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I am going to lose it! I need a hint

  1. Dec 7, 2007 #1
    1. The problem statement, all variables and given/known data
    just to get me started! This is killing me....

    I need to derive the formula for an work done W of an adiabatic expansion of an ideal gas: [tex]W=\frac{1}{1-\gamma}[p_2V_2-p_1V_1][/tex]

    Using the fact that [tex]p_1V_1^{\gamma}=p_2V_2^{\gamma}=constant[/tex]

    thanks
    RW
     
  2. jcsd
  3. Dec 7, 2007 #2
    I know that [tex]W=\int pdV[/tex] also.....
     
  4. Dec 7, 2007 #3
    This is going nowhere extremely fast...
     
  5. Dec 7, 2007 #4

    Doc Al

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    Staff: Mentor

    Here's a hint.

    Rewrite: [tex]pV^{\gamma}=C[/tex]

    As [tex]p=C/V^{\gamma}[/tex]
     
  6. Dec 7, 2007 #5
    Okay from now on Will use n in place of gamma for simplicity.

    So now if I have [tex] W=c\int\ V^{-n}dV[/tex]

    [tex]\Rightarrow W=c\frac{V^{-n+1}}{-n+1}[/tex].....

    Now I am stuck again. ....
     
    Last edited: Dec 7, 2007
  7. Dec 7, 2007 #6

    Doc Al

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    Staff: Mentor

    Redo that integration more carefully. (Check by taking the derivative of your answer.)
     
  8. Dec 7, 2007 #7
    Gotcha......now I am having trouble putting that in terms of p1 and p2...I now have it in terms of c.....and I still have 1-n up in th exponent....do I need a natural log somewhere?

    Thank you
    RW
     
  9. Dec 7, 2007 #8

    Doc Al

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    Staff: Mentor

    You can replace c with what it equals.
     
  10. Dec 7, 2007 #9
    But which do I use....I know [tex]p_1V_1^n=p_2V_2^n[/tex]

    oh..wait.... how about I distibute c first and then replace it...it's clear now.

    Thanks Doc!!!
    You're the best!

    RW
     
  11. Dec 7, 2007 #10
    So to actually evaluate this expression when I am given pressure in atms and volume in cubic centimeters do i need to convert these units into SI.

    I see that the coefficient of 1/(1-n) is dimensionless. And I had assumed that since p1v1-p2v2 had the SAME dimensions it would be okay to NOT convert.......

    i think I see my error though....I need Joules when all is said and done. I may have answered my own question.

    RW
     
  12. Dec 7, 2007 #11

    Doc Al

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    Staff: Mentor

    Use standard SI units.
     
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