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Thermo question, two methods of changing volume/temp

  1. Aug 30, 2012 #1
    n2fgqb.jpg

    does this look right?

    for the quasistatic case you can use thermodynamics to find the temperature at any time,

    for the other case you have to use dU=dW and so on since its not quasistatic
     
  2. jcsd
  3. Aug 30, 2012 #2

    ehild

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    It is correct for the constant external pressure. But what is the final temperature in the quasi-static case? It is true that Tf=PVf/(NKB) but P changes during the process. How do you get the final pressure Pf?


    ehild
     
  4. Aug 31, 2012 #3
    Oh right,

    so the quasistatic case should be:

    Tf=PfVf/(NKB)

    for some reason i forgot that the pressure was changing in that case as well,

    this makes more sense, in the quasistatic case you can find the state of the system at any point in terms of the state variables, and P is one of those too!

    thanks ehild =]
     
  5. Aug 31, 2012 #4

    ehild

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    Can you show what you got? Just to complete the solution, so as other people learn from it. :smile:

    ehild
     
  6. Aug 31, 2012 #5
    quasistatic case :

    Tf=PfVf/(NKB)

    under constant pressue:

    Tf=2Pex(Vi-Vf)/3NkB + Ti
     
  7. Aug 31, 2012 #6

    ehild

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    Well, it is not the solution yet. You need to give Tf in terms of the initial and final volumes and the initial temperature. Do you know the equation that governs a quasistatic adiabatic process?

    ehild
     
  8. Aug 31, 2012 #7
    Oh,

    do you mean something like

    43344f9db2f081da94da515e703efed7.png

    where gamma is the ratio of heat capacities under constant pressure / volume?
     
  9. Aug 31, 2012 #8

    ehild

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    Yes. But you can combine it with the ideal gas law to give an equation between V and T.

    ehild
     
  10. Aug 31, 2012 #9
    yea theres afew like

    TV^(gamma - 1)=constant

    PV^gamma = const

    but wouldnt mucking around with these and introducing them into my problem make it like

    Tf=PfVf/(NKB)=PiVf1-λViλ/(NKB)

    using λ as gamma

    PV^gamma = const = C say



    Tf=CVf1-λ/NKB

    then i have this unknown constant C
     
  11. Aug 31, 2012 #10

    ehild

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    That is a good start. You can find that constant as you know the initial volume and temperature:
    TfVfγ-1=TiViγ-1.
    From the equation for the internal energy you can see that it is a mono-atomic gas, so you know Cv, and you also know the relation between Cv and Cp. What is γ-1 then?

    ehild
     
  12. Aug 31, 2012 #11
    C=NKb Tf Vf^(γ-1)=NKb TiVi^(γ-1)


    Tf= Ti(Vf Vi)^(γ-1)

    for γ=Cp/Cv

    Tf= Ti(Vf Vi)^(Cp/Cv-1)

    is that kind of what you mean?
     
  13. Aug 31, 2012 #12

    ehild

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    Yes, but you made a little error:

    Tf=Ti(Vi/Vf)γ-1.
    And you know the numerical value of Cp/Cv what is it?

    ehild
     
  14. Aug 31, 2012 #13
    oh whoops,


    Tf=Ti(Vi/Vf)5/3-1.



    Tf=Ti(Vi/Vf)2/3.

    thanks ehild
     
  15. Aug 31, 2012 #14

    ehild

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    It is the SOLUTION now:smile:

    ehild
     
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