Assumptions to be taken when rearranging terms in first law of thermodynamics Assuming a closed system, show how the first law of thermodynamics.
$$\mathrm dU = \mathrm dQ + \mathrm dW$$
becomes,
$$\mathrm dQ = \frac{C_V}{R}V\,\mathrm dP + \frac{C_p}{R}P\,\mathrm dV$$

I have an answer, but am hesitant to say it is a final answer or even the correct one at that.
I am concerned with the partial derivatives when rearranging this equation. How do certain parts cancel out or how does one approach the problem without them? Is it safe to assume that since you know how the state variables will act in a certain process, that you can use that knowledge to provide a better or more exact answer to the problem?
Attached you will find a picture of the problem that I have done to the best of my ability (which is limited I must say).
How do I make my assumptions?
Do I assume holding one part constant to solve the equation in one way - that is via an adiabatic process, isothermal process - do the same for another path and compare?
This equation can not be applied in general because it can't describe the amount of heat transferred to an incompressible solid or liquid at constant pressure.
For these cases, it would predict that $\mathrm dQ=0$ (which is not necessarily correct).  If it is not generic, it also cannot describe real gases beyond the ideal gas region.  Therefore, the equation must apply exclusively to an ideal gas. The presence of the ideal gas constant $R$ in the equation is a dead giveaway.
For a reversible path,
$$\mathrm dQ=T\mathrm dS=C_v\mathrm dT+P\mathrm dV$$
But, for an ideal gas,
\begin{align}
&amp;\mathrm dT =\frac{P\mathrm dV+V\mathrm dP}{R}\\
&amp;C_v+R = C_p
\end{align}
Therefore,
\begin{align}\mathrm dQ&amp;=C_v\frac{P\mathrm dV+V\mathrm dP}{R}+P\mathrm dV \\ &amp;=\frac{C_v}{R}V\mathrm dP+\frac{(C_v+R)}{R}P\mathrm dV \\
\implies\mathrm dQ&amp;=\frac{C_v}{R}V\mathrm dP+\frac{C_p}{R}P\mathrm dV
\end{align}
This equation applies only to an ideal gas and a reversible path.