The temperature at which NOx is in equilibrium with its liquid and vapor phase at 1 atm is about –84 °C. Does that mean NOx exists as a liquid at that temperature? What about it's vapor phase then, because it's vapor and liquid phase should be at equilibrium, right?
Yes, the vapor and liquid can coexist, provided P and T are on the coexistence (!) line (as already explained in another answer).
Also, is that the reason why compressed NOx when released from its closed and pressurized container (e.g. 50 bar) at 20 °C becomes a gas while its temperature drops? Therefore, the compressed NOx was in it's liquid state at 50 bar, right?
No, the reason it cools when depressurized has to do with the adiabatic nature of the expansion. For an adiabatic expansion of an ideal gas, the final and initial pressures and temperatures can be related using the following expression: $$PT^{-C_{pm}/R}=\text{constant}$$
By the way, 50 bar is not quite enough to bring NOx to the coexistence line at $20^\circ C$, at least according to data tables in reference 1, which say 50.6 bar is required.
If that's the case, then what is the relationship between the vapor pressure curve and the liquid-vapor dome?
The region under the dome defines the liquid-vapor coexistence region, and is usually represented as a 2D projection, say in P-V coordinates, of the higher dimensional phase diagram. The coexistence line in a typical 2-D P-T diagram represents a projection of the dome in 3D P-T-V space in which the V-dimension is collapsed.
How would you use the saturated liquid-vapor tables to predict the vapor pressure for your substance?
At a temperature of interest, the (saturation) vapor pressure is given by the saturated vapor pressure indicated in the table.
And which specific volume would you chose to define your substance: its liquid specific volume or its vapor specific volume?
As another answer explains, the observed specific volume (on a point within the dome) is an average of the saturated vapor and saturated liquid $V_g$. The specific volume of the gas is likely more responsive to the position on the coexistence line, since liquids are typically only slightly compressible. According to the ideal gas law, $$V_g=\frac{RT}{MP}$$
References
- Thermophysical properties of nitrous oxide (2004). ESDU 91022; ESDU Series on Physical Data, Chemical Engineering and on Physical Data, Mechanical Engineering. ESDU International plc, 27 Corsham Street,London N1 6UA.
