# Silly doubt about thermodynamics: molar representation

1. Jan 29, 2012

### Telemachus

Well, I have a doubt about something that I've found in the book. It's really silly, but it's been bothering me for a while, so perhaps you can help me to understand this.

As you should know, the fundamental equation of a system can be represented in the entropic representation as a function S(U,V,N). And this function is a homogeneous first order function of the extensive parameters. That is, if all the extensive parameters of a system are multiplied by a constant , the entropy is multiplied by this same constant.

$$S(\lambda U,\lambda V, \lambda N)=\lambda S(U,V,N)$$

Then one can take $$\lambda=1/N$$ to obtain $$S(U,V,N)=NS(U/N,V/N,1)$$
U/N is the energy per mole, and is denoted by u. Similarly V/N is the volume per mole, denoted by v, and S/N is the entropy per mole.

S(U/N,V/N,1) is the entropy of a system of a single mole, denoted s(u,v).
S(U,V,N)=Ns(u,v).

I think that the previous ideas are clear to me. But then I found some difficulties with some examples.

I have that the entropy per mole of a system is:
$$s=4A^{-1/2}u^{1/4}v^{1/2}+s_0$$
A is just a constant.
Then the entropy for all the moles in the system according to the book is:

$$S=4A^{-1/2}U^{1/4}V^{1/2}N^{1/4}+Ns_0$$

I don't know where the N1/4 comes from. I know I'm not understanding this clearly, because as I think it, there would be no N's at all, and that's just wrong. As I'm reasoning this, I just multiply by N the left side of the equation to obtain the big S, and then in the other side I can do the same, or multiply each U and V by N. If I just multiply everything by N I would get:
$$S=4NA^{-1/2}u^{1/4}v^{1/2}+Ns_0$$
and if I multiply U and V by N I would get:
$$S=4A^{-1/2}U^{1/4}V^{1/2}+s_0$$

I don't know what I'm doing wrong, and I don't know how to get that N1/4 on it's place. What's so wrong with my reasoning?

Last edited: Jan 29, 2012
2. Jan 30, 2012

### Päällikkö

N u^1/4 v^1/2 = N (U/N)^1/4 (V/N)^1/2 = U^1/4 V^1/2 N^1/4, right? Sorry for not using latex, on mobile.

3. Jan 30, 2012

Thank you :)