Are textbooks sloppy with the entropy change of an irreversible process?

[atyy]

But how would P_gas be increased to be equal to P_ext?

It starts off lower and is compressed by a pressure, decreased by friction, so it can never reach P_ext.

If we allow a frictionless piston then, as Andrew says, the gas would be subject to a sudden adiabatic compression, again violating the conditions of the question.

Yes, it seems I'd need something complicated like a pressure dependent friction.

 

[Studiot]

Yes, it seems I'd need something complicated like a pressure dependent friction.

Did you say friction or fiction?



Sorry, couldn't resist that

 

[atyy]

Did you say friction or fiction?



Sorry, couldn't resist that

Both obviously

 

[atyy]

@Studiot, I tried searching for a discussion on whether the explicit form of the friction is known for this textbook example. These are the only leads I have so far that mention friction. Both mention irreversible iosthermal compression (not necessarily due to constant applied force), with the irreversibility due to friction, but neither gives an explicit form.

Bizarro, Entropy production in irreversible processes with friction, Phys. Rev. E 78, 021137 (2008).

Thomsen, Thermodynamics of an Irreversible Quasi-Static Process, American Journal of Physics 28 p119 (1960).

 

[Studiot]

I have no problem with friction being included in a thermodynamic system in general.

I just can't see how a frictional system can meet the conditions of 5.8 (or 5.7 for that matter)

That is why I suggested the hydraulic idea - it has no issues in this regard.

If you were to say this is not a well framed question, I think everyone would agree - perhaps it is just an exercise in symbolic manipulation for the posers.

I have not seen any similar questions in any thermodynamics books in the British orbit, physical chemistry or otherwise and since I do not have access to the actual book

I'll try to look up Engel and Reid's solution later.

Here is an extract from Rogers and Mayhew, one of the most famous (British) engineering thermo books:

Constant volume, constant pressure, polytropic and isothermal processes: when these are irreversible we must know either Q or W in addition to the end states before the process is completely determined.

go well

 

[atyy]

I have no problem with friction being included in a thermodynamic system in general.

I just can't see how a frictional system can meet the conditions of 5.8 (or 5.7 for that matter)

That is why I suggested the hydraulic idea - it has no issues in this regard.

If you were to say this is not a well framed question, I think everyone would agree - perhaps it is just an exercise in symbolic manipulation for the posers.

I have not seen any similar questions in any thermodynamics books in the British orbit, physical chemistry or otherwise and since I do not have access to the actual book

Here is an extract from Rogers and Mayhew, one of the most famous (British) engineering thermo books:

Constant volume, constant pressure, polytropic and isothermal processes: when these are irreversible we must know either Q or W in addition to the end states before the process is completely determined.

go well

Couldn't we see the specification of constant Pext in 5.8 as equivalent to Rogers and Mayhew's requirement that W be specified in addition to the end states, which in this case are the same as the reversible case of 5.7? It'd be very funny indeed if this were a transatlantic difference!

 

[atyy]

I just can't see how a frictional system can meet the conditions of 5.8 (or 5.7 for that matter)

That is why I suggested the hydraulic idea - it has no issues in this regard.

I have been thinking about how the conditions in the questions might be effected and here is what I have come up with.

The gas is confined in a thin rigid but thermally conductive bulb immersed in a liquid bath.
Leading to the bulb is a tube from a substantial header tank of a liquid that does not interact with the gas, via a tap. The level of the fluid is such that the pressure in the pipe is the desired end pressure.

At some time the tap is partially opened, allowing the liquid to compress the gas, at the sensibly constant pressure of the header tank, but at the same time controlling the rate of compression so that the temperature in the bulb remains sensibly constant.

Perhaps the carefully engineered friction that would be needed is equivalent to the carefully controlled rate of flow in the hydraulic set-up? The former is more notional, the latter more practical.

 

[juanrga]

Trying hard to understand a basic textbook model meant to illustrate that entropy (of the universe) increases for irreversible processes. Help me out please?

I get this part: A gas is compressed isothermally (constant T) and reversibly, getting worked on and expelling heat. To calculate ΔS for the system, you can use the formula ΔS=q/T since it is a reversible process. The same formula can be used for the entropy change in the surroundings, and of course, the entropy changes are equal and opposite since q_sys=-q_surr. Total entropy change of the universe is zero.

Now for the irreversible compression, with same initial and final system states. Since entropy is a state function, ΔS for the system is exactly the same value as above. However when you calculate the heat transfer q, that has a higher value now (more work was required for the irreversible compression, so more heat was expelled). To get entropy change of the surroundings, textbooks use ΔS_surr=q_surr/T=-q_sys/T, which is higher than ΔS_sys. But isn't it illegal to use that formula for this irreversible process? I'm totally confused. All of the textbooks brush right over this point, but it seems like an obvious objection.

Thank you for your help!

As you say the total entropy change of the universe ΔS_uni is zero for the reversible process

( ΔS_uni )_rev = 0

Now for the irreversible process just apply the second law

( ΔS_uni )_irrev >= ( ΔS_uni )_rev

and you get

( ΔS_uni )_irrev >= 0

or in words

the entropy of the universe increases or remain constant

.

 

[bbbeard]

If the temperature of the gas and the surroundings are the same it can be isothermal only if it is quasistatic.

I am saying that it can't be quasistatic, hence it can't be isothermal, if there is an initial finite difference between the external pressure on the gas and the internal pressure of the gas.

So in order for problem 5.8 to make any sense I think you have to replace "is isothermally compressed by a constant external pressure" by "is compressed by a constant external pressure" and add "At the end of the process, T=300K." where T refers to the temperature of the gas.

AM

I think you're making this problem a lot harder than it needs to be.

You are concerned that the assumption that the irreversible compression process is isothermal is somehow not valid, or inconsistent, or unrealistic (I'm not sure which). I will agree with "unrealistic". Of course, there are no isothermal processes in reality. "Reversible isothermal process" is an idealization like "steady flow" or "incompressible fluid". Objecting to this stipulation is like objecting when a pendulum problem tells you to ignore the mass of the string, or when a mass-spring problem tells you to ignore the mass of the spring, or when you are told to use the ideal gas law instead of including real-gas effects. It's just an idealization that is meant to allow the student to focus on other aspects of the model.

I think it is a conceptual mistake to bring these issues into a problem of thermodynamics. Your objections are essentially about heat transfer, not thermodynamics. In particular, it would seem that your objections about the isothermal assumption would vanish in the limit of large thermal diffusivity for the gas. The assumption doesn't introduce an inconsistency within thermodynamics. While you're not a student, I have found that many students get very confused when they try to reason about thermodynamics by resorting to reasoning about heat transfer. Conversely, one has to focus on the thermodynamics when learning thermodynamics and set aside any conceptions about the difficulties introduced by finite-rate heat transfer.

As I read the problem, the authors are merely stipulating that the student ignore any irreversibilities that are due to the finite rate of heat transfer, or the need for heat transfer to take place across a finite temperature difference. The problem is "about" figuring out the difference between reversible work and irreversible work. For the given conditions, the reversible work is NRT ln(V₂/V₁), while the irreversible work is P(V₂-V₁). Since both processes are stipulated to be isothermal (even though a real irreversible compression can't be isothermal), this allows the student to infer that at every step of the irreversible process, the internal energy of the gas is fixed, so that every increment of the work input becomes heat output to the surroundings at a known temperature. This enables the calculation of the entropy added to both the system and the surroundings.

A more detailed calculation could be more realistic. As you wrote, in actuality the finite rate of piston movement induces finite changes in the temperature of the gas, which results in irreversible heat transfer. But wouldn't this calculation necessarily require additional information about the gas, the geometry of the cylinder, the distribution of the work irreversibilities (friction? gas viscosity?) etc.? Isn't the student better off learning the basic thermodynamics first, before launching into a full transient 3d Navier-Stoke simulation?

The bottom line is that the simplifications in this problem, albeit unrealistic, introduce no inconsistency in thermodynamics and serve a valid pedagogical purpose. And it's really no more unrealistic than any of the dozens of other routine idealizations we encounter.

 

[Studiot]

I think an issue (under discussion) here is the poor wording of 5.8 as regards to 'constant pressure' and therefore determining exactly what the question meant.

As a matter of interest the reversible work does not contain N since only 1 mole is being considered, as I noted when I tried to kick off a collective effort solution to both parts of the problem.

go well

 

[bbbeard]

I think an issue (under discussion) here is the poor wording of 5.8 as regards to 'constant pressure' and therefore determining exactly what the question meant.

As a matter of interest the reversible work does not contain N since only 1 mole is being considered, as I noted when I tried to kick off a collective effort solution to both parts of the problem.

go well

In the formula for the reversible work the extensive quantity is the number of moles N. Since the process is isothermal one could just as easily have written P₁V₁ ln(V₂/V₁) or P₂V₂ ln(V₂/V₁), since all these quantities are equal to NRT ln(V₂/V₁). In the formula for the irreversible work the extensive quantity is the volume.

What is the problem with using the term "constant pressure"?

 

[atyy]

Objecting to this stipulation is like objecting when a pendulum problem tells you to ignore the mass of the string, or when a mass-spring problem tells you to ignore the mass of the spring, or when you are told to use the ideal gas law instead of including real-gas effects. It's just an idealization that is meant to allow the student to focus on other aspects of the model.

A more detailed calculation could be more realistic. As you wrote, in actuality the finite rate of piston movement induces finite changes in the temperature of the gas, which results in irreversible heat transfer. But wouldn't this calculation necessarily require additional information about the gas, the geometry of the cylinder, the distribution of the work irreversibilities (friction? gas viscosity?) etc.? Isn't the student better off learning the basic thermodynamics first, before launching into a full transient 3d Navier-Stoke simulation?

