So does the temperature change of the gas matter then?

yes it does, but we don't nee to worry about it, as no heat is leving the gas through any means other than the work done on the bullet.

other than the work done on the bullet.

 

Exactly. The heat is doing work on the projectile (actually not a bullet but no worries) - hence the term. I never said it was positive...

its not doing work the way you think it is

 

you could theoretically due the problem by using the formula nr delta t= w

 

however since knowledge of the temperature is beyond us we should use the formula delta pv=w however the two are exactly equivalent.

 

heat merely means the energy transferred, so technically you can state that work is a form of heat, however as people don't usually think of it in this fashion, it is better to say that the transferred energy is equal to Q+W where Q is the energy transferred in the form of temperature, out of the system. in this case you could say that the work was zero or that the heat is zero and get the same answer.

 

although you should see that the temperature changed in my equation to, by the amount

 

E0(ln(b)-E0)/nr