Reversible paths minimize the dissipation of mechanical energy to thermal energy, and maximize the ability of temperature differences to be converted into mechanical energy. In reversible paths, the pressure exerted by the surroundings at the interface with the system is only slightly higher of lower than the pressure throughout the system, and the temperature at the interface with the surroundings is only slightly higher or lower than the temperature throughout the system. This situation is maintained over the entire path from the initial to the final equilibrium state of the system.Thanks Chester.
Yes. I really did find that clear.
Which is not to say I understood it...
What is special about the reversible paths?
They integrate to different values < ΔS. The equal sign does not apply to irreversible paths. They are all less.Are all the other paths, the non-reversible ones, the same, or do some integrate to different values <= DeltaS than others?
Hi Jimster. You ask great questions.Thanks Chet, I hope it's okay if I keep asking you questions. It really is my favorite way to learn, and I can get enough of the second law, and I'm sure I will learn something - not the least of which will be precision of terms.
In the the case of a gas in a cylinder with a piston (or damper) why does the amount of dissipation vary with the amount of force per unit time? what does the time rate of force have to who with the efficiency of conversion to mechanical energy? Why does the difference at any given time between the system and surroundings, dictate the reversibility, as opposed to say the amount of energy transferred altogether?
I'll take a look and see if I can contribute. There are lots of pages and lots of posts, so it may take me a while to come up to speed. No guarantees.Thanks Chet. Your explanation of the "Clausius Inequality" and you answer on the difference between reversible and non-reversible paths were helpful and lucid, and It means a lot to know they are at least sensible questions.
I don't suppose @techmologist and I could get your help reading an old Galvin Crooks paper from 1999 on the "generalized fluctuation theorem"? We've got a thread going in the cosmology forum. PeterDonis has been helping us (humoring us more like it). It's under @techmologists question "why are there heat engines?" It's pretty rambly at this point so I would be more than happy to restart it focusing it back on Crooks' paper and handful of equations, and drill in with your guidance.
Thanks INFO_MAN. It's nice to be appreciated.Hello Chestermiller.
Entropy and the Second Law of Thermodynamics is not exactly an intuitive concept. While I think your article is basically a good one, it is obviously somewhat limited in scope, and my only critique is that you did not cover some of the most important aspects of entropy.
This is an example of one of those instances I was referring to in which the constraints on the equations is not spelled out clearly enough, and, as a result, confusion can ensue. The situation you are referring to here with the inequality (ΔS > 0) and equality (ΔS = 0) applies to the combination of the system and the surroundings, and not just to a closed system. Without this qualification, the student might get the idea that for a closed system, ΔS≥0 always, which is, of course, not the case.I agree that most people have a very hard time grasping entropy and the second law of thermodynamics. But I am not sure I understand why your article keeps referring to reversible processes and adiabatic idealizations. In natural systems, the entropy production rate of every process is always positive (ΔS > 0) or zero (ΔS = 0). But only idealized adiabatic (perfectly insulated) and isentropic (frictionless, non-viscous, pressure-volume work only) processes actually have an entropy production rate of zero. Heat is produced, but not entropy. In nature, this ideal can only be an approximation, because it requires an infinite amount of time and no dissipation.
I'm sorry that impression came through to you because that was not my intention. I feel that it is very important for students to understand the distinction between real irreversible processes paths and ideal reversible process paths. Irreversible process paths are what really happens. But reversible process paths are what we need to use to get the change in entropy for a real irreversible process path.You hardly mention irreversible processes. An irreversible process degrades the performance of a thermodynamic system, and results in entropy production. Thus, irreversible processes have an entropy production rate greater than zero (ΔS > 0), and that is really what the second law is all about (beyond the second law analysis of machines or devices). Every naturally occurring process, whether adiabatic or not, is irreversible (ΔS > 0), since friction and viscosity are always present.
This equation applies to the more general case of an open system for which mass is entering and exiting, and I was trying to keep things simple by restricting the discussion to closed systems. Also, entropy generation can be learned by the struggling students at a later stage.Here is my favorite example of an irreversible thermodynamic process, the Entropy Rate Balance Equation for Control Volumes:
As I said above, I was trying to limit the scope exclusively to what the beginning students needed to understand in order to do their homework.And here are are a couple of other important things you did not mention about entropy:
1) Entropy is a measure of molecular disorder in a system. According to Kelvin, a pure substance at absolute zero temperature is in perfect order, and its entropy is zero. This is the less commonly known Third Law of Thermodynamics.