Yeah, I always had problems with the massless string. If it's made of photons, won't they fly away? Then how would they support the mass? But the worst were the massless pulleys!

 

[Andrew Mason]

I think you're making this problem a lot harder than it needs to be.

You are concerned that the assumption that the irreversible compression process is isothermal is somehow not valid, or inconsistent, or unrealistic (I'm not sure which). I will agree with "unrealistic". Of course, there are no isothermal processes in reality. "Reversible isothermal process" is an idealization like "steady flow" or "incompressible fluid". Objecting to this stipulation is like objecting when a pendulum problem tells you to ignore the mass of the string, or when a mass-spring problem tells you to ignore the mass of the spring, or when you are told to use the ideal gas law instead of including real-gas effects. It's just an idealization that is meant to allow the student to focus on other aspects of the model.

I think it is a conceptual mistake to bring these issues into a problem of thermodynamics. Your objections are essentially about heat transfer, not thermodynamics. In particular, it would seem that your objections about the isothermal assumption would vanish in the limit of large thermal diffusivity for the gas. The assumption doesn't introduce an inconsistency within thermodynamics. While you're not a student, I have found that many students get very confused when they try to reason about thermodynamics by resorting to reasoning about heat transfer. Conversely, one has to focus on the thermodynamics when learning thermodynamics and set aside any conceptions about the difficulties introduced by finite-rate heat transfer.

I don't want to belabour the point, but it was an issue that seemed to cause a lot of trouble with the OP.

Massless ropes and pulleys represent a limit that can be approached. At some point, the mass can be considered negligible for the purposes of the problem. I think that there is a difference between making an assumption that a process is isothermal when it can be arbitrarily close to isothermal and makng such an assumption when it cannot be arbitrarily close to isothermal. It is not really about heat transfer rates. It is about putting energy rapidly into a gas (an unbalanced pressure of 25:10) while maintaining that heat flows out of it while there is an arbitrarily small temperature difference between it and the surroundings.

All I was suggesting is that the word "isothermal" in relation to the process is not only unnecessary (which it is) it is not even approachable here. So to avoid confusion to students, we should ignore the details of the process and just concern ourselves with the beginning and end states.

AM

 

[atyy]

I don't want to belabour the point, but it was an issue that seemed to cause a lot of trouble with the OP.

Massless ropes and pulleys represent a limit that can be approached. At some point, the mass can be considered negligible for the purposes of the problem. I think that there is a difference between making an assumption that a process is isothermal when it can be arbitrarily close to isothermal and makng such an assumption when it cannot be arbitrarily close to isothermal. It is not really about heat transfer rates. It is about putting energy rapidly into a gas (an unbalanced pressure of 25:10) while maintaining that heat flows out of it while there is an arbitrarily small temperature difference between it and the surroundings.

All I was suggesting is that the word "isothermal" in relation to the process is not only unnecessary (which it is) it is not even approachable here. So to avoid confusion to students, we should ignore the details of the process and just concern ourselves with the beginning and end states.

AM

Anyway, regardless of the specific example, would you agree that although in general one needs the initial and final states to compute the entropy change of an irreversible process, in the exceptional case when the irreversible process is quasi-static, the "lost work" can be treated as "heat added" and the entropy change calculated by integrating "dQ/T" as in a reversible process?

 

[bbbeard]

I don't want to belabour the point, but it was an issue that seemed to cause a lot of trouble with the OP.

Massless ropes and pulleys represent a limit that can be approached. At some point, the mass can be considered negligible for the purposes of the problem. I think that there is a difference between making an assumption that a process is isothermal when it can be arbitrarily close to isothermal and makng such an assumption when it cannot be arbitrarily close to isothermal. It is not really about heat transfer rates. It is about putting energy rapidly into a gas (an unbalanced pressure of 25:10) while maintaining that heat flows out of it while there is an arbitrarily small temperature difference between it and the surroundings.

Well, as I pointed out, if you increase the thermal diffusivity (α=k/ρc_p) of the gas, the temperature rise can be made arbitrarily small. [It is the diffusivity, not the conductivity, that shows up in the http://highered.mcgraw-hill.com/sites/dl/free/0073129305/314124/cen29305_ch04.pdf"]. So in my view it is exactly analogous to assuming that the mass of a pendulum string can be made negligibly small. Unless I'm mistaken, the gas thermal diffusivity can be specified independently of the parameters that dictate how rapidly the piston will accelerate under the unbalanced pressure given in the problem.

Another way to achieve the same limit is to make the gas cylinder taller and skinnier. The non-dimensional time in the transient heat conduction problem is the Fourier number Fo = αt/R², where R is the radius of the cylinder. So for any given characteristic time of the piston movement (1 second?) we can make the Fourier number arbitrarily large by making R smaller (the previous paragraph was about making α larger). The effect of making Fo larger is to push the solution of the transient conduction farther out in time, i.e. more toward thermal equilibrium, i.e. driving the gas temperature closer to the heat bath temperature. Imagine the piston movement happens in some time t for a given experimental setup, and suppose that the gas temperature surges to, say, 310 K instead of staying isothermal at 300 K. Not negligible enough for you? Cut the cylinder radius in half and make it four times taller. Then the temperature surge should be roughly 302.5 K compared to 300 K. Not negligible enough? Repeat the radius reduction until you're happy that the temperature overshoot is negligible.

