# Reversible Vs Irreversible Gas Compression/Expansion Work - Comments

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DrDu
I also find it helpful to consider the example of the compression of a gas when the gas is viscuous and remains homogeneous. This is approximately the case in sound waves and is the main mechanism for the damping of sound waves in gases.

There, one distiguishes between the mechanical pressure ##p_m## and the usual thermodynamical pressure ##p_t##. The two are related as ##p_m=p_t +\kappa \frac{1}{\rho} \frac{d\rho}{dt}##, where ##\kappa## is the coefficient of bulk viscosity and ##\rho## is the density. While the mechanical pressure is constant and equal to the external pressure, the thermodynamic pressure is smaller. It is clear that the friction represents a source of entropy. It can also be seen that the extra work done ## -\kappa \frac{1}{\rho}\frac{d\rho}{dt}dV ## vanishes when the process is done quasi-statically.

Drakkith
Staff Emeritus
Good job, Chet. I wish I was far enough along in my schooling to understand it though.

Mentor
I also find it helpful to consider the example of the compression of a gas when the gas is viscuous and remains homogeneous. This is approximately the case in sound waves and is the main mechanism for the damping of sound waves in gases.

There, one distiguishes between the mechanical pressure ##p_m## and the usual thermodynamical pressure ##p_t##. The two are related as ##p_m=p_t +\kappa \frac{1}{\rho} \frac{d\rho}{dt}##, where ##\kappa## is the coefficient of bulk viscosity and ##\rho## is the density. While the mechanical pressure is constant and equal to the external pressure, the thermodynamic pressure is smaller. It is clear that the friction represents a source of entropy. It can also be seen that the extra work done ## -\kappa \frac{1}{\rho}\frac{d\rho}{dt}dV ## vanishes when the process is done quasi-statically.
Hi DrDu,

Great minds must think alike. I had originally planned to do exactly what you describe, and had even worked out the solution for the volume as a function of time, but, in the end, I decided to leave the analysis out of the article. My starting equation was essentially the same as yours:
$$\frac{nRT}{V_f}=\frac{nRT}{V}-{\kappa}\frac{1}{V}\frac{dV}{dt}$$
One reason for omitting the analysis was that I determined that, for typical values of the parameters (including the viscosity) in the irreversible case, the velocity of the piston would exceed the speed of sound in the gas at time zero. This is because the distributed mass of the gas is omitted from the model, and also because turbulence, which would increase the effective value of gas viscosity, is also omitted. I just didn't want to have to deal these points in detail in the article. However, as you say, if I had continued the analysis to the end anyway, I would have obtained the same final result for the viscous work as in the article.

Thanks for bringing this up. It's a great point.

Chet

DrDu
Thank's Chet.
What is also a pedagogical point is the fact that many students conclude from the equilibrium states of a gas being describable in terms of, say, p and T, that there can only be p-V work being done on a gas.
As your analysis already showed, there are many more possibilities. A specifically simple example would be stirring.
If one really wants to consider only p-V work, I think the model with bulk viscosity is one of the simplest. It has the additional bonus that the viscosity can be understood relatively easily, especially if you are a chemical enigneer to be.
The main effect behind bulk viscosity is the non-equilibrium distribution of the distances of the particles of the gas. This can be shown for e.g. a van der Waals gas, but, even more intuitively for gases whose non-ideal behaviour is due to chemical reactions.
E.g. Nitrogen dioxide is partially dimerized at room temperature and the degree of dimerization depends on pressure and temperature : 2 NO2 <--> N2O4
If the system is compressed rapidly, than there will be more NO2 than what would correspond to equilibrium and the pressure would be higher than the equilibrium pressure. At expansion, there would be more N2O4 and the pressure correspondingly lower. Given we reaction constants of this reaction, we could even calculate the coefficient of viscosity from this.