2) "A system will select the path or assemblage of paths out of available paths that minimizes the potential or maximizes the entropy at the fastest rate given the constraints." This is known as the Law of Maximum Entropy Production. "The Law of Maximum Entropy Production thus has deep implications for evolutionary theory, culture theory, macroeconomics, human globalization, and more generally the time-dependent development of the Earth as a ecological planetary system as a whole." http://www.lawofmaximumentropyproduction.com/
Suppose you compress a gas isothermally and reversibly in a closed system. To hold the temperature constant, do you have to add heat or remove heat? After you compress the gas to a smaller volume at the same temperature, are the number of quantum states available to it greater or fewer?That was great Chet. It helps to know the purpose and scope. Hey, can you explain to a confused student why the change in entropy in a closed sytem is not always greater than or equal to 0? I think I know (Poincare' recurrence?) but I also think I'm probably wrong.
No sir, I was not clear on that precise difference of terms! Now I am. I believe you need to remove heat. Hmm, the quantum states. That one really makes me think, with great confusion, which is not good, since the answer should probably be obvious. In the closed sytem that has been isothermically compressed (heat removed), I would say the number of states is fewer? But it 's basically a guess. I don't know how to decompose that question, with any confidence. I think I know something about the parts, but probably have way too many questions and misconceptions tangled up in it. Please do illuminate!Suppose you compress a gas isothermally and reversibly in a closed system. To hold the temperature constant, do you have to add heat or remove heat? After you compress the gas to a smaller volume at the same temperature, are the number of quantum states available to it greater or fewer?
You are aware that, in thermodynamics, there is a difference between a closed system and an isolated system, correct?
Both your answers are correct. You remove heat from the system in an isothermal reversible compression, so ΔS < 0 (q is negative). The number of states available to the system is fewer, so, by that criterion also, ΔS < 0.No sir, I was not clear on that precise difference of terms! Now I am. I believe you need to remove heat. Hmm, the quantum states. That one really makes me think, with great confusion, which is not good, since the answer should probably be obvious. In the closed sytem that has been isothermically compressed (heat removed), I would say the number of states is fewer? But it 's basically a guess. I don't know how to decompose that question, with any confidence. I think I know something about the parts, but probably have way too many questions and misconceptions tangled up in it. Please do illuminate!
I say fewer because the volume is less, and so the available "locations" are reduced. But this does not seem very satisfactory, right, or clear.
What you're missing is that, at the interface, the local temperature of the system matches the temperature of the surroundings. There is no discontinuity in temperature (or in force per unit area) at the interface. However, in an irreversible process, the temperature within the system varies with distance from the interface.I find your "temperature at the interface with the surroundings" confusing in that to me it implies an average temperature between the system and the surroundings at that point. Would it be more clear to say "temperature of the surroundings at the interface" or am I missing something?
Yes. With an irreversible process, there is a temperature difference between the average temperature in the system and the temperature at the interface. However, at the very interface, the local system temperature matches the local surroundings temperature.So you're saying there is a temperature gradient between the "bulk" system and the interface
Not necessarily. I've tried to get us focused primarily on the system. I'm assuming that we are not concerning ourselves with the details of what is happening within the surroundings, except at the interface, where we are assuming that either the heat flux or the temperature is specified. (Of course, more complicated boundary conditions can also be imposed, and are included within the framework of our methodology). Thus, the "boundary conditions" for work and heat flow on the system are applied at the interface., but no temperature gradient between the "bulk" surroundings and the interface?
I realize why the quantum states question confuses me. Probably it is an issue of specific terms.Both your answers are correct. You remove heat from the system in an isothermal reversible compression, so ΔS < 0 (q is negative). The number of states available to the system is fewer, so, by that criterion also, ΔS < 0.
A closed system is one that cannot exchange mass with its surroundings, but it can exchange heat and mechanical energy (work W). An isolated system is one that can exchange neither mass, heat, nor work.
My goal was to emphasize the classical approach to entropy in my development, and to generally skip the statistical thermodynamic perspective.I realize why the quantum states question confuses me. Probably it is an issue of specific terms.
If I picture the piston and cylinder made of graph paper cells, containing 1's and 0's, with the volume of the uncompressed cylinder as an area of zeros 0's mixed with 1's (representing the uncompressed gas in the cylinder) this area then surrounded by some more 1's representing the boundaries of the cylinder, including the piston. If I then compress the gas, by changing some of the cylinder volume cells to 1's, I haven't changed the number of states in the system (the graph paper hasn't shrunk or lost cells) I have just added information assigning some of the cells of the cylinder volume with specific values. So I guess by "available QM states" you mean those that are uncertain, or "free" to be randomly set to 1or 0.
Maybe it's a bad metaphor, because I get even more confused when I think that to expand the "cylinder" I still have to add information, changing a set of "fixed cells" to be "free".
It is an integral sign. Apparently, you haven't had calculus yet. You are not going to be able to understand and apply much of thermodynamics without the basic tool of calculus.What does the ∫ symbol in the thermodynamics equations mean?
Brackets are a kind of parenthesis.(i know what + - * / mean :) Oh and brackets)