Intensive and extensive thermodynamic quantities

An extensive quantity measures how much of a thermodynamic system there is; an intensive quantity describes its state independently of that amount. The clean comparison is replication: take independent identical systems at the same thermodynamic equilibrium, with the same substance, composition and thermodynamic phase. Combining their accounting boundaries multiplies internal energy $U$, entropy $S$, volume $V$, mass $m$ and amount of substance $n$, while leaving temperature $T$, pressure $p$ and chemical potential $\mu$ unchanged.^[ref]

For $\lambda$ copies the ledger isAn integer $\lambda$ counts actual copies; a positive real $\lambda$ describes the corresponding bulk scaling law. Copies can remain separated: replication need not involve removing a partition, changing a container or performing any work. Distinguish $n$, the amount of substance measured in moles, from $N=n{N}_A$, the dimensionless count of specified entities, where $N_A$ is the Avogadro constant in ${\mathrm{mol}}^{-1}$. For example, $2\mathrm{mol}$ contains $1.204428152\times 10^{24}$ specified entities; tripling it gives $6\mathrm{mol}$ and $3.613284456\times 10^{24}$ entities. This uses the exact Avogadro constant $N_A=6.02214076\times 10^{23}{\mathrm{mol}}^{-1}$, which does not triple. In the differential $dU=TdS-pdV+\mu dn$, $\mu$ is a molar chemical potential in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$; the coefficient of $dN$ instead has units of energy per entity.

As a direct check, the ideal gas law $pV=nRT$ givesFor a monatomic ideal gas with only translational degrees of freedom active, internal energy of an ideal gas gives $U=3nRT/2$, so $U^{\prime }=\lambda U$ at unchanged temperature. Here $R$ is the molar gas constant. The scaling distinction lets a small sample supply an equation of state for a much larger amount: material state and inventory become separate questions.

Replication does not mean multiplying every linear dimension by $\lambda$: a geometrically similar container then has ${\lambda }^3$ times the volume. Nor does it mean combining samples at different states. Equal temperature establishes thermal equilibrium but does not imply equal pressure, composition or mass density. Density of a composite system makes this distinction concrete.

Two related properties of an extensive quantity must be distinguished. Additivity says that a quantity $X$ for independent subsystems A and B obeysThe subsystems may be in different states. Extensivity expressed through a homogeneous function says that the equilibrium fundamental relation obeyswith composition held fixed. This is degree-one homogeneity along a replication ray; it does not say that internal energy is a linear function of arbitrary independent changes of entropy, volume and amount of substance.

Additivity gives integer replication by repeated addition. Subdividing a uniform bulk sample into $b$ equivalent parts gives rational scaling $a/b$; continuity then extends it to positive real factors. This argument requires negligible boundary contributions and a consistently prepared state. Conversely, a degree-one homogeneous function need not be additive between different rays: $f(x,y)=\sqrt{x^2+y^2}$ has $f(\lambda x,\lambda y)=\lambda f(x,y)$, but $f(1,0)+f(0,1)=2\ne \sqrt{2}=f(1,1)$. Thus a scaling identity alone cannot justify summing arbitrary states.

For finite samples, the surface correction to thermodynamic free energy can matter. Under geometric enlargement with volume multiplied by $\lambda$, surface area grows as ${\lambda }^{2/3}$, so a surface energy term is not a bulk extensive quantity. Replicating separated droplets retains every original surface, whereas merging them changes the total surface area. Likewise, appreciable mutual gravitational potential energy makes interacting subsystems fail the independent-copy premise. The thermodynamic limit explains when bulk scaling becomes accurate; it does not remove these qualifications for every small or strongly interacting system.

For a single-component equilibrium system with extensive internal energy $E(S,V,N)$, scaling its entropy, volume and particle number together gives $E(\lambda S,\lambda V,\lambda N)=\lambda E(S,V,N)$. The Euler theorem for homogeneous functions therefore gives $E=S{\partial }_{S}E+V{\partial }_{V}E+N{\partial }_{N}E$. The first law of thermodynamics identifies these derivatives as $T,-P,\mu$, respectively, proving $E=TS-PV+\mu N$. Dividing by volume yields the entropy density formula $s=(\rho +P-\mu n)/T$. This step fixes the entropy normalization rather than merely identifying its differential. Differentiating the Euler relation and subtracting $dE=TdS-PdV+\mu dN$ gives the Gibbs-Duhem equation $SdT-VdP+Nd\mu =0$. The argument requires extensivity; it is not an assumption about arbitrary nonadditive systems.

