Archived RMS Speed of a Gas Molecule

1. Feb 5, 2008

ChopChop

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=$$\sqrt{}((3RT)/M)$$

3. The attempt at a solution
Vrms, O2=$$\sqrt{}((3*8.31J*293K)/(32g/mol))$$ =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. Mar 7, 2016

LukeEWilliams

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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