in this thread is based on the same number of moles of monatomic ideal gas in each side. That is what I worked out. The $\Delta T$ of the two sides will not be equal and opposite if there are different quantities of gas in the two compartments, which is what those values indicate.

AM

OK. I'll check your solution and compare with mine using the data in this thread.

What am I missing. The problem statement says the initial temperatures are 525 and 275. This is not a factor of 2. The ratio is 1.9091.

Why do you say that, with your method, the change in entropy is almost all on the left side. According to you approximation, it should be exactly all on the left side.

I don't see any of your results for the left side.

 

[Chestermiller]

Using your method, for the final volume on the right, I get 0.7841V. For the final volume on the left, I get 1.2159 V. For the final temperature on the right, I get RT=1.1761 VP_B or, equivalently, T=1.1761 T_B. For the final temperature on the left, I get RT = 1.8239 VP_B, or equivalently, T = 1.8239 T_B = 0.912 T_A. These are close to the values you obtained. In my next post, I will give the corresponding values using the approximation that I am recommending.

 

[Chestermiller]

Using my approximation, for the final temperature on the right, I get $T=1.2T_B$ and, for the temperature on the left, I get $T=0.9T_A$. For the final volume on the right, I get 0.8 V and, for the final volume on the left, I get 1.2 V. Notice how closely our results resemble one another.

 

[Andrew Mason]

Why do you say that, with your method, the change in entropy is almost all on the left side. According to you approximation, it should be exactly all on the left side.

It is a close approximation. Entropy change would be zero on the right if the process occurred with the right side in perfect equilibrium at all times.

I don't see any of your results for the left side.

$\Delta T, \Delta V and \Delta P$ are the same magnitude, opposite sign.

AM

 

[Chestermiller]

It is a close approximation.

I can make a case for the approximation I suggested being a close approximation (even though the two approximations give quantitatively very similar results). Any interest in seeing the analytic development?

 

[Andrew Mason]

I can make a case for the approximation I suggested being a close approximation (even though the two approximations give quantitatively very similar results). Any interest in seeing the analytic development?

Sure. It is difficult to analyse a dynamic process. You might be interested in this paper I found which is similar to an analysis of the left side of the present problem.

AM

 

[Chestermiller]

Sure. It is difficult to analyse a dynamic process. You might be interested in this paper I found which is similar to an analysis of the left side of the present problem.

AM

For a reversible quasi-static process, the compressive normal stress exerted by the gas on the piston face is given by the ideal gas law: $$\sigma=\frac{RT}{v}$$where v is the molar volume of the gas. For an irreversible non-quasistatic process, Newton's law of viscosity adds the basic viscous contribution to the normal stress (see Transport Phenomena by Bird, Stewart, and Lightfoot) such that, at the piston face, $$\sigma=\frac{RT}{v}-\frac{4}{3}\frac{\mu}{v}\frac{d v}{d t}$$where $\mu$ is the gas viscosity, and where all the properties are evaluated at the piston face.

If the piston velocity is small compared to the speed of sound in the gas, the molar volume of the gas will be essentially uniform, and v in these equations can be replaced by the total gas volume V divided by the number of moles of gas n. In addition, if, as expected, the rapid irreversible deformation of the gas is turbulent, the gas viscosity in this equation can be replaced by the turbulent viscosity $\mu_T$ which is relatively independent of temperature. Under these circumstances, the above equation reduces to:
$$\sigma=\frac{nRT}{V}-\frac{4}{3}\frac{\mu_T}{V}\frac{d V}{d t}$$
So, as a result of these considerations, the compressive normal stresses exerted by the gases on the left and right faces of the partition in the present problem are given by:$$\sigma_L=\frac{n_LRT_L}{V_L}-\frac{4}{3}\frac{\mu_T}{V_L}\frac{d V_L}{d t}$$and $$\sigma_R=\frac{n_RRT_R}{V_R}-\frac{4}{3}\frac{\mu_T}{V_R}\frac{d V_R}{d t}=\frac{n_RRT_R}{V_R}+\frac{4}{3}\frac{\mu_T}{V_R}\frac{d V_L}{d t}$$where use has been made here of the fact that the sum of the two gas volumes is constant.

