I've asked a number of chemists why the ideal gas constant is in the equation for Gibbs free energy, and all say the same thing: "The name is a relic of the argon gas chamber used to determine energy and temperature at different pressures and temperatures, but it applies to every thermodynamic system." This sounds great and I understand names of constants can be relics, but why can a conversion factor that only appears to pertain to ideal gases be used for other substances? R assumes elastic collisions, where energy is not lost during molecular collisions, but only ideal gases have this property. Wouldn't you use a different conversion factor for something much denser and fluid like water or even more dense like lead to convert from temperature to energy? I've heard that R is used for free electrons as well, which are pretty different from atoms much less gases. Would "R-specific" be best to use everywhere, as I've heard they do in engineering, so maybe every substance has its own temperature-energy conversion factor?

  • $\begingroup$ Every substance does indeed have its own temperature-energy conversion factor. It is different for different substances, and generally has nothing to do with R. $\endgroup$ – Ivan Neretin Apr 9 at 20:45
  • $\begingroup$ YOu should do some research on "kinetic theory of gases" and the "Dulong-Petit Law" $\endgroup$ – A.K. Apr 10 at 0:55
  • $\begingroup$ Thank you. Ivan, I see R listed in Gibbs equation, namely the expression RTlnQ. Maybe “this R” is different from “that R” you’re suggesting. Should I take the ideal gas constant and divide by the molecular weight of the substance? I heard that’s “specific R”. Or is there a table of R’s for lots of elements and molecules in various phases? Let’s assume I’m not working with gases at all, like a deep-sea sediment (pure silica fine sand). Could I take R and divide by silica’s MW to get an adjusted energy-temperature conversion factor? $\endgroup$ – onnea Apr 11 at 4:54
  • $\begingroup$ @IvanNeretin I commented below, didn’t hit reply directly. (New to this site.) $\endgroup$ – onnea Apr 11 at 21:55
  • $\begingroup$ No, R from RTlnQ is really the same everywhere, and there are profound reasons for that. I wouldn't call it a temperature-energy conversion factor, though. $\endgroup$ – Ivan Neretin Apr 11 at 22:48

It is a very good question: Why does R appear everywhere in equations even though it is apparently related to or derived from the ideal gas law? Yes historically R was derived from the limiting gas law in the 1830s.

The beauty of R is that it is related to two physical constants called the Boltzmann constant and Avogadro's constant. Read here


The Boltzmann constant appears everywhere in thermodynamics, semiconductors and statistical mechanics: https://en.wikipedia.org/wiki/Boltzmann_constant

"The name is a relic of the argon gas chamber used to determine energy and temperature at different pressures and temperatures, but it applies to every thermodynamic system."

I feel this is nothing but fiction. Not sure why people make up stories. William B Jensen traces the history of R in the Journal of Chemical Education.

  • $\begingroup$ Thank you! I suppose the next question is does the Boltzmann constant apply to all molecules in all phases? I heard that a mathematical constant is a non-variable value (so any number not a specific number), whereas a physical constant is an immutable value no matter the context. Could a gas like argon at room temp have one Boltzmann constant and a solid like lead at room temp have a different Boltzmann constant? Avagadro’s number sounds like a physical constant whereas the Boltzmann constant sounds like a mathematical constant, subject to the constext, then R responds in kind as their ratio. $\endgroup$ – onnea Apr 11 at 5:00

The profoundly universal significance of $\rm\bf R$ ultimately stems from Boltzmann's equipartition theorem, which states that (roughly speaking) you are guaranteed to have $\rm kT/2$ of energy per one degree of freedom of whatever nature, or, consequently, $\rm{\bf R}T/2$ per mole of those. It is only in this sense that $\rm\bf R$ may be perceived as a some kind of intermediary between temperature and energy. Real compounds all have different number of degrees of freedom, and hence different temperature-to-energy coefficients (heat capacities, as we call them).

So you see that $\rm\bf R$ is not derived from ideal gases at all, to the point that calling it the "ideal gas constant" might be considered a misnomer. Their relation is of different nature. Like all ideal things, ideal gases are simple; they only have 3 well-defined degrees of freedom per particle, hence the simple formula for their heat capacity. No wonder it is hinged on a universal constant.

Long story short, your bewilderment is misplaced. It is as if we happened to name $\rm CO$ "the Ford gas", because of its presence in the exhaust of Ford cars, and then marveled at its widespread appearance in nature, all the way up to deep space. How could old man Ford put it there? He didn't.

So it goes.

  • $\begingroup$ This makes a little more sense. It sounds like the ideal gas law shows gases use R and the Dulong-Petit law shows solids use something closer to 3R. I presume this is because the solid’s molecules have more degrees of freedom (or is it fewer?). I’ve never seen this adjustment in a thermodynamics calculation. Would that mean a liquid like water or a gel have a energy-temperature conversion not of R but between R and 3R? What would the conversion be for an ion in water, maybe around 1.5R because solvated ions move somewhat freely like a gas but bump into water and don’t fly about in a vacuum? $\endgroup$ – onnea Apr 12 at 15:34
  • $\begingroup$ A little? 😠 Thanks for that! Now to the point. You'd better stop using vague terms like "use R" (what does that even mean?) or "energy-temperature conversion" (which is technically OK, but this is not the way people in the field call it). Yes, an ideal gas has heat capacity $c_v={3\over2}R$, which translates to $c_p={5\over2}R$, which grows to ${7\over2}R$ for diatomic gases and then even more. And yes, a solid within a certain idealized model would have $c_v=c_p=3R$, which is pretty much what Dulong and Petit say, and which is more or less true for pretty many elements. $\endgroup$ – Ivan Neretin Apr 12 at 19:21

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