Equations for internal energy and their relevance to the derivation of the equation for entropy for perfect gases

My lecture handout proves derives an expression for entropy thus:



> 
> For a perfect gas $U=U(T)$ (i.e. is not dependent on volume; only temperature). Therefore, $(\frac{\partial U}{\partial V})\_T=0$
> 
> 
> Therefore: 
> 
> 
> Equation 1) $$dU=(\frac{\partial U}{\partial V})\_TdV+(\frac{\partial U}{\partial T})\_VdT=C\_VdT$$ Since $(\frac{\partial U}{\partial T})\_V=C\_V$
> 
> 
> Now, for a reversible change: 
> 
> 
> Equation 2) $$dU=dq\_{rev}+dw\_{rev}=dq\_{rev}-PdV$$
> 
> 
> Equation 3) $$dq\_{rev}=PdV+dU=\frac{nRT}VdV+C\_VdT$$
>  Then divide by $T$ to make the equation equal to entropy rather than heat:
> 
> 
> $$dS=\frac{nR}{V}dV+C\_VdT$$
> 
> 
> 


I have a couple of issues with this. Mainly stemming from the fact that $U=U(T)$ and therefore that $(\frac{\partial U}{\partial V})\_T=0$. I understand it on a physical level why this is the case (no intermolecular forces). I am not asking for a proof of this; I just have an issue with the implications of this. Let me explain:


An equation for dU is:


$$dU=TdS-PdV$$


This is understandable. $TdS$ accounts for heat transfer in/out of a system and $PdV$ accounts for work done by/on the system. However, surely this equation implies that $U=U(S,T)$ since both $dS$ and $dT$ appear in the expression.


The crux of it is: how can $dU=C\_VdT$ and $dU=TdS-PdV$. Please help me understand what's going on here.




Your equation $dU = TdS - pdV$ is valid for all pure substances, not just ideal gases. It is telling you that if you knew an equation that explicitly defined $U$ as a function of $S$ and $T$, then you could calculate all other thermodynamic properties of that substance, i.e. if you knew that $U=f(S, V)$, then you could easily calculate $T$ and $p$ because $T=\frac{\partial f(S, V)}{\partial S}$ and $p=\frac{\partial f(S, V)}{\partial V}$. This is because $S$ and $V$ are [canonical variables](https://physics.stackexchange.com/questions/158182/what-makes-a-thermodynamic-canonical-variable-canonical) for $U$.


Now let's apply the ideal gas equation $dU=C\_V dT$. You can integrate this to yield $U=C\_V T + K$. Unfortunately, this does not express $U$ as a function of $S$, but rather of $T$. So how do we proceed? This [physics.se](https://physics.stackexchange.com/questions/158182/what-makes-a-thermodynamic-canonical-variable-canonical) question has a good general answer.


But in this specific case things are bit easier. Let's find $dS$ from the equation you were given.


$dS =\frac{1}{T}(dU + p dV)=\frac{1}{T}(C\_V dT + \frac{RT}{V}dV)=C\_V \frac{dT}{T} + R \frac{dV}{V}$


Since you now know $dS$ and from the ideal gas law you also know $p dV$, you have an expression for $dU$ from which you can derive all thermodynamic quantities.


 1