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First law of thermodynamics

  1. Dec 8, 2015 #1
    1. The problem statement, all variables and given/known data
    In my lecture notes(beginner thermodnyamics), we just got introduced to the first law(DeltaU=q+w) and two scenarios. One at constant volume which yields the equation Delta U = qv. I understand that
    The second scenario is at Constant pressure and it says that At constant pressure q subcript p = DeltaH. DeltaH being change in enthalpy. I don't understnad that bit at all. How is being at constant pressure tell you that the heat energy is the change in enthalpy?

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Dec 8, 2015 #2
    If the pressure is constant, what is w equal to?
     
  4. Dec 8, 2015 #3
    w = -p times deltaV. note that p is the external pressure
    So if pressure is constant, the formula is the same.
     
  5. Dec 8, 2015 #4
    OK. What do you get if you substitute that into your first law equation?
     
  6. Dec 8, 2015 #5
    w = -p times deltaV. First law is DeltaU = q+w. So subbing it in would give me:
    DeltaU = q-pDeltaV
     
  7. Dec 8, 2015 #6
    OK. Now how is ΔH defined in terms of ΔU and Δ(pV)?
    What does this defining equation reduce to if p is constant?

    Chet
     
  8. Dec 8, 2015 #7
    Well DeltaH is change in enthalphy which is change in heat energy, so shouldnt DeltaH always equal to q no matter if its at constant pressure or not since the only other energy transfer is work and that doesnt involve heat. So at all times DeltaH =q? Ive just confused myself somehow.
     
  9. Dec 8, 2015 #8
    No, because the work is not always -pΔV, it is more generally the integral of -pdV. And ΔH is not always ΔU+pΔV, it is more generally ΔU+Δ(pV). Try it using that information and see what you get.
     
  10. Dec 8, 2015 #9

    Mark Harder

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    Gold Member

    Remember the multiplication rule from differential calculus: d(f(x)*g(x))/dx =g(x)* df(x)/dx + f(x)* dg(x)/dx. Or, in the differential form (shortcut):
    d(f(x)*g(x)) = g(x)* df(x) + f(x)* dg(x). and so, d(p*v) = ?
    Now integrate that, given p=constant.
     
  11. Dec 9, 2015 #10
    Do you find an error in what DiamondV and I have done so far?

    Chet
     
  12. Dec 9, 2015 #11
    Thanks. I think ive gotten a grasp of it anyways. had my final thermodynamics exam, never seeing it again thank god.
     
  13. Dec 14, 2015 #12

    Mark Harder

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    Gold Member

    No, not at all. Reading over the posts in this thread, I couldn't find an explanation for why d(PV) = PdV + VdP, so I provided one, just in case the OP didn't know that. It's a necessary step in the derivation for delta-H at constant pressure.
    Mark
     
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