Relationship between the ideal gas constant and entropy

So it dawned on me the other day that the ideal gas constant $$R = \pu{8.31 J mol-1 K-1}$$ has the same units as molar entropy. Is there some deeper meaning behind this or is it just a coincidence that this occurs?




The gas constant is equal to Avogadro's constant times Boltzmann's constant, the latter serving as a proportionality constant between the average thermal (kinetic) energy of the particles in an ideal gas and the temperature:


$$\left(\frac{\partial \bar U}{\partial T}\right)\_p=\frac{3}{2}k\_\mathrm{B}$$


The entropy can be regarded as a proportionality constant between the change in free energy $G$ with change in temperature of a system at constant pressure, since


$$\left(\frac{\partial G}{\partial T}\right)\_p=-S$$


Note also that the average entropy of a particle can be written as


$$S=k\_\mathrm{B} \log\Omega$$


So both $R$ and $S$ can be regarded as proportionality constants between energy and temperature.


The Wikipedia page on the [equipartition theorem](https://en.wikipedia.org/wiki/Equipartition_theorem#Relation_to_the_virial_theorem) may provide some enlightenment on the origin of these proportionalities.


 6

The fundamental equation is Boltzmann's for the entropy $S =k\_B\ln(\Omega)$ where $k\_B$, Boltzmann's constant, has units J/molecule/K. When we use $R$ instead of $k\_B$ it is trivially because molar units are used to define $S$. 


In the equation $\Omega$ is the number of arrangements or configurations (or 'complexions' to use an old word) of distinguishable 'particles' among all the available energy levels. In that sense entropy is a measure of the uniformity of population in these levels.


(If there are $N$ distinguishable particles then $\displaystyle \Omega =\frac{N!}{n\_1!n\_2!n\_3!\cdots}$ where there are $n\_i$ particles in level $i$. The log is evaluated by assuming all the numbers $N,\, n\_i$ etc. are large numbers so that Stirling's approximation for $\ln(N!) $ etc. is valid.)


 6

In the ideal gas, $3R/2$ is the thermodynamic entropy of one Mol.


The factor $\frac{3}{2}$ comes from $ST=N\bar{E}=N\frac{3k}{2}T=\frac{3R}{2}T$.


The energy of 1 Mol of ideal gas is either given by $3pV/2$ or $3RT/2$ or $ST$.


(In the ideal gas) *the entropy does not depend on temperature*.
That is why one writes the $T$ separate in $TS$.
For non-ideal gases $S$ changes with temperature: $E=\iint dSdT$.


For the ideal gass $C\_V = S$ (always per Mol here).
$ST = \int dQ\_{\text{rev}} = C\_V \int dT = S \int dT$.
Because of constant volume, all the energy goes into $T$.
$C\_V$ and $S$ stay constant.
On the $pV$ side this amounts to an increase in pressure.


 1