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Calculating work from graph

  1. Jul 22, 2007 #1
    1. The problem statement, all variables and given/known data

    a sample of ideal gas is expanded to twice its original volume of 1.00m^3 in a quasi-static process for which P=alphaV^2, with alpha=5.00 atm/m^6. How much work is done on the expanding gas.

    2. Relevant equations

    w=-intergral sign Pdv from intial to final volume( i dont know how to type this out)

    3. The attempt at a solution

    w=-aplha(v^3/3) from 2m^3 to 1m^3
    =(-5.065 x10^5/3)(2^3-1^3)=-1.2x10^6

    this actually the correct answer, however i dont understand why -5.065x10^5 is being divided by3 i thought it would be : -5.065x10^5((2^3/3)-(1^3/3)=w/e could someone explain why your suppose to set it up like that....
     
  2. jcsd
  3. Jul 22, 2007 #2
    Hint:

    What is

    [tex]\int\alpha x^2 dx[/tex]
     
  4. Jul 22, 2007 #3
    alpha)(x^3/3).... did i forget something
     
    Last edited: Jul 22, 2007
  5. Jul 22, 2007 #4
    What is the work done in a reversible quasistatic process? It is given by

    [tex]W = \int P dV[/tex]

    Thats how the factor of 1/3 comes about. Maybe the answer is wrong?

    (PS--Which book is this?)
     
  6. Jul 22, 2007 #5
    the text is principles of physics serway and jewett 4th edition, is the correct setup to the anser then -5.065x10^5(2^3/3-1^3/3) or the way they put it? Should the 1/3 still be factored out?
     
    Last edited: Jul 22, 2007
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