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Justification of generalisation from reversible processes to all procesess

  1. Jan 26, 2012 #1
    Hi all,

    I've been having some difficulty understanding the derivation of the Fundamental Theorem of Thermodynamics, [itex]dU=T \ dS-P \ dV[/itex].

    The derivation, which can be found at Wikipedia (http://en.wikipedia.org/wiki/Fundamental_thermodynamic_relation) first starts with the universal First Law of Thermodynamics,

    [itex]dU=dQ-dW[/itex]

    and then supposes that the situation is reversible, allowing the substitution of [itex]T \ dS[/itex] and [itex] P \ dV [/itex] for [itex]dQ[/itex] and [itex]dU[/itex] respectively. I stress again these individual substitutions are only valid for the reversible case.

    However, at this point, both Understanding Thermodynamics by Van Ness and Wikipedia do something I don't quite understand. They say that, even through we derived the equation for the reversible case, because this equation only contains functions of state (or 'properties of the system' in Van Ross' terms) this equation must apply for all processes, both reversible and irreversible.

    I do not understand this at all. These substitutions were only valid for reversible processes and, as far as I can see, from the moment you make these substitutions and implicitly assume that the process is reversible, the equation can only be applied (with certainty) to the reversible case, irrespective of its final form.

    As a 'counter-example' to this logic, consider the equation [itex]dS=0[/itex]. This is applicable for all adiabatic, reversible processes - any reversible processes that take place within an 'insulated' environment. Now why can't we apply this same logic and say that, because this equation only contains 'functions of state' (just [itex]S[/itex]) then, logically, all processes, both reversible and irreversible, which occur in this insulated environment must obey this 'law'. Because of course this final conclusion would be false.

    I would appreciate it immensely if anyone could lead me in the right direction, thanks!
     
  2. jcsd
  3. Jan 27, 2012 #2
    The thermodynamic relation is [itex]TdS \geq dU + PdV[/itex] but is only valid for closed simple system.

    The expression [itex]TdS = dU + PdV[/itex] is only valid for reversible processes, as you correctly notice, and closed simple system. It cannot be "fundamental".
     
    Last edited: Jan 27, 2012
  4. Jan 27, 2012 #3
    Thanks for your help! What you say makes a lot more sense. But given that, what does wikipedia mean when they say:

    "...This equation has been derived in the case of reversible changes. However, since U, S, and V are thermodynamic functions of state, the above relation holds also for non-reversible changes."

    and similarly, 'Understanding Thermodynamics' when it says

    'now we derived this equation for reversible processes, but once derived we see that it contains just properties of the system, and so it must not depend on the kind of process considered. What we have really done is to derive an equation for a special case and then conclude it must be general'

    Thanks
     
  5. Jan 28, 2012 #4
    For irreversible process between an initial equilibrium state (0) and a final equilibrium state (F) the equation is [itex]TdS \gt dU + PdV[/itex]. Now if you substitute the real irreversible process by an idealized reversible process [itex]T'dS' = dU' + P'dV'[/itex] and integrate between the same initial (0) and final state (F) you can obtain the same variation of entropy [itex]\Delta S=S_f - S_0[/itex] because the initial and final state are the same. But the two processes are different.

    Another way to see this. Think of a real process for which [itex]dU = δQ + δW[/itex] so that [itex]dU \gt δQ[/itex]. You can imagine an alternative process [itex]dU = δQ'[/itex] and obtain the change in internal energy between an initial (0) and a final state (F), since U is state function, but evidently [itex]δQ'[/itex] is not the heat [itex]δQ[/itex] in the original process.

    This kind of tricks (where a real process is substituted by another process) are common in classical thermodynamics where you are only interested in initial and final equilibrium states. Those tricks cannot be used in modern thermodynamics dealing with nonequilibrium states and with the real processes.
     
    Last edited: Jan 28, 2012
  6. Jan 29, 2012 #5
    Thanks, I think that finally all makes sense, it all seems so simple now :) . It all works because the equation is only referring to changes of state functions within the system. Irrespective of what process is used to get from A to B, the changes in entropy, etc. will be the same. However, for the irreversible process, you will get more entropy generated in the surroundings, which we don't care about here.

    I also see why my 'counterexample' is wrong. In an open system, it is possible to construct a reversible process between any two equilibrium states (0) and (F), so any irreversible process in the system can be modeled as a reversible process between (0) and (F), giving the desired ΔS, ΔV, etc. However, for a closed system, if ΔS > 0 between 2 points (0) and (F), then there is no reversible process that can be used to get between these 2 points, so we can't imagine a reversible process to justify the application of the reversible equation.

    I have to admit I struggled for a while to understand why [itex]T \ dS > dU + P \ dV[/itex] would ever hold for a process between 2 equilibrium states. Surely, as [itex]\Delta T, \Delta P ,\Delta S, \Delta U[/itex] and [itex]\Delta V [/itex] are all going to be the same for an imagined reversible process between these two states, it is contradictory to have two process between the same two states, one obeying [itex]T \ dS > dU + P \ dV[/itex] and the other obeying [itex]T \ dS = dU + P \ dV[/itex]. I think the solution to this comes from the fact that in the irreversible process excess heat is inevitably leaked into the surroundings, so the heat (which is [itex]T \ dS[/itex] in the reversible case) required for the internal energy change and work done by the system ([itex]dU + P \ dV[/itex]) is inevitably greater for the irreversible case than for the reversible case (so [itex]dQ_{real}=T_{real} \ dS_{real}>T_{reversible} \ dS_{reversible}=dU+P \ dV[/itex])

    Again, thanks, you've cleared my mind up significantly :)
     
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