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

Using enthalpy when pressure is not constant

  1. Sep 27, 2011 #1
    I started another post about this but I made things needlessly complicated by talking about non-ideal gasses. My question is equally valid for ideal gasses. Sorry for double-posting, but I can't figure out how to delete the old post.

    If you do an energy balance for a control mass where there is no heat input, potential energy effects are negligible, and the only work interaction is work done by the systems on the surroundings through the expansion of the gas, you get
    Q-W=dU+dKE+dPE
    0-intPdV = dU + dKE + 0
    0 = dU + int PdV + dKE

    I see how if P is constant, the integral from state 1 to state 2 of PdV is
    P*V(2)-P*V(1), and that
    0 = P*V(2)-P*V(1) + U(2)-U(1) + KE(2) - KE(1)

    which, since H=U+PV, reduces to
    0 = H(2)-H(1) + KE(2)-KE(1)

    But what happens when P is not constant? For example, when a gas escapes from a pressurized tank through a throttling valve? For this situation, if you look at some mass of air that escapes the tank, you can do an energy balance between state 1, when the air is motionless in the tank, and state 2, when the air is in motion and moving away from the valve. In this case P(1) does not equal P(2). Can you still use enthalpy the same way we did above?
    Q-W=dU+dKE+dPE
    0-intPdV = dU + dKE + 0
    0 = dU + int PdV + dKE

    can we say that dh=du+intPdV and write
    0 = dH + dKE = H(2)-H(1) + KE(2) - KE(1)?

    I don't understand the mathematics behind this, but it makes sense that you can look up enthalpy in tables and compare values at different pressures the same way you would under constant pressure.
     
  2. jcsd
  3. Sep 29, 2011 #2

    Bill_K

    User Avatar
    Science Advisor

    You're trying to use enthalpy in a situation where it is not particularly meaningful. You can always define H = U + PV and get dQ = dH + V dP. When P = const then ΔQ = ΔH, and so you can say the change in enthalpy is equal to the heat required for an isobaric process. This is useful.

    But it sounds like you want to talk instead about an adiabatic process where P ≠ const, in which case ΔH = - ∫V dP. Sure you can calculate this quantity, or look it up in a table, but it has no simple interpretation.
     
  4. Sep 29, 2011 #3

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Enthalpy is a state function It can be a state function because it is defined as the change in internal energy and change in PV in a reversible process between two states:

    [tex]\Delta H = \Delta U + \Delta (PV) = \Delta U + \int PdV + \int VdP = \Delta Q_{rev} + \int VdP = \int TdS + \int VdP[/tex]

    In a non-reversible adiabatic process, such as this, you cannot use [itex]\Delta Q_{rev} = \int TdS = 0[/itex] in determining the change in enthalpy because this assumes that there is no change in entropy. In a free expansion of a gas, there is an increase in entropy.

    AM
     
  5. Sep 30, 2011 #4

    Bill_K

    User Avatar
    Science Advisor

    H = U + PV and ΔH = ΔU + Δ(PV) have nothing to do with reversibility. As you say, they are state functions, defined by the initial and final conditions, not by any kind of process, reversible or not.

    And I said Q = 0, which holds for any adiabatic process. Not ∫T dS = 0, which implies reversibility.
     
  6. Oct 1, 2011 #5

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    You are quite right that U and PV are state functions. If they were not, H could not be a state function. What I intended to say was that if you express entropy change in terms of heatflow, you have to use reversible heatflow. Since [itex]dH = dU + PdV + VdP[/itex], if you separate out the dU and PdV terms and equate their sum to dQ you can say [itex]dH = dQ +VdP => \Delta H = \Delta Q + \int VdP[/itex] ONLY IF [itex]\Delta Q = \Delta Q_{rev}[/itex].

    I agree. But if you want to calculate [itex]\Delta H [/itex] in an irreversible adiabatic process you cannot use [itex]\Delta Q = 0[/itex] in the expression for enthalpy: [itex]\Delta H = \Delta Q + \int VdP[/itex]. This would mean that [itex]\Delta H = \int VdP[/itex] which is not correct. [itex]\Delta H = \int TdS + \int VdP[/itex] and for an irreversible adiabatic process, [itex]\int TdS \ne 0[/itex].

    AM
     
  7. Oct 2, 2011 #6
    LS: as long as no distinction is made between the internal pressure an the external pressure this discussion will go the same way any piston would: nowhere.
    The subject is adressed in the attachment zeppos10 placed elsewhere in this forum:
    https://www.physicsforums.com/showthread.php?t=88987&referrerid=219693.
    Bill_K states: You're trying to use enthalpy in a situation where it is not particularly meaningful. I think it is particularly usefull, because p in H=U+pV is the external pressure.
     
  8. Oct 2, 2011 #7

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    The P in PV is the pressure of the gas, not the external pressure.

    AM
     
  9. Oct 3, 2011 #8
    Enthalpy is also defined for solids and liquids, indeed for any system: how do you define p as in H=U+pV there ?
     
  10. Oct 3, 2011 #9
    use p=nRT/V the intPdV will be nRT(ln(V2/V1))
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook