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

I have a case to contradict my own question but in few books, it is given that x/y is always intensive. So I got confused. My case is as follows :

Let a wire be our system whose resistance is R,
area of cross section is A,
length is L
and constant of resistance be ρ.
V,R are extensive
V(volume) = AL
R = ρ
(L/A)

So R/V = [ρ(L/A)]/AL = ρ/A^2
ρ is a constant but, A depends on the shape and size of system. So ρ/A^2 depends on shape and size of system. So it is an extensive property.

Therefore x/y can be extensive as well.

But in the below picture which is of G.R.B Op Tandon physical chemistry, it is given x/y is always intensive. Where did I make a mistake ?

Please note that I did not understand this is related more to physics or chemistry, so I asked it here as chemistry is where I have first come across this problem. If I should ask this in the physics website, please tell me and I will do so.

• Perhaps the book is right. It depends on what is written before the assertions that you have posted... Nov 1 '20 at 8:38

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: $$R \propto A^{-1} \propto n^{-2}$$.

• Thanks for the reply, you helped me a lot. Nov 3 '20 at 6:30
• @RyugaGod Happy to help. And thanks for the question -- it was fun to think about! Nov 3 '20 at 6:52

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).