Since the piston is massless and frictionless, we have that $$\sigma_L=\sigma_R=\sigma$$and thus
$$\sigma=\frac{n_LRT_L}{V_L}-\frac{4}{3}\frac{\mu_T}{V_L}\frac{d V_L}{d t}=\frac{n_RRT_R}{V_R}+\frac{4}{3}\frac{\mu_T}{V_R}\frac{d V_L}{d t}$$Eliminating $dV_L/dt$from these relationshiips yields $$\sigma=\frac{n_LRT_L+n_RRT_R}{(V_L+V_R)}\tag{1}$$ But, since the internal energy of the combined system is constant, we have $$n_LT_L+n_RT_R=n_LT_{Li}+n_RT_{Ri}$$and Eqn. 1 becomes:$$\sigma=\frac{n_LRT_{Li}+n_RRT_{R_i}}{(V_{Li}+V_{Ri})}=\frac{(P_{Li}V_{Li}+P_{Ri}V_{Ri})}{(V_{Li}+V_{Ri})}=P_f\tag{2}$$where $P_f$ is the final pressure. Eqn. 2 indicates that, at all times during the irreversible process, the normal stress on both sides of the partition is approximately equal to the final pressure.

 

[Chestermiller]

I've also calculated the entropy changes for the two chambers and the two approximations, expressed as $\Delta S/C_v$. I found that, even though the predicted temperature changes and volume changes are very similar for the 2 approximations, the entropy changes for the two chambers and the total entropy changes are quite different.

Andrew approximation:
Left = 0.0382
Right = 0.0
Total = 0.0382

Chet approximation:
Left = 0.0162
Right = 0.0338
Total = 0.0498

As you can see, with the viscosity-based approximation that I am proposing, twice as much entropy is generated in the right compartment as in the left, compared to your approximation which features zero entropy generation in the right compartment. And the total entropy generation is greater with my proposed approximation by about 25 %.

 

[Andrew Mason]

I've also calculated the entropy changes for the two chambers and the two approximations, expressed as $\Delta S/C_v$. I found that, even though the predicted temperature changes and volume changes are very similar for the 2 approximations, the entropy changes for the two chambers and the total entropy changes are quite different.

Andrew approximation:
Left = 0.0382
Right = 0.0
Total = 0.0382

Chet approximation:
Left = 0.0162
Right = 0.0338
Total = 0.0498

As you can see, with the viscosity-based approximation that I am proposing, twice as much entropy is generated in the right compartment as in the left, compared to your approximation which features zero entropy generation in the right compartment. And the total entropy generation is greater with my proposed approximation by about 25 %.

There are two things I am struggling with in your analysis. 1. since it is an ideal gas, I don't understand where the viscosity comes from. 2. I don't see how there could be more entropy generated on the compressed right side. Because the right side is compressed, the right side pressure effectively determines the work that is done, which is very close to a reversible/quasi-static adiabatic compression. The left side, however, has undergone a partial free expansion so entropy has increased.

Suppose that in the exact middle of the cylinder we added a fixed massless adiabatic panel that has a small aperature and put the panel a very small distance, say 1 mm, to the left of the moveable adiabatic panel (the moveable panel can not move farther to the left). Initially the space between the panels is a vacuum. We then open the aperture and gas from the left flows into the space. When the pressure in the 1mm space is a tiny bit greater than the pressure in the right side, the right panel moves. It keeps moving until the pressures in all three compartments are equal. In that case, the work done is very close to that of a quasi-static reversible adiabatic compression of the right side. So very little entropy increase on the right side. The left side, however, in expanding has done less work than $\int P_{internal}dV$ so it has more internal energy than it would have if it had followed a reversible path from its initial to final volume. Therefore, the reversible path is not adiabatic but one in which there is positive heat flow. So the entropy of the left side must increase.

AM

 

[Chestermiller]

There are two things I am struggling with in your analysis. 1. since it is an ideal gas, I don't understand where the viscosity comes from.

