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

[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