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Gibbs free energy and enthalpy relationship

  1. Dec 28, 2013 #1
    *to moderators: i apologize about the double post, but i didn't know how to get rid of the other post and i wanted a get an answer asap and no one was answering the other post so i reposted here in the appropriate section*

    enthalpy is the energy available if a system with some defined volume were to be annihilated and have the atmosphere collapse inward where the atmospheric pressure remains constant. hence..

    dH= dU + PdV (i like sticking to differentials)


    Gibbs: this one takes place at constant temperature and pressure. I don't understand why we do the following for the gibbs:

    G=U+PV-TS ----> dG=dU+d(PV)-d(TS) ----> dG=dU+PdV+VdP-TdS-SdT

    Question 1: the cases for both the enthalpy and gibbs free energy involve the environment being at constant pressure hence the PV term right? (if i'm wrong please correct me) so why do i use PdV for one of them (enthalpy case) and PdV+VdP for the other (the gibbs case)?


    Question 2: VdP looks like the case where we are shoving/compressing air into a fixed volume (perhaps a container or something...), so i was wondering, could this term could be somehow combined with the μdN term? it doesn't seem necessary to keep both terms around since they seem like different expressions for the same thing.
     
  2. jcsd
  3. Dec 30, 2013 #2

    stevendaryl

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    The differential form of thermodynamics equations are:

    [itex]dU = T dS - P dV + \mu dN[/itex]

    [itex]H = U + PV \Rightarrow dH = T dS + V dP + \mu dN[/itex]

    [itex]G = H - S T \Rightarrow dG = -S dT + V dP + \mu dN[/itex]

    The significance of these various kinds of energy can be understood a little better by considering the system interacting with a much larger system in limited ways.

    Imagine that you have some gas in an insulated container whose volume is allowed to change. For example, a piston, which is a cylindrical where the top is allowed to slide up and down. This piston is not allowed to exchange heat or particles with the environment, but it can change its volume. If the volume of the piston expands, then the volume in the room contracts, so that the total volume is constant. In that case, the piston will expand (or contract) until the pressures inside and outside the piston are equal. Minimizing the total energy of system + room is equivalent to minimizing the enthalpy [itex]H[/itex] of the system in the piston alone.

    Now change the scenario so that the piston is not insulated. The system inside the piston is allowed to exchange heat with the room. In that case, the system inside the piston will absorb or release heat until the temperatures inside and outside the piston are the same. Minimizing the total energy of system + room in this case is equivalent to minimizing the Gibbs free energy [itex]G[/itex] of the system in the piston alone.
     
  4. Dec 31, 2013 #3
    if the gibbs free energy takes place in a const. pressure and const. temp environment, then wouldn't the
    $$-SdT +VdP$$ terms disappear? if not, then why not?
     
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