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Homework Help: Why no Vdp?

  1. May 17, 2007 #1
    1. The problem statement, all variables and given/known data
    The books often always present pdV but never Vdp. Why is that? surely it is present in expressions like dE or dG where G is the grand potential. Is it because we don't treat p as a variable? If so why not? For each of these function we can only have 3 variables? So having V and p both as variables would be a bit redundant? If so why?
    Last edited: May 17, 2007
  2. jcsd
  3. May 17, 2007 #2
    In a grand canonical ensemble is pressure constant for any system? Hence dp=0? The grand potential dosen't appear to have dp in it.
  4. May 18, 2007 #3
    Is it also because P is a state function so dP can always be expressed as other variables just like dU can. However temperture is also a state function? but it is treated as a variable.
  5. May 19, 2007 #4

    Andrew Mason

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    They mention VdP. VdP is related to the change in internal energy of the gas at constant volume. PdV is the work done by/on the gas. Together, VdP + PdV = nRdT. Since nCvdT is the heat flow at constant volume and nCpdT is the heat flow at constant pressure, and Cp-Cv = R, VdP + PdV = nRdT = n(Cp-Cv)dT

  6. May 19, 2007 #5
    Why don't they have Vdp in the expression of the grand potential?

    G = grand potential = -pV

    So dG = -pdV - Vdp = -SdT -pdV - Ndu

    So -Vdp = -SdT - Ndu?

    or Vdp = SdT + Ndu which must be true by defintion would it? It's more convineint to use the right hand side instead of the left hand side?
  7. May 19, 2007 #6

    Andrew Mason

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    dQ = TdS
    dU = nCvdT
    dW = PdV

    So the first law dQ = dU + dW can be written:

    (1) TdS = nCvdT + PdV

    Since d(PV) = VdP + PdV = d(nRT) = nRdT,

    PdV = nRdT - VdP

    So substituting this for PdV in (1):

    (2) TdS = nCvdT + nRdT - VdP

    (3) VdP = nCvdT + nRdT - TdS = nCpdT - TdS = nCpdT - dQ

    So the first law can be written:

    (4) dQ = nCpdT - VdP

  8. May 19, 2007 #7
    You haven't disproved Vdp = SdT + Ndu have you?
  9. May 19, 2007 #8

    Andrew Mason

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    I am not sure what you mean by Ndu. I think it refers to added molecules.

    From the Euler equation:

    U = TS - PV

    dU = d(TS) - d(PV) = (TdS + SdT) - (VdP + Pdv)

    But the first law states that dU = dQ - PdV. Since dQ = TdS:

    dU = TdS - PdV so

    SdT - VdP = 0

  10. May 19, 2007 #9
    u stands for the chemical potential constant. If you have dU=dQ-Pdv you have not accounted for the chemical potential of particles entering and leaving the system. If you did then dU=TdS-pdV+udN. So

    We have
    U = TS - PV + uN

    dU = d(TS) - d(PV) + d(uN)= (TdS + SdT) - (VdP + Pdv) + (Ndu + udN)

    We have

    So SdT - Vdp + Ndu = 0
    hence Vdp = SdT + Ndu

    Although why is U = TS - PV + uN?

    Or even why is U = TS - PV when not accounting for chemical potential?

    Is it more the fact that we first know the expressions for dU then work out U from those expressions?
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