Mentor
Thank's Chet.
What is also a pedagogical point is the fact that many students conclude from the equilibrium states of a gas being describable in terms of, say, p and T, that there can only be p-V work being done on a gas.
As your analysis already showed, there are many more possibilities. A specifically simple example would be stirring.
If one really wants to consider only p-V work, I think the model with bulk viscosity is one of the simplest. It has the additional bonus that the viscosity can be understood relatively easily, especially if you are a chemical enigneer to be.
In the equation I wrote, the viscosity parameter κ also includes the extensional viscosity of the gas which, for the kinematics of this deformation, is equal to twice the shear viscosity.
The main effect behind bulk viscosity is the non-equilibrium distribution of the distances of the particles of the gas. This can be shown for e.g. a van der Waals gas, but, even more intuitively for gases whose non-ideal behaviour is due to chemical reactions.
E.g. Nitrogen dioxide is partially dimerized at room temperature and the degree of dimerization depends on pressure and temperature : 2 NO2 <--> N2O4
If the system is compressed rapidly, than there will be more NO2 than what would correspond to equilibrium and the pressure would be higher than the equilibrium pressure. At expansion, there would be more N2O4 and the pressure correspondingly lower. Given we reaction constants of this reaction, we could even calculate the coefficient of viscosity from this.
I don't quite follow what you are saying here. It appears you are talking about an adiabatic expansion/compression, but are you referring to the situation when the final equilibrium state is reached or to the situation immediately after the volume change is complete, but before the system has reached final thermodynamic equilibrium?

Chet

Thank's Chet.
What is also a pedagogical point is the fact that many students conclude from the equilibrium states of a gas being describable in terms of, say, p and T, that there can only be p-V work being done on a gas.
As your analysis already showed, there are many more possibilities. A specifically simple example would be stirring.
If one really wants to consider only p-V work, I think the model with bulk viscosity is one of the simplest. It has the additional bonus that the viscosity can be understood relatively easily, especially if you are a chemical enigneer to be.
In the equation I wrote, the viscosity parameter κ also includes the extensional viscosity of the gas which, for the kinematics of this deformation, is equal to twice the shear viscosity.
The main effect behind bulk viscosity is the non-equilibrium distribution of the distances of the particles of the gas. This can be shown for e.g. a van der Waals gas, but, even more intuitively for gases whose non-ideal behaviour is due to chemical reactions.
E.g. Nitrogen dioxide is partially dimerized at room temperature and the degree of dimerization depends on pressure and temperature : 2 NO2 <--> N2O4
If the system is compressed rapidly, than there will be more NO2 than what would correspond to equilibrium and the pressure would be higher than the equilibrium pressure. At expansion, there would be more N2O4 and the pressure correspondingly lower. Given we reaction constants of this reaction, we could even calculate the coefficient of viscosity from this.
I don't quite follow what you are saying here. It appears you are talking about an adiabatic expansion/compression, but are you referring to the situation when the final equilibrium state is reached or to the situation immediately after the volume change is complete, but before the system has reached final thermodynamic equilibrium?

Chet
The concept of reversibility in thermodynamics is very important. The difficulty students face in understanding the concept is also genuine, especially in view of the fact that there exist many a way in which the concept is sought to be explained to them in simple and more simple ways. Unfortunately, not all of them are equivalent or reinforce one another.

The easiest way the correct concept of reversible process can be obtained is from the definition given by Max Planck, in his Treatise on Thermodynamics, which says, 'A process is said to be irreversible, if it is impossible with the assistance of all agents in nature, to bring the system back to its original state without leaving any changes elsewhere in the universe, once the process has occurred.' A process which is not irreversible is reversible.

In view of this definition, it would be always useful to conceive of a cyclic process that consists of the process of interest as a part of the cycle and then see if there are any changes left in the surroundings at the end of the cycle, that it is impossible to bring the surroundings back to its original state.

With this brief background, let us see the difference in the work output (or work input) of a reversible and irreversible process connecting two equilibrium states i and f of a closed system of a given mass of ideal gas.

During the course of a reversible process, the system is in a state of equilibrium (within as well as with the surroundings) at any point of the process along the path i-f. That is, properties of the system have unique values for any state j on the path i-f. In contrast to this, for an irreversible process connecting the states i and f, the system will not be at equilibrium and will not have unique values for its properties at any intermediate point along the irreversible path i-f.

