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Difference between enthalpy, Helmholtz free energy,& Gibbs free energy

  1. May 11, 2013 #1
    I am study chapter 5 of An Introduction to Thermal Physics by Schroeder and I am having trouble understanding his explanation of the differences between enthaply, Helmholtz free energy, and Gibbs free energy.

    Schroeder defines enthalpy of a system as its energy plus the work needed to make room for it in an environment with constant pressure P : H = U + PV.

    Then he explains that if the environment is one of constant temperature, the system can extract heat from its environment for free, so all that is needed to create the system from nothing is any additional work needed. (Helmholtz free energy, F is defined by F = U - TS).

    My main question is why can the system extract heat for free from an environment at constant temperature. Can someone explain this in more detail?

    Also, helmholtz free energy, unlike enthalpy, does not account for the work needed to make room for the system in the environment? Is this because Helmholtz free energy is measured when the environment is at constant temperature and not at constant pressure?

    So overall, is it true that the main difference between H, F, and Gibbs free energy is that

    Enthalpy is the energy needed to create a system out of nothing in an environment at constant pressure only.

    Helmholtz free energy is the energy needed to create a system out of nothing in an environment at constant temperature only.

    Gibbs free energy is the energy needed to create a system out of nothing in an environment at both constant pressure and constant temperature.

    Am I understanding this correctly?

  2. jcsd
  3. May 12, 2013 #2


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    Everything is based on the 1st and 2nd law of thermodynamics. For reversible processes you can write it in the form
    [tex]\mathrm{d} U=T \mathrm{d} S-P \mathrm{d} V,[/tex]
    where I assumed the canonical ensemble with a fixed number of particles to keep things simple.

    According to this equation the work done on the gas, [itex]-P \mathrm{d} V[/itex] equals the change in internal energy, [itex]\mathrm{d} U[/itex] for adiabatic changes, i.e., for [itex]\mathrm{d} S=0[/itex].

    If you want convenient thermodynamic potentials for other circumstances, it's usefull to perform Legendre transformations from the above total differential. While the natural independent variables for [itex]U[/itex] are [itex]S[/itex] and [itex]V[/itex] and because of
    [tex]T=\left (\frac{\partial U}{\partial S} \right)_{V}, \quad P=-\left (\frac{\partial U}{\partial t} \right )_{S}[/tex]
    to get a potential with the natural independent variables to be, e.g., [itex]T[/itex] and [itex]V[/itex] you perform the Legendre transformation to the Helmholtz free energy
    [tex]F=U-T S.[/tex]
    Indeed taking the total derivative of this, using again the above law for [itex]\mathrm{d} U[/itex], you indeed get
    [tex]\mathrm{d} F=\mathrm{d}U-T \mathrm{d} S-S \mathrm{d} T=-S \mathrm{d} T-P \mathrm{d} V.[/tex]
    Thus the change of the free energy is the work done on the gas for isothermal processes, [itex]\mathrm{d} T=0[/itex]. You have the relations
    [tex]S=-\left (\frac{\partial F}{\partial T} \right)_{V}, \quad P=-\left (\frac{\partial F}{\partial V} \right )_{T}.[/tex]
    This you can do for any pair of independent variables you like. This leads to the different potentials like enthalpy, Gibbs free energy and other potentials. Which one to use depends on the process you consider. All potentials are equivalent, but for a given process one might be more convenient to handle than the other.
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