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Thermodynamics: H2 - H1 and U2 - U1

  1. Sep 1, 2007 #1
    Hi. Could anyone clear my doubts?


    U_2-U_1= C_v(T_2 - T_1) and H_2 - H_1 = C_p (T_2 - T_1):
    Do these two equations always valid, regardless the process is adiabatic, isobaric ect?
    At the first place, I thought one works only for isobaric and another for isochloric, because I see C_v and C_p in the equations. I read an example, both the equations are used , even though the process is adiabatic!

    Especially the second eqn, how can it be applied to a adiabatic process?
    dQ = d(U + PV) =dH, and dH is not 0 by the second eqn, due to different temp.
    but for adiabatic, dQ=0!

    where is my mistake?

    Another question, how to get W=R_0 x T x ln(P' / P) for a isothermal process?

    There are so many formula in thermodynamics, I wonder how you guys master this topic?


    thanks in advance.
     
  2. jcsd
  3. Sep 2, 2007 #2
    U_2-U_1=C_v(T_2-T_1) ALWAYS holds for an ideal gas because internal energy is only dependent on temperature, and not on pressure or volume etc. I'm not sure what H_2-H_1 means in the second equation.

    W=R_0 x T x ln(P' / P) can be obtained by integrating P*dv and using PV=nRT.
     
  4. Sep 2, 2007 #3
    mune: thermodynamics, or more accurately analytic thermodynamics, is actually a very nice and self-contained theory. Practise is the only way to learn it properly. Get to the point where you can see past the equations and understand the underlying relations! Also, be careful about the fact that some things only apply to special systems like ideal gases, whilst others apply all the time. And one last hint: the laws of thermodynamics are richer than they first seem! When doing problems, try stating the entire logical argument, right from first principles.
     
  5. Sep 2, 2007 #4

    Andrew Mason

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    [itex]dH = C_pdT[/itex] is true only where dp = 0 (constant pressure). On the other hand, [itex]dU = C_vdT[/itex] is always true - it does not depend on the thermodynamic path.

    If the VdP term is non-zero then [itex]dH \ne dQ[/itex]:

    [itex]dH = d(U + PV) = dU + PdV + VdP[/itex] and [itex]dQ = dU + PdV[/itex]

    AM
     
  6. Sep 2, 2007 #5

    olgranpappy

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    ...but isn't that is what the subscript 'p' *means*.

    Um... at constant volume (which is what the subscript 'v' means). I have to be holding *something* else constant since U is a function of two variables...
     
  7. Sep 2, 2007 #6

    olgranpappy

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    Oh, nevermind, I see that you were just explaining that dH = dQ only at constant pressure... I really should read throught all the posts before I reply...
     
  8. Sep 3, 2007 #7
    thanks everyone.

    I can understand d(U+PV) = dU + PdV + VdP, but why dQ is not equal to d(U + PV)?

    Q= U + W
    => dQ = d(U + PV), where W=PV?

    but dH = d(U + PV) makes sense to me too!
     
  9. Sep 3, 2007 #8

    olgranpappy

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    because you already know that dU = dQ - PdV, or rather, dQ=dU+PdV.

    you are forgetting the fact that d(PV)=PdV+VdP again.

    No. W is *not* PV. Similarly Q is *not* TS. All we can say is that dW is PdV and dQ is TdS. This does not mean that W=PV or that Q=TS.

    This is a definition:

    H=U+PV
     
  10. Sep 4, 2007 #9
    can I say dW=Vdp?

    as the work done is the area under under the PV-curve, we also can use VdP to find the area right? just like finding area under a curve by integrating xdy.
     
  11. Sep 4, 2007 #10

    olgranpappy

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    No.

    yes

    no. say you have a single curve for P as a function of V and that the function can be inverted to give V as a function of P. Then (assuming the axes are drawn as usual) the integral of PdV gives the area under the curve, but the integral of VdP gives the area "to the left of" the curve. Those areas aren't the same...

    On the other hand, if you are calculating the area *enclosed* by some P-V cycle, then you could split the cycle up into two functions of P and integrate VdP for each and subtract the answers to obtain the work over the cycle...
     
  12. Sep 5, 2007 #11


    thanks. you clear my head:smile:
     
  13. Mar 22, 2011 #12
    I'm posting a reply here to this old question since some of the earlier answers are not correct. Both the statements (du = Cv dT and dh = Cp dT) are true under all conditions for ideal gases. They relate different properties (u, T, h - what are called state functions) and do not depend on the process (or path). Whether process is isobaric or isothermal or adiabatic is not of consequence.

    I think what is being confused is whether δq = Cv dT or δq = Cp dT. These two relations hold for specific professes - the first one here is true only for a constant volume process, and the second is true for a constant pressure process only as shown below.

    The first law for a system is δq = du + pdv (ignoring all other forms of energy other than internal energy, and all forms of work other than displacement work). For a constant volume process, pdv=0, and hence δq=du (=CvdT). For a constant pressure process, du + pdv = d(u+pv) (e.g. dx + 5dy = d(x+5y)). Since h=u+pv, therefore δq=dh (=CpdT).

    p.s.: I teach thermodynamics to grad students
     
  14. Mar 22, 2011 #13

    Andrew Mason

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    Thanks for the correction. You are quite right that dH = nCpdT is true by definition for an ideal gas:

    dH = dU + d(PV) = nCvdT + nRdT = n(Cv+R)dT = nCpdT

    AM
     
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