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Gas law problem

  1. Dec 11, 2016 #1
    1. The problem statement, all variables and given/known data
    http://i.imgur.com/wSvUEU2.png
    wSvUEU2.png

    2. Relevant equations
    (written in the photo)

    3. The attempt at a solution
    (written in the photo)
    Thanks
     
  2. jcsd
  3. Dec 11, 2016 #2
    It all seems to be there, so I'm not sure what you are missing.

    The area is clear from the 10cm cylinder. If you know the area then the pressure is clear from the force of gravity on the weight divided by the area (and + 1 atmosphere for the external pressure on the cylinder.) n is given, the Ts are given. What else do you need?
     
  4. Dec 11, 2016 #3

    TSny

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    As Cutter Ketch noted, the pressure of the gas is not atmospheric pressure. The weight of the piston must be taken into account.
     
  5. Dec 11, 2016 #4
    Do I really need to consider the weight of the piston? because it seems to be a constant factor... no change in the case...
    do you mean "F=PA=mg" to be simple? however, whats going on with it?
     
  6. Dec 11, 2016 #5
    Oh, I see. You aren't getting the right answer. So I take it the hand written answers are correct and the red is what you got?
     
  7. Dec 11, 2016 #6
    Words in red are written by me
    and the circle of the option is the right answer
    i got c, while the right answer is a...
    thanks
     
  8. Dec 11, 2016 #7
    You don't need it because it changes. You need it because you can't define the initial condition without it. Since h1 isn't given you have to figure it out. Your formula for h1 is correct, and in that formula is pressure. You need the absolute pressure to determine h1, and that pressure includes holding up the piston.
     
  9. Dec 12, 2016 #8
    so you mean i should eliminate the factor for h1 first in order to find out the pressure if piston is not present?
    how should i relate PV=nRT to force?
     
  10. Dec 12, 2016 #9

    TSny

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    Your calculation in red is correct except for the value of the pressure P1. P1 does not equal atmospheric pressure.

    To find the correct pressure, switch your attention to the piston and consider all the forces acting on the piston.
     
  11. Dec 12, 2016 #10
    by F=PA=mg, P should be equal to mg/A=50*9.81/(0.05^2*pi)=62452Pa?
    I still can't get the written answer...
     
  12. Dec 12, 2016 #11

    TSny

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    That's not the pressure of the gas inside the container.

    How many forces act on the piston? Describe each force.
     
  13. Dec 12, 2016 #12
    weight and pressure by gas
    am i right?
     
  14. Dec 12, 2016 #13

    TSny

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    There's one more force. The problem states that the top of the piston is "open to the air".
     
  15. Dec 12, 2016 #14
    pressure of the air?
    mg+P(a)A=P(g)A
    P(g)=163689Pa ?
     
  16. Dec 12, 2016 #15

    TSny

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    That's it. Good!
     
  17. Dec 12, 2016 #16
  18. Dec 12, 2016 #17

    TSny

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    I'm off to bed. :sleep: Will check back tomorrow.
     
  19. Dec 12, 2016 #18
    Thanks you~
    Then I will come back here at night of my local time~
    because i am going to have exam tomorrow, hope all problems in the exercises can be settled by today:)
     
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