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1 US gallon = 0.133680556 cubic foot (Google) (80 gallon)*(0.133680556 cubic foot/gallon) = 10.69444448 cubic feet The current pressure is meaningless as to the volume of the tank. If you have the temperatrue and pressure of the tank, you can determine the mass and hence density of the air inside the tank, and also determine how many cubic feet of air you have if you were to release the pressure into another fixed volume. The temperature is a tricky one though. If you pump the tank up, the air inside the compressor is hotter than ambient. This is due to the compressor adding internal energy and the conservation of energy principles. The air has a quantity of energy to it, capturing a large volume and reducing the volume (compressing) puts all that energy into a smaller container. More energy per mass means hotter gas. The gas will eventually cool to ambient, and lower the tank pressure slightly. Point being, when you release the gas at a cooled stage, the air re-expands and drops temperature due to the conservation of energy (the inverse of before). In essence this is how AC works. The conversion of energy and all that good stuff like entrophy makes it a challenge, so here'e what I think your asking, in a nut shell. If you released all the air into a giant balloon or something, the general equation, and I mean very very general, P1V1 = P2V2 V2 = P1V1/P2 P1 = 175 PSIG + 14.1 PSI = 189.1 PSIA P2 = 14.1 PSIA V1 = 10.69444 ft^3 V2 = 189.1/14.1*10.69444 = 143.426 ft^3143.426/.133680556 = 1072.903 gallons of Air in its released state at standandard pressure. Again, this is assuming an adiabatic isentrophic process. It is only a vague estimation. Chris Morrison Aero Student OSU

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