Since the system is in equilibrium with surroundings at any point j along the reversible path i-f, the values of the intensive properties of the system, such as temperature, pressure are equal to the values of the corresponding external variables. Therefore, for a reversible process of expansion (or compression) of an ideal gas, Pext = Pgas, Work output =integral(Pext x dV). For an irreversible process, Pext is not equal to Pgas.

Whether the process is reversible or irreversible, the equation to calculate work out put is one and the same: Work output =integral(Pext x dV). Work output and work input have opposite signs since dV has opposite signs in the two cases.

For an irreversible process we cannot do that substitution for Pext, and hence, cannot obtain an equation but only an inequation. For expansion process we get lower work output and for a compression process we get more work input compared to that of a reversible process connecting the same end points. We cannot go into an elaboration of this point here in the discussion.

Bringing 'time' to play For reversible processes, we can substitute the known value of Pgas for Pext and get: Wr = nRTln(Vf/Vi).
a role in explanation of thermodynamic process deprives thermodynamics of its vital strengths. Time has no role to play in thermodynamics (Classical or Equilibrium).

While mechanical processes are covered by laws of thermodynamics, thermodynamics is more general and covers the processes that involve heat which is not a part of mechanics. All mechanical processes, in principle, are reversible. Therefore, there is no scope to introduce irreversible processes through mechanical analogs.

Drakkith
Mentor
The concept of reversibility in thermodynamics is very important. The difficulty students face in understanding the concept is also genuine, especially in view of the fact that there exist many a way in which the concept is sought to be explained to them in simple and more simple ways. Unfortunately, not all of them are equivalent or reinforce one another.

The easiest way the correct concept of reversible process can be obtained is from the definition given by Max Planck, in his Treatise on Thermodynamics, which says, 'A process is said to be irreversible, if it is impossible with the assistance of all agents in nature, to bring the system back to its original state without leaving any changes elsewhere in the universe, once the process has occurred.' A process which is not irreversible is reversible.

In view of this definition, it would be always useful to conceive of a cyclic process that consists of the process of interest as a part of the cycle and then see if there are any changes left in the surroundings at the end of the cycle, that it is impossible to bring the surroundings back to its original state.

With this brief background, let us see the difference in the work output (or work input) of a reversible and irreversible process connecting two equilibrium states i and f of a closed system of a given mass of ideal gas.

During the course of a reversible process, the system is in a state of equilibrium (within as well as with the surroundings) at any point of the process along the path i-f. That is, properties of the system have unique values for any state j on the path i-f. In contrast to this, for an irreversible process connecting the states i and f, the system will not be at equilibrium and will not have unique values for its properties at any intermediate point along the irreversible path i-f.

Since the system is in equilibrium with surroundings at any point j along the reversible path i-f, the values of the intensive properties of the system, such as temperature, pressure are equal to the values of the corresponding external variables. Therefore, for a reversible process of expansion (or compression) of an ideal gas, Pext = Pgas, Work output =integral(Pext x dV). For an irreversible process, Pext is not equal to Pgas.

Whether the process is reversible or irreversible, the equation to calculate work out put is one and the same: Work output =integral(Pext x dV). Work output and work input have opposite signs since dV has opposite signs in the two cases.

For an irreversible process we cannot do that substitution for Pext, and hence, cannot obtain an equation but only an inequation. For expansion process we get lower work output and for a compression process we get more work input compared to that of a reversible process connecting the same end points. We cannot go into an elaboration of this point here in the discussion.
I completely agree with everything you have said up to here. In fact, the very points you have emphasized, for the most part, are covered in my earlier Insights article at the following link: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/. I would be pleased to receive feedback from you on that article as well.
Bringing 'time' to play For reversible processes, we can substitute the known value of Pgas for Pext and get: Wr = nRTln(Vf/Vi).
a role in explanation of thermodynamic process deprives thermodynamics of its vital strengths. Time has no role to play in thermodynamics (Classical or Equilibrium).

