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I used this resource (http://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1151&context=physicskatz) to derive the ideal gas law from the kinetic model of gases:

$$PV=n(2/3)N_oT$$ with T being equal to $(1/2)mv_{average}^2$.

And now I'm super confused. Does this mean that the universal gas constant R is simply equal to (2/3)*avogadro's number?

Nova
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1 Answers1

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$T$ here does not refer to temperature but rather the average kinetic energy of a molecule.

I can see why there is some confusion. Sadly we only have so many symbols and kinetic energy also happens to be commonly denoted with $T$.

To relate the average kinetic energy to the temperature, you need to bring in equipartition theory

$$\frac{1}{2}m\langle v^2\rangle = \frac{3}{2}kT$$

(assuming three translational degrees of freedom only). If you substitute that into your above equation

$$pV = \frac{2}{3}nN_\mathrm{A}\cdot m\langle v^2\rangle$$

then you will find that $pV = n(N_\mathrm{A}k_\mathrm{B})T \Longrightarrow R = N_\mathrm{A}k_\mathrm{B}$, hopefully a familiar expression. The question of how the value of $R$ is derived then becomes a question of how equipartition is derived - and that deserves another question itself.

orthocresol
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    As to the equipartition theorem a derivation is given here http://chemistry.stackexchange.com/questions/61255/derivation-of-mean-kinetic-energy/61332#61332 which while it looks very hard mathematically it is quite easy to follow. – porphyrin Feb 16 '17 at 09:44