Single species
At the surface of a liquid, molecules are constantly being ejected from the liquid surface and gas molecules rejoining it. When those two rates are equal there's no net change and the system is in equilibrium.
The rate at which liquid water will eject a molecule is roughly independent of pressure and is based solely on temperature. The temperature determines how much kinetic energy the molecules have while the pressure just determines how tightly packed they are and they're already very tightly packed so the density doesn't change much. There are always water molecules right on the surface because of this high density.
The rate at which gaseous water molecules rejoin the water depends on how often one hits the surface which depends on the density. By the ideal gas law, the density is related to pressure and temperature, so if we've already set the temperature, it only depends on pressure. Thus the higher the pressure, the higher the density, the more gaseous water molecules there are to hit the surface and join it.
Locally, the density and pressure of the gaseous water molecules will increase until the rate of rejoining is equal to the rate of evaporation.
Multi gas species
Other gasses can feel free to bombard the surface of the water, compressing it a little bit, and increasing the pressure significantly. However, this doesn't increase the rate at which water molecules escape, so the vapor will still equilibrate at the same partial pressure.
Multi liquid species
This is where Raoult's law comes in. On the liquid surface there are now two species of molecules but the rate at which molecules are ejected is still the same. Now those ejections must be split between species, and according to Raoult's law, it's based on the concentration. If 30% of the liquid's molecules are water then 30% of the ejections will be water, so the rate of ejection is reduced to 30%. So in equilibrium the rate of rejoining must also be 30% so there must be 30% as many collisions thus 30% the density and thus 30% the vapor pressure.
Non Equilibrium
Examining the gas directly above the surface would show a partial pressure of water that approaches the vapor pressure of water for that temperature very quickly. These water molecules would then diffuse through the nitrogen which would slowly lower the local partial pressure if it was not being replenished by additional evaporation from the surface. This evaporation requires energy to overcome the latent heat and as such it will lower the local temperature of the water. Water is a much better heat conductor than air so this heat is drawn from the water.
In some evaporation processes the slow diffusion of the gas dominates, in others, the temperature drops until the diffusion process dominates. This is why liquid nitrogen gets and stays so cold, and why wet bulb temperatures are lower than dry bulb temperatures.
If you'd like to learn more about the steady state solution I would recommend reading about relative/absolute humidity, condensation etc.
If you'd like to learn more about the rate limiting effect look into gas diffusion.
If you'd like to learn more about the evaporation condensation process learning about statistical thermodynamics would help.
Hmm, you're right. High-quality phase diagrams don't seem to float around for free on the web. I suppose I shouldn't be as surprised by this as I am.
Reading from this random page it looks like at 6 atm the boiling point for nitrogen is around 85 K. Oxygen is a little harder to find, since there has been news in the past few years about metallic and superconducting oxygen at very high pressures. It looks like you can buy the data to generate your own phase diagrams from Wolfram|Alpha for a few dollars; squinting at the tiny diagram that's available for free, I guess that the boiling point at 0.6 MPa is around 100–110 K.
Probably the behavior of a nitrogen-oxygen mixture (assuming that water and CO2 would freeze out) would be comparable to their behavior at one atmosphere, as shown here: a boiling that varies roughly linearly from that of nitrogen to that of oxygen as the oxygen concentration changes.
This may be more of a cryogenic engineering question.
Best Answer
This problem has certainly been solved in practice in a number of applications - cryosurgery, cryogenic treatment of metals, and cryogenic cooling of x-ray crystallography samples, but the approach will depend on the application.
The flow of liquid nitrogen is always always unstable and fluctuating, because it is usually impossible to prevent small amounts of boiling from generating unpredictable amounts of gas. If this is unacceptable, you will need to boil the liquid first in a vessel or phase separator, and then pass it through the nozzle.
If you intend for the liquid to be boiled in the hose by external heating or contact with ambient air, you may be disappointed. Droplets of liquid will remain in the stream because of the Leidenfrost effect - poor thermal contact between the droplets and the wall of the hose. The result will be a spray of droplets mixed with gas.
You might therefore need a proper heat exchanger - using a long thin pipe or perhaps a metal mesh, to ensure the liquid is fully boiled. Ambient heat can be quite adequate, and is routinely used in boil-off heat exchangers, provided you don't mind large amounts of ice and water condensing on the outside. I wouldn't worry about condensation of oxygen - as long as this occurs in an open environment the oxygen will not build up in dangerous amounts. Oxygen buildup can certainly be dangerous if it occurs inside an enclosed container.
If you need a controlled temperature, then I agree that a heater and temperature sensor are required. A pipe exposed to the environment will gradually cool down as it becomes insulated by ice.
The dewar will have two outlets - one for liquid (unpredictable as I explained above) and one for gas. The gas outlet is very useful, but is not fully insulated, so there will be a minimum temperature below which it will not go. You say that the container has a pressure regulator - but you need to clarify this. Does it have a built-in boiloff device to maintain the pressure? If not, you will find that extracting liquid or gas depressurises the dewar.
Before you do any of that, however, you need to say how big a room you are going to be doing this in, what size of dewar you are using, and what rate of flow you are hoping to use. I would like to carry on discussing with a StackExchanger who is still breathing air with at least a bit of oxygen in it.
Sorry if this answer is disappointingly unmathematical - you need to define the physical regime first, before equations can be usefully applied.