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How to derive pv^gamma=constant

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data
    The relationship between pressure and volume of an ideal gas is expressed as pv=constant in a reversable isothermal condition. Show that the relationship between pressure and volume of the same gas is expressed as pV^gamma=constant in a reversible adiabatic condition where gamma=Cp,m/Cv,m.




    2. Relevant equations

    gamma=Cp,m/Cv,m.
    pV^gamma=constant, rev. adiabatic
    pv=constant, rev. ideal



    3. The attempt at a solution
    p1v1^(Cpm/Cvm)=p2v2^(Cpm/Cvm) take ln and mult both sides by the Cvm/Cpm
    ln p1v1 = ln p2v2 e to both sides
    p1v1=p2v2
    p1v1/p2v2 = 1
     
    Last edited by a moderator: Feb 26, 2013
  2. jcsd
  3. Feb 26, 2013 #2
    please answer it if you have better than these.
    your's thankfully milan talaviya
     
  4. Feb 26, 2013 #3

    ehild

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    Hello Milan, welcome to PF.

    You have to show that pVγ= constant assuming a reversible adiabatic process on an ideal gas.

    What is an adiabatic process? How does the internal energy change in an adiabatic process? What are Cp and Cv?

    ehild
     
  5. Feb 26, 2013 #4

    vela

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    You haven't proved anything here. For one, you've made an algebra mistake. The more serious error is that you started with what you're supposed to be proving. That sort of argument isn't valid logically.
     
  6. Feb 26, 2013 #5

    Andrew Mason

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    The adiabatic condition that you have to prove is not:

    [itex](PV)^{\frac{C_p}{C_v}} = \text{constant}[/itex]

    Rather, it is :

    [tex]PV^{\frac{C_p}{C_v}} = \text{constant}[/tex]

    which means:

    [tex]\frac{P_1}{P_2} = (\frac{V_2}{V_1})^{\frac{C_p}{C_v}} [/tex]

    1. Start with the first law and find an expression for dU in terms of PdV (hint: what is dQ if it is adiabatic?).

    2. Then express dU in terms of dT and substitute your answer in 1. for dU.

    3. Finally express dT in terms of d(PV). (hint: use R = Cp-Cv).

    AM
     
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