All I was suggesting is that the word "isothermal" in relation to the process is not only unnecessary (which it is) it is not even approachable here. So to avoid confusion to students, we should ignore the details of the process and just concern ourselves with the beginning and end states.

Actually, the idealization that the irreversible compression process is isothermal is necessary if you need to ignore the irreversibilities generated during the heat transfer process. The closer the process is to being isothermal, the less entropy is generated by the heat conduction through the cylinder walls. Suppose you send an amount of heat Q through the walls. If the process has a uniform temperature T, then the gas loses entropy in the amount -Q/T and the surroundings pick up entropy in the amount Q/T, so there is entropy convection but no net entropy generation. But if the gas is at a temperature Tgas > T, then the gas entropy change is -Q/Tgas and the heat bath entropy change is Q/T. The total entropy has increased by an amount Q(1/T-1/Tgas). So the authors of the problem want you to ignore this contribution to the entropy generation by assuming that Tgas is arbitrarily close to T.

BBB

 

[Studiot]

What is the problem with using the term "constant pressure"?

Nothing except that it applies the the surroundings not the system.

Your formula

while the irreversible work is P(V2-V1).

is the work done by the surroundings, not the work done on or by the system, when P is P_ext

The question makes a big song and dance about this.

Some, if not most, of the work done by the surroundings remains in the surroundings and is dissipated as heat due to friction or whatever.

The question specifies the impossible but sensible stance that the heat capacity of the surroundings is so large that the process is sensibly isothermal, something you are taking Andy and atyy to task for doing in a different direction.

Both questions actually ask for the entropy change to the system, the surroundings and the combination.

I agree the implication is that all work done on the system passes back to the surroundings as heat so the exercise can simply be a bit of algebraic manipulation, but I still reckon it is poorly draughted.

In particular Rogers and Mayhew are correct in saying that you need to know the work done or heat transferred to solve the problem - since it is impossible (even theoretically) for it to be zero.

 

[Andrew Mason]

Actually, the idealization that the irreversible compression process is isothermal is necessary if you need to ignore the irreversibilities generated during the heat transfer process. The closer the process is to being isothermal, the less entropy is generated by the heat conduction through the cylinder walls. Suppose you send an amount of heat Q through the walls. If the process has a uniform temperature T, then the gas loses entropy in the amount -Q/T and the surroundings pick up entropy in the amount Q/T, so there is entropy convection but no net entropy generation.

Only if the process is reversible. The change in entropy of the gas is always -\int |dQ_{rev}|/T whereas the change in entropy of the surroundings is +\int |dQ_{actual}|/T.

But if the gas is at a temperature Tgas > T, then the gas entropy change is -Q/Tgas and the heat bath entropy change is Q/T. The total entropy has increased by an amount Q(1/T-1/Tgas). So the authors of the problem want you to ignore this contribution to the entropy generation by assuming that Tgas is arbitrarily close to T.

In problem 5.8 (above) when determining change in entropy it makes absolutely no difference what the temperature of the gas is between the beginning and end points. What matters is the total heatflow out of the gas, which is necessarily the amount of work done on the gas. That heatflow divided by the temperature of the surroundings (assuming it to be constant) determines the change in entropy of the surroundings.

AM

 

[atyy]

I agree the implication is that all work done on the system passes back to the surroundings as heat so the exercise can simply be a bit of algebraic manipulation, but I still reckon it is poorly draughted.

In problem 5.8 (above) when determining change in entropy it makes absolutely no difference what the temperature of the gas is between the beginning and end points. What matters is the total heatflow out of the gas, which is necessarily the amount of work done on the gas. That heatflow divided by the temperature of the surroundings (assuming it to be constant) determines the change in entropy of the surroundings.

Yes, I think that was the OP's question - why can one use dQ/T to calculate entropy change even though this process is irreversible? Would you agree that the basic idea is that in a quasi-static process in which the irreversibility is due to dissipative work, the surroundings that accept the heat from the dissipative work cannot distinguish whether the heat is due to "reversible heating" or to "irreversible work", so dQ/T can still be used in the exceptional case of a quasi-static irreversible process (ie. this trick should not work for a non-quasi-static irreversible process, in which case one has to know the initial and final states of the environment to find a corresponding reversible process)?

 

[bbbeard]

Nothing except that it applies the the surroundings not the system.

Your formula P(V2-V1) is the work done by the surroundings, not the work done on or by the system, when P is P_ext

The question makes a big song and dance about this.

Some, if not most, of the work done by the surroundings remains in the surroundings and is dissipated as heat due to friction or whatever.

