In retrospect,
\(\cfrac{V}{n}=\cfrac{1}{\rho_n}\ne \cfrac{dn_s}{df}\)
of is it?
Not,
\(PV=nRT\)
but,
\(PV=[T_{boom}]R\)
\(n=1\)
where for any instance of \(T_{boom}\) as part of a distribution,
\(\cfrac{E_a}{kT_{boom}}=\cfrac{1}{m}\)
\(T_{boom}\) is associated with each particle and with
\(mT_a=T_{boom}\)
where \(T_a\) is associated with each particle without aggregation. Energy states is separately accounted for by \(m\), such that,
\(\cfrac{E_a}{kT_{boom}}=\cfrac{1}{m}\)
but,
\(\cfrac{E_a}{kT_a}=\cfrac{1}{m}m=1\)
please refer to the post "An Old Friend For New Reasons" 17 Nov 2017. With \(E[m]=n_{\small{E}}\)
\(PV=n_{\small{E}}T_a.R\)
the number of mole, \(n=1\). \(T_a\) measures the kinetic energy of a single, not clustered particle.
OK?