The reason you are struggling with this is that you have never had a course in Newtonian fluid dynamics. Newtonian fluid dynamics deals with the behavior of liquids and gases (including gases in the low-density limit of ideal gas behavior) at finite rates of deformation. According to Newton's law of viscosity, expressed in proper tensorial form, the state of stress in a Newtonian fluid (even in the ideal gas region) is described in Cartesian coordinates by (see Bird, Stewart, and Lightfoot, Transport Phenomena, Chapter 1):
$$\Pi_{xx}=p-2\mu\frac{\partial v_x}{\partial x}+\frac{2}{3}\mu\left(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}\right)$$
$$\Pi_{yy}=p-2\mu\frac{\partial v_y}{\partial y}+\frac{2}{3}\mu\left(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}\right)$$
$$\Pi_{zz}=p-2\mu\frac{\partial v_z}{\partial z}+\frac{2}{3}\mu\left(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}\right)$$
$$\Pi_{xy}=-\mu\left(\frac{\partial v_x}{\partial y}+\frac{\partial v_y}{\partial x}\right)$$
$$\Pi_{xz}=-\mu\left(\frac{\partial v_x}{\partial z}+\frac{\partial v_z}{\partial x}\right)$$
$$\Pi_{yz}=-\mu\left(\frac{\partial v_y}{\partial z}+\frac{\partial v_z}{\partial y}\right)$$
where the $\Pi's$ are the components of the compressive stress tensor in a deforming liquid or gas, p is the local pressure calculated from the equation of state in terms of the local temperature and specific volume (e.g., the ideal gas law), the v's are the components of the local velocity vector (field), and $\mu$ is the fluid viscosity.
Please note that, with all due respect to what you learned about ideal gases, without viscosity, there is no mechanism for non-quasistatic mechanical entropy generation in gases. So to understand what is happening in a non-quasistatic deformation of a gas, viscosity is an essential consideration. For more details on this, please see Example 11D.1. Equation of change for entropy in Chapter 11, Transport Phenomena.

2. I don't see how there could be more entropy generated on the compressed right side. Because the right side is compressed, the right side pressure effectively determines the work that is done, which is very close to a reversible/quasi-static adiabatic compression.

This is just your own personal opinion. If what you are saying is correct, then, for a massless frictionless piston, since the normal stress exerted by the two gases on the opposite faces of the partition must always be equal, the normal stress exerted by the gas to the left on the partition starts out at $2P_B$, but then drops immediately to $1P_B$ and then rises during the process to $1.5P_B$ (so that it matches what the stress is doing on the right of the partition) while, at the same time, the gas on the left is expanding. In my judgment, it defies logic to say that the force on the left increases while the gas expands.

The velocity of the partition is the same for both gases, and the magnitude of the rate of change of volume is also the same for both. From fluid dynamics, we know that the rate of volumetric strain is equal to the rate of volume change divided by the volume. Initially the magnitudes of the rates of volumetric strain are the same for both gases because the initial volumes are equal. This means that their rates of viscous dissipation are equal. However, as time progresses, the left volume becomes larger and the right volume becomes smaller. This means that the magnitude of the rate of volumetric strain on the right becomes larger than than the rate of volumetric strain on the left. This means that the rate of viscous dissipation on the right becomes larger than the rate of viscous dissipation on the left. So we expect a larger entropy increase on the right than on the left.

 

[Andrew Mason]

This is just your own personal opinion. If what you are saying is correct, then, for a massless frictionless piston, since the normal stress exerted by the two gases on the opposite faces of the partition must always be equal, the normal stress exerted by the gas to the left on the partition starts out at 2PB, but then drops immediately to 1PB and then rises during the process to 1.5PB
(so that it matches what the stress is doing on the right of the partition) while, at the same time, the gas on the left is expanding. In my judgment, it defies logic to say that the force on the left increases while the gas expands.

Well, the force on the gas on the right has to increase as the gas on the left expands. So what is there to provide that force other than the left side? It is the resistance to the applied pressure that results in a force. As that resisitance increases (i.e. pressure on the right) the left side applies more force.

AM

 

[Chestermiller]

Well, the force on the gas on the right has to increase as the gas on the left expands. So what is there to provide that force other than the left side? It is the resistance to the applied pressure that results in a force. As that resisitance increases (i.e. pressure on the right) the left side applies more force.

AM

Ask yourself this: how can the gas on the left possibly apply more force if the gas on the left is expanding? On the other hand, consider this:

The force on the right remains relatively constant as a result of (a) the equation of state compressive stress contribution $p_R$ increasing in conjunction with (b) the viscous compressive stress decreasing toward zero (as a result of the partition slowing down). The force on the left side also remains relatively constant as a result of (a) the equation of state compressive stress contribution $p_L$ decreasing in conjunction with (b) the viscous compressive stress (which, on this side of the partition, is negative, i.e. tensile) increasing toward zero (as a result of the partition slowing down).