While mechanical processes are covered by laws of thermodynamics, thermodynamics is more general and covers the processes that involve heat which is not a part of mechanics. All mechanical processes, in principle, are reversible. Therefore, there is no scope to introduce irreversible processes through mechanical analogs.
These last two paragraphs are where we differ in our perspectives. Regarding the first paragraph, certainly time is not very important in discussing reversible processes. However, for irreversible processes, the speed at which a deformation occurs determines the rate at which mechanical energy is dissipated into thermal energy, and certainly, time is relevant to that. In any event, unlike you, I see no compelling reason why time should be excluded from the discussion of irreversible processes.

With respect to the second paragraph, I disagree with the contention that mechanical processes, in principle, are reversible. Any mechanical process that involves frictional dissipation or viscous dissipation of mechanical energy to thermal energy cannot be regarded as inherently reversible. Certainly processes like these cannot be considered reversible within the context of the definition you attributed to Max Planck (and which I fully agree with). So, whereas you feel that there is no value in introducing irreversible processes through mechanical analogs (at least mechanically irreversible processes), my judgement is that there is great pedagogical value in doing this. I guess we are just going to have to disagree on this.

Chet

DrDu
Whether the process is reversible or irreversible, the equation to calculate work out put is one and the same: Work output =integral(Pext x dV). Work output and work input have opposite signs since dV has opposite signs in the two cases.

For an irreversible process we cannot do that substitution for Pext, and hence, cannot obtain an equation but only an inequation. For expansion process we get lower work output and for a compression process we get more work input compared to that of a reversible process connecting the same end points. We cannot go into an elaboration of this point here in the discussion.
Of course we can write down an equation for Pext also for irreversible processes using irreversible thermodynamics.
Even though irreversible thermodynamics is not in the scope of introductory thermodynamics classes, it seems necessary to give students an idea in what respect non-equilibrium systems differ from equilibrium ones.
In my experience, students have great difficulties to imagine how a difference between equilibrium pressure and pressure away from equilibrium can arise.
My point was to find an easy example where Pext is still spatially constant but not equal to the equilibrium pressure.

Nidum
Gold Member
Time has no role to play in thermodynamics (Classical or Equilibrium).
Steam takes a finite time to transition from one condition of dryness/wetness to another during an expansion .

Otherwise identical expansions should give slightly different work outputs and have slightly different end conditions depending on rate of expansion ?

Mentor
Of course we can write down an equation for Pext also for irreversible processes using irreversible thermodynamics.
Even though irreversible thermodynamics is not in the scope of introductory thermodynamics classes, it seems necessary to give students an idea in what respect non-equilibrium systems differ from equilibrium ones.
In my experience, students have great difficulties to imagine how a difference between equilibrium pressure and pressure away from equilibrium can arise.
My point was to find an easy example where Pext is still spatially constant but not equal to the equilibrium pressure.
Hi Dr. Du. Thank you for your perceptive comment.

I'd like to elaborate a little on your response. It is not usually necessary to use irreversible thermodynamics to solve problems with irreversible gas dynamics, and to take into account the fact that the compressive stress which each parcel of gas exerts on its neighbors (as well as on solid boundaries) is not equal to the thermodynamic pressure (determined by the local temperature and density). Computational Fluid Dynamics (CFD) can adequately handle quantification of these effects, even if the deformation is turbulent. In both our comments within this thread, we correctly noted that, during an irreversible deformation, the difference between the compressive stresses within a gas and the thermodynamic pressure is attributable to the viscous stresses, which depend on the rate of deformation. I strongly agree with you that "it seems necessary to give students an idea in what respect non-equilibrium systems differ from equilibrium ones," and this answers that question (at least for mechanical irreversibilities).

In my judgement, the assumption that the compressive stress within the deforming gas is nearly uniform and equal to the externally applied force per unit area at the boundary, Pext, is usually a good one. This is because compression waves travel through a gas at the speed of sound, and the time scale for the expansion or compression is typically much larger than the time scale for a wave to travel across the system. However, of course, there can be situations in practice where this is not the case, such as in internal combustion engines.

