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Force exerted by a gas

  1. Dec 2, 2007 #1

    sun

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    [SOLVED] Force exerted by a gas

    1. The problem statement, all variables and given/known data

    A sealed cubical container 30.0 cm on a side contains three times Avogadro's number of molecules at a temperature of 23.0°C. Find the force exerted by the gas on one of the walls of the container.

    2. Relevant equations
    F=Nm/d(v^2)
    P=F/A


    3. The attempt at a solution
    There always seems to be a missing variable to plug into the equations. Are there any hints that can get me pointed in the correct direction?

    thanks
     
  2. jcsd
  3. Dec 2, 2007 #2

    Astronuc

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    Staff: Mentor

  4. Dec 2, 2007 #3

    sun

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    V of one mole of ideal gas=22.4x10^-3m^3
    V of container=2.7x10^-5
    T=296K
    R=8.31j/mol*k
    A=.54m^3
    n=?
     
    Last edited: Dec 2, 2007
  5. Dec 2, 2007 #4

    sun

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    This is what i've done:
    V=m(He)/density(He))
    m=.004Kg/(6.02x10^-23*3)=2.21x10^-27kg
    V(He)=2.21x10^-27kg/1.79x10^-1Kg/m^3=1.23x10^-26m^3

    Am i on the correct path? If so, all i need to figure out is how many moles there are, correct? Then i can solve for P.
     
  6. Dec 2, 2007 #5

    sun

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    does this essentially mean there are 3 moles?
     
  7. Dec 2, 2007 #6

    sun

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    any help would be appreciated :)
     
  8. Dec 3, 2007 #7

    sun

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    I'm not sure how to calculate the moles.

    I tried V(f)/V(i)=n(f)T(f)/n(i)T(i)

    when i solved for n(i) i got .0011mol. But i guess that isn't correct.
     
  9. Dec 3, 2007 #8

    sun

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    I also tried PV=NkT

    would N equal 3*(6.02*10^23)?
    If so, what i did was P(.027m^3)=1.81*10^24(1.381*10^-23)(296)
    i got 2.74*10^5

    Then i multiplied that by the area .54 and got 1.47*10^5N, but thats wrong.
     
  10. Dec 3, 2007 #9

    sun

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    argg still lost and confused
     
  11. Dec 3, 2007 #10

    Kurdt

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    Yes 3 times Avogadro's number means there are 3 moles of the gas. You want to use the ideal gas law with the molar constant R. Thus:

    [tex] PV = n R T [/tex]

    If you work out the pressure you can find the force exerted on one wall.
     
  12. Dec 3, 2007 #11

    sun

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    this is what i've done:
    P(.027m^3)=(3mol)(8.31J/mol*K)(296K)

    P=2.73*10^5

    F=2.73*10^5(.54)=1.47*10^55 N, but that is incorrect. any idea as to what i may be doing wrong.

    And thank you for your help.
     
  13. Dec 3, 2007 #12

    sun

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    do i divide 1.47*10^5 by 6, b/c there are 6 sides to the cube?
     
  14. Dec 3, 2007 #13

    Kurdt

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    The question asks for the force on one side, and the 0.54 you've used is the total area of the box.

    EDIT: or you could divide your answer by 6.
     
  15. Dec 3, 2007 #14

    sun

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    YES!!!!!!!!!!!!!!!!!!!!
    Thank you !!!
    I got it :)
    <3
     
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