Why is there a constant in the ideal gas law?

Question

Why do we have constants?

Consider, for example, the ideal gas law,

$$PV = nRT \, . \tag{ideal gas law}$$

Sometimes I believe that the constant is there in order to make the equation work (make the units line up per se), but other times I feel like such assumptions are unnecessary.

I don't entirely understand why that constant is used, besides the fact that it is necessary for the units.

NB/ This is not intended to stir philosophical debate. I am purely curious of the nature of constants in cases such as $R$ (not $c$ as I understand that the speed of light is uniformly constantly) I am simply asking whether these constants are necessary for our equations and understandings or if they are universally constant.

Answer 1

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.

Answer 2

1. Constants are used to convert between quantities of different dimensions.

2. Take the case of $I(t) = I_0\sin(\omega t)$, for example. The argument of the $\sin$-function must be dimensionless. Therefore, if $t$ has dimensions of time, we need to multiply it by a constant with dimensions of inverse time so that the argument is dimensionless. Thus $\omega$ is defined such that $\omega t$ is dimensionless. Similarly, if $I(t)$ has dimensions of current, we need another constant, $I_0$ to make the right hand side also have dimensions of current. Moreover, if the amplitude of the current is, say, 5 Amps, we express that in the constant $I_0$.

3. In the case of the ideal gas law we want $P$, $V$, and $T$ to have different dimensions. The constant $R$ (or $k_B$), scales and relate the dimensions on the right hand side with the dimensions on the left hand side: namely temperature to pressure (force per area).

4. Note that for the case of the ideal gas law, it would be perfectly okay to write $PV = NT$; you would just have to understand that $T$ now means something different, i.e. temperature would have dimensions of energy, which is perfectly reasonable as described in this other post.

Answer 3

Constants have two important role in any mathematical equations . 1- They make dimensions equal on both sides of equation. 2- They multiply or add up to give the correct value of the expression ,and this value is determined by experiments.

For example F=Gm1m2/r^2

Here the G has both the purpose by taking the value 6.674 08 x 10-11 m3kg-1 s-2 it is giving up the exact force which when two masses of 1 kg each will exert on each other when kept 1 m apart.

And secondly by having dimension of m3kg-1 s-2 it is making the dimension of the whole expression equal to dimension of force.