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Why does the enthalpy equation include work (PV term) twice?

  1. Mar 11, 2013 #1
    Hello. I am a thermodynamics novice trying to gain a better understanding of state functions, particularly enthalpy.

    I understand that enthalpy is defined as

    "A measure of the total energy of a thermodynamic system, including internal energy, which is the energy required to create a system, and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure."

    The equation:

    ΔH=ΔU(internal energy) + ΔPV

    confuses me b/c

    ΔU= q(heat added) - w(work done by system on environment)


    ΔH really means:

    ΔH=q - w + ΔPV

    There are two terms of work (w and ΔPV) and b/c of the opposite sign, they cancel out, leaving only q. This means ΔH= q which is at odds with the accepted definition of enthalpy. Where did I mess up?
  2. jcsd
  3. Mar 12, 2013 #2


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    That not at odds with the usual definition. Mostly enthalpy is determined by measuring heat.
    However ##\Delta(PV)=P\Delta V=w## only if p is constant and volume work is the only kind of work the system is performing.
  4. Mar 12, 2013 #3
    so you are saying that enthalpy is only equivalent to heat if pressure is held constant?


    ΔH=q - PΔV + PΔV= q (only when pressure is constant and only PV work is being exerted)

    Otherwise, when pressure is not constant the equation should like this, perhaps?:

    ΔH=q - w + PΔV

    And the work defined by the 'w' above includes all forms of work, whether PV or mechanical, etc?

    If so, that makes a little more sense. It's just that all the example problems I've encountered with ΔU only use PV work and no other form.
  5. Mar 12, 2013 #4


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    In general, ##d(PV)=PdV+VdP##. The second term will not vanish when P is not constant while the first term gives the volume work done in an infinitesimal step.
    Hence ## \Delta H=q-w+\int PdV +\int V dP## in general.
    If there is no work done other than volume work, this reduces to
    ## \Delta H=q+\int V dP##.
    An example of non-volume work is e.g. the work done when stirring a viscous fluid.
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