If I read your post right, here's my 2 cents.
If you can find the force, which you already know, you just multiply that by 2 to get the spring constant, "k". Then get a spring with a slightly lower "k" than what you calculated, and have the spring compress the piston enough to close the valve, but still have a force difference between the air and spring compression that is high.
Example:
If you have a piston with 1" of surface area for the air to push on (to open), and you want 1/2" of piston travel at 110 psi, your spring constant, "k", will be 220 lb./in. Equations: (110 lbs./in^2 * 1 in^2. = 110 lbs.) Then you take the force, 110 lbs., and divide by the change in distance, in this case, 1/2 in. , and you get a spring constant of 220 lb./in. (110 lbs. / 1/2 in. = 220 lb./in.)
So then you find a spring with slightly less than that constant, probably on the order of 210 - 215 lb./in. That way, you can put the spring in your gun in such a way that when the system is unpressurized, there is 5 - 10 lbs. of force acting on the piston to close it. When you've got 110 lbs. of force due to the air trying to open the piston at firing time, and only 5 - 10 lbs. of spring force acting the opposite way, you still get a force that will push the piston very close to 1/2 in. If you factor in the mass of the piston, which should ideally be low, you probably get around a 1/2 in. of travel.
It should be noted that mass and friction were assumed to be negligible.
Give us your piston's diameter, and the outer diameter of your barrel, and one of us could crunch the numbers for you!
I'm going to assume that your piston is 2 in. in diameter, if you are using the barrel / 4 piston travel principle.
Am I right?