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Units problem

  1. Oct 9, 2004 #1
    What does (kg)(m/s^2) = what unit?

    also does 1 cm^3= 1 mL?

    i forgot all these units stuff its really bugging up my mind
     
  2. jcsd
  3. Oct 9, 2004 #2

    Pyrrhus

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    (kg)(m/s^2) = Newton

    1 cm^3= 1 mL Yes
     
  4. Oct 9, 2004 #3
    ok, im confused, ive been stuck on this problem for about an hour

    So far i got this in my calculations:

    (5.316 L*atm) / ( (211.68 kg*m/s^2)+(.0073m^2*atm))

    the answer is suppose to be in meters, and from these units i have no idea how it is possible..anyone have any ideas?
     
  5. Oct 9, 2004 #4

    Pyrrhus

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    Show your work, by the way, L is liters right?
     
  6. Oct 9, 2004 #5
    oh yeah and 1 more question, if u add units, then what happens?

    like for example, if u have 8 atm + 3 liters, will it become 11 (atm+liters)?
     
  7. Oct 9, 2004 #6
    k here is my work:

    the equation is this:

    h=n*R*T/(m*g+Po*A)

    n= 1.85 moles
    R=gas constant
    T= 350 K
    A= .00730 m^2
    m= 21.6 kg

    and Po is atmospheric pressure (1 atm or 1.0135e5 kPa)
     
  8. Oct 9, 2004 #7
    Absolutely Not!
     
  9. Oct 9, 2004 #8

    Pyrrhus

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    I don't believe you can do that, look up atm basic units, i mean Newton is kg*m/s^2, what is atm equal to? and use the relation above cm^3 = mL, and see if you can follow from that. This problems can be solved by converting to the same types of units, and simple elimination of analogous units.
     
  10. Oct 9, 2004 #9
    that last post was directed at the previous post...about adding units
     
  11. Oct 9, 2004 #10

    Pyrrhus

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    a Pascal is N/m^2, use that. Also 1 atm = 101325 pascal.
     
  12. Oct 9, 2004 #11
    ok, so i ended up with this:

    5.315975 L*atm / (211.68 N +739.6725 m^2*Pa)

    but i am stuck here, how can i cancel out further units? since the bottom is addition, i have no idea what to do from here...am i still allowed to convert the m^2 to liters and cance it out with the top, as well as the Pa with the atm on top?
     
  13. Oct 9, 2004 #12
    well, a pascal times m^2 is just Newtons!

    so you have all newtons on the bottom....then instead of using that gas constant...use the one that is J * mol^-1 * K^-1

    then you have J / N.....which is meters (change them to their base forms if you want to double check it :) )
     
  14. Oct 9, 2004 #13

    Pyrrhus

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    nicely done :smile:
     
  15. Oct 9, 2004 #14
    gas constant:

    8.314 J * mol^-1 * K^-1

    instead of

    8.2057 L * atm * mol^-1 * K^-1

    there are a few other forms of the gas constant too, but these are the most commonly used ones.
     
  16. Oct 9, 2004 #15
    thx!! i finally got the right answer..this units are really bugging

    does anyone know a website that has all these info?
     
  17. Oct 9, 2004 #16

    Pyrrhus

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    It should be on your textbook, if not try google.
     
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