Your argument completely ignores one of the most impostant factors, transport of water into the fuel. If the tank is full the surface area available to adsorb water from the air is very small (basically the area of the vent vose in cross section). One the fuel absorbs water from the air in the vent hose more water vapor must diffuse down the hose to the fuel water interface. That process is relatively slow, so the absolute rate of adsorption of water by the fuel is slow if the hose is long and the surface arae of fuel exposed to air is small. In contrast, if the tank is half full, the entire free fuel surface in the tank can adsorb water from the air. Consider a 24"x15" tank. It has a free fuel surface area os 360 square inches when not full. If the tank is completely full, the free surface area is the area of the vent hose. For a typical 5/8" diameter hose that area is 0.3 square inches. The mass transfer is directly proportiona to the contact area between the two phases (air and fuel). Thus a partly full tank of the dimensions describer will be able to absorb over 1,000 times more water in unit time than a full tank. In addition the volume of air availabe from which to adsorb water is also much larger for a partly full tank than for a full tank. More air means more available water vapor. Furthermore a partly full tank will pump more air into the tank with temperature change than a full tank which will only move fuel up and down the vent hose. That transports more water vapor into the tank for the partly full tank. It is a very simple diffusion problem. It is NOT a simple chemical equilibrium problem. Take my word for it. My background is in physical chemistry with an emphasis on chemical kinetics and diffusion.

Todd D, Ph.