Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Change in enthalpy at constant VOLUME

  1. Aug 30, 2011 #1
    Hi, I know enthalpy is used a lot to calculate isobaric processes, but what about isochoric ones?
    For example, if I have gas in a piston and I add heat at constant VOLUME.
    I know that [itex]Q=\Delta U[/itex]
    But since [itex]H=U+PV[/itex] or [itex]\Delta H=\Delta U + \Delta PV[/itex]
    I see that although V is constant P is not, it increases and so my entropy increase is bigger than my internal energy increase.
    That is ok, I've searched this forum and found the same results, but my question is. Since enthalpy is energy, and energy is conserved from where does this energy come from?
    Since the heat supplied to the system is smaller than the change in enthalpy.
    Let's say I give 1000 Joules to the system in form of heat, the internal energy increases by 1000J, but the Enthalpy increases more...
    Any help appreciated.
  2. jcsd
  3. Aug 30, 2011 #2
    by how much? do you have any idea?
  4. Aug 30, 2011 #3


    User Avatar
    Science Advisor

    Enthalpy is not the energy, U is the energy. Enthalpy only has the dimensions of energy. Enthalpy is a mathematical potential useful in processes that take place at constant pressure. The change in enthalpy during an isobaric process is equal to the heat that is transferred. In the fields of engineering and chemistry enthalpy is more useful than internal energy, since most processes occur isobarically.
  5. Aug 30, 2011 #4

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Yes. Exactly. I was in the middle of responding but this is a much more succinct and clearer answer than I was about to give.

  6. Aug 30, 2011 #5
    @Bill K, but the question is under isochoric process.
  7. Aug 30, 2011 #6
    It feels weird that in an ISOCHORIC process the enthalpy(the potential to do work? Can I say that way?) increases more than the heat supplied...so I get more work(energy) than I gave in(in the form of heat)
    Yes, I know that I'm wrong but I was having trouble seeing this.
    But thanks for the clarification.

    What I'm thinking is that this isn't a perpetual energy machine(duh) because to keep the constant volume I need to have something holding the piston, exerting some force on it to keep it from expanding. And also because I would need to compress it again and in the end the delta H would be zero.
    Going to try some calculations later.
  8. Aug 30, 2011 #7

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    In an isochoric process, the change in enthalpy differs from the change in internal energy by the amount [itex]\Delta PV[/itex]. But it is just a mathematical book-keeping thing. The only actual heat flow/energy change is [itex]\Delta Q = \Delta U[/itex] in a constant volume process. In fact, the VdP term is already incorporated into [itex]\Delta U[/itex] so, in effect, it is being counted twice in the expression for [itex]\Delta H[/itex] in a constant volume process.

  9. Aug 30, 2011 #8
    Aha, now everything is clear in my mind.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook