I have found it mentioned in various sources that the value of R depends on the amount of substance $n$ in the ideal gas equation: $PV=nRT$ If this is true, it would imply that $R$ is a function of $n$ and we can replace $R$ by that function. I want to know what is the relationship between $R$ and $n$. If $R=f(n)$ then what is $f(n)$?

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    $\begingroup$ The universal gas constant is given by $R=k_\text{B}N_\text{A}$ where $k_\text{B}$ is the Boltzmann constant and $N_\text{A}$ is Avogadro's constant. $\endgroup$ – Paul Nov 16 '16 at 15:05
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    $\begingroup$ I do not see how R and n being on one side of an equation implies that one is the function of the other. That is simply not true. $\endgroup$ – Jon Custer Nov 16 '16 at 15:16
  • $\begingroup$ @JonCuster I am not saying that them being on the same side of the equation implies that they are related(Read the question before commenting), secondly if you are saying that if two variables on one side of the equation cannot be related, you are again wrong. For example: If someone says that the area of a circle =pi*r*(d/2), then they are not wrong. It may be possible that the value of d=2r may not be substituted in the equation. $\endgroup$ – Aniansh Nov 16 '16 at 15:27
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    $\begingroup$ The answer is that R is a constant (the universal gas constant) and has no relation to the number of moles. I still do not see anything that would imply that R = f(n) from the ideal gas equation. $\endgroup$ – Jon Custer Nov 16 '16 at 15:29
  • $\begingroup$ Your book states that for a given amount of gas $PV/T$ is a constant which they define as $K$. When you take twice the amount of gas, $K$ becomes two times bigger. $K$ depends linearly on the number of moles $n$, thus you can write $K=nR$ where $R$ is a new constant that defines the increase in the constant $K$ per mole of gas. $\endgroup$ – Paul Nov 16 '16 at 19:18

I think you are misunderstanding something badly here. Let’s take a sentence that may be the cause of your misunderstanding:

Where $R$ is a constant of proportionality and is found to be independent of the nature of the gas and depends only on the amount of gas taken. For 1 mole of any gas the value of $R$ is the same. (emphasis mine.)

Proportionality constants are the $m$ in a typical linear equation $(1)$. With them and a value $t$, which is potentially present (but physics and chemistry attempt to choose variables in such a way that $t = 0$) it is possible to calculate $x$ from $y$ or vice-versa assuming they are in a linear relationship. As mentioned, it is a constant and thus neither a function of $x$ nor one of $y$.

$$y = mx + t\tag{1}$$

The second sentence I highlighted also includes a potential source of confusion. You might have understood this to mean ‘the value of $R$ is the same for any given gas assuming that exactly $1~\mathrm{mol}$ of it is present.’ However, that is the wrong interpretation. It should be taken to mean: ‘Whichever the chosen gas may be, the $R$ in the equation above will be the same — whether we are calculating with $1~\mathrm{mol}$ or $2~\mathrm{mol}$ or $10~\mathrm{mol}$.’

Indeed, $R$ happens to be a physical constant invariable of the chosen gas. It can be related to other physical constants by equation $(2)$, wherein $k_\mathrm{B}$ is the Boltzmann constant and $N_\mathrm{A}$ is Avogadro’s constant; both of which also invariable with respect to the chosen gas.

$$R = k_\mathrm{B} N_\mathrm{A}\tag{2}$$


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