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RMS Speed of a Gas Molecule

by ChopChop
Tags: molecule, speed
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ChopChop
#1
Feb5-08, 01:20 AM
P: 6
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
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