Local equilibrium
If the assumption of local equilibrium (usually exploited to use thermodynamics in fluid mechanics), you can use thermodynamic relations for every fluid particle.
Solution of the problem
It seems to me that you did it right. Let's revisit the problem using integral equations for a fluid in a control volume:
- mass:
$\dfrac{d}{dt}\displaystyle \int_V \rho + \oint_S \rho \mathbf{u} \cdot \mathbf{\hat{n}} = 0$
- momentum (not needed here): ...
- total energy:
$\dfrac{d}{dt}\displaystyle \int_V \rho e^t + \oint_S \rho e^t \mathbf{u} \cdot \mathbf{\hat{n}} = \int_V \rho \mathbf{g} \cdot \mathbf{u} + \int_V (-p\mathbf{\hat{n}} + \mathbf{s}_n) \cdot \mathbf{u} - \oint_S \mathbf{q} \cdot \mathbf{\hat{n}}$
to be simplified using some assumptions, basically
- steady flow $\frac{d}{dt} \equiv 0$;
- no volume force $\mathbf{g} = \mathbf{0}$;
- negligible viscous stress $\mathbf{s}_n = 0$, negligible heat transfer $\mathbf{q} = 0$, without shock waves and thus isentropic conditions (don't make me prove it here. If needed, look for balance equation for entropy, or Clausius-Duhem condition);
- perfect gas law $P = \rho R T$; for isentropic conditions you get $\dfrac{P}{\rho^\gamma} = const$;
- uniform properties on each section of the nozzle.
The simplified equations read
mass:
$\oint_S \rho \mathbf{u} \cdot \mathbf{\hat{n}} = 0 \qquad \rightarrow \qquad \rho_1 A_1 V_1 = \rho_2 A_2 V_2$
total energy, introducing the definition of total enthalpy $h^t = e^t + \frac{P}{\rho} =\dfrac{1}{2}|\mathbf{u}^2| + e + \frac{P}{\rho}$:
$\oint_S \rho h^t \mathbf{u} \cdot \mathbf{\hat{n}} = 0 \qquad \rightarrow \qquad \rho_1 A_1 V_1 h^t_1 = \rho_2 A_2 V_2 h^t_2 \qquad \rightarrow \qquad h^t_1 = h^t_2$
and you can use them along with the condition for isentropic transformations $P_1/\rho_1^\gamma = P_2/\rho_2^\gamma$ and the constitutive equation for perfect gases.
Differential form of quasi-1D isentropic equations
Since mass flux is constant through the sections of the nozzle, you can write
$0 = d \dot{m} := \rho A V + \rho dA V + \rho A dV$.
Since $h^t$ is constant, you can use one of the equivalent forms of the following equation
$0 = d h^t = V dV + dh = V dV + de + d\left(\dfrac{P}{\rho}\right) = V dV + \underbrace{de - \left(\dfrac{P}{\rho^2}\right)d\rho}_{= de + P dv = Tds \quad (1^{st} Pr.)} + \dfrac{dP}{\rho} = VdV + Tds + v dP$.
For isentropic flows, where $ds=0$, you get
$VdV + v dP = 0$.
