As in the title, I'm trying to establish the approximate velocity (sans friction and other losses) of a pellet propelled by compressed gas from a tank. Below is what I have came up with myself, I would appreciate if someone could review this as say whether the end values are reasonable. I have looked for a formula that would give me the energy of the pellet (for which I know the mass and diameter), and I've found the formula for work done in volume on the wikipedia page for the adiabatic process, which goes like this: W=P1 * V1^γ * ( V2^(1-γ) - V1^(1-γ) ) / (1-γ) Where P1 is the pressure (in Pascals) in the tank of volume V1 (cubic meters), V2 is the total volume after work has been done (so tank+barrel), γ is the heat capacity ratio (~1.4 for air for this purpose). Assuming that all of the work is put towards the kinetic energy of the pellet (please tell me if the assumption is unreasonable), that would make the result be v = sqrt(2*W/m) Where W is the work from the previous equation and m is the mass (kg) of the pellet After that, I've run a calculation for what reasonably could be a real air gun with the following values, for a simple scenario: The starting pressure is 10 bar = 1000000 Pa Volume V1 of the tank is half volume of the barrel. The barrel has the diameter of the pellet and lenght of 0.5m. The pellet has a diameter of 6mm and mass of 0.25 g (0.00025 kg) So that's V1 ~ 0.00000706858 m3 (~7068.58 mm3) V2 is the barrel+tank, which is 2 * V1 + V1 = 3 * V1 For γ = 1.4, that gives us V1^γ ~ 6.15 *10^-8 V2^(1-γ) - V1^(1-γ) ~ 74.032 - 114.886 = -40.866 Therefore W ~ 10^6 * 6.15*10^-8 * -40.866 / -0.4 ~ 6.3 J From that v = sqrt(2 * 6.3 / 0.00025) = sqrt(50400) ~ 224.5 m/s Does that seem like a number that's anywhere close to what should be true for a 6mm 0.25 g plastic BB pellet shot with 10 bar? Are there any considerable errors in either the assumptions or calculations?