Irrespective of whether the compressive stress within the system is uniform or not, the parameter Pext is usually used to refer to the external force per unit area applied at the boundary (interface) between the system and the surroundings. In his comment, Radhakrishnam seems to imply with regard to Pext that, if the process is irreversible, then all bets are off and Pext can not be established. However, this is not the case. We are perfectly free to impose any external force loading history that we choose at the boundary of our system. In many cases, once this loading history is established, the system can be solved deterministically for the final equilibrium state. This is certainly the case for a constant external force per unit area imposed in an irreversible expansion or compression within a cylinder.

Chet

The concept of reversibility in thermodynamics is very important. The difficulty students face in understanding the concept is also genuine, especially in view of the fact that there exist many a way in which the concept is sought to be explained to them in simple and more simple ways. Unfortunately, not all of them are equivalent or reinforce one another.

The easiest way the correct concept of reversible process can be obtained is from the definition given by Max Planck, in his Treatise on Thermodynamics, which says, 'A process is said to be irreversible, if it is impossible with the assistance of all agents in nature, to bring the system back to its original state without leaving any changes elsewhere in the universe, once the process has occurred.' A process which is not irreversible is reversible.

In view of this definition, it would be always useful to conceive of a cyclic process that consists of the process of interest as a part of the cycle and then see if there are any changes left in the surroundings at the end of the cycle, that it is impossible to bring the surroundings back to its original state.

With this brief background, let us see the difference in the work output (or work input) of a reversible and irreversible process connecting two equilibrium states i and f of a closed system of a given mass of ideal gas.

During the course of a reversible process, the system is in a state of equilibrium (within as well as with the surroundings) at any point of the process along the path i-f. That is, properties of the system have unique values for any state j on the path i-f. In contrast to this, for an irreversible process connecting the states i and f, the system will not be at equilibrium and will not have unique values for its properties at any intermediate point along the irreversible path i-f.

Since the system is in equilibrium with surroundings at any point j along the reversible path i-f, the values of the intensive properties of the system, such as temperature, pressure are equal to the values of the corresponding external variables. Therefore, for a reversible process of expansion (or compression) of an ideal gas, Pext = Pgas, Work output =integral(Pext x dV). For an irreversible process, Pext is not equal to Pgas.

Whether the process is reversible or irreversible, the equation to calculate work out put is one and the same: Work output =integral(Pext x dV). Work output and work input have opposite signs since dV has opposite signs in the two cases.

For an irreversible process we cannot do that substitution for Pext, and hence, cannot obtain an equation but only an inequation. For expansion process we get lower work output and for a compression process we get more work input compared to that of a reversible process connecting the same end points. We cannot go into an elaboration of this point here in the discussion.
I completely agree with everything you have said up to here. In fact, the very points you have emphasized, for the most part, are covered in my earlier Insights article at the following link: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/. I would be pleased to receive feedback from you on that article as well.
Bringing 'time' to play For reversible processes, we can substitute the known value of Pgas for Pext and get: Wr = nRTln(Vf/Vi).
a role in explanation of thermodynamic process deprives thermodynamics of its vital strengths. Time has no role to play in thermodynamics (Classical or Equilibrium).

While mechanical processes are covered by laws of thermodynamics, thermodynamics is more general and covers the processes that involve heat which is not a part of mechanics. All mechanical processes, in principle, are reversible. Therefore, there is no scope to introduce irreversible processes through mechanical analogs.
These last two paragraphs are where we differ in our perspectives. Regarding the first paragraph, certainly time is not very important in discussing reversible processes. However, for irreversible processes, the speed at which a deformation occurs determines the rate at which mechanical energy is dissipated into thermal energy, and certainly, time is relevant to that. In any event, unlike you, I see no compelling reason why time should be excluded from the discussion of irreversible processes.

