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Ideal gas law, Find the Temperature of the container?

  1. Oct 19, 2016 #1
    1. The problem statement, all variables and given/known data

    There is a lid on a .25m diameter, .30m tall cylindrical container enclosing .021kg of air. The lid is held in place solely by atmospheric pressure. It take 220N of force to pull of the lid at an atmospheric pressure of 101kPa. What is the Temperature of the enclosed gas.

    2. Relevant equations
    PV =nRT

    3. The attempt at a solution
    So got the volume of the container by doing .3m * pi * (.25m/2)^2 which is .0147m^3 than I got the density of the container doing (.021kg/.0147m^3) = 1.43kg/m^3

    At this point I am lost I don't understand what to do with the 220N force on the lid or how to get the pressure of the inside of the container.
     
  2. jcsd
  3. Oct 19, 2016 #2

    TSny

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    Gold Member

    Hello and welcome to PF!

    It will help to identity all forces acting on the lid (including the directions of these forces).
     
  4. Oct 19, 2016 #3
    The only force on the lid is the atmospheric pressure which makes a 220N force downwards on the container. I don't understand how that helps find the temperature of the gas.
     
  5. Oct 19, 2016 #4
    There can't be just one force on the lid. Otherwise, it would be accelerating . Actually, there are three forces acting on the lid, and the gas is at a pressure below atmospheric pressure. Can you identify the 3 forces from a free body diagram?
     
  6. Oct 19, 2016 #5
    How would there be three forces Ii the lid is held in place solely by atmospheric pressure, wouldn't the only the only to force be the atmospheric pushing down on the lid and the pressure inside pushing upward to keep it in place.
     
  7. Oct 19, 2016 #6
    The problem statement isn't very clear. What they mean is that the lid is like a piece of rigid cardboard sitting on top of the rim of the cylinder, and held in place by atmospheric pressure. The pressure inside the cylinder is less than atmospheric, so the rim of the cylinder is pushing upward on the lid. So there are initially three forces acting on the lid: atmospheric pressure pushing down, the gas pressure inside the cylinder (less than atmospheric) pushing up, and the rim of the cylinder pushing up.

    When you apply an additional upward force of 220 N, you are just able to separate the lid from the rim. This tells you that the rim force on the lid initially was 220 N.
     
  8. Oct 19, 2016 #7
    But, how is the force on the lid connect to temperature of the gas inside the container? I can't use PV=nRT because I don't know the moles or pressure of the gas. I don't see a way to get the pressure of the gas and getting to moles is probably not possible.
     
  9. Oct 19, 2016 #8
    You can get the pressure from the information I discussed. If the net downward force exerted by the atmosphere on the lid is equal to the atmospheric pressure times the cross sectional area of the cylinder and the upward force exerted by the rim of the cylinder on the lid is 220 N, what is the upward force exerted by the gas on the lid (neglecting the weight of the lid, and assuming it is in force equilibrium)? What is the pressure exerted by the gas on the lid?

    Regarding the number of moles of gas, what do the words ".021kg of air" mean to you?
     
  10. Oct 19, 2016 #9
    So the cross sectional area of the cylinder is pi * (.25 m/2)^2 = .0491 m^2
    So the net downwards force is .0491 m^2 * 101000 Pa = 4959 N
    So would the upwards force = 4959 N - 220 N = 4739 N ?
    If that is the case then the pressure by the gas would be 4759N / .0491 m^2 =96920 Pa

    For the moles should i just assume the molar mass of the air is 28.9 g/mol
    So the moles= 21/28.9 = .727 mol

    And answer to the problem would just be plugging it in to the formula which is (96920)(.0147) = (.727)(8.314)T
    T = 235.7 K
     
  11. Oct 20, 2016 #10
    I haven't checked your arithmetic, but your methodology is definitely correct.
     
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