 

[Andrew Mason]

Ask yourself this: how can the gas on the left possibly apply more force if the gas on the left is expanding?

The expansion results in a directed flow of gas to the right. That flow carries momentum. That momentum is transferred to the gas molecules on the right. The rate of change of momentum is a force. That force compresses the right side.

AM

 

[Chestermiller]

The expansion results in a directed flow of gas to the right. That flow carries momentum. That momentum is transferred to the gas molecules on the right. The rate of change of momentum is a force. That force compresses the right side.

AM

I'm talking exclusively about the gas on the left. Its volume is increasing and its temperature is decreasing. How can the force that it is exerting on the partition be increasing?

There is momentum transfer on both sides of the partition. These momentum transfer effects are captured by the turbulent viscosity contributions to the total compressive stresses. The viscous contributions give rise to entropy generation both in the left and the right compartments.

 

[Andrew Mason]

I'm talking exclusively about the gas on the left. Its volume is increasing and its temperature is decreasing. How can the force that it is exerting on the partition be increasing?

The internal energy of the gas is converted into kinetic energy of the directed flow to the right. As the pressure differential between left and right is largest initially, kinetic energy of the gas flow and temperature is greatest initially and pressure, therefore, force, is least (Bernoulli principle) in the flowing part of the gas (which is in the volume of gas nearest the partition). As the gas flow slows, kinetic energy is reduced and pressure, therefore, force increases.

There is momentum transfer on both sides of the partition. These momentum transfer effects are captured by the turbulent viscosity contributions to the total compressive stresses. The viscous contributions give rise to entropy generation in both the left and the right compartments.

There will be some oscillation but these oscillations will die down due not to friction but to randomization of the molecular motion. The paper that I mentioned earlier discusses this.

AM

 

[Chestermiller]

The internal energy of the gas is converted into kinetic energy of the directed flow to the right. As the pressure differential between left and right is largest initially, kinetic energy of the gas flow and temperature is greatest initially and pressure, therefore, force, is least (Bernoulli principle) in the flowing part of the gas (which is in the volume of gas nearest the partition). As the gas flow slows, kinetic energy is reduced and pressure, therefore, force increases.
There will be some oscillation but these oscillations will die down due not to friction but to randomization of the molecular motion. The paper that I mentioned earlier discusses this.

AM

In your paper, the "dynamic pressure" is given by the equation $$\hat{P}=\frac{nRT}{xA}\left[1-v\sqrt{\frac{8M}{\pi RT}}\right]=\frac{nRT}{xA}\left[1-\frac{dx}{dt}\sqrt{\frac{8M}{\pi RT}}\right]\tag{3}$$This equation can be re-expresssed as $$\hat{P}=\frac{nRT}{V}-\frac{1}{V}\frac{dV}{dt}\frac{n}{A}\sqrt{\frac{8MRT}{\pi}}$$The quantity $\frac{n}{A}\sqrt{\frac{8MRT}{\pi}}$ in this equation is proportional to the gas viscosity approximation determined by Maxwell in 1860 using kinetic theory. Note the similarity of this expression for the "dynamic pressure" to the expression I have presented for compressive stress $\Pi$. This is how the continuum viscosity theory connects to your kinetic theory approach. In my development, all I have done is replace your gas viscosity with turbulent viscosity to take into account the fact that the flow will be turbulent.

The net result of the approach indicated in your paper actually supports my contention that there will be significant entropy generation in both compartments.

 

[Chestermiller]

@Andrew Mason

So, if I understand correctly, you accept the equation for the "dynamic pressure" given in the paper you have referenced, right? Because, if that is the case, I will solve the equations for our partitioned gas system (numerically), subject to this dynamic pressure representation, to obtain the gas temperatures, volumes, and dynamic pressure at the partition as a function of time. We can then see once and for all whose approximation is a better representation for predicting the final state of the system and the entropy changes. Are you willing to accept these results if I go to the trouble to obtain them?

Chet

 

[Andrew Mason]

@Andrew Mason

So, if I understand correctly, you accept the equation for the "dynamic pressure" given in the paper you have referenced, right? Because, if that is the case, I will solve the equations for our partitioned gas system (numerically), subject to this dynamic pressure representation, to obtain the gas temperatures, volumes, and dynamic pressure at the partition as a function of time. We can then see once and for all whose approximation is a better representation for predicting the final state of the system and the entropy changes. Are you willing to accept these results if I go to the trouble to obtain them?

