# Relation between the Ideal gas constant and Boltzmann's constant

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.

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

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, $$M=\textrm{mass of your 1-mole sample}$$ $$m=\textrm{molecular mass of your microscopic species}$$
• 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? – Pedro Nogueira Feb 22 at 18:13
• @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. – Jakob Feb 22 at 18:43