Adiabatic Expansion: Solve for Final Volume of Gas

[Andrew Mason]

What happens when the barrier is released, or suddenly vaporizes is that the particles already moving toward that barrier no longer encounter reflection and continue moving into the vacuum until they reach the new wall. There is no sudden movement of every gas particle toward the vacuum. There is a gradual (on some time scale) migration of particles into the vacuum analogous to the evaporation of a liquid into the air, except there is no energy threshold to overcome.

It certainly wouldn't be gradual. It would be extremely fast.

Consider xanthym's box but with two plates with the gas confined between them. The plates are lifted at the same time and the gas expands into a vacuum on both sides. The ends of the container, some distance from the centre are connected to ratcheted springs with large spring constants. As the gas hits the ends, the springs are compressed by the translational kinetic energy of the gas which store all that translational energy by compressing only a very small distance. What is the temperature of the gas? I would say it has to be:

$$T_f = T_i(\frac{V_i}{V_f})^\gamma$$

AM

 

[OlderDan]

It certainly wouldn't be gradual. It would be extremely fast.

Consider xanthym's box but with two plates with the gas confined between them. The plates are lifted at the same time and the gas expands into a vacuum on both sides. The ends of the container, some distance from the centre are connected to ratcheted springs with large spring constants. As the gas hits the ends, the springs are compressed by the translational kinetic energy of the gas which store all that translational energy by compressing only a very small distance. What is the temperature of the gas? I would say it has to be:

$$T_f = T_i(\frac{V_i}{V_f})^\gamma$$

AM

I agree with you if the ends of the box are free to move. The gas particles would then transfer energy to the walls. Work would be done and the temperature would drop. I just don't think the idea of an oscillating center of mass works. There would certainly be a wave front as the faster particles moved ahead into empty space, and they are certainly going to hammer the ends when they get there. But a wave front does not mean there is going to be a net movement of the center of mass past the middle of the box.

I'm not arguing against you conclusion. In fact I have some some serious heartburn with the concept of a massless barrier confining the gas in the first place. Such an idea is preposterous on its face. The notion is a violation of conservation of momentum. What it is saying is that the particles in the middle of the box are magically reversing direction without interacting with anything to absorb their lost momentum. Can't happen; never will happen; and such a concept probably has no place in defending any argument about how even an ideal gas will actually behave. The only thing that can confine a gas is a real barrier, and expansion against that barrier is going to move it and do work.

 

[Clausius2]

I think I know where are you getting at. But to say the truth we are lost in semantic issues. I have discovered you're very difficult to get convinced.

That is the change in entropy of the system only, not system + surroundings. As long as that is clear, I agree.

As $$dS=\delta Q_R/T$$ we understand the entropy of a system. Of course there is a change of surroundings entropy, so that the change of universe (surroundings+system) entropy is $$dS_{universe}=dS_{system}+dS_{surroundings}$$. We will assume $$dS_{surroundings}=0$$ hereinafter, therefore the process is so-called adiabatic. This is the definition of an adiabatic process. You have mentioned processes of internal chemical reactions (combustion), or electrical heat dissipation in your example. Both of them had an insulated (adiabatic) container. After thinking of it for a while:

I) we agree that the example of electrical-heat dissipation is not an adiabatic process. It is because $$dS_{surroundings}<0$$ from the electrical generation system.

II) the example of chemical reaction and internal viscous dissipation is not as clear as the last one. Although $$dS_{surroundings}=0$$ the release of chemical reaction heat such as in a combustion process, or the heat released by internal fluid friction, both of these effects have the same behavior of an inflow of external heat. Semantically, we would say they are systems with adiabatic walls, but thermodynamically their behaviours are not similar to an adiabatic process, so that we can conclude that they are not adiabatic processes. But they have an essential difference with an usual non-adiabatic process: there is no heat inflow and so none impact on the surroundings. The irreversibility doesn't come from external heat inflow issues $$dS_{surroundings}$$, but it comes from internal entropy generation. The combustion process is an irreversible one which is enclosed to an increasing of entropy due to chemical reaction, while internal viscous dissipation is a source of entropy due to the heat released.

