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What equation to use

  1. Sep 7, 2007 #1
    1. The problem statement, all variables and given/known data

    If 1.0 x 10^23 carbon dioxide molecules strike 1.0 cm2 of wall per second at a 90o angle to the wall when moving toward it with a speed of 45,000 cm/s, what pressure (in atm) do they exert on the wall?

    does anybody know what equation to use for this problem ??
     
  2. jcsd
  3. Sep 7, 2007 #2

    robphy

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    Start at the beginning....

    what is the definition of pressure?
     
  4. Sep 7, 2007 #3
    well iit can't be P=nRT/V
     
  5. Sep 7, 2007 #4

    robphy

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    That's correct... that is not the definition of pressure.. but merely an important equation that is applicable in certain situations.
     
  6. Sep 7, 2007 #5
    well cant i just convert the molecules into grams, then into kilograms then use F=MA
    to get force then convert it into atm, but what do i do with the 1.0cm2 wall, those that even matter
    oh and the definition of pressure is P=F/A
     
  7. Sep 7, 2007 #6

    Dick

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    F/A is right. I think robphy would now ask you if you remember that force is also the time rate of change of momentum. Since momentum=mv and F=ma and a=dv/dt. What's the time rate of change of momentum for your collection of molecules?
     
  8. Sep 11, 2007 #7
    ok well i cant use F=ma since the problem has a velocity and not acceleration, but momentum can be used in this way, i hope iam right :)

    so F = P/t
    So for P =(CO2 in Kg * 450m/s) then F= (CO2Kg*450m/s)/(1sec) then when i get Force put it bake into P=F/A
    so then

    P= (CO2Kg*450m/s)/(1sec)/(1.00cm^2) those this seem right ??
     
  9. Sep 11, 2007 #8

    Dick

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    Does the CO2 bounce off the wall or stick to it? There's a factor of two in the momentum change between the two cases since after the bounce they are now going 450m/sec AWAY from the wall. I'm assuming you mean them to bounce.
     
  10. Sep 11, 2007 #9
    well since its a gas it would bounce of the wall and this is the question again

    "If 1.0 x 10^23 carbon dioxide molecules strike 1.0 cm2 of wall per second at a 90o angle to the wall when moving toward it with a speed of 45,000 cm/s, what pressure (in atm) do they exert on the wall?"

    and this is what i got when i sloved it

    P=(.007308Kg of CO2)(450m/s)
    P= 3.28869
    so then

    F= 3.288869/1sec
    F= 3.28869
    and fianlly

    P= 3.28869/.0001m^2
    i converted the area of 1.00cm^2 to .0001m^2 and used this as area instead does this seem right
    and my finaly answer in Pressure in Pascal = 32886.91
     
  11. Sep 11, 2007 #10

    Dick

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    Your numbers seem right, except you weren't listening. F is not equal to P/t. It's equal to delta(P)/delta(t). P is +mv before it hits the wall and -mv after it hits the wall. What's the delta P?
     
    Last edited: Sep 12, 2007
  12. Sep 12, 2007 #11

    Dick

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    And before someone stomps me here, yes, the real expression is F=dP/dt. It's a rate.
     
  13. Sep 12, 2007 #12
    well then iam just confued now
     
  14. Sep 12, 2007 #13

    Dick

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    The force exerted by the gas on a patch of the wall is equal to the rate at which that patch of wall is changing the momentum of the gas molecules. Your answer is off by a factor of two, because to change P to -P I need a delta(P) of 2*P. I seem to be just repeating myself...
     
  15. Sep 12, 2007 #14
    so would be something like this if iam understanding right

    and this is what i got when i sloved it

    P=(.007308Kg of CO2)(450m/s)*2
    P= 6.57738
    so then

    F= 6.57738/1sec
    F= 6.57738
    and fianlly

    P= 6.57738/.0001m^2
    i converted the area of 1.00cm^2 to .0001m^2 and used this as area instead does this seem right
    and my finaly answer in Pressure in Pascal = 65773.8
     
  16. Sep 13, 2007 #15

    Dick

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    Yes. That looks better.
     
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