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Work done during expansion of a gas

  • Thread starter tanzerino
  • Start date
  • #1
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[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
 
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Answers and Replies

  • #2
48
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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
 
  • #3
48
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any help?
 
  • #4
Andrew Mason
Science Advisor
Homework Helper
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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
 
  • #5
48
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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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