I have $n$ mole of an ideal gas with pressure $p$, volume $V$, temperature $T$ and constant heat capacity $C_v$. The question is to calculate the inner energy $U$.
Solution: $$ \left( \frac {\partial U}{\partial T} \right)_V=C_v \to U(T,V)=\left( \frac {\partial U}{\partial T} \right)_V T+\left( \frac {\partial U}{\partial V} \right)_T V=C_v T+u(V) $$
where $u(V)$ is a function of the volume. To calculate $u(V)$, we use
$$dU=TdS-pdV \to \left( \frac {\partial U}{\partial V} \right)_T=T \left( \frac {\partial S}{\partial V} \right)_T-p$$ $$ \left( \frac {\partial S}{\partial V} \right)_T = \left( \frac {\partial p}{\partial T} \right)_V$$ $$pV=nRT \to \left( \frac {\partial p}{\partial T} \right)_V=\frac {nR} {V} $$
So finally we get
$$ \left( \frac {\partial U}{\partial V} \right)_T = T \frac {nR} {V}-p=\frac {nRT} {V} - \frac {nRT} {V} = 0 \to u'(V) = 0 \to u(V)=Constant$$
Final answer is: $U=C_V T+constant$
I don't quite understand the concept of putting a variable outside of my paranthesis, I thought that it meant the variable was constant in the given system.
I'm confused by how we can write $dU=TdS-pdV \to \left( \frac {\partial U}{\partial V} \right)_T=T \left( \frac {\partial S}{\partial V} \right)_T-p$, are we not setting the temperature as constant here? Why is that OK? I understand that we do this so it's easier to work with and I can use the maxwell-relation later on, but is that enough of an argument when setting $T$ as constant?
I'm also confused about $ \left( \frac {\partial S}{\partial V} \right)_T = \left( \frac {\partial p}{\partial T} \right)_V $. If $T$ is constant, then the right side should be $0$, and if $V$ is constant, then the left side should be $0$? What am I missing?
