# Chemistry - If x and y are 2 extensive properties of a system then is x/y always intensive?

## Solution 1:

Extensive properties are those that are directly proportional to the mass of the system, under the restriction that all intensive properties are held constant.

The restriction that "all intensive properties are held constant" means we are not changing the inherent nature of the system; rather, we are only changing its size.

For instance, if we keep all the intensive properties constant, and double the mass of the system, all the extensive properties will double.

Thus extensive properties all have the form $$X=nZ$$, where n is the number of particles in the system, and Z is some intensive property. [We can use n in place of mass because the restriction that all intensive properties are held constant means that the composition doesn't change.]

Consequently, if we have two extensive properties, X1 and X2, their ratio is:

$$\frac{X1}{X2}=\frac{nZ1}{nZ2}=\frac{Z1}{Z2}$$

Since the n's cancel out, the ratio of two extensive properties becomes the ratio of two intensive properties. And since intensive properties are independent of n, their ratio is also independent of n.

Based on the above definition of extensivity, volume is only extensive in 3-dimensional systems, because it's only in 3-D systems that $$V\propto n$$. Area is only extensive in 2-D systems, because it's only in 2-D systems that $$A\propto n$$.

I have never before considered resistance in the context of extensivity, but I believe the reason you are running into trouble is that resistance depends both on the size of the system and its geometry. Given that, I would conclude that resistance is neither intensive nor extensive.

You could make resistance extensive by constraining the geometry, specifically by constraining A to be constant, in which case resistance would be a 1-D extensive property that depends upon L: $$R \propto L \propto n$$.

However, I don't believe you could turn resistance into an extensive property by constraining L to be constant, since then you would have: $$R \propto A^{-1} \propto n^{-2}$$.

## Solution 2:

The conductance (inverse resistance) describes a non-equilibrium property, the current transported through a medium divided by the applied voltage:

$$\frac{1}{V}\frac{dQ}{dt} = R^{-1} = \Lambda$$

where $$\Lambda$$ is the conductance.

You might say that an applied voltage is an intensive property independent of geometry, and could write instead

$$V = R\frac{dQ}{dt} = \frac{dQ/dt}{\Lambda}$$

Both conductance (or resistance) and current depend on the geometry (dimensions) of the sample, but clearly something cancels when dividing these extensive properties because we obtain an intensive property (voltage).

On the other hand, conductivity is often assumed to be an intrinsic property of a material independent of geometry, defined as

$$\rho = \frac{E}{J}$$

where $$E$$ and $$J$$ are the magnitude of the local electric field and the current density in the material (the vector properties $$\vec E$$ and $$\vec J$$ are here assumed collinear, more generally $$\rho$$ is a tensor).