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How much work?

  1. Jan 8, 2016 #1
    1. The problem statement, all variables and given/known data

    zjgw49.jpg


    2. Relevant equations

    no

    3. The attempt at a solution

    no

    Since the problem asks how much work was done by OR on the gas, I did not understand why the book's answer is 162 J instead ±81 J that I've found. (sorry my bad english)

    Sorry, the correct question on the problem is how much work was done from b to c instead from a to b as it's in the image.
     
    Last edited by a moderator: Jan 8, 2016
  2. jcsd
  3. Jan 8, 2016 #2
    Show us how you got your answer, then we can point out what you did wrong. Also, be careful with your units. I don't know all the unit conversions but to use Joules the units for P and V are pascals and meters cubed, not atmospheres and liters.
     
  4. Jan 8, 2016 #3
    Okay.

    dW = dV p

    In this case we have the initial and final values of V and p. So, W = (Vc - Vb) x 10-3m³ x (Pc - Pb) x 1.013 x 105Pa, wich gives W = 81.04 J.
     
  5. Jan 8, 2016 #4

    haruspex

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    You have calculated ##\Delta V\Delta p##. That is not the same as ##\int p.dV##.
     
  6. Jan 8, 2016 #5
    and how can I solve the integral for T?
     
  7. Jan 8, 2016 #6

    haruspex

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    In the graph, p is the y ordinate and V the x ordinate. So the integral is equivalent to ##\int y.dx##. what's a geometric interpretation of that integral?
     
  8. Jan 8, 2016 #7
    oh yes, I see that and I solve the problem by this way. But I wonder if there's anyway to solve this integral for T using only calculations without the graph. Is there a way?
     
  9. Jan 8, 2016 #8

    haruspex

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    Yes, but you first have to turn the graph into an equation relating p to V. Then plug that function into ##\int p.dV##.

    Edit: When you say you solved the problem that way, are you referring to your solution in post #3? That solution was wrong.
     
  10. Jan 8, 2016 #9
    Ok. Thank you.
     
  11. Jan 8, 2016 #10

    Mister T

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    They do not want an answer with a ± sign in front of it. They want a positive number and they want you to determine whether it's "on" or "by".
     
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