As a first conclusion, I think you're right when you said internal viscous dissipation is not an adiabatic process. Maybe it is the key of this whole discussion. Let's try to summarize it:

i) the irreversible adiabatic process can be described as an adiabatic one because the walls are adiabatic, there is no entropy exchange with the surroundings.

ii) such process can be viewed as a non-adiabatic one, because it is a heat internal release and such heat release behavior is mathematically similar to a heat external inflow. The main difference between both behaviors is that the first one doesn't make any effect on the surroundings, all the entropy generated is done inside the system; while the second one has an effect on the surroundings, the whole irreversibility is generated by means of the sum of the surroundings effect and system effect.

Maybe the two points of view are right. The fact is that I have always believed that the quantity $$\delta Q$$ in entropy was referred to a heat exchanged through the boundaries, and that it is the true and original sense of the entropy definition. My opinion is that such effects of internal viscous dissipation are so difficult to treat them with "thermodynamic usual equations" that physicists and engineers model them as some kind of additional entropy generation ad hoc, as I made in my last post. After thinking of this, I realize your point of view is quite right, but I think my interpretation of the event is right too. I mean, it could be said the irreversibility comes from a non effective adiabatic behavior as you said, but it could be said to that the irreversibility comes from an internal entropy generation due to a heat release although the whole system can be considered adiabatic from the usual point of view of $$dS_{surroundings}=0$$.

I don't see why not. Can you show me some actual data on this? It seems to me that since the decrease in internal energy of the gas is converted entirely to external work, so long as that work is not being sent back to the gas in the form of heat, the rapidity of the expansion does not affect entropy. The only way the gas can fall off the isentropic curve is if some of the external work done by the gas is converted to heat and let back into the system.

Sorry, I don't have any piston filled with gas at hand to check it .
I have said to you that the irreversibility (and so the not coincidence with an isentropic trajectory) comes from the proper fluid field. If the process is too fast, sure there will be internal dissipation and viscous effects. Such entropy increasing (which we agree it exists) has the effect of taking away the final equilibrium point from the same isentropic curve. That's because $$dS_{surroundings}=0$$ and then $$dS_{system}>0$$. There must be a loose of Availability somewhere, and in order to reinstaurate the original state there must be employed more work than the exerted in the expansion.

As a second conclusion, any rapid (non-cuasiestatic) process is irreversible due to the proper flow field produced. Such flow field might enhance internal viscous dissipation, which could be modeled as an external heat inflow but taking into account that $$dS_{surroundings}=0$$ (which right now seems to me puzzling and without any sense), or it can be modeled too as a volumetric heat release which provokes an additional source of internal entropy generation (which I called $$d\sigma$$).


What do you think? Have we come to some agreement?

 

[Andrew Mason]

I think I know where are you getting at. But to say the truth we are lost in semantic issues.

Perhaps. We seem to disagree over the meaning of adiabatic. But it is an interesting point.

I have discovered you're very difficult to get convinced.

Yeah. part of my legal training.

As $$dS=\delta Q_R/T$$ we understand the entropy of a system. Of course there is a change of surroundings entropy, so that the change of universe (surroundings+system) entropy is $$dS_{universe}=dS_{system}+dS_{surroundings}$$. We will assume $$dS_{surroundings}=0$$ hereinafter, therefore the process is so-called adiabatic. This is the definition of an adiabatic process. You have mentioned processes of internal chemical reactions (combustion), or electrical heat dissipation in your example. Both of them had an insulated (adiabatic) container. After thinking of it for a while:

I) we agree that the example of electrical-heat dissipation is not an adiabatic process. It is because $$dS_{surroundings}<0$$ from the electrical generation system.

In my example, the electrical generation system and heating coil is inside the insulated container. Why would the entropy of the surroundings increase?

As a first conclusion, I think you're right when you said internal viscous dissipation is not an adiabatic process. Maybe it is the key of this whole discussion. Let's try to summarize it:

i) the irreversible adiabatic process can be described as an adiabatic one because the walls are adiabatic, there is no entropy exchange with the surroundings.

If that is your definition of adiabatic, then ok. But this is not how adiabatic is normally defined. As I would understand and use the term, "Adiabatic" refers to an absence of heat inflow/outflow into/out of the system. ie. a process that occurs without heat entering or leaving the system.

Heat can be added to the system in at least two ways:

1. heat being transferred from surroundings to system;
2. Energy in a form other than heat being converted to heat inside the system (whether or not the energy is transferred from the surroundings).

In the first case (1.) there is an exchange of heat with the surroundings causing a decrease in entropy of the surroundings and an increase of the entropy of the system. (1.) is NOT adiabatic. We both agree.

