I didn't think the information I looked at did a very good job of explaining the Boltzmann Constant, kB, but once I did this next simple calculation, it seemed clear that what I didn't understand were the units. So I simplified the units only:

$8.31... = R$ = energy per degree/mole

NA = molecules/mole, Avogadro's constant

kB = R/NA = (energy per degree/mole)/(molecules/moles)

kB = energy per degree/mole * moles/molecules

kB = energy per degree per molecule

Since kB is in units of energy per deg K per molecule, multiplying it by T in units of deg K will arrive at total energy per molecule. Multiplying that by the number of molecules gives total energy of the mass of air, same as PV. Energy per degree per molecule x degrees x molecules = energy.

But every reference I can find states that the Boltzmann constant is in units of energy per temperature, or joules/deg. Kelvin. What am I missing?

It seems to me that kB is joules per degree per molecule, while R is joules per degree per mole.


3 Answers 3


The correct unit is joules per kelvin oscillator. A molecule might have several modes of vibration. The theory runs numeric constants for oscillators / molecule.

SI is not a coherent system here, because everyone else derives moles in the same way as electron volts are: a unit and constant. The constant is a dalton (currently a unified mass unit: dalton has been derived for the generic name), the unit is M, and a M-mole is M/(1 dalton). SI uses M in grams here :o.

Since $k=R/N_A$ one sees the unit is (J/kg-mol. K) / (dalton/kg) = (J/Da-mol K) A dalton-mole is by definition, 1, so one gets J/K. If one starts to add in a unit "molecule" for dalton-mole, one gives the impression that there is only one oscillator per molecule, or that $k_B$ applies only to molecular matter.

The theory of black body radiation does not call down to molecular matter, but the boltzman constant is found there too. It is better to think of heat as being made of thermions or modes of energy, the actual energy is thermion * temperature, in much the same way that planck's constant gives energy = h * wavelength * photons.

Thermodynamics normally handles 'oscillators/molecule' as a numeric constant, often simply putting in the resolved value (eg 1.4). So the energy per molecule is not $kT$, but some numeric like $1.4kT$ etc. Oscillators are distributed over molecules, not over weight. So all di-atomic gasses, like H_2 or N_2 or O_2, have the same number of oscillators per molecule, or per lb-mole. In general, an equipartition rule holds, so each thermion tends to have a similar energy, but that is a subject of thermodynamics, not of metrology.

Likewise, 'h' is a measure per oscillator, and $N_A h$ molar planck constant is also known. h and its kith $\hbar$ are likewise missing units, (correctly, these are J.s/cycle = J/Hz, and J.s/rad respectively. Crossing the wavelength $\lambda$ converts that to a radian-length, so $E = hc/\lambda$ works when both are in cycles or in radians.

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    $\begingroup$ +1 for the beginnings of a very good answer - I like your turn of phrase "joules per kelvin oscillator": perhaps you could explain a bit more of what you clearly understand: e.g. equipartition of energy and so forth. $\endgroup$ May 28, 2014 at 2:13
  • $\begingroup$ So you're saying that my math is right, but because of other uses of Boltzmann constant, "per molecule" would generally be inappropriate? $\endgroup$
    – Luther
    May 28, 2014 at 3:37
  • $\begingroup$ The maths is in part wrong because SI is not coherent. A molecule might have 7 oscillators, and to suppose k=1.381 J/K would think that the energy is 1,381 J/molecule, when it is 1.381 J/thermion, and at 7 thermions/molecule, 9,667 J/molecule. This is why using molecules is a bad idea. $\endgroup$ May 28, 2014 at 3:42

Wendy Krieger's answer with her allusions to the principle of equipartition of energy principle is one take on your question. Here's another, information-theoretic / Jaynesian (see footnote) take on it.

The Boltzmann constant is simply a unit-dependent dimension-normalising factor that accounts for the fact that on the one hand (i) we have a "traditional" temperature scale in Kelvin, Celsius and so forth defined experimentally in terms of expansions of liquids and gasses in thermometers and then an arbitary assignment (fixed by the temperature unit definition) of scale to an experimentally reproducible temperature interval e.g. 100 units between the freezing and boiling points of water, whereas (ii) on the other hand, in a system comprising statistically independent "particles", the Shannon (information theoretic) entropy of the system is proportional to the total system energy which, in turn, is proportional to (i) the number of particles and (ii) the reciprocal of a certain Lagrange multiplier $\beta^{-1}$ (the latter only true in the "high temperature limit"; see for example my answer here where I show it is not true for low temperatures). $\beta$ is defined with the Boltzmann probability distribution, derived from the Canonical Ensemble, prevails at thermodynamic equilibrium and, in a system of statistically independent particles, one finds that the probability of a particle's being in the $\ell^{th}$ energy state is proportional to $\alpha\,e^{-\beta\,E_\ell}$, where $\alpha,\,\beta$ are simply a Lagrange multipliers explained in the Wiki artcile on the Canonical Ensemble. At low values of $\beta$ (more energetic systems containing more heat), $\beta^{-1}$ is proportional to the mean particle energy.

