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Reversible, equilibrium and quasi-static

  1. Jan 8, 2009 #1

    KFC

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    Hi there,
    I found it is too much concepts in my head after reading more and more about thermodynamics. In the very beginning of the text, it emphize that in many cases (at lease in beginning level), we only deal with the equilibrium state in thermodynamic system. To make sure every stage of the system is in equilibrium, we assume the process it undergoes is quasi-static, right?

    Later on, in the chapter about entropy, it read the definition [tex]dS = \frac{dQ}{T}[/tex] to calculate the entropy can be only applied to reversible system. Now, involving entropy, all of these concepts are quite confused. What's the basic connection between these three concepts? For equilibrium, as mentioned before, we have to make sure the process is quasi-static so each stage is in equilibrium. so,

    1) if the process is not quasi-static so that the system is not in equilibrium, does it mean the system is not reversible?

    2) if the system is reversible, can I say it must be in equilibrium?

    3) if the process is not in equilibrium, can I still apply [tex]dS = \frac{dQ}{T}[/tex] to calculate the change entropy?

    4) Reversibility is quite confusing. For Carnot cycle, we consider it is reversible because after a cycle it goes back to the inital state exactly. Here we assign 'reversibility' to the cycle. But can I say in each stage (for example in the first isotherm expansion), it is reversible?

    It is not clear how to tell if a system is reversible or not. For instance, if I start from a initial state and after some process I go back to the inital state again. Can I say this process is reversible? Or I have to add an additional conditon about the total heat change and total work change is zero? For free expansion, all gas will expand to a larger space spontaneously. Obviously, this is not a reversible process because the gas will not *spontaneously* going back to its inital state. But if I find some way to compress the gas so that it complete going back to its inital state, and I INCLUDE the gas and that compress process as a whole system, can I say this is reversible?
     
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  3. Jan 8, 2009 #2

    Mapes

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    I found this confusing too when learning thermo. In particular, I didn't understand how an adiabatic, quasi-static process could exist, since adiabatic processes are often conceived to occur quickly to limit heat transfer to the surrounding environment and quasi-static processes are conceived to occur slowly to eliminate gradients. Eventually I realized that these processes assume a happy medium--not too slow, not too fast. In reality, there is no such thing as a truly reversible process, and all thermo models are somewhat inaccurate to allow ease of conception and computation.

    Your questions:

    1) Right, not quasi-static = irreversible, because of the appearance of gradients in material, temperature, momentum, charge, etc.

    2) Yes, although we idealize some processes as being arbitrarily close to reversible.

    3) There will be some error in taking [itex]dS=\frac{dQ}{T}[/itex] because a state of non-equilibrium will produce entropy as the system tends towards equilibrium. Again, though, you might conceive of a situation in which the error is arbitrarily small.

    4) Yes, each stage is reversible because no stage generates entropy. Reversibility implies no gradients and no entropy creation.
     
  4. Jan 9, 2009 #3

    atyy

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    3) S is a state function, a property of the system independent of its history (dS is an exact differential; Q is history dependent, dQ is not an exact differential). If the process is not reversible, the entropy change can be calculated using dS=dQrev/T using a different reversible process that has the same initial and final states.
     
  5. Jan 12, 2009 #4
    A "thermodynamic process" is an ordered sequence of states of a system, which normally starts and ends in equilibrium states.

    If all the intermediate states are also equilibrium states (or "close enough"), the process is called quasistatic. In a quasistatic process, the quasistatic heat flux is dQ = TdS and the quasistatic work is dW = pdV.

    A process is called adiathermal if there is no heat flux from the surroundings during it. A process is called adiabatic if it is both adiathermal and also \Delta S = 0.

    A process can be called "reversible" only if the system under consideration is the whole universe, and Delta S = 0.

    A Carnot Cycle is called reversible because it is an ordered sequence of states of the universe ( two heat baths, a work reservoir and a piston) which has no change in entropy.

    The word "reversible" applies to processes, not systems. The reversibility of the carnot cycle is not the fact that it is a cycle which gets back to its starting point, it is the point that you can run it in reverse and use it as a heat pump rather than a heat engine.

    The change of a state function during a process only depends on the end points (definition of state function). This is why it is is possible to use a quasistatic process to calculate the change in a state function, even when the real process under consideration is not quasistatic, eg a Joule Expansion.
     
  6. Jan 12, 2009 #5

    Mapes

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    Just for completeness, I want to note that most texts define "adiabatic" to correspond to an absence of heat transfer through the system boundaries (but not necessarily systems undergoing reversible processes) and use "reversible adiabatic" (= "isentropic") for systems with [itex]\Delta S=0[/itex].
     
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