Constants in physics are not just unit matching things. They are actually very fundamental. Yes, it is an heuristic and easy way to explain constants as unit keepers and I have nothing against that; but constants represent a sort of privileged group in nature. They are like symmetry points were everything moving around most do so in a way to keep their values the same.
Now for gas constant ($R$): it is an experimental constant.
Imagine that you have a thermos bottle filled with a gas having a piston at its top which you can pull/push, an electric resistance inside that you can use to heat the gas, a thermometer and a barometer. The thermometer and the barometer are placed in such a way they can give the temperature and the pressure of the gas inside the bottle.
At a certain moment you make a measurement of all these three parameters $p, V$ and $T$. Let’s say you get the values $p_0, V_0, T_0$. Now do any of the following:
Heat up the gas or pull/push the piston up/down. You can do all of that at once. After that perform a new measurement of the above parameters. Let’s say you get $p_1, V_1, T_1$.
You will realize that no matter what you do, in an isolated system, the values of the parameters $p, V$ and $T$ will always change in such a way that the ratio between the product $pV$ by $T$ is constant, i.e.,
$$φ=\frac{p_0 V_0}{T_0}=\frac{p_1 V_1}{T_1}=\frac{pV}{T}=constant \tag{1}$$
This means that, once you make an initial measurement and get a value for $φ$, in the future you’ll be required to measure just 2 of the parameters, and the third will be established using an equation of the form
$$pV=φT \tag{2}$$
The problem is, you cannot make any assumption about the general validity of equation (2). By this time, it is just and ad hoc equation which serves the purpose of your current setup or experiment. What if you increase/reduce the amount of gas inside the bottle? Or you change the gas type?
In the case of increasing/reducing the amount of gas inside, just as expected, the value of $φ$ will increase/reduce by the same proportion $n$ as the amount of gas added/removed. Or
$$φ =\frac{pV}{T}= nφ_0 \tag{3}$$
where $φ_0$ is the value of $φ$ for a unit amount of gas.
The big leap here is a discovery by Amadeo Avogadro known as Avogadro’s law, which in other words, says that, if one uses the amount of substance $n$ in terms of the number of moles instead of $\mathrm{kg}$ or $\mathrm{lbs}$, then, under the same conditions of $p$ and $T$ all gases occupy the same volume, i.e., the values of the $φ$’s are the same. He discovered that, for 1 mole of any gas under $1 \, \mathrm{atm}=101.325•10^5 \, \mathrm{ \frac{N}{m^2}}$ and $0 \, \mathrm{°C}= 273.15 \, \mathrm{K}$ the gas occupy $V_0=22.4•10^{-3} \, \mathrm{m^3}$.
Now we can generate an universal value for $φ_0$ as
$$φ_0=R=\frac{p_0 V_0}{T_0}=\frac{101.325 •10^5×22.4•10^{-3} \, \mathrm{\frac{N}{m^2}×m^3}}{273.15 \, \mathrm{K}}=8.3 \, \mathrm{J/K} \tag{4}$$
Now (2) can be written as
$$pV=nRT \tag{5}$$
and if we do so, we get a compact and universal form to describe the thermodynamic system.
But there is more in (5) then just a compact form of describing the thermodynamics system. As you can see in (4) the units of $pV$ turns out to be $J$. It actually represents total work done by an isolated thermodynamic system. Deriving (3) for the same amount of substance, we get
$$p \mathrm{d} V+V \mathrm{d} p=nR \mathrm{d}T \tag{6}$$
$p \mathrm{d} V$ is the so called expanding reversible work and $V \mathrm{d} p$ is the so called shaft work. Since in the right side of (4) the only variable is $T$ it gives a new meaning for temperature as some form of energy (or energy potential) of some sort, and we can understand heat as energy and not some kind of substance as it was thought in past.
