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deltaH and q

Discussion in 'MCAT Study Question Q&A' started by Sylvee, Aug 14, 2015.

  1. Sylvee

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    Hi everyone,

    I'm having trouble wrapping my head around the deltaH = deltaU + PdeltaV equation
    My main issue is this: doesn't deltaU already include PV work in its equation deltaU = q + w = heat + PV work? So why is it considered twice in the enthalpy equation, and why is deltaH = q when there is no change in volume valid?

    This concept has been driving me crazy. Thanks!
     
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  3. Neuroplasticity

    2+ Year Member

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    1. The equation for enthalpy you give is already assuming constant pressure. Enthalpy is actually defined as dH=dU+d(pV), the change of pressure and volume, not dH=dU+P*dV which is change only in volume (http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Enthalpy)
    This does not include a redundant term since dU=dq+dW and dW=-p*dV so dU=dq-p*dV

    2. dH=q when pressure is constant
    with the 2 definitions above -> dH=dU+d(pV) dU=dq-p*dV
    plug dU into dH -> dH=(dq-p*dV)+d(pV)
    product rule (https://en.wikipedia.org/wiki/Product_rule) -> dH=(dq-p*dV)+(p*dV+V*dp)
    cancel terms -> dH=dq+V*dp
    so when there is no change in pressure, dp=0 so V*dp=0 -> dH=q
     
  4. Sylvee

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    Whew! So that's where it comes from...thanks!
     
    Neuroplasticity likes this.

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