# Working out the mean velocity of particles in a gas

I'm trying to answer the following question:

Air consists of molecules Oxygen (Molecular mass = 32$amu$) and Nitrogen (Molecular mass = 28$amu$). Calculate the two mean translational kinetic energies of Oxygen and Nitrogen at 20($^\circ C$)

To solve it I have done:

Use $E = \frac{3}{2}kT$

Energy = $\frac{3}{2} \times (1.38 \times 10^{-23}) \times (20+273) = 6.07 \times 10^{-21}$

Use $KE = \frac{1}{2}mv^2$:

For oxygen: $\sqrt{\frac{6.07 \times 10^{-21}}{2 \times (32 \div 6.02\times 10^{23})}} = 15.11$

However, 15.11 isn't the answer in the textbook (the answer is 480m/s)

For nitrogen: $\sqrt{\frac{6.07 \times 10^{-21}}{2 \times (28 \div6.02\times 10^{23}) \div 32}} = 16.12$

16.12 isn't the answer either (it's 510m/s)

I know that my answers are wrong (gas molecules don't move as slow as I calculated at room temperature) but I can't see why my method doesn't work. Any help?

-
Could you please try to add the units into your calculations which will help you to see why your formulae are wrong? At the end, the error boils down to a multiplicative factor of $\sqrt{1,000}$ because your method to treat Avogadro's constant and the amu unit of mass assumes that one gram is the "desired" SI unit of mass (working with amu means assuming 1 g/mole en.wikipedia.org/wiki/Atomic_mass_unit ) while it should be one kilogram. –  Luboš Motl Dec 26 '12 at 8:21
I see, if you put that as an answer I'll accept it. Thanks! –  Todd Davies Dec 26 '12 at 8:26

It may be a good idea to add the appropriate units in your calculations. Doing so will help you to localize the mistake. At the end, your results are only wrong because of a multiplicative factor of $\sqrt{1,000}\sim 31.7$ that must be added to your result to obtain the right one.

It is actually not hard to see where this wrong factor comes from.

http://en.wikipedia.org/wiki/Atomic_mass_unit

In your denominators, you used a value for the mass of the nuclei that is based on the ratio of the type $32$ divided by Avogadro's constant (number of particles per mole). However, in this way, you obtain the value that assumes the natural conversion factor $1\,{\rm g/mole}$: Avogadro's constant was originally defined as the number of molecules in one gram-molecule. However, you want to get the masses in kilograms – and the proton mass is about $1.66\times 10^{-27}\,{\rm kg}$, to proceed in the SI units.

So effectively, the right easiest fix of your formulae is either to substitute the explicit masses in kilograms or to replace your Avogadro's constant by $6.023\times 10^{26}/{\rm mole}$ whose numerical value is the number of molecules in one kilogram-molecule (note the kilo) or, equivalently, keep Avogadro's and replace $32$ by $0.032$ etc. Then you get the right results within some tiny error margins (the masses 32 and 28 amu aren't quite accurate: proton and neutron masses differ and there are additional corrections from electrons and from nuclear binding energies).

-