An air rifle shoots a lead pellet by allowing high - pressure air to expand, pro

Question

An air rifle shoots a lead pellet by allowing high - pressure air to expand, propelling the pellet down the rifle barrel. Because this process happens very quickly, no appreciable thermal conduction occurs and the expansion is essentially adiabatic. Suppose the rifle starts with 13.0 cm^3 of compressed air, which behaves as an ideal gas with lambda = 1.40. The expanding air pushes a 1.20 - g pellet as a piston with cross - sectional area 0.0275 cm^2 along the 59.5 - cm - long gun barrel. What initial pressure is required to eject the pellet with a muzzle speed of 140 m/s? Ignore the effects of the air in front of the bullet and friction with the inside walls of the barrel.

Explanation / Answer

for the system of the bullet

Wbullet = dK (change in kinetic energy)

system of gas undergoing an adiabatic process , Q = 0

Wgas = dEint

Wbullet = -Wgas

dEint = -dK

nCvdT = -dK

n(5/2*R)*dT = -dK

(5/2)*nR (Tf-Ti) = -dK

in adiabatic process Tf

Ti(Vi/Vf)^y-1 - Ti

5/2*nR[Ti(Vi/Vf)^y-1 - Ti ] = -dK

5/2*nR*Ti[(Vi/Vf)^y-1 - 1 ] = -dK

here use PV = nRT

5/2PiVi[Vi/Vf)^y-1 - 1 ] = -(1/2mv^2 - 0)

Pi = mv^2/5Vi[1-(Vi/Vf)^y-1]

The final volume of the gas is equal to the initial volume plus the volume of the rifle barrel

Vf = Vi + Ah = 14.64 cm^3

substitue values

Pi = (1.2 x 10^-3 kg) ( 140 ^2) / 5*13 x 10^-6 m^3)[1-(13/14.64)^0.400]

Pi = 7.8 x 10^6 Pa = 76.94 atm

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