Prove that $G=\mu N$ and independence of $\mu$ on $N$

Question

I have read somewhere that $G(T,P,N)=\mu N$ and I tried to prove it.

But by proving this I ended up with the results that $\mu(T,V)$ doesn't depend on $N$.

I would like to check my proof and to understand if it is true why we a

We have:

$$dG=VdP-SdT+\mu dN ~(1)$$

and:

\begin{equation} G(T,P,\alpha N)= \alpha G(T,P,N) ~(2) \end{equation}

Thus, using $(1)$:

$$ \frac{\partial[G(T,P,\alpha N)]}{\partial N}=\alpha\frac{\partial G}{\partial N}(T,P,\alpha N)=\alpha \mu(T,P,\alpha N) $$

Also using $(2)$:

$$ \frac{\partial[G(T,P,\alpha N)]}{\partial N}=\alpha\frac{\partial G}{\partial N}(T,P,N)=\alpha \mu(T,P,N) $$

Then : $\mu(T,P,N)=\mu(T,P,\alpha N)$, so $\mu(T,P)$.

And then if we integrate over $N$ we have:

$$G(T,P,\alpha N)=\alpha N \mu(T,P)$$

And so : $G(T,P,N)=\mu(T,P) N$

Is what I did right ? In fact I thought that when working with thermodynamic variables the conjugates variable always depends on all their conjugates variables but it seems not true?

First question : Is my derivation ok?

Second question : So in general the conjugate intensive variable never depends on the extensive variable associated to? Because this result seems quite general as soon as we have an extensive dependence on the potential.

Answer 1

First off, $G$ is actually a function of $P$, not $V$, so you've shown $$G(T, P, N) = \mu(T, P) N.$$ That is, if you take a system and make it twice as big, keeping the temperature and pressure the same, $\mu$ shouldn't change. That makes perfect sense: you can take a box of gas and place it next to a second, identical box of ideal gas to double $N$, and there's no way that should change $\mu$.

More generally, $\mu$ is intensive, so it should be expressible in terms of only other intensive quantities, which is exactly what's done here. That doesn't mean it's the only way to write it; you could e.g. have $\mu(V(T, P), S(T, P))$ if you used a different potential.