With respect to the second paragraph, I disagree with the contention that mechanical processes, in principle, are reversible. Any mechanical process that involves frictional dissipation or viscous dissipation of mechanical energy to thermal energy cannot be regarded as inherently reversible. Certainly processes like these cannot be considered reversible within the context of the definition you attributed to Max Planck (and which I fully agree with). So, whereas you feel that there is no value in introducing irreversible processes through mechanical analogs (at least mechanically irreversible processes), my judgement is that there is great pedagogical value in doing this. I guess we are just going to have to disagree on this.

Chet
Introduction of time in the analysis of irreversible process does not pay any dividends, on the other hand there is a possibility that it could be counterproductive, leading to misunderstanding of the simple concepts. Let us consider the spring-damper system. Let us take the system through a cycle ABA. Certain amount of net work is done on the system by the external force. As a result, the input work and output work differ, which could be in some form of energy ( such as electrical work, for eg) in the surroundings. The damper exerts a time dependent force. But for that property, there is no indication that the difference in the works for the forward and reverse portions of the cyclic processes, can not be transformed into some energy such as electrical energy. You did mention that the difference in works is dissipated in the form of heat; but it sounds like a request to the students to accept it as an extra fact. Introduction of friction on the other hand would lead to a natural feeling of dissipation into heat. If you still see the spring damper system (that doesn't have the properties associated with friction, I leave it to you).

The second point is more important! The pedagogical advantage you see in mechanical analogues to explain irreversible thermodynamic processes could mislead the students, indeed!

For, it is always possible for us to take a mechanical system through a cycle (whether reversible or irreversible), at the end of which, the system reaches its initial state. However, it is not the case with thermodynamic cyclic processes. For example, it is impossible to take a closed system of ideal gas through an irreversible adiabatic cyclic process! The system, simply will not reach the initial state A by an adiabatic process, after you take it from an equilibrium state A to an equilibrium state B by an irreversible adiabatic process.

I am yet to see the URL you sent for pressure of time. I will see it at the earliest. People are addicted to Statistical thermodynamics, these days and it is difficult to find people who discuss classical thermodynamics. From this point of view I like your discussion.

Whether the process is reversible or irreversible, the equation to calculate work out put is one and the same: Work output =integral(Pext x dV). Work output and work input have opposite signs since dV has opposite signs in the two cases.

For an irreversible process we cannot do that substitution for Pext, and hence, cannot obtain an equation but only an inequation. For expansion process we get lower work output and for a compression process we get more work input compared to that of a reversible process connecting the same end points. We cannot go into an elaboration of this point here in the discussion.
Of course we can write down an equation for Pext also for irreversible processes using irreversible thermodynamics.
Even though irreversible thermodynamics is not in the scope of introductory thermodynamics classes, it seems necessary to give students an idea in what respect non-equilibrium systems differ from equilibrium ones.
In my experience, students have great difficulties to imagine how a difference between equilibrium pressure and pressure away from equilibrium can arise.
My point was to find an easy example where Pext is still spatially constant but not equal to the equilibrium pressure.
Introduction of time in the analysis of irreversible process does not pay any dividends, on the other hand there is a possibility that it could be counterproductive, leading to misunderstanding of the simple concepts. Let us consider the spring-damper system. Let us take the system through a cycle ABA. Certain amount of net work is done on the system by the external force. As a result, the input work and output work differ, which could be in some form of energy ( such as electrical work, for eg) in the surroundings. The damper exerts a time dependent force. But for that property, there is no indication that the difference in the works for the forward and reverse portions of the cyclic processes, can not be transformed into some energy such as electrical energy. You did mention that the difference in works is dissipated in the form of heat; but it sounds like a request to the students to accept it as an extra fact. Introduction of friction on the other hand would lead to a natural feeling of dissipation into heat. If you still see the spring damper system (that doesn't have the properties associated with friction, I leave it to you).

The second point is more important! The pedagogical advantage you see in mechanical analogues to explain irreversible thermodynamic processes could mislead the students, indeed!

For, it is always possible for us to take a mechanical system through a cycle (whether reversible or irreversible), at the end of which, the system reaches its initial state. However, it is not the case with thermodynamic cyclic processes. For example, it is impossible to take a closed system of ideal gas through an irreversible adiabatic cyclic process! The system, simply will not reach the initial state A by an adiabatic process, after you take it from an equilibrium state A to an equilibrium state B by an irreversible adiabatic process.

