Is the pressure the same if I heat to the same temperature different closed containers with distinct ratios of water to air? Say I have 3 closed containers of 1L each. By volume, the 1st one is 50/50 air and water, the 2nd is 20% water and 80% air, and the third is 80% water and 20% air. If I heat all of them to the same temperature of 120 C. Are they all under the same internal pressure?
EDIT: They all start at atmosphere pressure (1 atm) and room temperature (20 C) before I close the lid and start heating them. 
 A: No in general they're not. When you start the experiment you have liquid water. The total number of particles is not the same in the three different containers. If you have to consider the values given in your question , than you basically have three pressure cookers. Even the container containing less water, about 200 g, given that you have 1 l room and stop heating at 120 °C,** provides enough molecules to attain the vapour P. It will be about 2 atm, you may look at the water PT diagram.
(**PV = nRT gives 358 atm) 
Let us assume the volume of the containers be much larger and/or the amount of water much less. Then consider that the molecular weight (averaged) of air is almost twice that of water and especially that air in the initial conditions is hundreds time less dense than water. You can then simplify your calculations or at least your reasoning by assuming that your system contains just water, in terms of number of moles (number of particles) and specific heat.
A: The containers all start at the same pressure $P_0$, volume $V_0$ and temperature $T_0$. They each consist of two gas species, consisting of a number of particles $N_1$ and $N_2$. These exert partial pressures $P_1$ and $P_2$. By Dalton's law of partial pressures, the total pressure in the container is given by
$$P = P_1+P_2$$
which, when we apply the ideal gas law, becomes
$$P = \frac{N_1kT}{V_0}+\frac{N_2kT}{V_0} = (N_1+N_2)\frac{kT}{V_0}$$
which means that the total pressure in the container as a function of temperature depends only on the total number of particles $N=N_1+N_2$ in the container. Since each gas mixture is at the same temperature, volume, and pressure, we can see that
$$N = \frac{P_0V_0}{kT_0}$$
which is the same in each case. Therefore, the pressure in each container is the same as a function of temperature.
A: What you are describing is a pressure cooker.
Probably_someone's answer is right if all water evaporates or boils and you just have a mix of two gases. As he said, pressure will depend on the amount of gas inside of the container.
However, as long as there is liquid water in the container, it will stop boiling when pressure in the container reaches its vapour pressure at its temperature.
According to this steam pressure calculator vapour pressure at 120ºC is 198.665 kPa (about two atmospheres) and that will be the pressure in the container at 120ºC as long as some water remains in it. Since if the container originally had 20%, 50% or 80% of water, some water will remain in liquid state, and therefore in the conditions of the question, the container will be at this pressure.
Only if the container contained very little water (let's say less than 1/20 of the container) or if you heat it more than 120ºC, all water will be converted to steam and probably_someone's answer will apply. Please notice that in this "dry" situation pressure will be lower. 
A: I'll show you how to do the calculation for 20% water and 80% air, and you can then repeat the calculations for the other mixtures.  We will neglect the initial amount of water vapor in the air at 20 C.  So there are 200 cc of liquid water and 0.8 l of air.  The specific volume of liquid water at 20 C is 1.002 cc/gm, so the mass of water is 199.6 gm and the number of moles of air is determined from the ideal gas law as [(0.9869)(0.8)]/[(0.0821)(293)]=0.0328 moles = 0.952 gm.  
From the steam tables, at 120 C, the equilibrium vapor pressure of water vapor is 1.985 bars, the specific volume of liquid water is 1.060 cc/gm and the specific volume of water vapor is 892 cc/gm.  So, if x represents the number of grams of water vapor in the gas space, we have (199.6-x) grams of water in the liquid space.  So,we have, $$892x+1.060(199.6-x)=1000$$This gives 0.885 gm of water in the gas phase and 199.6-0.885=198.7 gm of liquid water.  So the volume of liquid water at 120 C will be  (198.7)(1.060)=211 cc and the volume of gas will be 1000-211=789 cc.  Neglecting any air dissolved in the water, the partial pressure of the air in the gas space will be $[(1.0)(800)(273+120)]/[(789)(293)]=1.36\ bars$. So the total pressure in the container will be 1.985+1.36 = 3.35 bars.
