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Work done during expansion of a gas

  1. Apr 3, 2010 #1
    [PLAIN]http://img101.imageshack.us/img101/8176/24851738.png [Broken]

    it is definetaly negative as Eint=Q+W
    and Q is smaller than the area under curve taking care of the scaling of pressure in atm.
    but what i get is -1.5x10^6
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 3, 2010 #2
    Einternal=Q+W
    Q he said is given so +ve and work from the graph is -ve
    calculated work was much greateer thats why i think the answer is -ve then wen i tried the numbers i get -1.5 which isnt an answer
     
  4. Apr 3, 2010 #3
    any help?
     
  5. Apr 3, 2010 #4

    Andrew Mason

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    You should show your work. But I get the same answer as you do.

    [tex]\Delta Q = \Delta U + W[/tex]

    where W is the work done by the gas.

    That work is the area under the graph. There appears to be an increase in volume of 5 m^3 and a drop in pressure from 7 to 3 atm. So the work done is (7+3)*5/2 = 25 atm-m^3 of work = 2.525e6 Joules.

    So:

    [tex]\Delta U = \Delta Q - W = 1.02e6 - 2.525e6 = -1.5e6 J[/tex]

    AM
     
  6. Apr 4, 2010 #5
    thanks for your help.i will check for any mistakes with the teacher this is cengage and the second mistake already.shame!
     
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