I have an equation of state as a function of $p$ (pressure), $T$ (temperature) and $V$ (volume) and need to find an expression for the internal energy $U$. If I use the First Law, $$dU=TdS-pdV$$ and set $dS = 0$, then I can find how $U$ depends on $V$, but not how it depends on $S$. Can I then set $dV = 0$ and integrate $T dS$?
Generally, is it possible to integrate $dU$ above given just an equation of state? Can we get a unique answer for $dU$, or is it ambiguous, and if so, how?
Internal Energy $dU$ can be claculated from
\begin{equation} dU=\left(\frac{\partial U}{\partial V}\right)_{T}dV+\left(\frac{\partial U}{\partial T}\right)_{V}dT \end{equation}
We know that \begin{equation} \left(\frac{\partial U}{\partial T}\right)_{V}=C_{V} \end{equation}
One can use Maxwell's equation (which can be seen from $dU=TdS-PdV$):
\begin{equation} \left(\frac{\partial T}{\partial V}\right)_{S}=-\left(\frac{\partial P}{\partial S}\right)_{V} \end{equation}
and can calculate
\begin{equation} \left(\frac{\partial U}{\partial V}\right)_{T}=T\left(\frac{\partial P}{\partial T}\right)_{V}-P \end{equation}
Therefore,
\begin{equation} dU=\left(T\left(\frac{\partial P}{\partial T}\right)_{V}-P\right)dV+C_{V}dT \end{equation}
In the case of ideal gas, equation of state is $PV=nRT$,
\begin{equation} \left(\frac{\partial U}{\partial V}\right)_{T}=T\left(\frac{\partial P}{\partial T}\right)_{V}-P=0 \end{equation}
and internal energy can be integrated which gives \begin{equation} \Delta U=C_{V}(T_{f}-T_{i}) \end{equation} .
In the case of adiabatic expansion $dS=0$, which gives \begin{equation} dU=-PdV \end{equation}
Using, $PV^{\gamma}=\text{constant}$ .If constant is k we can integrate and find \begin{equation} \Delta U=k\left(\frac{V_{f}^{-\gamma-1}-V_{i}^{-\gamma-1}}{\gamma+1}\right) \end{equation}
It was simple in the above two cases because we could reduce integration in one variable.
Now if we consider a toy model \begin{equation} PVT^{3}=k \end{equation}
In this case
\begin{equation} \left(\frac{\partial U}{\partial V}\right)_{T}=\left(T\left(\frac{\partial P}{\partial T}\right)_{V}-P\right)=-4P=-4\frac{k}{VT^{3}} \end{equation}
If we want to integrate $dU$ , $\left(\frac{\partial U}{\partial V}\right)_{T}dV=-4\frac{k}{VT^{3}}dV$ has 2 independent variables and cannot be integrated.
Hence, integration of the internal energy depends on the equation of state. In some cases, it is possible to integrate and in other cases it won't be possible to integrate internal energy.
Given a function $U=U(x_1, \dots, x_n)$ of class $\mathcal{C}^1$, for any $a=(a_1,\dots,a_n)$ and $b=(b_1,\dots,b_n)$,
$$\begin{align} U(b) - U(a) &= \sum\limits_{i=1}^{n}U(b_1, \dots, b_{i}, a_{i+1}, \dots, a_n) - U(b_1, \dots, b_{i-1}, a_{i}, \dots, a_n) \\ &= \sum\limits_{i=1}^{n}\int\limits_{a_i}^{b_i}\frac{\partial U}{\partial x_i}(b_1, \dots, b_{i-1}, t_i, a_{i+1}, \dots, a_n)\textrm{d}t_i \end{align}$$
So, in your case $U = U(S, V)$, and let $U_0 = U(S_0, V_0)$ a reference state, then
$$ \begin{align} U(S_f,V_f) &= U_0 + \int\limits_{S_0}^{S_f}\frac{\partial U}{\partial S}(S, V_0)\textrm{d}S + \int\limits_{V_0}^{V_f}\frac{\partial U}{\partial V}(S_f, V)\textrm{d}V \\ &=U_0 + \int\limits_{S_0}^{S_f} T(S, V_0)\textrm{d}S -\int\limits_{V_0}^{V_f}p(S_f,V)\textrm{d}V \end{align}$$
The First Law gives you the partial derivative of $U$, and the equation of state will give you the relation between the useful variables, so it is in fact possible to find the value of $U$ for any state.
You typically hold one thing constant and get a differential equation in terms of the other quantities. For instance,
$$\left(\frac{dU}{dV}\right)_T=p$$
This is a helpful reference: http://authors.library.caltech.edu/25018/6/TOE05.pdf
