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I Adiabatic Process, Internal Energy vs. Enthalpy.

  1. Mar 1, 2016 #1
    I'm struggling to understand a concept which I assume is basic, but I can't seem to fit the pieces together. When speaking about an ideal gas, I understand that

    ΔH = ΔU + Δ(PV) = ΔU + RΔT

    So far so good. I also understand the relationship:

    ΔU = Q + W... (W here is work being done on the system)

    In an adiabatic reversible process, Q = 0, which also makes sense. So,

    W = ΔU = nCvΔT

    Now, where my confusion lies is in the next part. My book works out a problem, and says:

    ΔH = W = nCpΔT

    How can both ΔU and ΔH equal W? This doesn't make sense to me, unless Δ(PV) from the first equation was 0. I don't see how this could be 0 unless we were talking about an isothermal case. It also doesn't make sense because W can't equal both nCvΔT and nCpΔT simultaneously, since Cp = Cv + R.

    Am I missing something?
     
  2. jcsd
  3. Mar 1, 2016 #2
    Never mind. Just answered my own question. The first instance refers to a closed adiabatic system, and the second refers to an open steady-state adiabatic process.
     
  4. Mar 7, 2016 #3
    How nCvΔT will equal to W.It is the heat absorbed/released at constant volume only(no work is done at constant volume,the expression
    Cv).i think instead of work it is heat,ie ΔU=ΔH=qv at constant volume and ΔH=qp at constant pressuer
    actually it is the proof for Cp-Cv=R
     
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