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Entropy Qrev/T?

  1. Jun 16, 2008 #1
    Hey can neone tell me why this is so? I have not got the ans in any book i read...
    Thnx 4 replying
  2. jcsd
  3. Jun 16, 2008 #2


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    Hi aniketp,

    One way of writing the First Law is


    which means that the differential change in energy equals the heat transferred to a system plus the work done on a system. But another way is


    where the work is expressed as a generalized force (pressure) times a generalized displacement (change in volume). The heat transfer term is expressed in an analogous way: a generalized force (temperature) times a generalized displacement (change in entropy).

    If you haven't heard the terms generalized force and generalized displacement before, they're just ways to classify variables. Generalized forces are intensive, and they drive processes; a change in temperature drives heat flow, and a change in pressure drives mass flow. Generalized displacements, which are extensive, are the "stuff" that is transferred: entropy, volume.

    [itex]S=Q_\mathrm{rev}/T[/itex] arises because at constant volume and temperature, and if the process is reversible (no excess entropy generated) we can integrate [itex]q=T\,dS[/itex] to give [itex]Q_\mathrm{rev}=T\Delta S[/itex].

    Does this make sense?
  4. Jun 16, 2008 #3
    Hey mapes,
    But isn't dU=TdS-PdV actually derived from dS=dQ/T?
  5. Jun 17, 2008 #4


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    Entropy is defined as S = k*log(omega)
    Temperature is defined as 1/T = dS/dU at constant V
    Pressure is given by P = T * dS/dV at constant U

    From the previous two relationships it follows that
    dU = TdS - PdV, and S = Q/T is a special case.

    For an intuitive discussion see, for example, An Introduction to Thermal Physics by Schroeder, sections 2&3.
  6. Jun 17, 2008 #5


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    I should mention that these expressions for T and P are derived (in Schroeder) by consideration of equilibrium conditions.
  7. Jun 17, 2008 #6
    Hi nicksauce, thnx 4 the help. So T & P are actually defined on the basis of entropy....
    But in k*log(omega) what do 'k' n 'omega represent?
  8. Jun 17, 2008 #7


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    k is the boltzmann constant, and omega is the "multiplicity" of the system, or the number of possible microscopic configurations the system can have.

    Simple example:
    Say I have a system of 3 "oscillators" (that are quantized), with a total energy of "q" units. Say q = 0, then there is only one possible arrangement (Omega=1). Say q = 1, then there are three possible arrangements (Omega = 2). Say q =2, then there are 6 possible arrangements, ie (2,0,0) three times, and (1,1,0) 3 times. So Omega = 6. Say q = 3, then there are 10 possible arrangements, (3,0,0) 3 times, (2,1,0) 6 times, and (1,1,1) once. So Omega = 10.
  9. Jun 17, 2008 #8
    Oh ,ok. got it now...thanks once more !
  10. Jun 17, 2008 #9
    http://en.wikipedia.org/wiki/Fundamental_thermodynamic_relation" [Broken]
    Last edited by a moderator: May 3, 2017
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