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Thermodynamic potential question,help!

  1. Mar 1, 2007 #1
    i have studied my lecture notes back to front and looked high and low for a clue how to do this question but i tried in a) constructing an expression for u and came up with DU= +TDS -PDV - FDL BAUT I COULD BE MILES OUT. could anyone help.

    thanks

    here it goes:


    A piece of rubber of length L is subject to work by a hydrostatic pressure P and also
    by a tensional force F.

    a)Construct the expression for dU.
    [Hint: the usual expression will have an extra term added to it corresponding
    to the increment of work performed by F

    b)Generate (i.e. invent new) thermodynamic potentials that have as proper vari-
    ables (S; V; F); (S; P; F) and (T; P; F).
    [Note: the potential R has (S; V; F) as proper variables if dR = xdS+ydV +zdF
    where x; y; z are to be found.]
     
  2. jcsd
  3. Mar 1, 2007 #2
    you can't think of three thermo equations with variables (S; V; F); (S; P; F) and (T; P; F)?
     
  4. Mar 1, 2007 #3
    for example could it be DR =TDS + PDV + LDF FOR (SVF), AND FOR (SPF) DN = TDS + VDP + LDF ? I am just not sure what question means thats all, can anyone help?
     
  5. Mar 1, 2007 #4
    i suppose they want you to derive new (invent new?) potentials, but I would go ahead and use internal energy change and say
    dU = dQ - dW
    dQ = TdS
    dW = PdV +ldf
    now dU = TdS - (PdV + LdF)

    stick to something basic and use the four potentials (internal, hemholtz, gibbs, and enthalpy) as a basis.
     
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