Thermodynamics problem- paint-ball being launched.

[Urmi Roy]

Homework Statement

Some paint ball rifles use compressed carbon dioxide to launch the paint ball.The system consists of a gun barrel with an inner diameter of 2 cm and 40 cm long. A chamber, 15 cm
long, is charged with CO2 gas at 300 K and a pin holds the ball in place. When the pin is pulled, the paint ball is launched.


f.)If the expansion process is reversible, please find the initial pressure required for the paint
ball to exit the muzzle at 100 m/s. Please assume the CO2 can be modeled as an ideal gas where the gas constant and specific heat at constant volume for CO2 are R = 189 J/kg-K and cv = 662 J/kg-K,respectively. Since the expansion process is fast and the thermal conductivity of carbon dioxide is low, you may assume the heat transfers to the muzzle and ball are zero. In addition, please assume (for the sake of simplicity) that the pressure of the environment is zero and that the ball’s contribution to the gas volume(s) can be neglected.

Homework Equations

Work done in an adiabatic process= (P1V1-P2V2)/(1.4-1), where 1.4=Cp/Cv=polytropic index for adiabatic process.

The Attempt at a Solution

I was thinking if we could use the above formula...since the external pressure is 0, P2=0...so we now equate P1V1/(1.4-1) to the kinetic energy of the paint ball...we know the volume V1, acc. to the problem, so P1 comes out...the answer I get is near the one they provided on the soln too...but is my procedure correct?

 

[rude man]

Are you sure p2 = 0?

BTW I hope they mean the chamber has the same cross-sectional area as that of the barrel.

How did you get p1 BTW?

(Ignore my first equation perviously. ΔU doesn't directly enter the problem.

 

[Urmi Roy]

Are you sure p2 = 0?

In the question it says "pressure of the environment is zero and that the ball’s contribution to the gas volume(s) can be neglected."

BTW I hope they mean the chamber has the same cross-sectional area as that of the barrel.

ya, from what its seems, i think we can safely assume that.

How did you get p1 BTW?

So we have p2=0, so P1V1/(1.4-1)=work done= 1/2 (mvsquared) (kinetic energy of the ball)

Knowing the other terms, we get P1..thats what we're supposed to find here.

 

[rude man]

Pressure p2 is not zero. It's the pressure just before the ball leaves the barrel. After that, mass is not conserved and all your beautiful idea-gas equations go down the drain ... the reason they specify atmospheric p = 0 is so that all the work ∫pdV accumulated during passage in the barrel is translated into k.e., and p can be the pressure in the barrel. Otherwise there would be a Δp = p(barrel) - p(atmospheric) which would confuse matters, albeit in my opinion not excessively.