We define the thermodynamic temperature to be proportional to $\beta^{-1}$ and, experimentally, our traditional temperature measurements are found to agree with this notion if we choose the right "offset" to define zero temperature. The constant $k_B$ simply lets us keep our traditional temperature scales, or something near to them in practical cases, whilst being aware of the statistical mechanical interpretation of $T \propto \beta^{-1}$. In "natural" (Planck) units, $k_B$ is set to unity, by definition.

Not all "particles" though behave "atomically" in a thermodynamic sense: experimentally we find that we must treat them as though they behave as though they are made up of several particles if we are to reconcile statistical mechanics with e.g. observations of heat capacity. This is to do with the number of degrees of freedom, and this number can change with temperature as different modes of vibration are "frozen" in or out. So you are right in the sense that entropy is an extensive property, but it becomes a very messy notion if you insist on interpreting it as per "particle" as we would think of it.

So we simply think of $k_B$ as a scaling constant between "experimental" and "fundamental", information theoretic notions of temperature: the "hotter" something is, the greater the number of bits needed to endcode its state fully. The latter are important of course because microscopically Nature's laws are reversible, so an isolated system's state at any time is mapped bijectively to its state at any other time. Nature "does not forget" how it gets into a certain state, and a necessary condition for this to hold is that a system's information content cannot shrink, which of course is the second law of thermodynamics.

Footnote: Edwin T. Jaynes was a physicist, mathematician and philosopher who thought very deeply about the theoretical groundings of probability theory, inspired by an information theoretic / symmetry (group) theoretic take on thermodynamics and the wish to unite Shannon's information theory with Boltzmann / Gibbs thermodynamics.


The other answers are more complicated and right, and you learn more, but a simpler answer is just that "number of molecules" isn't a proper unit -- it is a count with an absolute interpretation, not a standard against which other units are measured. If they figure out how to speak english, aliens and us could agree on numbers of molecules, or counts of things, even if they never exchange a single apparatus with us. Our units, however, are simply standards that we've adopted due to the Enlightenment bent of the French Revolution, the conquest of Europe by Napoleon, and Western Imperialism. A resident of Gliese 583c would have no reason to understand what a meter is, or to use it.

  • $\begingroup$ As I suspected, the simple answer seems to be that molecules aren't units, they are objects. That is physics, but I haven't seen an answer to what's wrong with my algebra, so I assume my algebra is correct, aside from not following the practices of physics. So there must be a right way to say we're talking about molecules (Boltzmann constant is 1.38 x 10^-23 j/K, but wikipedia and all its copiers stop there and leave the molecules hanging as if they aren't there. But 1.38 etc does refer to molecules, so what is the right way to say this, so beginners like myself would not get so confused? $\endgroup$
    – Luther
    May 28, 2014 at 4:15
  • $\begingroup$ Not everything exists in molecules. NaCl is a lattice of alternating atomc, not di-atomic molecules. When one uses moles, You get a "chemical mass" which equates to a number of atom-packages. In an equation like NaOH+HCl-> H2O+ NaCl. You can deal with the equation if it's NaCl or Na2Cl2 or whatever. A lb-mole of NaCl is 1/2 lb-mole of Na2Cl2. A molecule, in the metrological sense, is then a weight, divided by a sum of daltons, to give a weighed number of packages as used in the formula. Not all of them are free floating packets of atoms. $\endgroup$ May 28, 2014 at 4:44
  • $\begingroup$ @wendy.krieger: but the key point remains -- N is a pure number that has no true unit. Different rules can change the count in situations, but it's still just a count, which is fundamentally different than, say, lengths, which are measured relative to an arbitrary standard (even appealing to a planck length doesn't change this -- one can define a $g_{0}$ such that $g_{0} = \frac{1}{4\pi G}$, which is no fundamentally different than the version using $G$, but $g_{0}, c$ and $\hbar$ give a different planck length than $G, c$ and $\hbar$, and there is no universal rule to say which is right.) $\endgroup$ May 28, 2014 at 4:58
  • $\begingroup$ @JerrySchirmer Quanties do not have dimensions or units. Quantities have scales, and scales have dimensions. Even though N is a number, the correct dimension for it is 'M'. That is, you meausre chemical-mass by weighing it and dividing it by a number that you want it to appear as. The tables remain the same, so water is 18 daltons, but the weights go by country, eg lb-mol = lb/Da, kgmol = kg/Da, gmol = gram / Da. SI is different here, by using kg and g-moles. You can measure mass, for example, in $L^2M/T^2Q$, which is done all the time in atomic physics. $\endgroup$ May 28, 2014 at 7:16
  • $\begingroup$ I am working with air and compressed air, thus my interest in PV = nRT etc. So I'd think molecules are a good place for me to stop, the heavy physics doesn't answer (for me) the question I was asking. If n is pounds or kilograms of air in the macroscopic view, then moles (a count of molecules) from my mechanistic viewpoint is as small and abstract as I want to get. My question only applies to where Boltzmann constant is referring to a molecule count. $\endgroup$
    – Luther
    May 28, 2014 at 8:27

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