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

[atyy]

I am having difficulty understanding how the compression could be isothermal. How does the rapid initial compression (which, it seems to me, has to occur) not increase the temperature of the gas to more than an infinitessimal amount higher than 300K?

It seems to me that the process must involve a rapid compression until Pgas = Pext followed by a slow compression as the heat flows out of the gas to the surroundings.

AM

I'm guessing that there is friction that has some maximum strength (analogous to static friction in classical mechanics). So most of the finite pressure difference simply balances the friction, and the friction is only exceeded by an infinitesimal amount. Because the excess is infinitesimal, the motion is still quasi-static, and because it is in thermal contact with a heat bath, it is isothermal. The irreversibilty is due to friction not changing sign if the external pressure is infinitesimally reversed. For the surroundings, the effect of work done in the forward direction against friction is the same as frictionless work plus reversible heating, since the environment can't "tell the difference" between heat from friction and heat from reversible heating.

There are other forms of irreversible compression which are not quasi-static, and so definitely not isothermal. I think it is only in the quasi-static irreversible case that one can use this trick of replacing irreversible work against friction with frictionless work plus heating.

 

[Andrew Mason]

I'm guessing that there is friction that has some maximum strength (analogous to static friction in classical mechanics). So most of the finite pressure difference simply balances the friction, and the friction is only exceeded by an infinitesimal amount. Because the excess is infinitesimal, the motion is still quasi-static, and because it is in thermal contact with a heat bath, it is isothermal. The irreversibilty is due to friction not changing sign if the external pressure is infinitesimally reversed. For the surroundings, the effect of work done in the forward direction against friction is the same as frictionless work plus reversible heating, since the environment can't "tell the difference" between heat from friction and heat from reversible heating.

There are other forms of irreversible compression which are not quasi-static, and so definitely not isothermal. I think it is only in the quasi-static irreversible case that one can use this trick of replacing irreversible work against friction with frictionless work plus heating.

So are you saying: (Pext-Pgas)xArea = Ffriction?

Would that not mean that the pressure applied to the gas was Pgas and not Pext?

AM

 

[atyy]

So are you saying: (Pext-Pgas)xArea = Ffriction?

Would that not mean that the pressure applied to the gas was Pgas and not Pext?

AM

Let me put my guess down again. I'll try to look up Engel and Reid's solution later. (Edit: I googled a similar problem in http://books.google.com/books?id=eH...ce=gbs_ge_summary_r&cad=0#v=onepage&q&f=false, Example 3.9 on p126.)

From the point of view of the gas, everything is the same as in the reversible case. However, compared to the reversible case, we put finitely more weights on the piston. So more weight has been lowered in the gravitational field. The extra work (compared to the reversible case) has been used to generate net heat (compared to the reversible case) via friction, which here has gone to the surroundings.

 

[Andrew Mason]

Example Problem 5.8

One mole of an ideal gas at 300 K is isothermally compressed by a constant external pressure equal to the final pressure in Example Problem 5.7. At the end of the process, P=P_ext. Because P≠P_ext at all but the final state, this process is irreversible. The initial volume is 25.0 L and the final volume is 10.0 L. The temperature of the surroundings is 300 K. Calculate the entropy change of the system, surroundings and their sum.

I have thought a little more about this problem and I conclude it is not possible to have an isothermal compression where Pext >> Pgas

As far as I can see, if you apply an external pressure to the gas that is so much greater than the internal pressure of the gas, there will be rapid compression that cannot be isothermal. It has to be practically adiabatic.

An example would be putting a balloon filled with air at 1 atm in a rigid box at the bottom of a lake where the ambient pressure is 2.5 atm, and then suddenly removing the top of the box. The gas will compress rapidly adiabatically, which will heat up the gas to a higher temperature than a reversible adiabatic compression. The gas will then compress further as it cools.

AM

 

[atyy]

I have thought a little more about this problem and I conclude it is not possible to have an isothermal compression where Pext >> Pgas

As far as I can see, if you apply an external pressure to the gas that is so much greater than the internal pressure of the gas, there will be rapid compression that cannot be isothermal. It has to be practically adiabatic.

An example would be putting a balloon filled with air at 1 atm in a rigid box at the bottom of a lake where the ambient pressure is 2.5 atm, and then suddenly removing the top of the box. The gas will compress rapidly adiabatically, which will heat up the gas to a higher temperature than a reversible adiabatic compression. The gas will then compress further as it cools.

AM

The compression can be quasi-static if there is friction.

Apart from Engel and Reid, it's also discussed in
http://books.google.com/books?id=eH_1dIZr-zMC&source=gbs_navlinks_s, Example 3.9
http://books.google.com/books?id=W0bpzXFK3lEC&dq=physical+chemistry&source=gbs_navlinks_s, p51
http://books.google.com/books?id=OdDyAy0xE3cC&dq=physical+chemistry&source=gbs_navlinks_s, p63
http://books.google.com/books?id=sTu-qT3jeP0C&dq=chen+thermodynamics&source=gbs_navlinks_s, p57-61

The last reference by Chen probably has the most explicit discussion of friction in the process.

 

[Andrew Mason]

The compression can be quasi-static if there is friction.

Apart from Engel and Reid, it's also discussed in
http://books.google.com/books?id=eH_1dIZr-zMC&source=gbs_navlinks_s, Example 3.9
http://books.google.com/books?id=W0bpzXFK3lEC&dq=physical+chemistry&source=gbs_navlinks_s, p51
http://books.google.com/books?id=OdDyAy0xE3cC&dq=physical+chemistry&source=gbs_navlinks_s, p63
http://books.google.com/books?id=sTu-qT3jeP0C&dq=chen+thermodynamics&source=gbs_navlinks_s, p57-61

The last reference by Chen probably has the most explicit discussion of friction in the process.

I don't see how you could have a quasistatic compression and still apply a pressure of Pext = 2.5 Pgas to the gas. You can make it quasistatic only if the pressure of the gas is equal to (infinitessimally less than) the pressure that is applied to the gas. You can reduce the external pressure through friction and make the compression quasistatic. But that means that the pressure applied to the gas is simply Pgas, not Pext.

AM

 

[atyy]

I don't see how you could have a quasistatic compression and still apply a pressure of Pext = 2.5 Pgas to the gas. You can make it quasistatic only if the pressure of the gas is equal to (infinitessimally less than) the pressure that is applied to the gas. You can reduce the external pressure through friction and make the compression quasistatic. But that means that the pressure applied to the gas is simply Pgas, not Pext.

AM

OK, but it's still isothermal compression then, right?

 

[Andrew Mason]

OK, but it's still isothermal compression then, right?

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

 

[Studiot]

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.

 

[atyy]

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

But couldn't it make sense if the term "external pressure" referred to the pressure applied without accounting for friction, ie. one puts a bunch of weights mg on the piston, and divides by its area A, so P_ext=mg/A. Then the gas is observed to compress quasi-statically. Knowing that P_ext>>P_gas, we infer that there's friction.

 

[Studiot]

But couldn't it make sense if the term "external pressure" referred to the pressure applied without accounting for friction, ie. one puts a bunch of weights mg on the piston, and divides by its area A, so Pext=mg/A. Then the gas is observed to compress quasi-statically. Knowing that Pext>>Pgas, we infer that there's friction.

If there was any significant friction this would not meet the terms of the question.

Not because of the work done by the surroundings overcoming friction, since this is returned to the surroundings as heat. But because when the piston comes to rest it is partly supported by the raised gas pressure and partly by the force of static friction around the piston perimeter. Since the piston at rest is in static equilibrium the force on one side must balance the force on the other so the weights can never compress the gas to P_ext, violating one end condition of the question.

 

[atyy]

If there was any significant friction this would not meet the terms of the question.

Not because of the work done by the surroundings overcoming friction, since this is returned to the surroundings as heat. But because when the piston comes to rest it is partly supported by the raised gas pressure and partly by the force of static friction around the piston perimeter. Since the piston at rest is in static equilibrium the force on one side must balance the force on the other so the weights can never compress the gas to P_ext, violating one end condition of the question.

How about a non-constant friction force (something like F≤uN of classical mechanics)?

 

[Studiot]

But isn't static friction greater than dynamic?

 

[atyy]

But isn't static friction greater than dynamic?

That might help, but I was just thinking that in the condition P_gas=P_ext, the forces trying to move the piston either way are balanced, so the friction is zero at that point. Just like static friction can be zero or μN.

 

[Studiot]

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.

 

[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.