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

[azaharak]

Isn't the correct term for a reversible adiabatic process isentropic?

 

[atyy]

Why is a finite pressure difference as in Example Problem 5.8 irreversible? First, if it is very large, then the gas will compress quickly, and things will not be quasi-static, hence irreversible. If the compression is not quasi-static, equilibrium thermodynamic variables are not defined throughout the process, which rules out "isothermal" as a description. However a finite pressure difference can still compress quasi-statically but irreversibly if there is the kind of static friction that increases up to a certain maximum value. In that case one has to add a finite amount of weight to match the maximal friction plus an infinitesimal amount to make the piston move quasi-statically. It is the second case that is assumed in 5.8, so that the compression can be isothermal.

Friction is irreversible because if you reverse the action (remove an infinitesimal amount of external weight), the friction does not reverse sign. For accounting purposes, we can treat the friction as an equivalent reversible process consisting of frictionless work plus reversible heat. The entropy of the surroundings cannot distinguish between heat from friction during a quasi-static isothermal process or heat from reversible heating. What makes the friction irreversible is that the reverse process does not correspond to a reverse accounting.

 

[bbbeard]

Isn't the correct term for a reversible adiabatic process isentropic?

A reversible adiabatic process is isentropic. I wouldn't say that's the "correct" terminology, as though "reversible adiabatic process" were somehow incorrect.

 

[bbbeard]

Thank you Andrew Mason and bbbeard for going at it like learned gentlemen...

I've read a lot of Andrew's posts and he is quite knowledgeable and thorough. I was lucky enough to catch him in a slight error so I jumped right in... ;-)

 

[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 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

 

[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