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...
