
We have repeatedly
stressed the fact that
whenever you encounter Boltzmann's constant k we deal with the particle
unit "atom" or "molecule", while whenever we encounter the
gas constant R, we deal with the the unit "mol". Here we will
quickly survey the connection. 


First we have the general law for ideal gases with volume
V, pressure p and temperature T 





This was first an empirical law that later became fully understood
by statistical thermodynamics. 

The next step is to realize that if
you increase the volume while keeping everything else constant, the
"const" in the law must increase in the same proportion. This leads
to the much more universal formulation that is generally used: 





With n = quantity of the gas, and
R = gas constant with a value depending on how you measure
n. 

This would still leave room for
R being different for different kind of gases. Avogadro enters, proposing that identical
volumina of gases under identical pressure and temperature contain identical
numbers of particles. 

This permits to define R for
all (ideal) gases and to measure it. We find 


R 
= 
p · V
n_{mol} · T 
= 
8.32441 J · K^{–1} ·
mol^{–1} 




if me measure the
quantity n of the gas in mols. º 


One mol of a substance,
per definition, contains just as many
particles, objects, or building blocks of that substance (i.e. atoms,
molecules, electrons, vacancies, ...), as there are carbon atoms in 1 g
of ^{12}C which gives 





Avogadros constant
then automatically is 


N_{A} 
= 
6.022 · 10^{23} mol^{–1} 




i.e. we have 6.022 · 10^{23} particles
per mol of a substance. 

If we set n = 1 we have
for the molvolume V_{m}, i.e. for the volume that 1 mol
of a gas occupies 


V_{m} 
= 
n · R · T
p 
= 22.414 
l
mol 




This is valid for for "old" standard
conditions (p = 1013 mbar = 101 325 Pa, and T =
0^{o}C ). 


For the "new" standard conditions
(p = 100 000 Pa, T = 298.15 K) we have
V_{m} = 24.789 l/mol 


Why the international
standards and units of measurements must change all the time is beyond me, but
that's the way it is. I have suffered through 4 changes in the units for
pressure by now, not to mention the big pain caused by the fact that the
Americans normally don't care and still stick to psi. 

If we now measure substance
quantities not per mol, but per particle, we must divide R by Avogadros
constant N_{A} and obtain 


p·V 
= 
n_{part} · 
R
N_{A} 
· T = n_{part}
· kT 







and n_{part} is now the
number of particles in
V. 


For R/N_{A}
=: k = Boltzmann's
constant we obtain 


k 
= 
8.32441
6.022 · 10^{23} 

J · K^{–1} 
= 8.616 · 10^{–5 }eV ·
K^{–1} 



Fine, we can see that as a definition of Boltzmann's constant k. But now
we have two questions: 


1. Why is the k from the gas law
the same number as in the famous entropy
equation S = k · ln P ? 


Not obvious 
and not exactly easy to prove. Essentially, you have to unleash the full power
of statistical thermodynamics to show that both k's are identical. So
either grab your thermodynamic textbook, or believe your professor at this
point. 


2. Is there a way to calculate the
numerical value of k from some more fundamental constants? Well, as far
as I know, it cannot be done. So k
is a basic constant of nature, in the same
league as other fundamental constants of nature, like the speed of light, the
gravitational constant, or the elementary charge. 

Finally something to make things
really complicated: 


Changing from mols to particle
numbers or densities, changes
the precise formulation of the mass action law. Consult
the link for
details. 


