Thank You Admiral Cheng!

If the universal gas constant is,

$R=2*\cfrac{4}{3}\pi$

then,

$pV=nRT=n.2*\cfrac{4}{3}\pi.T$

then,

$T$ is free within the space denoted by $4c^3$

when we compare quite arbitrarily with,

$A_{\small{D}}=\cfrac{4}{3}\pi(2c)^3$

equally arbitrarily if we let $R\rightarrow \cfrac{1}{2}R$ then,

$T$ is free within $(2c)^3$

$n$ multiplies one entangled particles sphere of radius $2c$, for $n$ many particle spheres in consideration.  This assumes that the sphere does not interact between themselves.  Which sets the limit on expression (*), for the  $n$ spheres do interact when they approaches light speed. So,

$T\lt(2c)^3$

When $T\rightarrow (2c)^3$, $n\rightarrow n_{c}$ where $n_c\lt n$.  When $T=(2c)^3$, $n=1$ because everyone is entangled and there is just one sphere.  ie.  Quantum effects on a "ideal" gas can be adjusted for by adjusting $n$

$n=n_c(T\rightarrow (2c)^3)$

that an ideal gas is not ideal when its constituents interact with each other, quantum effects or otherwise.

Does $T$ give energy per particle in the "Durian" sphere correctly.  No.  $T$ should be the estimated mean energy of the population in the sphere, such that,

$T.N_A= total\,\,energy$

given the distribution of $T$ in the population.

To scale $T$correctly,

 $R\rightarrow 1$.

The new smell like durian gas law is,

$pV=n.E[T]$

where $p$ is pressure, $V$  is volume, $n$ is the number of moles in consideration and $E[T]$ is estimated mean energy per particle given $T$ distribution among the population of the sphere.

Thank you Admiral Cheng Ho...