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Force and pressure done by molecules

  1. Jan 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Molecules of hydrogen, each has mass of 3.3 x 10-27 kg, move with speed 1.6 kms-1 hit a wall at angle π/3 rad to the normal. If there are 2.0 x 1020 molecules s-1 hit the area of 1.2 x 10-4 m2, find:
    a. the average normal force on the wall if all the molecules are absorbed by the wall
    b. the average normal force on the wall if all the molecules are reflected
    c. pressure exerted by case (a) and (b)


    2. Relevant equations
    P = F/A
    p = mv (momentum)
    Δp = F.Δt

    3. The attempt at a solution
    Δp = F.Δt
    Nmv = F cos (π/3) . t ; N = number of molecules
    F = (N/t . m . v) / cos π/3

    What is the difference when the molecules are absorbed and when reflected? Thanks
     
  2. jcsd
  3. Jan 23, 2012 #2

    ehild

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    The absorbed molecule stays in rest after the collision, so it loses the initial normal component of momentum. The normal component of momentum of the reflected particle changes to the opposite.

    ehild
     
  4. Jan 23, 2012 #3
    Oh I see. I think it will be like this:
    a.
    Δp = F.Δt
    m(v2 - v1) = F . t

    Because the molecules are absorbed, v2 = 0 and v1 = v cos π/3

    b. Because the molecules are absorbed, v2 = - v cos π/3 and v1 = v cos π/3

    Correct? Thanks
     
  5. Jan 23, 2012 #4

    ehild

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    You meant reflected in the second case, did you not? Then correct.

    ehild
     
  6. Jan 24, 2012 #5
    Ah yes. I meant reflected, not absorbed.

    Thanks :smile:
     
  7. Jan 24, 2012 #6

    ehild

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    You are welcome. Have you got the solution?


    ehild
     
  8. Jan 25, 2012 #7
    Haven't finished it, but this is my idea:

    a. all the molecules are absorbed
    Δp = F.Δt
    m(v2 - v1) = F . t
    -mv1 = F.t
    F = -N/t . m0 . v cos π/3 ,where N is number of molecules and m0 is the mass of one molecule

    b. all the molecules are reflected
    m(v2 - v1) = F . t
    m(- v cos π/3 - v cos π/3) = F.t

    The force for (b) is twice of (a)

    Am I correct? Thanks
     
  9. Jan 25, 2012 #8

    ehild

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    Yes, but the sign. You calculated the force the wall exerts on a molecule and multiplied it by N/t: You need to multiply the force a molecule exerts on the force, which is just of the opposite sign. And you need to give the numerical values.

    ehild
     
  10. Jan 25, 2012 #9
    For (a), the force the molecules exert on the wall is:
    F = N/t . m0 . v cos π/3
    = 2.0 x 1020 x 3.3 x 10-27 x 1.6 x 103 cos π/3
    = 5.28 x 10-4 N
     
  11. Jan 25, 2012 #10

    ehild

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    Right!

    ehild
     
  12. Jan 25, 2012 #11
    Thanks again :smile:
     
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