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Archived RMS Speed of a Gas Molecule

  1. Feb 5, 2008 #1
    1. The problem statement, all variables and given/known data
    The atmosphere is composed primarily of nitrogen N2 (78%), and oxygen O2 (21%). Find the rms speed of N2 and O2 at 293K


    2. Relevant equations
    Vrms=[tex]\sqrt{}((3RT)/M)[/tex]


    3. The attempt at a solution
    Vrms, O2=[tex]\sqrt{}((3*8.31J*293K)/(32g/mol))[/tex] =15.11m/s

    When I looked at the answer to the book, it was 478 m/s because instead of putting 32g/mol in the denominator, they converted it to 0.032 kg/mol. Can somebody explain to me why the authors of my book decided to do that? Is my first answer still correct? Or do I need to convert to kg every time I have to do a RMS problem?

    Thank you for your time
     
  2. jcsd
  3. Mar 7, 2016 #2
    It is used in terms of kg because Joules is a unit that involves kg (kg*m^2*s^-2). In order for it to work out, you have to divide by the same units.
     
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