Normalizing extensive quantities

Dividing two extensive quantities with the same replication factor produces an intensive quantity:Division by volume gives a density, such as mass density $\rho =m/V$ or energy density $U/V$. Division by mass gives a specific quantity, such as specific internal energy $u=U/m$ or specific volume $v=V/m$. Division by amount of substance gives a molar quantity, such as molar volume $V_m=V/n$. Their units are respectively $\mathrm{kg}{\mathrm{m}}^{-3}$, $\mathrm{J}{\mathrm{m}}^{-3}$, $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, ${\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}$ and ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. The denominator must be stated: “per particle”, “per mole” and “per unit mass” are different normalizations.

For a hypothetical uniform parcel with mass $2\mathrm{kg}$, volume $0.002{\mathrm{m}}^3$ and internal energy $40\mathrm{kJ}$ relative to a specified reference, $u=20\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$ and $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$. Three independent copies have $6\mathrm{kg}$, $0.006{\mathrm{m}}^3$ and $120\mathrm{kJ}$; both ratios remain unchanged. This permits comparison of materials without confusing the amount tested with the state tested.

If $X$ and $Y$ are additive but the parcels have different $r_i=X_i/Y_i$, their composite ratio is a weighted arithmetic mean:The denominator supplies the weights: specific internal energy uses mass weights, while mass density uses volume weights. This identity is accounting for the specified subsystems; predicting a new state after mixing also needs the relevant equation of state.