Broken Boltzmann Constant

What about absolute temperature, zero Kelvin?

The Kelvin temperature scale was defined so that there is no constant temperature term in the ideal gas equation as when the Celsius temperature scale is used.  Physically its suggests that if the ideal gas equation still applies at zero Kelvin, all molecular motions would stop.  The ideal gas equation does not apply when the temperature is low, which would then suggests that molecular motions will cease before zero Kelvin.  This is not true.  Molecular motion is still seen at temperature very close to zero Kelvin.

If a spinning positive temperature particle behaves like a partial positive charge, the electric potential associated with such a charge, $q_{T}$ is given by,

$V_{q_T}=\cfrac{q_{T}}{4\pi\varepsilon_o r_{q_{T}}}$

For a gas of $N$ molecules, the total electric potential due to all positive temperature particles is,

$V_{q_T}=n_{T}\cfrac{Nq_{T}}{4\pi\varepsilon_o r_{q_{T}}}$

where $n_{T}$ is the average number of positive temperature particles associated with each gas molecule.  $n_T$ may not be an integer.

$r_{q_T}$ is a hypothetical radius defining a sphere around each temperature particle that confines the partial charge $q_{T}$.

If $\psi$ around the temperature particle increases with temperature $T$ in Kelvin by the introduction of  a temperature particle ($\psi$s superimpose and sum to a resultant) with a corresponding small charge $\Delta q$ as given by,

$\Delta q=\cfrac{q_{T}}{K_{q_T}}\Delta T$

where $K_{q_T}$ is in Kelvin ($K$), then the change in work done, $\Delta W$, in introducing the change in temperature is,

$\Delta W=V_{q_T}.\Delta q=n_{T}\cfrac{Nq_{T}}{4\pi\varepsilon_o r_{q_{T}}}\cfrac{q_{T}}{K_{q_T}}\Delta T$

When $\Delta T$ is small,

$\cfrac{dW}{dT}=n_{T}\cfrac{Nq^2_{T}}{4\pi\varepsilon_o r_{q_{T}}}\cfrac{1}{K_{q_T}}$

the gradient of $W(T)$ at a specific point $T$.

In the simplest case when all parameters are independent of temperature, as in the case of the ideal gas equation,

$PV=W=Nk_BT$

where $k_B$ is the Boltzmann constant, we have

$W=\int{\cfrac{dW}{dT}}dT=n_{T}\cfrac{Nq^2_{T}}{4\pi\varepsilon_o r_{q_{T}}}\cfrac{1}{K_{q_T}}.T$

where we assume at $T=0$, $q_{T}=0$ and so, $W=0$.  Comparing this with the ideal gas equation,

$k_B=n_{T}\cfrac{q^2_{T}}{4\pi\varepsilon_o r_{q_{T}}}\cfrac{1}{K_{q_T}}$

If we further let,

$q_{T}=\alpha_{q_T}q$

where $\alpha_{q_T}$ is the fraction of the partial charge, and $q$ the unit charge,

$W=n_{T}\cfrac{N\alpha^2_{q_T}q^2}{4\pi\varepsilon_o r_{q_{T}}}\cfrac{1}{K_{q_T}}.T$

Furthermore,

$V_{T}=\cfrac{W}{q}=n_{T}\cfrac{N\alpha^2_{q_T}q}{4\pi\varepsilon_o r_{q_{T}}}\cfrac{1}{K_{q_T}}.T$

which makes sense when $V_{T}$ is the thermal voltage across a p-n junction encountered in semiconductors.  With $N=1$ when this voltage develops over a sharp boundary of a singular p type semiconductor beside a n type semiconductor, and so equivalently,

$N=1$

$k_B=n_{T}\cfrac{\alpha^2_{q_T}q^2}{4\pi\varepsilon_o r_{q_{T}}}\cfrac{1}{K_{q_T}}$

for

$V_T=\cfrac{k_B}{q}T$

We have just fractured $k_B$, the Boltzmann constant into four smaller pieces, $n_T$, the number of positive temperature particles associated with each gas molecule, $\alpha_{q_T}$, the partial charge on each temperature particle due to its spin,  $r_{q_{T}}$, the radius of the partial charge, and $\cfrac{1}{K_{q_T}}$, the fractional increase in partial charge due to increasing energy density $\psi$ per unit Kelvin increase in temperature.