The question specifies the impossible but sensible stance that the heat capacity of the surroundings is so large that the process is sensibly isothermal, something you are taking Andy and atyy to task for doing in a different direction.

Both questions actually ask for the entropy change to the system, the surroundings and the combination.

I agree the implication is that all work done on the system passes back to the surroundings as heat so the exercise can simply be a bit of algebraic manipulation, but I still reckon it is poorly draughted.

In particular Rogers and Mayhew are correct in saying that you need to know the work done or heat transferred to solve the problem - since it is impossible (even theoretically) for it to be zero.

That's not how I read the problem.

What I read is that the surroundings exert a constant pressure on a piston that is part of the thermodynamic system under consideration. I don't see any wording that suggests that part of the work goes back into the surroundings. You may be defining the "system" as excluding the piston, that is, you may be defining the piston as part of the surroundings. But it seems to me that if the (perhaps massive) piston starts at rest and winds up at rest, then in the end it simply passes all the P_ext(V₂-V₁) work into the gas. But I say that because, as I discuss below, I also dismiss the scenario where friction between the wall and the piston is the source of dissipation.

You can surely invent scenarios where the simple interpretation of the problem is untenable, where, for example, the piston has some heat capacity, or the walls have some heat capacity, or the friction between the wall and the cylinder is the source of the dissipation. But since there is no hint in the problem statement that suggests that these complications play a role in the thermodynamics, I see no reason to depart from the simplest interpretation.

Among the complications that have been discussed in this thread is the issue of dynamic consistency. Since the piston has different pressures acting on the gas side and on the heat bath side, unless it has a mass the acceleration will be infinite, at least whenever mechanical friction is negligible.

There are various ways of making the system dynamically consistent -- none of which have anything to do with the thermodynamics per se. The easiest thing to imagine is that the piston has finite mass, and starts to accelerate when it is released from rest with an unbalanced pressure force. But you could also imagine that the acceleration of the mass of the gas itself provides the extra term that makes F=ma balance. Or, conceptually at least, there could be a dashpot attached to the gas side of the piston, so that the friction in the dashpot provides a counterbalancing force (and standing in for dissipation in the gas itself). (I think some folks have tried to make the friction act between the piston and the cylinder walls -- this seems simpler mechanically, but it introduces conceptual difficulties with the distribution of the resulting internal energy). But for the purpose of doing this problem, the details of the mechanical assembly don't matter. All the information you need to do the problem is stated. If you assume that peculiar unmentioned boundary conditions prevent you from doing the problem -- if, for example, you assert that some of the work done by the surroundings on the piston mysteriously winds up as work done on the the surroundings by the surroundings -- you will get the problem wrong, I guarantee.

 

[bbbeard]

Only if the process is reversible. The change in entropy of the gas is always -\int |dQ_{rev}|/T whereas the change in entropy of the surroundings is +\int |dQ_{actual}|/T.

I'm not understanding your meaning here. Heat transfer across a finite temperature difference is by definition irreversible because entropy is generated. The paradigmatic example is essentially what I outline.

Perhaps it would be more revealing to consider a situation where two heat baths are put in contact with each other through a diathermal membrane. Suppose the bath temperatures are T_L (T_low) and T_H (T_high). For every increment of heat transfer ΔQ that is transmitted through the diathermal wall, the high temperature bath loses an amount of entropy |ΔS_H| = |ΔQ|/T_H, and the low temperature bath gains an amount of entropy |ΔS_L| = |ΔQ|/T_L. Because T_H > T_L, |ΔS_H| < |ΔS_L|, and therefore there is a net generation of entropy: ΔStotal = |ΔS_L|-|ΔS_H| > 0.

This is exactly the mechanism by which entropy is generated by heat transfer. For a temperature field with spatial gradients, you do this calculation for differential elements, but the process is the same.

In problem 5.8 (above) when determining change in entropy it makes absolutely no difference what the temperature of the gas is between the beginning and end points. What matters is the total heatflow out of the gas, which is necessarily the amount of work done on the gas. That heatflow divided by the temperature of the surroundings (assuming it to be constant) determines the change in entropy of the surroundings.

You are right. Any excess entropy comes out "in the wash" when the gas settles down to its final temperature.

 

[Andrew Mason]

Yes, I think that was the OP's question - why can one use dQ/T to calculate entropy change even though this process is irreversible? Would you agree that the basic idea is that in a quasi-static process in which the irreversibility is due to dissipative work, the surroundings that accept the heat from the dissipative work cannot distinguish whether the heat is due to "reversible heating" or to "irreversible work",

Yes, provided no work is done on the environment ie. there is only heat flow to/from the environment, it does not matter how the heat originates in the system when determining the entropy change of the surroundings.

so dQ/T can still be used in the exceptional case of a quasi-static irreversible process (ie. this trick should not work for a non-quasi-static irreversible process, in which case one has to know the initial and final states of the environment to find a corresponding reversible process)?

