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:

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


\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.


1 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.

  • $\begingroup$ Yes it depends on $P$ I wrote a little fast. Thanks $\endgroup$
    – StarBucK
    Feb 1, 2018 at 21:39

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