4C4 :  Muzzle Velocity of a Pellet Fired From an Air Gun  6 pts 

An airpowered pellet gun uses 1.5 mL of compressed air at 1.2 MPa and 29^{o}C in a small tank to propel a projectile. Assume the pellet, with a mass m_{p} = 0.02 kg, seals against the walls of the cylinder and behaves like a piston.  


The pellet is held by a pin until the trigger is pulled, removing the pin and releasing the pellet. Assume the air expands isothermally as the pellet moves along the barrel of the gun and the pressure in the barrel drops all the way to ambient  
pressure, 100 kPa, just as the bullet leaves
the gun barrel. Estimate… a.) The mass and volume of the air in the cylinder as the pellet reaches the end of the barrel b.) The work done by the air within the cylinder on the pellet and 

the work done ON the ambient air (outside the gun) by the bullet c.) The velocity of the bullet when it leaves the gun barrel (muzzle velocity) Boldly assume that the process is a quasiequilibrium process. 

Read :   We must assume that the process is a quasiequilibrium process. This is not a great assumption, but it does yield a reasonable 1st approximation of the muzzle velocity of the bullet.  
 We can use the Generalized Compressibility EOS to show that, despite the molar volume, the gas behind the bullet actually behaves as an ideal gas. This fact allows us to solve the problem using the Ideal Gas EOS.  
 We can determine the work done on the bullet by the air behind it using the relationship for boundary work done by an ideal gas as it expands isothermally.  
 The bullet does work on the surrounding air in a constant pressure process. So, we can evaluate this work term using the formula for isobaric compression of an ideal gas.  
 Finally, we can apply the 1st Law to the bullet. There is no heat exchanged and no change in the internal energy or potential energy of the bullet. The only remaining terms are the two work terms we already know how to determine and the change in kinetic energy. The initial velocity is zero, so the only unknown left in the 1st Law equation is the final velocity of the bullet as it leaves the barrel of the gun !  
Diagram:  The diagram in the problem statement is adequate.  
Given:  V_{1}  1.5E06  m^{3}  P_{2}  100  kPa  
P_{1}  1200  kPa  
T_{1}  29  ^{o}C  T_{c}  132.5  K  
m_{B}  0.02  kg  P_{c}  3770  kPa  
Find:  V_{2}  ???  m^{3}  W_{on surr}  ???  J  
W_{on bullet}  ???  J  v_{2}  ???  m/s  
Assumptions:  
 For purposes of computing the work done on the bullet, you may treat the air inside the cylinder as an ideal gas. This is not entirely accurate because the initial pressure is so high.  
 Assume that the air in the barrel is initially in an equilibrium state.  
 Assume that the air in the barrel as the bullet leaves the gun is also in an equilibrium state.  
 Assume the process is isothermal.  
 For estimation purposes, assume that the process is a quasiequilibrium process. This assumption will yield the maximum muzzle velocity that the bullet could attain.  
 Assume that the air within the system is a closed system until the bullet leaves the gun.  
Part c.)  
 If the temperature of the bullet remains constant, then its internal energy does not change.  
 Changes in the gravitational potential energy of the bullet are negligible, especially if the gun is fired horizontally !  
 Heat transfer to or from the bullet is negligible if the process is isothermal.  
Equations / Data / Solve:  
Part a.)  Begin by using the initial state to determine the number of moles of air inside the barrel. This remains constant until the bullet leaves the gun and that is the time interval in which we are interested.  
Use the Generalized Ccompressibility EOS : 

Eqn 1  or : 

Eqn 2  
Reduced temperature and pressure are required in order to use the compressibility charts to determine the compressibility, z :  

Eqn 3 

Eqn 4  
R  8.314  J/molK  z_{1}  1  (Because P_{c} is so high)  
T_{R1}  2.28  
P_{R1}  0.318  n  7.17E04  moles  
T_{R2}  T_{R1}  V_{1}  2.1E03  m^{3}/mol  
P_{R2}  0.027  z_{2}  1  (Because P_{c} is so high)  
Since Z = 1 throughout the process, it is safe to treat air as an ideal gas throughout this process.  
This is a surprise since the molar volume is 2.5 L/mol and that is less than 5 L/mol.  
The process is assumed to be isothermal and we discovered that the air could be treated as an ideal gas. Therefore:  

Eqn 5  or : 

Eqn 6  
V_{2}  1.80E05  m^{3}  
m = n MW  Eqn 7  MW_{air}  28.97  g/mole  
m  0.0208  g  
Part b.)  Next, we can calculate the work done by the air, on the bullet using the work equation derived for isothermal processes like this one:  

Eqn 8  
W_{on bullet}  4.473  J  
The bullet does work on the surrounding air against a constant restraining pressure, P_{atm}.  
Therefore:  

Eqn 9  
W_{on surr}  1.650  J  
Part c.)  In order to determine the muzzle velocity of the bullet, we must determine the change in the kinetic energy of the bullet as a result of the net amount of work done on it.  
We can do this by applying the 1st Law, using the bullet as our system.  

Eqn 10  
The net work is the difference between the work done on the surroundings and the work done by the bullet.  

Eqn 11  
W_{net}  2.823  J  
Solving for v_{2} : 

Eqn 12  
v_{2}  16.80  m/s  
Verify:  The ideal gas assumptions were shown to be valid using the Generalized Compressibility EOS.  
The remaining assumptions are reasonable, but cannot be verified using only the information given in the problem statement.  
Answers :  V_{2}  1.80E05  m^{3}  W_{on surr}  1.65  J  
W_{on bullet}  4.47  J  v_{2}  16.8  m/s 