# Will all thermodynamic equations have a linear dependence with moles?

I'm taking chemical thermodynamics this semester, and currently seeing things like the Maxwell's Relations.

My professor never uses explicitly the variable "n" and always works with molar quantities like $\bar{V}$ and stuff like that.

I think this simplifies the algebra a great deal, and can be useful if you have the temptation to get a "molar quantity". However, I don't like it right now.

I think theres nothing that can assure me that if I were to work with the variable $n$ in my derivations, I won't somehow get a quadratic dependence of my thermodynamic equation with $n$, or logarithmic, who knows.

But it seems that there is always a linear dependence with $n$. This makes physical sense to me, however, how can I prove this to myself? Maybe this is more of a question of mathematics, but it has scientific implications as well.

Thanks.

Extensive properties actually have a linear dependence with the common magnitudes used for amount of substance.

Mathematical speaking, you work in some representation in which one magnitude is represented with a function of some variables, and the later represent other physical variables. For example $S(U,V,N)$. When a magnitude represents a extensive property, the function is a homogeneous function of of order 1 of the extensive variables, that is: $S(\lambda E, \lambda V, \lambda N) =\lambda \,S(E,V,N)$

We choose $\lambda = 1/N$, so $S(\lambda E, \lambda V, \lambda N) =\lambda \,S(E,V,N)$, using lower case symbols for (say molar) properties,

$N \,S(e, v, 1) = S(E,V,N)$

So, taking in mind that your equations are formulated for finding values for your thermodynamic functions, they must be stated in the form (for the previous example):

$S(E,V,N) =$ $whatever$

If you work with molar properties you will find just:

$S(e, v, 1) =$ $whatever$ $for$ $a$ $mol$, but using the third equation:

$S(E,V,N) = N\times$ $whatever$ $for$ $a$ $mol$

So the homogeneous (order 1) property will assure you that there is linear dependence with $N$.

Note: Thermodynamic functions that also depend on intensive variables have a zero order dependence in the homogeneous sense with these variables.

EDITED:

After reading the @Curt F. answer, to avoid confusions I decided made a comment: The explaining above works in the scope of 'traditional' thermodynamics, where no surface effects are involved. There are another reasons that can leads to draw wrong conclusions about this linearity, for example, if external fields are involved. But these cases are not in the scope of any course on thermodynamics that I aware of.

One exception to the general rule that equations will be linearly dependent on moles is if surface tension is to be modeled. With surface tension, we have to specify the shape of a system as well as the extent. The fundamental equation for a single-component system including surface tension would be:

$$d\bar{G} = \bar{V}dP - \bar{S}dT + \gamma d\bar{A}$$

where $\gamma$ is the surface tension and $d\bar{A}$ is a differential change in surface area (i.e. $\bar{A}$ is surface area per mole).

If the shape of the system is constant (imagine say a sphere of water that grows bigger and bigger as we add more water molecules, but still stays perfectly spherical), then $A$ is proportional to $N^{2/3}$, so $\bar{A}$ is proportional to $N^{-1/3}$. The exponent of 2/3 is not one, therefore the free energy is not linear in the number of moles.

These kinds of cases are rarely studied in chemical thermodynamics at the undergraduate level, however.