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If $R=N_A\cdot k_B$ how can the following equation be true as well?

$$\frac{R}{M}=\frac{k_B}{m}$$

That would result in: $R=\frac{1}{n}\cdot k_B$, which is different from the first equation.

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    $\begingroup$ How can it be checked if the definition of $M$ and $m$ are not given ? $\endgroup$ – Frederic Thomas Feb 22 at 17:20
  • $\begingroup$ @FredericThomas I'm sorry: $m$ is the mass and $M$ is the molar mass; $n$ is the chemical amount. $\endgroup$ – Pedro Nogueira Feb 22 at 18:06
  • $\begingroup$ And $N_{A}$ is Avogadro's number. $\endgroup$ – Pedro Nogueira Feb 22 at 18:14
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How can it be checked if the definition of M and m are not given ?

– @Frederic Thomas

Indeed. You posit, $$ R=N_{A}k_{B} $$ and then, you posit, $$ R=\frac{M}{m}k_{B} $$ This only makes sense --I'm assuming the whole of chemistry and statistical physics makes sense--, if, $$ mN_{A}=M $$ So you must mean, $$ M=\textrm{mass of your 1-mole sample} $$ $$ m=\textrm{molecular mass of your microscopic species} $$

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  • $\begingroup$ I'm sorry for not specifying what everything was: $m$ is the mass and $M$ is the molar mass, as you correctly guessed. $n$ is the chemical amount and $N_{A}$ is Avogadro's number. The problem is $mN_{A}=M$ is not true, right? This is true instead: $mn=M$. How can that be? $\endgroup$ – Pedro Nogueira Feb 22 at 18:13
  • $\begingroup$ @PedroNogueira : You have to be careful with the notation. $m$ is the mass of what? In this answer it was shown that $m$ has to be the mass of a particle, and not the total mass. The total mass here is $m_T = N\, m$. You can relate the number of particles with the Avogrado constant and the amount of substance. $\endgroup$ – Jakob Feb 22 at 18:43
  • $\begingroup$ I have now come to realize that. That's the answer I was looking for. I just got it myself a minute ago. $\endgroup$ – Pedro Nogueira Feb 22 at 18:44

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