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Second thermodynamics

  1. Apr 23, 2009 #1
    1. The problem statement, all variables and given/known data
    A sample of 3 mol of a diatomic perfect gas at 200K is compressed reversibly and adiabatically until its temperature reaches 250K. Given that Cv,m=27.5 JK-1mol-1, calculate q, w, ΔU, ΔH and ΔS.


    2. Relevant equations
    dS = dq/T
    ΔU = n(Cv,m)ΔT
    ΔH = n(Cp,m)ΔT


    3. The attempt at a solution
    (skipped the part of q and ΔS)

    ΔU = q+w = w
    w = -∫PdV = -PΔV = -nRΔT = -1247.2 J
    ΔU = -1247.2 J
    ΔH = ΔU+Δ(PV) = -2494.4 J

    the model answer:
    w= ΔU = CvΔT = +4.1 kJ
    ΔH = CpΔT = +5.4 kJ


    here are my questions:

    what's wrong with my answer?
    i should not assume Pressure is constant?

    if the problem is that Pressure does vary in the process,
    why ΔH is still calculated by Cp,m (i know ΔH=q in an isobaric process) ?
    why can't i use ΔH = ΔU -w in this case?

    is ΔU always equal to CvΔT for isobaric and isochoric process,
    and equal to 0 for isothermal process?

    Thank You!
    i have read my physical chemistry book but i am still confused!
     
  2. jcsd
  3. Apr 23, 2009 #2
    Well, you'll first have to understand that an adiabatic process is a process where all 3 of the basic Thermodynamics variables change. This means that Temperature,Pressure and Volume are all subset to changes , yet since this process is adiabatic, there is no exchange of Heat Q between the gas and the rest of the system.

    You should also understand that Cp and Cv are constants, that you can use when the gas in your problem is ideal/perfect and with a single atom. Cp=5R/2 and Cv=3R/2. When your gas isn't that perfect, there are other values that they take.

    ΔU is given by nCvΔT for every reversible process, and you can easily see than in an isothermal process (ΔT=0) ΔU=0.

    I am not sure about Entropy or Enthalpy since my memories of that are quite blurry (+ we were not taught of them in school, so..) I hope I have helped you understand some things, if others find mistakes in my explanations feel free to correct me.
     
  4. Apr 23, 2009 #3
    thank you karkas.

    actually you remind me of another question,

    for isothermal process,
    ΔT=ΔU=0
    ΔH = ΔU + Δ(PV)
    = ΔU + Δ(nRT)
    = 0+0 = 0

    so for isothermal process, heat is not necessarily be zero but ΔH is always zero,
    while an adiabatic process, heat must be zero while ΔH can be nonzero,
    am i right?


    anyway i am really confused by the equations...
     
  5. Apr 23, 2009 #4
    Really, the Second Law of Thermodynamics explains it all fairly easily, no need to be confused :)

    1)Q=ΔU+W (in an isothermal ΔU=0) and so Q=W=nRT lnΔV. <-- Isothermal
    2)Q=ΔU+W (Q=0) => ΔU = - W = - Δ(PV)/1-γ <--- Adiabatic

    So indeed in the isothermal : ΔΤ=0 and Q isn't zero.
    And in the adiabatic : ΔΤ isn't zero and Q=0.

    Anything else?? :) (I hope I am not making mistakes, someone confirm!)
     
  6. Apr 23, 2009 #5
    why Q=ΔU+W

    i know ΔU=q+w and it should be q=ΔU-w
     
  7. Apr 23, 2009 #6
    http://en.wikipedia.org/wiki/First_law_of_thermodynamics

    Does that clear your queries?

    @wikiquoting@

    Notice that a lot of textbooks (e.g., Greiner Neise Stocker) formulate the first law as:

    dU=\delta Q+\delta W\,

    The only difference here is that δW is the work done on the system. So, when the system (e.g. gas) expands the work done on the system is − PdV whereas in the previous formulation of the first law, the work done by the gas while expanding is PdV. In any case, both give the same result when written explicitly as:

    dU=\delta Q-PdV\
     
  8. Apr 23, 2009 #7
    i see!
    thanks a lot

    and could anyone answer my first question posted above...thanks
     
  9. Apr 26, 2009 #8
    hey could anyone help?
     
  10. Apr 26, 2009 #9

    Mapes

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    Your first question being "what's wrong with my answer?"? It's calculated assuming an isobaric process, which this isn't.
     
  11. Apr 26, 2009 #10
    why would the model use ΔH = CpΔT if it is not an isobaric process?
    i thought q=CvΔT and ΔH=q=CpΔT in isobaric process!
     
  12. Apr 26, 2009 #11

    Mapes

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    [itex]\Delta U=C_V\Delta T[/itex] and [itex]\Delta H=C_P\Delta T[/itex] always hold for an ideal gas. But [itex]\Delta H=Q[/itex] only for reversible isobaric processes (because [itex]dH=T\,dS+V\,dP=q+V\,dP[/itex])
     
  13. Apr 26, 2009 #12
    thx!

    do u mean that ΔH=CpΔT even in an isochoric process??

    so,
    for an isothermal process, ΔH=CpΔT = 0
    for an isochoric process, ΔH=CpΔT
    for an isobaric process, ΔH = Q= ΔU+Δ(PV)=CpΔT in reversible isobaric process

    am i correct?
    oh my god
    so confusing
     
  14. Apr 26, 2009 #13

    Mapes

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    Yes, this is true (and confusing). It may help to note that enthalpy is a state variable and therefore process-independent. Thus, if ΔH=CpΔT holds in one process then it holds in all processes (ideal gas only, though).

    (Is LaTeX down on PF? I thought maybe things were screwed up on my end.)
     
  15. Apr 26, 2009 #14
    oh this really helps me a lot
    thank you!
     
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