I am not sure what you are saying here. If there is only heatflow into/out of the surroundings (the system does no work on the surroundings) AND if the surroundings have sufficiently high heat capacity so that there is no change in temperature of the surroundings, then \Delta S_{surr}= \Delta Q_{actual}/T. It does not matter whether the actual process is quasi-static or not.

AM

 

[Andrew Mason]

I'm not understanding your meaning here. Heat transfer across a finite temperature difference is by definition irreversible because entropy is generated. The paradigmatic example is essentially what I outline.

This is true, of course. But problem (5.8 from atyy's post #27) does not involve a finite temperature difference. In this problem the system and the surroundings are at the same temperature T. I am just saying that since the change in entropy of the system depends only on the reversible work needed to compress the gas from 25 litre to 10 litres, not the actual work done, it does not matter whether the heat actually flowed out of the system at temperature T or whether it flowed out at a much higher temperature. And it doesn't matter to the calculation of the change of entropy of the surroundings either. All that matters is the amount of heat flow and that, of course, is determined by the actual amount of work done on the system (which will be higher than the reversible work).

AM

 

[atyy]

I am not sure what you are saying here. If there is only heatflow into/out of the surroundings (the system does no work on the surroundings) AND if the surroundings have sufficiently high heat capacity so that there is no change in temperature of the surroundings, then \Delta S_{surr}= \Delta Q_{actual}/T. It does not matter whether the actual process is quasi-static or not.

Yes. But how is dQ determined? One needs to somehow know the dissipative work. For example, in problem 5.8, if the volume oscillates between initial and final states, then it will be hard to calculate the dissipative work from the given information.

 

[Andrew Mason]

Yes. But how is dQ determined? One needs to somehow know the dissipative work. For example, in problem 5.8, if the volume oscillates between initial and final states, then it will be hard to calculate the dissipative work from the given information.

You definitely need to know the amount of work done on the system. In this case W = Pext(V2-V1). The compression will be very violent initially, causing turbulent flows within the gas that will cause the temperature to rise as they settle down.

AM

 

[bbbeard]

Yes. But how is dQ determined? One needs to somehow know the dissipative work. For example, in problem 5.8, if the volume oscillates between initial and final states, then it will be hard to calculate the dissipative work from the given information.

If the system is lightly damped, assuming it starts from rest at V1, and assuming the piston does not have zero mass, then the piston will overshoot V2 and go to an even smaller volume, then the higher pressure will be inside the cylinder, pushing the piston back out to nearly the initial volume, and the piston will continue oscillating (asymmetrically, because the pressure imbalance is not linear) until it eventually settles down to V2. Once equilibrium is achieved the work done on the surroundings will be P(V₂-V₁), regardless of the oscillations.

 

[atyy]

You definitely need to know the amount of work done on the system. In this case W = Pext(V2-V1). The compression will be very violent initially, causing turbulent flows within the gas that will cause the temperature to rise as they settle down.

AM

If the system is lightly damped, assuming it starts from rest at V1, and assuming the piston does not have zero mass, then the piston will overshoot V2 and go to an even smaller volume, then the higher pressure will be inside the cylinder, pushing the piston back out to nearly the initial volume, and the piston will continue oscillating (asymmetrically, because the pressure imbalance is not linear) until it eventually settles down to V2. Once equilibrium is achieved the work done on the surroundings will be P(V₂-V₁), regardless of the oscillations.

That seems reasonable. Perhaps the quasi-static requirement can be weaker than what I stated. The general idea for putting the requirement in is that for a general irreversible process, the entropy change cannot be calculated by integrating along the actual path, and must be found from an appropriate reversible process connecting the same initial and final points. However, there may be some exceptions to this when the irreversible process is quasi-static, and the dissipative work properly accounted for as heat. In this case, it seems that it is enough for the environment to be quasi-static, as P_ext and T_ext constant imply.

 

[Studiot]

What I read is that the surroundings exert a constant pressure on a piston that is part of the thermodynamic system under consideration.

One mole of an ideal gas at 300 K is isothermally compressed by a constant external pressure

I see no piston or indeed any compression mechanism mentioned.

I think it is quite clear that the 'system' is intended to be 1 mole of gas.

Further the question clearly states 'isothermally' so all this business of overshoot, oscillation etc is irrelevant.

If the compression is rapid it is impossible for it to be isothermal.

I cannot see how a piston driven by a constant pressure P_ext can be anything but rapid, which is why I suggested the hydraulic solution.

I suggest the alternative is that the compression is not isothermal ie that the word crept in by mistake from 5.7.

Do you think 5.8 makes more sense this way?

 

[bbbeard]

I see no piston or indeed any compression mechanism mentioned.

I think it is quite clear that the 'system' is intended to be 1 mole of gas.

Further the question clearly states 'isothermally' so all this business of overshoot, oscillation etc is irrelevant.

If the compression is rapid it is impossible for it to be isothermal.

I cannot see how a piston driven by a constant pressure P_ext can be anything but rapid, which is why I suggested the hydraulic solution.

I suggest the alternative is that the compression is not isothermal ie that the word crept in by mistake from 5.7.

Do you think 5.8 makes more sense this way?

The point is that the details of the structure of this system are irrelevant to its thermodynamics. This is a general observation about thermodynamics. Nick Carnot was able to place an upper bound on the efficiency of a heat engine operating between two temperatures, and this limit applies no matter what hardware you place in what configuration between the heat baths. Similarly, for problem 5.8 you don't need to know that the piston was made of 2" thick steel, or the exact height and width of the cylinder, or the chemical composition of the gas, or its viscosity, or the thermal conductivity of the walls, or any of a plethora of other parameters you would need to actually design the system. You are given everything you need to know to do the problem.

Most of this thread has consisted of various good-natured attempts to complexify this problem by asking what real system could come close to being described by the parameters of the problem. Sure, no real gas would really be isothermal under rapid compression -- but I've already laid out the strategy for approaching the system with real hardware: use a massive piston to slow down the compression, a gas with a high thermal diffusivity, and a tall, skinny cylinder with thin walls. I think I've shown that this approximation is analogous to similar idealizations we use without flinching: steady flow, incompressible fluid, diathermal membrane, adiabatic walls... indeed, even the ideal gas law ultimately gives way to http://en.wikipedia.org/wiki/Compressibility_factor" .

I still think it is clearer to compute the work interaction with the surroundings if you "draw the dotted line" to include the piston with the gas. Clearly this work interaction works out to be P_ext(V₂-V₁), and this is easily computed if you draw the control boundary outside the piston. Ultimately this must equal the work interaction computed if you draw the control boundary on the gas side of the piston, provided that the piston is frictionless (i.e. the dissipation is in the gas). Can you demonstrate this with an explicit calculation, i.e. integrate PdV for the motion of the piston? I think it would help to give the piston a finite mass, but it's your strategy, so let's see how you would do it.

 

[Andrew Mason]

Most of this thread has consisted of various good-natured attempts to complexify this problem by asking what real system could come close to being described by the parameters of the problem.

No need to complicate things by making up new words:)

I still think it is clearer to compute the work interaction with the surroundings if you "draw the dotted line" to include the piston with the gas. Clearly this work interaction works out to be P_ext(V₂-V₁), and this is easily computed if you draw the control boundary outside the piston. Ultimately this must equal the work interaction computed if you draw the control boundary on the gas side of the piston, provided that the piston is frictionless (i.e. the dissipation is in the gas). Can you demonstrate this with an explicit calculation, i.e. integrate PdV for the motion of the piston? I think it would help to give the piston a finite mass, but it's your strategy, so let's see how you would do it.

Why the concern about the mechanism? The problem says that Pext is constant, which means it is 25/10 x initial gas pressure, and it is applied to the gas. Work = Pext x (V1-V2). My quibble is with the characterization of the process as being isothermal.

There is tremendous initial acceleration since the gas molecules would have a very small mass. So if you want to figure out what mechanism could produce this, I think you have to have a piston with a very small mass so that it can accelerate extremely rapidly. You could not do this with weights pressing down on a piston, for example, as that would provide a maximum acceleration (g) which is much too small.

AM

 

[bbbeard]

No need to complicate things by making up new words:)

Heh. In my lexicon, the opposite of "simplify" is "complexify"... it sounds more analytical than "complicate" (the opposite of which must be "simplicate"?) ;-)

Why the concern about the mechanism? The problem says that Pext is constant, which means it is 25/10 x initial gas pressure, and it is applied to the gas. Work = Pext x (V1-V2). My quibble is with the characterization of the process as being isothermal.

It sounds like you agree with me that the system volume must include the piston. It should just be a matter of bookkeeping. But if the piston has any mass at all, the force exerted on the gas by the piston is less than the force exerted on the piston by the external pressure, by an amount equal to the mass of the piston times its acceleration. As I see it, that is how you would match the apparently inconsistent pressures of the outside atmosphere and the cylinder gas. If you try to draw the "dotted line" so that it only includes the gas, then you have to compute the work done on the gas by using the pressure on the inside of the piston (which does not equal Pext).

There is tremendous initial acceleration since the gas molecules would have a very small mass. So if you want to figure out what mechanism could produce this, I think you have to have a piston with a very small mass so that it can accelerate extremely rapidly. You could not do this with weights pressing down on a piston, for example, as that would provide a maximum acceleration (g) which is much too small.

The problem does not specify that the acceleration is tremendous. That's why I think you can pick whatever piston mass you want (again, the piston mass, or the magnitude of the acceleration of the gas make no difference to the thermodynamics, only to the dynamics).

