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The different between the work (PV ) in ΔU and the work (PV ) ΔH

  1. Apr 13, 2013 #1
    By definition
    ΔU = q+PV
    ΔH = ΔU+PV
    Well, what are the different between the work (PV ) in ΔU and the work (PV ) ΔH then ?

    thank you
     
  2. jcsd
  3. Apr 14, 2013 #2
    Both equations are incorrect.

    ΔU=q+w
    ΔH=ΔU+Δ(PV)

    So, now, what is your question?
     
  4. Apr 14, 2013 #3
    w= -P ΔV
    in case of constant Pressure, ΔH= (q -PΔV) +PΔV >> ΔH= q

    So w in ΔU will be the same when constant pressure with Δ(PV) in ΔH ?

    if the pressure is not constant What is P in ΔU ? is it P external? and what is P in ΔH ? P in ΔH is Pressure of system? in case of ideal gas, Δ(PV) = nRT so where is P ?

    and if not ideal gas how can we calculate Δ(PV) in ΔH?

    Thank you
     
  5. Apr 14, 2013 #4
    Dear izen,

    These are all excellent questions, but I don't have time to respond right now. We are expecting some out-of-town relatives, and I'm pretty tied up for the rest of the day. I will get back with you as soon as possible to answer all your questions.

    Chet
     
  6. Apr 14, 2013 #5
    Hi Chet,

    I will be waiting for you here. Enjoy your evening. :)

    Izen
     
  7. Apr 15, 2013 #6
    You need to think of U and H as properties of a system characteristic of each state of equilibrium. ΔU and ΔH are the changes in U and H from one equilibrium state (say, state 1) to another equilibrium state (say, state 2). Even though there may be many paths that will take the system between the two equilibrium states, the changes in U and H are independent of the path.

    On the other hand, the parameters q and w are quantities that are totally dependent on the path between the two equilibrium states. Only their difference is independent of the path. q and w are sometimes called energy in transit.

    For expansion or compression, the work w is always determined by [tex]w = -\int{P_extdV}[/tex], not by w= -P ΔV. If the process is reversible, the pressure P within the system is virtually homogeneous, and, in addition, P = Pext (actually P differs from Pext by an infinitecimal amount). If the process is irreversible, the pressure within the system is not homogeneous, and so there is no one single value of pressure that characterizes the pressure within the system.

    For the change in enthalpy, we have
    [tex]\Delta H = q -\int{P_{ext}dV}+\Delta (PV)[/tex]
    where the pressure P and the volume V in the Δ(PV) correspond to the system values in the initial and final equilibrium states. If a process is carried out at constant external pressure, the change in enthalpy is given by:
    [tex]\Delta H = q -P_{ext}(V_2-V_1)+(P_2V_2-P_1V_1)[/tex]

    In such a process, for the final state to be an equilibrium state, one must have that
    P2=Pext. Under such circumstances:

    [tex]\Delta H = q -(P_2-P_1)V_1[/tex]

    If the final pressure of the system is equal to the initial pressure (constant pressure process),

    [tex]\Delta H = q[/tex]

    Examples of when this applies are (a) if you have an ideal gas, and you add heat to the chamber while the piston is moved in such a way as to hold the pressure constant and (b) a combination of saturated vapor and saturated liquid in the chamber, and you add heat while the piston is moved in such a way as to keep the pressure constant.

    In the case of an ideal gas, there is no effect of pressure on either U or H. Both parameters are functions only of temperature. For a real gas, if you wish to calculate the effect of pressure on U and H, the formulation is more complicated. For infinitecimal changes between equilibrium states, the changes in U and H are described by the differential equations:

    dU = TdS - PdV
    dH = TdS+VdP

    These equations apply only to differential changes between equilibrium states. The effect of pressure P on U and H are derived starting with these equations, and, by applying the so-called Maxwell Relationships.
     
  8. Apr 16, 2013 #7
    Hi Chet thanks for the answer :)
     
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