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Enthalpy=Internal Energy at Constant Pressure, and InternalEnergy=q at constant volume. What about..

Discussion in 'MCAT Study Question Q&A' started by manohman, 09.28.14.

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  1. manohman

    manohman 2+ Year Member

    Joined:
    07.27.14
    Messages:
    110
    What about Enthalpy at constant volume?

    H = U + dPdV

    U= I + W

    W= -PdV

    so 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?

    I think i'm confusing the dPdV in the Enthalpy equation!
     
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  3. amy_k

    amy_k

    Joined:
    09.24.14
    Messages:
    192
    Status:
    Non-Student
    You made a slight mistake when you multiplied out d(PV), remember we are dealing with differential change here.

    The following equation applies for both isobaric and isochoric processes:

    dH = dU + d(PV)
    dH = dU + PdV + VdP

    ***(d(PV) = PdV + VdP, and not dVdP, using the product rule for differentiation)***

    Since dU + PdV = dQ, we can simplify it as follows:

    dH = dQ + VdP

    So for an isobaric process when dP = 0:

    dH = dQ

    And for isochoric when V is constant, we cannot cancel out the VdP term, so

    dH = dQ + VdP

    But you are correct about internal energy under isochoric conditions:

    dU = dQ - pdV
    dU = dQ

    I hope that helps you out.
     
  4. manohman

    manohman 2+ Year Member

    Joined:
    07.27.14
    Messages:
    110
    Ah Awesome That clears it up entirely and makes complete sense. I was having so much trouble. Thanks!
     

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