Chet

Let me think about it. I wouldn't want to ruin the holidays for you... Merry Christmas!
AM

 

[Chestermiller]

This is an analysis of the present problem using the model of non-quasistatic gas behavior recommended by @Andrew Mason in posts #35 and #44 of the present thread. The starting equations are as follows:

Energy Balances on the two compartments:
$$nC_v\frac{dT_L}{dt}=-\hat{P}\frac{dV_L}{dt}\tag{1}$$
$$nC_v\frac{dT_R}{dt}=-\hat{P}\frac{dV_R}{dt}=+\hat{P}\frac{dV_L}{dt}\tag{2}$$where n is the number of moles of gas in each of the two compartments and $\hat{P}$ is the time-dependent "dynamic pressure" on each of the two opposite faces of the massless frictionless partition, given by the the non-quasistatic model of gas behavior in Andrew's reference:

$$\hat{P}=\frac{nRT_L}{V_L}-\frac{\mu(T_L)}{V_L}\frac{dV_L}{dt}=\frac{nRT_R}{V_R}-\frac{\mu(T_R)}{V_R}\frac{dV_R}{dt}=\frac{nRT_R}{V_R}+\frac{\mu(T_R)}{V_R}\frac{dV_L}{dt}\tag{3}$$where $\mu(T)$ is a temperature-dependent parameter with units of viscosity that is proportional to $\sqrt{T}$.

Initial conditions (t = 0) on the problem are: $$V_L=V_R=V$$
$$T_L=2T_B$$
$$T_R=T_B$$$$P_L=2P_B=2\frac{nRT_B}{V}$$
$$P_R=P_B=\frac{nRT_B}{V}$$

 

[Chestermiller]

This is a continuation of the analysis begun in the previous post.

We can solve Eqns. 3 (representing the non-quasistatic Mungan gas model recommended by @Andrew Mason) to obtain the time derivative of the left compartment volume $dV_L/dt$ and the "dynamic pressure" $\hat{P}$ solely in terms of the compartment temperatures, volumes, and gas viscosities:
$$\frac{dV_L}{dt}=nR\frac{(T_LV_R-T_RV_L)}
{(\mu_LV_R+\mu_RV_L)}\tag{4}$$and$$\hat{P}=nR\frac{(T_L\mu_R+T_R\mu_L)}{(\mu_LV_R+\mu_RV_L)}\tag{5}$$where $\mu_L=\mu(T_L)$ and $\mu_R=\mu(T_R)$.

Next, if we add Eqns. 1 and 2, we obtain:$$nC_v\frac{d(T_L+T_R)}{dt}=0$$The solution to this equation, subject to the prescribed initial conditions is $$\frac{T_L+T_R}{2}=\bar{T}=\frac{3}{2}T_B\tag{6}$$where $\bar{T}$ is the (constant) average of the two compartment temperatures throughout the deformation.
Next, if we subtract Eqn. 2 from Eqn. 1, we obtain:$$nC_v\frac{d(T_L-T_R)}{dt}=2\hat{P}\frac{dV}{dt}$$or, equivalently, $$nC_v\frac{d(T_L-T_R)}{dV_L}=-2\hat{P}\tag{7}$$At this juncture, we define the following substitutions: $$V_L=V(1+\xi_V)\tag{8a}$$
$$V_R=V(1-\xi_V)\tag{8b}$$
$$T_L=\bar{T}(1+\xi_T)\tag{8c}$$
$$T_R=\bar{T}(1-\xi_T)\tag{8d}$$
$$\mu_L=\mu(\bar{T})\sqrt{1+\xi_T}\tag{8e}$$
$$\mu_R=\mu(\bar{T})\sqrt{1-\xi_T}\tag{8e}$$where, in writing Eqns. 8e and 8f, we have made use of the fact that, in the Mungan model, $\mu$ is proportional to $\sqrt{T}$. If we substitute Eqns. 8 into Eqns. 4, 5, and 6, we obtain:$$\frac{d\xi_V}{d\tau}=\frac{2(\xi_T-\xi_V)}{\sqrt{1+\xi_T}(1-\xi_V)+\sqrt{1-\xi_T}(1+\xi_V)}\tag{9}$$
$$\hat{P}=\frac{nR\bar{T}}{V}\left[\frac{\sqrt{1+\xi_T}(1-\xi_T)+\sqrt{1-\xi_T}(1+\xi_T)}{\sqrt{1+\xi_T}(1-\xi_V)+\sqrt{1-\xi_T}(1+\xi_V)}\right]\tag{10}$$
$$\frac{d\xi_T}{d\xi_V}=-(\gamma - 1)\left[\frac{\sqrt{1+\xi_T}(1-\xi_T)+\sqrt{1-\xi_T}(1+\xi_T)}{\sqrt{1+\xi_T}(1-\xi_V)+\sqrt{1-\xi_T}(1+\xi_V)}\right]\tag{11}$$
where the dimensionless time $\tau$ is given by $\tau=\frac{nR\bar{T}}{V\mu(\bar{T})}t$. For the present problem, initial conditions of these ordinary differential equations are: $$\xi_V(0)=0$$
and $$\xi_T(0)=\frac{1}{3}$$
One will note from Eqn. 11 that, even though this is a time-dependent problem, the trajectory of the dimensionless temperature variation $\xi_T$ as a function of the dimensionless volume variation $\xi_V$ is unique, and independent of the time history. Therefore, Eqn. 11 can be solved once-and-for-all for $\xi_T$ (and the dynamic pressure $\hat{P}$) vs $\xi_V$. In the present analysis this has been done by integrating Eqn. 11 numerically from $\xi_V=0$ to the final equilibrium point where, according to Eqn. 9, $\xi_T=\xi_V$. The results of these numerical calculations using the recommended Mungan model will be presented in the next post, together with corresponding predictions using Chet Miller's approximation and using Andrew Mason's approximation (to determine which approximation provides a better match to the Mungan model).

 

[Chestermiller]

Andrew Mason (AM) APPROXIMATION
The AM approximation is based on the assumption that the gas in the right compartment suffers an ideal adiabatic reversible compression during this overall irreversible process, such that essentially all the entropy generation is confined to the left compartment (in which the gas experiences expansion). This assumption seems to be incompatible with the Mungan Model of non-quasistatitic gas deformation that AM has recommended using for more accurate calculations.

Under the AM Approximation, one has that $$T_RV_R^{(\gamma - 1)}=T_BV^{(\gamma-1)}$$and$$\hat{P}V_R^\gamma=P_BV^\gamma$$or, in terms of the dimensionless parameters defined in the present development, $$\xi_T=1-\frac{2}{3(1-\xi_V)^{(\gamma-1)}}$$and $$\frac{\hat{P}}{P_B}=\frac{1}{(1-\xi_V)^\gamma}$$The predictions from these equations will be compared with the results of the Mungan Model calculation.

Chet Miller (CM) APPROXIMATION
The CM approximation assumes that the dynamic pressure throughout the deformation is equal to the final pressure at equilibrium of the system (and also equal to the volume weighted average of the initial pressures in the two compartments). One can also show that this assumption is equivalent to neglecting the temperature-dependence of the viscosity parameter $\mu$ in the Mungan Model.

Under the CM Approximation, for the parameter values in the present problem, one has that $$\frac{\hat{P}}{P_B}=1.5$$and, from Eqns. 10 and 11 of the present development, $$\xi_T=\frac{1}{3}-(\gamma-1)\xi_V$$As with the AM model, the predictions from these equations will be compared with the results of the Mungan Model calculation.

RESULTS OF CALCULATIONS

The figure below shows the results of the Mungan Model calculations for the dimensionless temperature as a function of the dimensionless volume. Also shown in the figure, for comparison, are the predictions using the AM and CM approximations.

The results from the CM approximation are barely distinguishable for the Mungan Model results. On the other hand, there is a significant difference between the AM approximation and the Mungan Model. This suggests that the AM approximation is a poor representation of physical reality.

The figure below shows the results of the Mungan Model calculations for the dimensionless "dynamic pressure" as a function of the dimensionless volume. Also shown in the figure, for comparison, are the predictions using the AM and CM approximations.

Here again, the CM approximation provides a fairly good match to the AM-recommended Mungan Model results. On the other hand, the AM approximation deviates significantly from the Mungan Model.

The obvious conclusion here is that it is a poor approximation to assume that the gas in the low-pressure (right) compartment suffers a nearly reversible adiabatic compression in this system and that there is negligible entropy generation in the right compartment. According to the Mungan Model and the CM approximation, there is substantial irreversibility and entropy generation in both compartments.