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Gibbs and Helmholtz equations for thermodynamic processes

  1. Dec 12, 2011 #1
    For a thermodynamic process, what equations would be used to find the change in Gibbs and Helmholtz free energy when:
    a.)The process is adiabatic
    b.)The process is isothermic
    c.)The process is at constant volume
    d.)The process is at constant pressure

    I know ΔG=ΔH-TΔS and ΔA=ΔU-TΔS but do these equations apply to all four processes?
  2. jcsd
  3. Dec 13, 2011 #2
    I think you need to get into the thermodynamic partial derivative definition of these processes, i.e.

    Gibbs Energy:
    dG = -SdT + VdP
    Helmholtz Energy:
    dA = - SdT - PdV

    Now you set the appropriate terms to zero.

    adiabatic: is it a reversible process?
    isothermic: dT=0
    isobaric (const pressure): dP=0
    const vol: dV=0
  4. Dec 13, 2011 #3


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    Does the process include chemical reactions?
  5. Dec 13, 2011 #4
    Basically, I'm asking if you were given a PV graph similar to this:

    http://www.websters-online-dictionary.org/images/wiki/wikipedia/commons/thumb/d/dc/Stirling_Cycle.png/200px-Stirling_Cycle.png [Broken]

    How would you find ΔG and ΔA for the processes 1 to 2, 2 to 3, 3 to 4 and 4 to 1?
    Last edited by a moderator: May 5, 2017
  6. Dec 13, 2011 #5
    For 1 to 2:
    ΔG = ΔH - TΔS
    ΔA = ΔU - TΔS


    so ΔA = ΔG=-TΔS; you know T but need to find ΔS. There are two derivations; either will work for you. I'm going fast so may have mixed up negative signs & numerators/denominators, you need to double check the math when you do it yourself to make sure nothing is wrong.

    Derivation 1
    ΔU = Q + W = 0 b/c ΔU = 0.
    Q = -W
    dW = -PdV
    dW= -(NRT/V)dV (plugged in ideal gas law)
    W = -NRTln(V2/V1)
    so Q = NRTln(V2/V1)
    ΔS = Q/T = -NRln(V2/V1)

    Derivation 2 (my preference)
    Use Maxwell's relations.
    (dS/dV) at const T = (dP/dT) at const V. Plug in ideal gas law.
    So dS/dV = NR/V; integrate to get ΔS = NR ln (V2/V1).
    Last edited: Dec 13, 2011
  7. Dec 13, 2011 #6
    For 2 to 3:
    Isochoric Cooling - Const Volume.

    ΔU = NCvΔT = Q + W.
    W = PΔV so W = 0.

    ΔU = Q = NCvΔT.
    Last edited: Dec 13, 2011
  8. Dec 14, 2011 #7


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    I was asking whether chemical reactions are taking place (which, as I learned from your answer, is not the case) because the formulas above refer to the change of G or A due to a chemical reaction taking place at constant T and p. The "Delta" is not simply an abbreviation for a difference here but
    [itex] \Delta X=\sum_i \nu_i \partial X/\partial n_i |_{T, P}[/itex] which the nu_i being the stochiometric coefficients of the reaction taking place. So this formula is little helpful when you consider a process which does not involve chemical reactions.
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