In the second case (2.), there is NO EXCHANGE OF HEAT with the surroundings. There should be no decrease in the entropy of the surroundings while there is an increase in the entropy of the system. (2.) is adiabatic according to your definition $(dS_{surroundings} = 0)$ but is NOT adiabatic under my definition as there is a flow of heat into the system.

Maybe the two points of view are right. The fact is that I have always believed that the quantity $$\delta Q$$ in entropy was referred to a heat exchanged through the boundaries, and that it is the true and original sense of the entropy definition.

I think this may be the essential difference between us.

I don't think the dQ (in dS=dQ/T) refers to heat transferred through a boundary. dQ is the heat added to the system. It need not be transferred as heat from the surroundings. The heat can be added to the system from an internal supply of energy (eg. chemical, gravitational, nuclear) converted to heat and entropy will increase with no decrease in entropy of the surroundings. And if that occurs,the process cannot be adiabatic.

My opinion is that such effects of internal viscous dissipation are so difficult to treat them with "thermodynamic usual equations" that physicists and engineers model them as some kind of additional entropy generation ad hoc, as I made in my last post. After thinking of this, I realize your point of view is quite right, but I think my interpretation of the event is right too. I mean, it could be said the irreversibility comes from a non effective adiabatic behavior as you said, but it could be said to that the irreversibility comes from an internal entropy generation due to a heat release although the whole system can be considered adiabatic from the usual point of view of $$dS_{surroundings}=0$$.

I think we change the meaning of adiabatic if we say that an adiabatic process can result in a change in entropy. If there is a change in entropy, there must be a flow of heat into/out of the system. And that means, by definition, that $ds \ne 0$

What do you think? Have we come to some agreement?

Almost. The difference between us appears to be whether a process that results in a change of the entropy of a system without any change in the entropy of the surroundings can be adiabatic.

AM

 

[Q_Goest]

Adiabatic is a process without heat transfer, which is not the same as isentropic (no change in entropy).

For a process which is both adiabatic and reversible, such as the compression or expansion of a gas in a cylinder, the process is also isentropic.

For a process which is adiabatic and irreversible, such as an expansion of gas in a pipeline, the process is not isentropic. Note that as the gas flows through a pipeline and expands, there is no conservation of mechanical energy. In this case, the process would be isenthalpic.

 

[Clausius2]

We are nearer each other right now.

In my example, the electrical generation system and heating coil is inside the insulated container. Why would the entropy of the surroundings increase?
[\quote]

Ah! is the generator inside the container also?. Well, I forgot it. I thought it was outside.

If that is your definition of adiabatic, then ok. But this is not how adiabatic is normally defined. As I would understand and use the term, "Adiabatic" refers to an absence of heat inflow/outflow into/out of the system. ie. a process that occurs without heat entering or leaving the system.

That's the same definition as mine. If there is no heat inflow/outflow then $$dS_{surroundings}=0$$. I think it has been the traditional definition of an adiabatic process.

Heat can be added to the system in at least two ways:

1. heat being transferred from surroundings to system;
2. Energy in a form other than heat being converted to heat inside the system (whether or not the energy is transferred from the surroundings).

In the first case (1.) there is an exchange of heat with the surroundings causing a decrease in entropy of the surroundings and an increase of the entropy of the system. (1.) is NOT adiabatic. We both agree.

In the second case (2.), there is NO EXCHANGE OF HEAT with the surroundings. There should be no decrease in the entropy of the surroundings while there is an increase in the entropy of the system. (2.) is adiabatic according to your definition $(dS_{surroundings} = 0)$ but is NOT adiabatic under my definition as there is a flow of heat into the system.

I think this may be the essential difference between us.

I agree.

I don't think the dQ (in dS=dQ/T) refers to heat transferred through a boundary. dQ is the heat added to the system. It need not be transferred as heat from the surroundings. The heat can be added to the system from an internal supply of energy (eg. chemical, gravitational, nuclear) converted to heat and entropy will increase with no decrease in entropy of the surroundings. And if that occurs,the process cannot be adiabatic.

I think we change the meaning of adiabatic if we say that an adiabatic process can result in a change in entropy. If there is a change in entropy, there must be a flow of heat into/out of the system. And that means, by definition, that $ds \ne 0$

As I was hoping, you're stating the definition of an isentropic process right now. But for you isentropic=adiabatic.

Keep in touch with this thread, I will post later an example of such a problem, and we will play with the equations to extract the final conclusions, ok?

