1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?

  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

    User Avatar
    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?

  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

    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted