deltaH and q

Sylvee

Full Member
2+ Year Member
Jul 21, 2015
13
2
    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!
     

    Neuroplasticity

    Full Member
    2+ Year Member
    Jul 27, 2015
    102
    54
    1. Medical Student (Accepted)
      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
       
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