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Enthalpy vs Internal energy for first law

  1. Feb 1, 2010 #1
    U is internal energy, which is the sum of heat + work done.

    [tex]\Delta[/tex]U = q + w -----(1) where w = [tex]\Delta[/tex](PV)

    Enthalpy is the internal energy + work done

    [tex]\Delta[/tex]H = [tex]\Delta[/tex]U + P[tex]\Delta[/tex]V

    At constant pressure,

    [tex]\Delta[/tex]H = q - P[tex]\Delta[/tex]V + P[tex]\Delta[/tex]V
    [tex]\Delta[/tex]H = q[tex]_{p}[/tex]

    Where q[tex]_{p}[/tex] is used to denote heat energy under constant pressure.

    I'm confused about a few things here.

    1) What is the work that is done here?
    - P[tex]\Delta[/tex]V > What work is done?
    + P[tex]\Delta[/tex]V > What work is done?
    By canceling them out, does it mean that no work is done at all i.e no volume change?

    2) If there is a change in the number of moles of gas, + P[tex]\Delta[/tex]V is actually has an impact on the equation. I'm under the impression that [tex]\Delta[/tex]U consists of thermal energy only(i.e no PV work done) when we're talking about enthalpy.

    Similarly under constant volume, from equation 1,
    [tex]\Delta[/tex]U = q[tex]_{v}[/tex] because [tex]\Delta[/tex]V = 0, hence w = 0.

    Since [tex]\Delta[/tex]H = [tex]\Delta[/tex]U + [tex]\Delta[/tex](PV)
    and [tex]\Delta[/tex]V = 0,
    therefore [tex]\Delta[/tex]H = q[tex]_{v}[/tex]?

    1) My next question is why is there this [tex]\Delta[/tex](PV) term here that doesn't do anything at all?

    2) Like the previous case, I'm under the impression that[tex]\Delta[/tex]U = thermal energy when we're talking about [tex]\Delta[/tex]H = [tex]\Delta[/tex]U + [tex]\Delta[/tex](PV)?.

    Of course which is not true since [tex]\Delta[/tex]U = q + w. So what exactly is the difference between the w in [tex]\Delta[/tex]U = q + w and [tex]\Delta[/tex](PV) in enthalpy equation [tex]\Delta[/tex]H = [tex]\Delta[/tex]U + [tex]\Delta[/tex](PV)?
    Last edited: Feb 1, 2010
  2. jcsd
  3. Feb 1, 2010 #2


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    Enthalpy and other thermodynamic potentials are defined by carrying out a http://en.wikipedia.org/wiki/Legendre_transform" [Broken] on the Energy. This sort of thing always gives me a headache but let me try to clarify some points.

    Firstly there is the ugly (but necessary for brevity's sake?) convention of using the same symbol for both variables (quantities) and functions (relationships between quantities). This I think is the principle source of headaches in this sort of stuff.

    The (time averaged) internal energy U is the total kinetic energy of all the particles in the gas plus their internal chemical energies, plus also the potential energies due to particle-particle interactions of the gas. It does not however include potentials for gas-boundary interactions and so doesn't take into account the difference between allowing the gas to expand freely vs letting it push outward on a resisting boundary.

    It is better though to think of U as the function giving the energy as a function of the entropy S, the volume V, and the number of particles N. (if you have many particle types you have separate N's and chemical potentials for each.)

    E = U(S,V,N)

    The volume takes into account the inter-particle forces, e.g. if you had a gas of electrons their mutual repulsion as well as random collisions would contribute to the energy. So in a sense U is the total energy if the particles were allowed suddenly to expand in free space as if you instantly removed walls. As they spread out far enough to ignore interactions you can then add up the final kinetic energies of each molecule plus any internal stored energy you want to account for.

    Now given these independent variables U=U(S,V,N) we can take the differential:
    [tex]d U = T d S - pd V + \mu d N[/tex]
    (the - p can be thought of as the external pressure holding in the gas so increasing volume does negative work on system).
    This defines T, p, and chemical potentials in terms of derivatives of the internal energy function as a function of (S,V,N). Note in this case T=T(S,V,N), and p = p(S,V,N) and mu = mu(S,V,N).

    Note that the Volume and particle number are extensive quantities and entropy can be treated as extensive for the most part. (Gibb's paradox shows entropy is not perfectly extensive.) The (total) internal energy U is likewise extensive so ratios (and thus derivatives) of these yield the intensive quantities p, T, and mu.

    Now the other thermodynamic potentials, Enthalpy, Gibbs free energy, Helmholtz free energy come into play when we wish to express an energy type quantity but as functions of the intensive variables.

    They are a bit abstract in physical meaning so it is best to understand them first purely in terms of the differential relationships you want them to have:

    [tex]dU = TdS - pdV + \mu dN[/tex]
    [tex]dH = TdS + Vdp +\mu dN[/tex]
    in other words, we want the Enthalpy H to be as much like U as possible but to depend on pressure as an independent variable and to define volume as the rate of change of enthalpy per unit change of pressure.

    This is simple enough, look at the difference of the differentials:
    [tex] dH - dU = Vdp + pdV = d(pV)[/tex]

    and thus we define:
    [tex] H = U + pV[/tex]

    Pressure times volume can be though of as the extrapolated energy required to expand to a given volume at constant pressure. Think of it as a linearized potential when in reality the pressure at different volumes may change in complex ways.

    When changing to independent variables (S,p,N) to define enthalpy as:
    [tex]H = U + pV[/tex]
    we should also be careful to understand that firstly we treated E = U(S,V,N) but we (assume that we may) solve that equation and the equation of state for the given system for V in terms of S, N, and p. So the defining equation should really read:
    [tex]H(S,p,N) = U(S,V(S,p,N),N) + pV(S,p,N)[/tex]
    Of course that reads terrible so it is shortened and the experienced reader eventually makes the translation implicitly.

    Now what does enthalpy mean? Well that's difficult to pin down exactly... it is effectively the total internal energy of the system plus the work that would be done on the walls of a container as it increases its volume from zero to the current volume at constant pressure. This is not an "out there" quantity but rather a type of mathematical relationship. Then again total energy also is that type of quantity since potential energy is also the work required to move a system from some zero point to the given configuration. So you can view the Enthalpy as the total energy with a different kind of potential energy definition (using pressure as the independent variable).

    The key point is that increasing pressure at constant volume does no volumetric work in and of itself. But it requires some type of work be done, either pumping in particles from outside, boiling a liquid, or simply heating the gas). Enthalpy keeps track of this change better than does U.

    Now as to your questions:

    Note that the differential work done is pdV not d(pV).

    And the sign is a matter of defining that work as done on the gas by the walls of the container (-pdV since that's the outside pressure pushing in on the gas while volume increases outward) or as work done on the walls by the gas (+pdV since the gas pushes outward in direction of increasing volume).

    If you vary particle number you generally need to take into account the chemical potential.
    [tex] \mu =\left. \frac{\parital U}{\partial N}\right|_{S,V\text{ const.}}=\left. \frac{\parital H}{\partial N}\right|_{S,p\text{ const.}}[/tex]

    Finally you should look at the other thermodynamic potentials http://hyperphysics.phy-astr.gsu.edu/HBASE/thermo/thepot.html" [Broken].

    Also see http://en.wikipedia.org/wiki/Fundamental_thermodynamic_relation" [Broken].
    Last edited by a moderator: May 4, 2017
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