I am yet to see the URL you sent for pressure of time. I will see it at the earliest. People are addicted to Statistical thermodynamics, these days and it is difficult to find people who discuss classical thermodynamics. From this point of view I like your discussion.

DrDu
The second point is more important! The pedagogical advantage you see in mechanical analogues to explain irreversible thermodynamic processes could mislead the students, indeed!
I did never claim any pedagogical advantage in mechanical analogues!

Whether the process is reversible or irreversible, the equation to calculate work out put is one and the same: Work output =integral(Pext x dV). Work output and work input have opposite signs since dV has opposite signs in the two cases.

For an irreversible process we cannot do that substitution for Pext, and hence, cannot obtain an equation but only an inequation. For expansion process we get lower work output and for a compression process we get more work input compared to that of a reversible process connecting the same end points. We cannot go into an elaboration of this point here in the discussion.
Of course we can write down an equation for Pext also for irreversible processes using irreversible thermodynamics.
Even though irreversible thermodynamics is not in the scope of introductory thermodynamics classes, it seems necessary to give students an idea in what respect non-equilibrium systems differ from equilibrium ones.
In my experience, students have great difficulties to imagine how a difference between equilibrium pressure and pressure away from equilibrium can arise.
My point was to find an easy example where Pext is still spatially constant but not equal to the equilibrium pressure.
Introduction of time in the analysis of irreversible process does not pay any dividends, on the other hand there is a possibility that it could be counterproductive, leading to misunderstanding of the simple concepts. Let us consider the spring-damper system. Let us take the system through a cycle ABA. Certain amount of net work is done on the system by the external force. As a result, the input work and output work differ, which could be in some form of energy ( such as electrical work, for eg) in the surroundings. The damper exerts a time dependent force. But for that property, there is no indication that the difference in the works for the forward and reverse portions of the cyclic processes, can not be transformed into some energy such as electrical energy. You did mention that the difference in works is dissipated in the form of heat; but it sounds like a request to the students to accept it as an extra fact. Introduction of friction on the other hand would lead to a natural feeling of dissipation into heat. If you still see the spring damper system (that doesn't have the properties associated with friction, I leave it to you).

The second point is more important! The pedagogical advantage you see in mechanical analogues to explain irreversible thermodynamic processes could mislead the students, indeed!

For, it is always possible for us to take a mechanical system through a cycle (whether reversible or irreversible), at the end of which, the system reaches its initial state. However, it is not the case with thermodynamic cyclic processes. For example, it is impossible to take a closed system of ideal gas through an irreversible adiabatic cyclic process! The system, simply will not reach the initial state A by an adiabatic process, after you take it from an equilibrium state A to an equilibrium state B by an irreversible adiabatic process.

I am yet to see the URL you sent for pressure of time. I will see it at the earliest. People are addicted to Statistical thermodynamics, these days and it is difficult to find people who discuss classical thermodynamics. From this point of view I like your discussion.
Hi Dr Du

Of course we can write down an equation for Pext also for irreversible processes using irreversible thermodynamics.

We can only write equations for the properties such as pressure, temperature etc. of the system, because system is well defined. On the other hand, surroundings is not well defined and doesn't have unique properties. Having known the pressure of the system, we apply or manipulate the Pext in a way we like. For example for reversible process, we maintain Pext equal to Psys. For irreversible processes Pext is not equal to Psys.

Students should easily be able to imagine that a well defined system at an equilibrium state has a unique value of pressure. A system in a state away from equilibrium doesn't have a unique values for its properties! Therefore, when a system is away from equilibrium, it is not in a well defined state and as such doesn't have unique values (or calculable values) for its properties.

I am unable to appreciate your statement: 'My point was to find an easy example where Pext is still spatially constant but not equal to the equilibrium pressure.'

What is the meaning of spatially constant ( but not equal to the equilibrium pressure), Pext?

DrDu
Oh, my fault , I meant that the system pressure of the gas does not vary with position and equals the external pressure.

