I have the following equation :
$$\frac{aT^3}{P^2} = \frac{-1}{V}\left(\frac{\partial V}{\partial P} \right)_T$$ Where $a$ is a constant, $T$ and $P$ are the temperature and pressure, with $T$ being held constant. I need to find an expression for the volume $V$.
My question is can I proceed as in a separable ODE ? I mean writing :
$$\frac{-1}{V}dV = \frac{aT^3}{P^2}dP$$ and integrate to find $V$ ?
Without converting it to an ODE, we can say the following
$$\frac{aT^3}{P^2} = - \frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T = - \left(\frac{\partial \log V}{\partial P}\right)_T $$
then integrate both sides with respect to $P$, but treating $T$ as an independent variable instead of another constant like $a$
$$\frac{aT^3}{P} = \log V +f(T) \implies V = f(T)\exp\left(\frac{aT^3}{P}\right)$$
where the constant of integration is a function of $T$