It is not clear from your question whether the gas in the bottle starts out as air, or the only substance in the bottle is water, either in liquid or gas form. I'll assume the latter since it's easier to answer.
Once the bottle is closed, ambient temperature doesn't matter. The pressure of the gas part in the bottle will be stricly a function of temperature, with that function being solely a property of water. You can find what that function is in what is commonly referred to as a steam table.
As you heat the bottle, and presumably everything inside gets to the same temperature, the vapor pressure increases. This causes a little more of the liquid to boil and thereby make less liquid and more gas. Eventually for any one temperature, a new equilibrium is reached where the gas is at the pressure listed in the steam table for that temperature.
If you let the system reach equillibrium at 200°C, for example, then open a valve at the bottom, water at the bottom will be forced out under pressure. As the volume of water in the bottle decreases, the volume available to the gas increases, which decreases its pressure. The liquid that was at equillibrium now no longer is, since its pressure is reduced but its temperature (for the short term) remains the same. This will cause the liquid to boil and make more gas. This cools the liquid, and if done slowly enough the liquid will track the ever decreasing boiling temperature for the ever decreasing pressure.
Meanwhile, the gas is expanding, so it will cool. It won't condense because there is no place for the heat to go. We are making the assumption that the bottle is insulated at this point. Boiling liquid will keep the gas "topped off" at whatever temperature it is according to the steam table.
Once all the liquid is gone, the steam will still be at pressure above ambient. If the valve stays open, the steam will vent until its pressure is the same as ambient. If the bottle is truly insulated, nothing more should happen. If heat is lost thru the bottle, then the temperature will decrease and the pressure of the steam decrease, again according to the steam table. Outside air is then sucked into the bottle. If this air is at 20°C, for example, it will significantly cool the steam, which will cause much of it to condense, which creates lower pressure, which sucks in even more cool air, etc. The proceeds until most of the water is condensed to liquid, with the only remaining water gas being whatever maximum partial pressure of water that air at that temperature and pressure can support.
If the valve is opened so that the 200°C water is expelled "quickly", then it gets a lot more complicated because there is not enough time for the system to track equillibrium as pressure is abruptly release. In that case, I think all we can say is that a bunch of super-heated water will be violently released and quickly undergo decompression, which will cause some of it to boil quickly, making a lot of steam, with the left over liquid being at boiling temperature for ambient pressure, which would be 100°C. After the water is gone from the bottle, a rush of steam will come out, condensing in the (relatively) cold ambient air and adding to the already large saturated water vapor cloud. Once enough steam is expelled so that ambient pressure is reached the remaining steam acts as above.
For most uses, some minimum pressure is required all the time, not just when the tank is totally full. The usual solution is to place the tank at some elevated position. As JMLCarter commented, using a narrow and tall tank would cause pressure to change a lot depending on the amount of water in the tank. That's the reason for water towers to be a tank on top of a tower instead of being a narrow tank as tall as a tower.
However, when building tanks there are other constraints, as ease of transport -as is pointed in the question- or available space which may lead to narrow and tall tanks. In such cases, either even pressure is not a requirement or the tank is placed high enough that its dimensions are negligible compared to its position.
Best Answer
You can answer this question yourself with a simple thought experiment. If you removed all the air, what would happen? The answer is, that regardless of how much you pressurize the tank, only a tiny bit of water could escape before the pressure has dropped (because water is essentially incompressible).
At the other extreme, if there was so much air in the tank that only a tiny bit of water remained, then the air could expel all the water, but that would not be very much water.
Right there, you can see that the answer to your question is "yes". Somewhere between these two extremes lies the amount of air that optimizes how much water can come out. What that optimum is depends on the maximum pressure your tank can withstand, and the minimum pressure needed to expel water. If we call these $p_{max}$ and $p_{min}$ respectively, and the pressurized volume is $V_p$ while the tank volume is $V_t$, then you can write down the following equations:
If $$\frac{p_{max}}{p_{min}} < \frac{V_t}{V_p}$$
then there is enough water in the tank, and the volume of air that ends up being expelled is
$$V_{out}=V_p \frac{p_{max}}{p_{min}}$$
From these two equations you should be able to determine the optimum amount of air in your pressure tank.