Mentor
Oh, my fault , I meant that the system pressure of the gas does not vary with position and equals the external pressure.
Oh, my fault , I meant that the system pressure of the gas does not vary with position and equals the external pressure.
Hi DrDu.

I think you meant here the "system compressive stress" rather than "the system pressure". The system compressive stress includes the combination of both the thermodynamic pressure and the viscous stress. This is what is equal to the external pressure.

Chet

DrDu
Ok, but the trace of the stress tensor is usually included in the total pressure.

Mentor
Ok, but the trace of the stress tensor is usually included in the total pressure.
The trace of the stress tensor is equal to 3x the thermodynamic pressure (assuming the sign convention that compressive stresses are positive) plus a contribution from the volumetric viscous deformation (Fluid mechanics is my primary area of expertise). The stress tensor is equal to the thermodynamic pressure times the identity tensor, plus the contributions of the viscous stress components.

Chet

DrDu
So isn't this in the end the "hhdrodynamic pressure"?

Mentor
So isn't this in the end the "hhdrodynamic pressure"?
In cases where the fluid is deforming and viscous stresses are present, if you are looking at the deformation involved in expansion or compression in a cylinder, the deformation is not isotropic, and the compressive stresses are different in different directions.

For incompressible flow only, the pressure is equal to 1/3 the trace of the compressive stress tensor. But, if the flow is incompressible, 1/3 the trace of the compressive stress tensor includes a volumetric contribution from viscous stresses , and, in gas expansion within a cylinder, because of the anisotropy of the deformation, this will not be equal to the force per unit area on the piston.

Chet

DrDu
Ok, i never talked about a cylinder with a piston. Consider isotropic stress. Isn't even for a compressible fluid the hydrodynamic pressure equal to one third of the trace of the stress tensor?

Mentor
Ok, i never talked about a cylinder with a piston. Consider isotropic stress. Isn't even for a compressible fluid the hydrodynamic pressure equal to one third of the trace of the stress tensor?
If the deformation is isotropic (purely a volume change), then the stress tensor is isotropic, and the stress is the same in all directions. It is equal to the thermdynamic pressure plus a viscous contribution. If you want to call that the hydrodynamic pressure, that's fine. Even for a non-isotropic deformation, it's perfectly OK to define the term hydrodynamic pressure as 1/3 the trace of the compressive stress tensor. But, in the expansion or compression of a gas within a cylinder (which is the focus of this thread), the deformation is 1 dimensional (i.e., anisotropic), and the force per unit area exerted on the piston is not equal to the hydrodynamic pressure (as described by this definition).

Chet

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Introduction of time in the analysis of irreversible process does not pay any dividends, on the other hand there is a possibility that it could be counterproductive, leading to misunderstanding of the simple concepts.
I'm beginning to get the feeling that you really don't like the idea of introducing time into the analysis of irreversible processes. If you had a valid reason other than the fact that you just plain don't like it, it might have more impact. The Continuum Mechanics people always use time in their analyses of irreversible processes. Have you been submitting papers in the open literature criticizing them for doing this, with valid arguments for why they must be getting the wrong answers. If so, please share the references. Why single my analysis out as the one you object to most strongly?
Let us consider the spring-damper system. Let us take the system through a cycle ABA. Certain amount of net work is done on the system by the external force. As a result, the input work and output work differ, which could be in some form of energy ( such as electrical work, for eg) in the surroundings. The damper exerts a time dependent force. But for that property, there is no indication that the difference in the works for the forward and reverse portions of the cyclic processes, can not be transformed into some energy such as electrical energy.
Incorrect. During the irreversible expansion step, the dissipative work done by the piston on the damper is always (a) positive or (b) negative? During the irreverisble compression step, the dissipative work done by the piston on the damper is always (a) positive or (b) negative? So for the cycle, the dissipative work done by the piston on the spring-damper system is always (a) positive or (b) negative?

The second point is more important! The pedagogical advantage you see in mechanical analogues to explain irreversible thermodynamic processes could mislead the students, indeed!