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Describe the thermodynamic system of a wire.

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data

    http://i.imgur.com/2WKCR.jpg

    2. Relevant equations

    Potentially: L(final) = L[1 + tau/(YA) + alpha(T - To)]

    3. The attempt at a solution

    Frankly I have no idea where to begin.
     
  2. jcsd
  3. Mar 28, 2012 #2

    rude man

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    What is F?
     
  4. Mar 28, 2012 #3
    Hi! In my opinion the answer is the following: the work done on the wire in order to stretch the wire of [itex]\delta l[/itex] is [itex]\delta W=t \delta l[/itex] where t is the tension of the wire (notice that \delta W is positive if [itex]\delta l[/itex] is positive, opposite behaviour with respect to the pressure and analogous behaviour with respect to the surface tension); indeed, one should reasonably make a positive work in order to stretch the wire; the first law of the thermodynamics in this situation is
    [itex]\delta U= T \delta S+ t \delta l[/itex]

    Since F=U-TS (I mean with F the Helmoltz free energy) i obtain

    [itex]\delta F=- S \delta T+ t \delta l[/itex]


    and so


    [itex]\frac{\partial F}{\partial l}|_T=t[/itex]

    [itex]\frac{\partial F}{\partial T}|_l=-S[/itex]
     
  5. Mar 28, 2012 #4

    rude man

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    Well, you left out quite a few steps but it's right.

    BTW my textbook uses "A" instead of "F" for the Helmholz function. And, if I may suggest, don't use t for temperature. Use T. t is almost always time.
     
  6. Mar 28, 2012 #5
    t is the tension and T is the temperature :)

    Anyway, can you tell me please what are the steps I left out? Thanks,
    Francesco
     
  7. Mar 28, 2012 #6

    rude man

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    Sorry, that got by me completely.

    You gave an equation, whence did it come? How about starting from the 1st law?
     
  8. Mar 28, 2012 #7
    I'm sorry, but I think that I don't understand: let me go into the details of the derivation, also covering well-known thermodynamics result; is the equation you are talking about
    [itex]\delta U=T\delta S+t\delta l[/itex] ?

    If so, then I think that this is just the first law for the system we are talking about (with [itex]\delta[/itex] I mean an infinitesimal change; well, to be rigorous, an infinitesimal reversible change); for every thermodynamic system we are supposed to have a function of state U (the internal energy) and to be able to define an ordered work ([itex]t\delta l[/itex]) and a disordered work ([itex]T\delta S[/itex]); then, by definition, I define
    [itex]F=U-TS[/itex] and then
    [itex]\delta F=\delta U-(\delta T)S-T(\delta S)=-S\delta T+t\delta l [/itex].

    Did I understand what you have asked or did you mean something else?
     
  9. Mar 28, 2012 #8

    rude man

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    [itex]\delta F=\delta U-(\delta T)S-T(\delta S)=-S\delta T+t\delta l [/itex].

    Maybe I'm being picky but
    dU - SdT -TdS = -SdT + tdL was not obvious to me. Admittedly I did not identify 'orderd' and 'disorderd' work as tdL and TdS respectively. In fact I don't remember coming across those terms at all except that certainly they make sense, qualitatively at least.

    You can do without that concept by
    dU = dQ + dW 1st law, dw > 0 here.
    dU = TdS + dW
    dW = tdL
    so dU = TdS + tdL
    dF = dU - TdS - SdT
    dF = TdS + tdL - TdS - SdT
    dF = tdL - SdT
    and the partials fall out as you indicated.
     
  10. Mar 29, 2012 #9
    Hello.

    dQ=TdS for every thermodynamic system that undergoes an infintesimal reversible transformation, as it is well known form basic thermodynamics.

    dW>0 doesn't make sense by itself; you should specify if dl is greater or smaller than 0.

    dW=tdl is analogous to -pdV in a usual gas system; if you want to build a simple (the simplest?) model where you can see this, you can imagine the wire as composed by N+1 very small masses disposed on a row, each distant d from the next ones. The fact that the wire has a tension t means that every particle A exerts on the next one B a force t in the direction B "with the arrow pointing from B to A". we want to stretch this wire of a length dl; then, we can move the particle identified by the integer n of the following length : dl((n-1)/N). You can now easily conclude.


    take two generic functions f(x,y), g(x,y) (sufficiently regular): then, by some mathematical analysis course,
    [itex]d(f g)=g(\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy)+f(\frac{\partial g}{\partial x}dx+\frac{\partial g}{\partial y}dy)[/itex]

    then we obtain the wantedd result if x=S, y=T, f(S,T)=S, g(S,T)=T.

    dU - SdT -TdS =TdS+tdl - SdT -TdS=-SdT + tdL

    Best,
    Francesco
     
    Last edited: Mar 29, 2012
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