Intensive variable

An intensive variable of a substance is such that its value does not depend on the amount of the substance.

It is the counterpart of an extensive variable.

Let there be one piece of substance whose quantity is n and another piece of substance whose quantity is m. Let V be an intensive variable. The value of variable V corresponding to the first substance is V(n) and the value of V corresponding to the second substance is V(m). Then put the two pieces together forming a substance with quantity n+m, then the value of their intensive variable should be

$V(n+m)={nV(n)+mV(m) \over n+m}\qquad \qquad (1)$

which is a weighted mean. If V(n)=V(m) then

$V(n+m)=V(n)=V(m)$

so the intensive variable is independent of the quantity.

Examples of intensive variables are: density, temperature, pressure, specific heat, voltage (electric potential).

Theorem

Generally, if

$V_{1}(n)={V_{2}(n) \over V_{3}(n)}\qquad \qquad (2)$

where V₂ and V₃ are extensive variables, then V₁ is an intensive variable.

Proof

$V_{1}(n+m)={V_{2}(n+m) \over V_{3}(n+m)}\qquad \qquad$

(hypothesis)

$={V_{2}(n)+V_{2}(m) \over V_{3}(n)+V_{3}(m)}$

(extensivity of V₂ and V₃)

$={V_{3}(n)\left({V_{2}(n) \over V_{3}(n)}\right)+V_{3}(m)\left({V_{2}(m) \over V_{3}(m)}\right) \over V_{3}(n)+V_{3}(m)}$

$={V_{3}(n)V_{1}(n)+V_{3}(m)V_{1}(m) \over V_{3}(n)+V_{3}(m)}$

(hypothesis)

$={nV_{3}(1)V_{1}(n)+mV_{3}(1)V_{1}(m) \over nV_{3}(1)+mV_{3}(1)}$

(extensivity of V₃)

$={nV_{1}(n)+mV_{1}(m) \over n+m}$

which is equation (1), therefore V₁ is an intensive variable.

Corollary

If an intensive variable V₁ is a ratio of two extensive variables, as in equation (2), then

Failed to parse (unknown function "\mobx"): {\displaystyle V_1(n) = {n V_2(1) \over n V_3(1)} = {V_2(1) \over V_3(1)} = \mobx{constant} = V_1(1) }

and

$V_{1}(0)={V_{2}(0) \over V_{3}(0)}={0 \over 0},\ {\mbox{which}}\ {\mbox{is}}\ {\mbox{undefined}}.$