 

[Clausius2]

Adiabatic is a process without heat transfer, which is not the same as isentropic (no change in entropy).

For a process which is both adiabatic and reversible, such as the compression or expansion of a gas in a cylinder, the process is also isentropic.

For a process which is adiabatic and irreversible, such as an expansion of gas in a pipeline, the process is not isentropic. Note that as the gas flows through a pipeline and expands, there is no conservation of mechanical energy. In this case, the process would be isenthalpic.

That's our discussion subject. I am agree with you, but Andrew argues that in the sudden expansion of a gas in pipeline the process is not adiabatic. Why? because the irreversibility comes from internal viscous dissipation originated inside the gas. Such dissipation is heat transformed further in internal energy. So that it seems it has been an inflow of "some kind" of heat inside the system. Do you understand his point?.

 

[Andrew Mason]

As I was hoping, you're stating the definition of an isentropic process right now. But for you isentropic=adiabatic.

Exactly:
http://scienceworld.wolfram.com/physics/Adiabatic.html

Keep in touch with this thread, I will post later an example of such a problem, and we will play with the equations to extract the final conclusions, ok?

Ok.

AM

 

[Q_Goest]

I guess we're all in agreement pretty much, but thought I'd toss out a few thoughts here to make sure...

the sudden expansion of a gas in pipeline the process is not adiabatic.

A real gas which expands as the pressure drops traveling down a pipeline can be modeled as adiabatic if
1) the pipe is well insulated
2) the temperature of the gas does not increase or decrease due to this expansion but stays constant and is equal to the temperature of the environment.

Note that when a gas expands isenthalpically, it can either increase in temperature (ex: Helium), decrease in temperature (ex: Nitrogen), or stay the same (ex: any gas at it's inversion temperature such as hydrogen at -183 C). It all depends on how the molecules react to the expansion.

Adiabatic only refers to heat transfer from the environment, not any kind of viscous dissipating forces induced by flow which can affect the temperature. Hope that helps.

 

[Andrew Mason]

I guess we're all in agreement pretty much, but thought I'd toss out a few thoughts here to make sure...


A real gas which expands as the pressure drops traveling down a pipeline can be modeled as adiabatic if
1) the pipe is well insulated
2) the temperature of the gas does not increase or decrease due to this expansion but stays constant and is equal to the temperature of the environment.

Note that when a gas expands isenthalpically, it can either increase in temperature (ex: Helium), decrease in temperature (ex: Nitrogen), or stay the same (ex: any gas at it's inversion temperature such as hydrogen at -183 C). It all depends on how the molecules react to the expansion.

Adiabatic only refers to heat transfer from the environment, not any kind of viscous dissipating forces induced by flow which can affect the temperature. Hope that helps.

Here is the conundrum:

1. What would you call an expansion of gas that:

a) does work without any heat exchange with its surroundings,
b) in which the work is in the form of electrical energy,
c) in which that electrical energy is passed through an electrical resistance inside the gas container and used to heat the gas?​

2. If you say that 1. is adiabatic, how would you characterize and expansion of the same gas, doing the same work where: that electrical energy is passed through an electrical resistance outside the gas container and converted to heat and the heat is then allowed to flow back to raise the temperature of the gas?

The second is clearly non-adiabatic. I am having trouble differentiating between the two.

AM

 

[Q_Goest]

AM,
Put a control volume around the gas, and around the gas only. This could also be called a "control mass" since there is no mass entering or exiting the volume.

For case 1, if there is no heat flux across the control surface, the process is adiabatic. If the gas is doing work, and assuming the work it does is completely reversible (which will not be perfectly true in a real world situation, but a close approximation) then the process is also isentropic.

For case 2, I'm not sure what you're getting at regarding the use of that work to create electric energy. The gas expands to do work, and that's a mechanical process, so if it expands through a turbine, or pushes on a piston, it must do that first, and it is that motion which is used to create electrical energy. Regardless of how the electrical energy is created though, the process is still adiabatic and isentropic.

For case 3, there is heat flux passing through the control surface and entering the control volume. For the case you created, the work done is equal to the heat flux. So energy in the form of work leaving the control volume is equal to the energy in the form of heat entering the control volume. If this is a closed volume, you could use the first law to describe this process and what you'd find is that, assuming there is no mass flow into or out of the control volume, the change in internal energy of the control volume will be zero. As you point out, this is not an adiabatic process since there is heat flux across the control surface.