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Enthalpy (H=U+PV) What is P, really?

  1. Sep 8, 2009 #1
    I'm studying for the Physics GRE and going over my thermo notes.
    Ok, so for enthalpy change

    (delta)H = (delta) U + P * (delta) V

    H is enthalpy, U is just proportional to T
    What pressure are they talking about?
    If the volume and temperature are changing, then the internal pressure is definitely changing, but in this equation it seems to be constant through out the process. So I assume that this is supposed to mean the pressure of the surroundings.
    But this is tricky.
    If you have a box of gas in space, the surrounding pressure is zero. So it seems like the only enthalpy change would be due to U. But this doesn't seem right. So let's look at how it expands.
    Some latch could get thrown during the process and allow the piston to slide out frictionlessly. Or the piston could be connected to a spring that would get compressed as the gas expanded. Or the piston could just be rusty and release energy as heat when it moves.

    I feel like all of these scenarios would have a different enthalpy change, but the equation doesn't seem to account for this. I'm guessing there's something more interesting in the P factor. What does it really mean?
  2. jcsd
  3. Sep 8, 2009 #2


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    The P is system pressure, and the equation you give assumes a constant system pressure (which is not always the case). This form is useful when a system is at equilibrium with the pressure of its surroundings.

    (Also, I don't agree that if the volume and temperature of a system are changing, then "the internal pressure is definitely changing.")
  4. Sep 8, 2009 #3
    P is system pressure then?
    I know that T and V changing does not imply that P changes. It just usually does when the system is not in mechanical contact with its surroundings.

    So I'm guessing that (delta)H is only useful in a setup where the system is held at constant pressure? This means that (delta)H is not very useful when it comes to engines/refrigerators or any system that is not held at constant pressure.
  5. Sep 8, 2009 #4


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    Your equation above assumes constant P, so not surprisingly it's only useful for constant-pressure situations. For more general situations, we can calculate [itex]\Delta H[/itex] by

    [tex]\Delta H=\Delta U+\int P\,dV+\int V\,dP[/tex]


    [tex]\Delta H=\int C_P\,dT[/tex]
  6. Oct 14, 2009 #5

    Can u please explain why there is both Delta P * V and Delta V * P?

    Would this not take Work into account twice?

    When P is constant, the Delta P * V term = 0?

    When V is constant, the Delta V * P term =0? Im just guessing
  7. Oct 14, 2009 #6


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    Work is [itex]\int P\,dV[/itex] (magnitude), not [itex]V\Delta P[/itex] or [itex]V\Delta P+P\Delta V[/itex], and not necessarily [itex]P\Delta V[/itex] if the pressure isn't constant.

    I agree with your guesses.
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