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Enthalpy of Reaction under Constant Volume?

  1. Sep 28, 2014 #1
    1. The problem statement, all variables and given/known data
    So we know Enthalpy under constant Pressure and Internal Energy under Constant Volume. By

    H = U + dPdV

    U= I + W

    W= -PdV

    under Isobaric conditions


    H = I -PdV + dPdV
    = U

    Enthalpy = Internal Energy
    _________________________________________
    Under Isochoric Conditions
    W= O because change in V =o

    so U = q + 0
    Internal Energy = q

    But wouldnt Enthalpy also = q under constant volume since dPdV = 0 as well if volume isnt changing?

    2. Relevant equations
    H = U + dPdV

    U= I + W

    W= -PdV

    3. The attempt at a solution
    ( Outlined above)

    Am i confusing the dPdV specific to enthalpy and PdV in work?
     
  2. jcsd
  3. Sep 28, 2014 #2

    rude man

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    Gold Member

    Looks confusing. what's the question?

    H = U + PV so dH = dU + pdV + Vdp
    dH = dU + Vdp under isochoric
    but dU = dQ - pdV = dQ under isochoric
    so dH = dQ + Vdp under isochoric.

    Is that something like what you're looking for?
     
  4. Sep 28, 2014 #3
    Yeah. I see. So Enthalpy = Internal Energy + Change in Pressure times volume.
    While change in internal energy (dU) = q + Change in Volume times pressure.

    is this correct?
     
  5. Sep 29, 2014 #4

    rude man

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    enthalpy = internal energy plus pressure times volume ... basic statement, always true
    change in internal energy = heat added minus pressure times change in volume. "Change" should read "differential change". To get change you integrate differential changes.
     
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