Thermodynamics, finding the fundamental equation of ideal gases

[fluidistic]

Homework Statement

An ideal monoatomic gas is characterized by the two equations $PV=NRT$ and $U=\frac{3NRT}{2}$ in which R is a constant.
Find the fundamental equation corresponding to a monoatomic ideal gas.

Homework Equations

$S=\left ( \frac{1}{T} \right ) U+\left ( \frac{P}{T} \right ) V- \left ( \frac{\mu }{T} \right ) N$.

The Attempt at a Solution

I'm given 2 equations of state, one is missing in order to get the fundamental equation. In the entropy representation the variables of the fundamental equation are $\frac{1}{T}$, $\frac{P}{T}$ and $\frac{\mu }{T}$.
The missing one is $\frac{\mu }{T} (U,V,N)=\frac{\mu}{T} (u,v)$ where u and v in lower script are the molar energy and volume respectively.
So I have that $\frac{1}{T}=\frac{3R}{2u}$ and $\frac{P}{T}=\frac{R}{v}$.
I use Gibbs-Duhem's relation $d \left ( \frac{\mu }{T} \right ) =ud \left ( \frac{1 }{T} \right ) +v d \left ( \frac{P }{T} \right )$. Replacing the variables in parenthesis and integrating both sides, I reach (like the book): $\frac{\mu}{T}=-\frac{3}{2}R \ln \left ( \frac{u}{u_0} \right ) -R \ln \left ( \frac{v}{v_0} \right ) +\left ( \frac{\mu }{\mu _0} \right ) _0$.
When I replace this equation into the expression for S given above and using the fact that u=N/U, $u_0=\frac{N}{U_0}$, I reach that $S =\frac{3U^2 R}{2N}+\frac{RV^2}{N}+ N \left [ \frac{3R}{2} \ln \left ( \frac{U_0}{U} \right ) +R\ln \left ( \frac{V_0}{V} \right ) - \left ( \frac{\mu }{T} \right ) _0 \right ]$. Using some log property this simplifies to $S=\frac{3U^2R}{2N} +\frac{RV^2}{N} +NR \ln \left [ \left ( \frac{U_0}{U} \right ) ^{3/2} \left ( \frac{V_0}{V} \right ) \right ] -N \left ( \frac{\mu }{\mu _0} \right ) _0$.
However this differs from the answer given in the book: $S= \frac {5NR}{2}-N \left ( \frac {\mu }{T} \right ) _0 +NR \ln \left [ \left ( \frac {U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac {N}{N_0} \right ) ^{-5/2} \right ]$.
I do not see, for the life of me, what I've done wrong.

 

[I like Serena]

Hi fluidistic.

Let's start with your integration constant.
It should be $(\frac{\mu}{T})_0$ instead of $(\frac{\mu}{\mu_0})_0$.

Then you write u=N/U, but that is not right.
Since you defined u to be the molar volume, u=U/N.

Perhaps you can redo it?

 

[fluidistic]

Thanks a lot for helping me!
You are absolutely right, I've redone everything with more care. I reach however $S=\frac{5RN}{2}-N \left ( \frac{\mu}{T} \right ) _0+NR \ln \left [ \left ( \frac{U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \right ]$.
Therefore it seems like I'm missing a term in the logarithm.
It seems it comes from the integration of the Gibbs-Duhem relation: $d \left ( \frac{\mu }{T} \right ) =ud \left ( \frac{1 }{T} \right ) +v d \left ( \frac{P }{T} \right )$. It seems that when I integrate I should get an extra term like "$-R \ln \left ( \frac{N}{N_0} \right )$. I don't really know how to get that.

 

[I like Serena]

Maybe it's simpler than you think.
What did you fill in for $u_0$?

 

[fluidistic]

Maybe it's simpler than you think.
What did you fill in for $u_0$?

Hmm I still don't see it. I took $u_0=\frac{U_0}{N}$.

 

[I like Serena]

Hmm I still don't see it. I took $u_0=\frac{U_0}{N}$.

Note that N is also a variable that changes.
It has an initial value too, just like $u_0$ and $U_0$.

 

[fluidistic]

Note that N is also a variable that changes.
It has an initial value too, just like $u_0$ and $U_0$.

Yes I know this but when should I make use of this?
Here are my steps:
1)$\frac{1}{T}=\frac{3NR}{2U}=\frac{3R}{2u}$
2)$\frac{P}{T}=\frac{NR}{V}=\frac{R}{v}$
I use Gibbs-Duhem's relation $d \left ( \frac{\mu }{T} \right ) =ud \left ( \frac{1 }{T} \right ) +v d \left ( \frac{P }{T} \right )$.
3)$ud \left ( \frac{1 }{T} \right ) =-\frac{3Rdu}{2u}$
4)$v d \left ( \frac{P }{T} \right ) =-R\frac{dv}{v}$.
Integrating the GD relations gives me $\frac{\mu}{T}=-\frac{3}{2}R \ln \left ( \frac{u}{u_0} \right ) -R \ln \left ( \frac{v}{v_0} \right ) +\left ( \frac{\mu }{T} \right ) _0$. The book (Callen's page 52) reaches exactly the same result so far.
He then says that inserting this into the Euler equation $S=\left ( \frac{1}{T} \right ) U+\left ( \frac{P}{T} \right ) V- \left ( \frac{\mu }{T} \right ) N$ one reaches $S= \frac {5NR}{2}-N \left ( \frac {\mu }{T} \right ) _0 +NR \ln \left [ \left ( \frac {U}{U_0} \right ) ^{3/2} \left ( \frac{V}{V_0} \right ) \left ( \frac {N}{N_0} \right ) ^{-5/2} \right ]$.
So my error is when I insert what I obtained into the Euler equation or it's in integrating GD relation (in which case the book would also be wrong) or it's in either of the 4 steps I denoted by 1), etc.?
I really don't see it. :/

 

[I like Serena]

I got to your book result substituting $u_0=\frac{U_0}{N_0}$ and $v_0=\frac{V_0}{N_0}$ in the integrated GD result.

 

[fluidistic]

I got to your book result substituting $u_0=\frac{U_0}{N_0}$ and $v_0=\frac{V_0}{N_0}$ in the integrated GD result.

Ah! That explains everything In my case those were variables, I didn't even notice. Now I reach the desired result... thank you very much for all.

 

[I like Serena]

Just out of curiosity, were my previous hints too cryptic?
It is not my intention to make a thread a cryptic puzzle...
Perhaps you have an idea what kind of hint I might have given that would have helped you better?

 

[fluidistic]

Just out of curiosity, were my previous hints too cryptic?
It is not my intention to make a thread a cryptic puzzle...
Perhaps you have an idea what kind of hint I might have given that would have helped you better?

You mean post 6? No, not cryptic. In fact now I understand your "hint" that I should have used N_0 instead of N. I did not realize that with my definition of u_0, it wasn't even a constant.
So in my case pointing out in post 6 that my definition of u_0 was wrong and that it's not even a constant would have made me understand where the problem lies. I did not realize it, because I already knew that N was a variable and I thought that was what you wanted me to know.
But I'm maybe a "weird learner" :) Anyway, thank you very much for having went through all of this with me.

 

[I like Serena]

Okay. Thanks.