I was thinking about a related problem. Suppose you have a rigid adiabatic reservoir with two chambers that are separated by a rigid diathermal membrane. Each chamber contains 1 mole of air at 300 K. But in Chamber A the pressure is 1 atm, while in Chamber B the air is at 2 atm (obviously the volumes have to be in the inverse ratio of the pressures). Scenario 1: suppose the membrane is moveable (like a piston) so that the volumes can change, and that the movement is frictionless. If the membrane is released, where does it wind up? What are the work and heat transfers between the two chambers? What are the final temperature and pressure? Scenario 2: suppose the membrane is "popped" so that air can move freely between the two chambers. What is the final temperature and pressure? Are these different from scenario 1? What happens to the entropy in each case? We can also consider case 3: suppose that the moveable diathermal membrane is attached to a rod that penetrates the wall of the reservoir, and thereby allows the extraction of work from the system as the high pressure chamber expands. Let the equilibrium be approached quasi-statically (i.e. isentropically). What are the final temperature and pressure, and how much work was extracted?

BBB

 

[Andrew Mason]

It sounds like you agree with me that the system volume must include the piston.

Why? There is no requirement that there be a piston. As I said previously, the situation could be achieved by having the gas in a balloon inside a rigid container at the bottom of a lake and then quickly removing the lid.

If you have a piston compressing the gas, the piston would have to be of negligible mass compared to the mass of the gas. Otherwise, as you say, the external pressure on the gas would decrease with the acceleration of the piston and you do not meet the conditions stated in the problem.

The problem does not specify that the acceleration is tremendous. That's why I think you can pick whatever piston mass you want (again, the piston mass, or the magnitude of the acceleration of the gas make no difference to the thermodynamics, only to the dynamics).

But if you did that, the pressure on the gas would not be Pext. Assuming the kinetic energy of the piston eventually ends up as thermal energy in the gas, the result would be the same. But it would be a different problem.

AM

 

[bbbeard]

Why? There is no requirement that there be a piston. As I said previously, the situation could be achieved by having the gas in a balloon inside a rigid container at the bottom of a lake and then quickly removing the lid.

In that case the balloon material is the piston. That wouldn't be a desirable formulation of the gedanken, in my view, because there is obviously elastic energy in the balloon that should be accounted for -- plus there is no reason to think that the compression will be spherically symmetric. I think it's "cleaner" to talk about a solid piston in a cylindrical geometry. Recall that for a given amount of stress, more energy is stored in the more elastic (smaller Young's modulus) material. (Just think of a spring: F = -kx, PE=(1/2)kx² = (1/2)F²/k.)

But on the other hand I am perfectly capable of ignoring the elastic energy of the balloon (just as I am capable of ignoring departures from isothermality!)

As I said in my graduate quantum mechanics class: if they can get you to believe in negative kinetic energy [during tunneling], they can get you to believe in anything.

If you have a piston compressing the gas, the piston would have to be of negligible mass compared to the mass of the gas. Otherwise, as you say, the external pressure on the gas would decrease with the acceleration of the piston and you do not meet the conditions stated in the problem.

But you would know exactly what the acceleration of the piston would have to be to match the pressures. As I see it, you have to have something to match the pressure at the gas-external boundary.

But if you did that, the pressure on the gas would not be Pext. Assuming the kinetic energy of the piston eventually ends up as thermal energy in the gas, the result would be the same. But it would be a different problem.

But the gas pressure is not equal to P_ext. The external pressure is P_ext. The gas pressure is some lower value, starting with P₁. I suggest a piston to match the pressures. It seems like what you are envisioning is closer to what you would see in a http://en.wikipedia.org/wiki/Shock_tube" , except with an infinite high pressure reservoir and a movable massless diaphragm. Which is okay in terms of the problem (the end states and total work & heat transfer are the same) -- except there is no way for the compression process to be isothermal inside a shock tube. What I suggest at least has a plausible limit which is isothermal. Of course, you have said (multiple times now) that the isothermal condition is not a requirement and was just sloppy problem writing on their part. I don't think that is necessarily true -- although I would bet the authors didn't spend 1/10 the amount of time thinking about it as we have...

 

[Andrew Mason]

In that case the balloon material is the piston. That wouldn't be a desirable formulation of the gedanken, in my view, because there is obviously elastic energy in the balloon that should be accounted for -- plus there is no reason to think that the compression will be spherically symmetric.

The balloon is a limp bladder inside the box. The pressure of the gas is exerted on the box not the balloon, so there is negligible elastic energy.

But the gas pressure is not equal to P_ext. The external pressure is P_ext. The gas pressure is some lower value, starting with P₁.

Precisely my point. The pressure ON the gas from outside is 2.5 times the internal pressure of the gas. So, there is an unbalanced inward force on the gas, initially. This unbalanced inward force gives the gas molecules in contact with the membrane to which Pext is applied (mass = \Delta m) acceleration a = (P_{ext}-P_{int})A/\Delta m.

Since \Delta m is very small to begin, the acceleration is very high. The piston would have to have that same acceleration. So it seems to me that if you are going to use a piston, you would want to give it a mass much less than the mass of the gas

AM