An intensive quantity is unchanged under replication at the same thermodynamic equilibrium. Temperature, pressure, chemical potential and the mass density of a uniform substance are examples. Independence from sample size does not mean independence from state: a material's mass density can change with temperature or pressure.

The relation to extensive quantities is particularly elegant for a differentiable bulk fundamental relation. Put $\mathbf{x}=(S,V,n)$ and writeTaking a partial derivative with respect to $x_i$ and cancelling $\lambda$ givesThe equilibrium differential $dU=TdS-pdV+\mu dn$ identifies these derivatives with temperature, minus pressure and molar chemical potential. They are degree-zero homogeneous functions. The same degree-one assumption yields the Euler relation in thermodynamics, but a surface energy contribution requires extra care.

An intensive quantity generally has no meaningful sum across samples: two equal temperatures do not add to a doubled temperature. Neither is the final temperature after thermal contact generally an unweighted average; heat capacities and the first law of thermodynamics determine the balance. Ratios of additive inventories lead to the useful weighted rules in 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.

Consider two hypothetical parcels at the same temperature $300\mathrm{K}$, kept separated with negligible partition mass and excluded container volume. Parcel A has mass $2\mathrm{kg}$ and volume $0.002{\mathrm{m}}^3$; parcel B has mass $3\mathrm{kg}$ and volume $0.006{\mathrm{m}}^3$. They may be different materials. Their individual mass densities and the composite mass density areThe composite mass density is total material mass divided by total occupied volume; it need not be the local mass density anywhere. Equal temperature does not make these parcels replicas. In particular, the unweighted average $750\mathrm{kg}{\mathrm{m}}^{-3}$ is wrong. The volume fractions are $1/4$ and $3/4$, givingWith mass fractions $2/5$ and $3/5$, the equivalent rule is a weighted harmonic mean:This follows directly by writing $V_i=m_i/{\rho }_i$ and dividing total volume by total mass; it is also the mass-weighted average of specific volumes. An unweighted average of mass densities is valid when the parcel volume values are equal, not merely when their temperatures are equal.

If a partition is removed and different substances mix, the final volume need not equal the sum of initial volume values. A new temperature, thermodynamic phase or composition may also change the local mass density. Use the actual final volume in $\rho =m/V$; the pre-mixing weighted identity cannot supply it. Even unchanged temperature alone does not guarantee unchanged volume.

A pycnometer makes the distinction between mass and mass density experimentally tangible. NIST's 1994 report describes a glass vessel with filling capillaries, wire plugs and a covered overflow bowl in a controlled water bath. The operators soaked a pure-water filling to establish the bath temperature, dried the outside, then left it overnight in the balance enclosure to reach thermal equilibrium before weighing. Empty and full force-restoration balance readings, together with known water mass density, calibrated its capacity at the bath temperature. A tin-oxide coating suppressed stray electric charge; an electrometer detected none during use. The authors distinguish bath and weighing temperatures and correct for buoyancy and changes of displacement volume. Capacity and water-reference uncertainties carry into later liquid mass density determinations. Their pycnometer data were not yet analyzed, so this account provides no measured density or achieved precision.^[ref]

The ideal accounting behind this measurement calibration can be derived without numerical observations. Let $m_0,m_w,m_x$ denote corrected masses of the empty vessel, water-filled vessel and sample-filled vessel. At the same filling temperature, with known reference water mass density ${\rho }_w$ and reproducibly filled volume $V_c$,The empty-vessel subtraction removes vessel mass; the water filling sets volume. Doubling $V_c$ for the same uniform sample doubles the sample mass increment but preserves ${\rho }_x$. Uncorrected balance indications, trapped bubbles, incomplete filling or unequal filling temperatures break the simple identification with corrected mass and a common $V_c$. This explains why replication is a statement about comparable states rather than arbitrary instrument readings.