Ballistics: The Theory and Design of Ammunition and Guns (2026 Edition) – Solutions Manual Complete Chapters Verified

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Ballistics: The Theory and Design of
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Ammunition and Guns 3rd Edition = = = = =

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Solutions Manual Part 0 = = = =




Donald E. Carlucci = = =




Sidney S. Jacobson
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** Immediate Download
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** Swift Response
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** All Chapters included
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2.1 The Ideal Gas Law
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Problem=1=-=Assume=we=have=a=quantity=of=10=grams=of=11.1%=nitrated=nitrocellulose=
(C6H8N2O9)=and=it=is=heated=to=a=temperature=of=1000K=and=changes=to=gas=somehow=wit
hout=changing=chemical=composition.=If=the=process=takes=place=in=an=expulsion=cup=with
3
=a=volume=of=10=in ,=assuming=ideal=gas=behavior,=what=will=the=final=pressure=be=in=psi?=
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Answer=p===292 =lbf= =


  in=2==  =
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Solution:=
=

This=problem=is=fairly=straight-
forward=except=for=the=units.=We=shall=write=our=ideal=gas=law=and=let=the=units=fall=out=dir
ectly.=The=easiest=form=to=start=with=is=equation=(IG-4)=
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pV===mg RT= = (IG-4)=
=


Rearranging,=we=have=
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mg RT=
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p===
V=

,= uytrewuytrew= =




=


Here=we=go=



( )
=
(10) g  =

1000 =1==   = g=kg  = 8.314   =kgmol=
=




( ) ( ) (
kJ== K=      ===2521==      = =kgkgmol=C=H=N==O=   737.6=
= = = ft=  =−=kJlbf==  12 = in  =ft= =   1000 =




) K =




 = = = =


== p== = =
3=  6= 8= 2====9== =


(10) in =



p===292 =lbf= =


  in=2==  =
=


You=will=notice=that=the=units=are=all=screwy=–
=but=that’s=half=the=battle=when=working=these=problems!=Please=note=that=this=result=is=unl

ikely=to=happen.=If=the=chemical=composition=were=reacted=we=would=have=to=balance=the=
reaction=equation=and=would=have=to=use=Dalton’s=law=for=the=partial=pressures=of=the=gas
es=as=follows.=First,=assuming=no=air=in=the=vessel=we=write=the=decomposition=reaction.=

C6 H8 N=2 O9 →=4H=2 O=+=5CO=+=N=2 +=C(s)=
= = = = = =


=


Then=for=each=constituent=(we=ignore=solid=carbon)=we=have=
p=i=== NiV= =T=
=


So=we=can=write=


=  10)g=C=H=N=O= = ==1==   =

= (4)   =  ===kgmol6===H=8=2====O===2======9===  =(8.314)  ==kgmol= kJ==-

= K=   =(=1000) K   ===252=1====     ==kgmo==kg=Cl=6C===H==68==H====N==O8=2=N=====2=9O= 9===     =(===  =6= 8= 2




9====   1,000 ==     ==g=kg=C=6=C===H=6==8H=====N=8==N2====O=====2=O9====9===    ====
kgmol=C=H==N==O=== = =

,= uytrewuytrew= =




p= == = = 

H=2O= (10 )  in
= =  
==3= === 1= =   == kJ =
 =  1  =   ft=  ==
=
=
=
=
=
=





=
=  =737.6=  ft=−=lbf=  =12=  in= =

= = 1,168 =lbf= =

pH2O==   in=2==  =



= (5) = kgmolCO==  (8.314) == kJ= (1000) K
 1  === ===
= kgmolC6=H8=N2O9=== (10) g ==




 == 1==  =kg=C6=H8=N2O9= =
=




 =kgmol= =  =kgmol=-=K= =  =252=  =kg= = C6=H8=N2O9====  =1,000=  =g= =

= = = =
= = = = =
C = H= N= O=== C = H==N= O= C = H==N==O===
pCO=        =


=  = = =


= == = 6====8=====2====9=== = ()
=   =   6====1=8======  2===  ===9=ft=   ===  =
6====8=====2====9===== =
10=in=3== 

1=   = kJ=

p 
CO= ==1,460 =lbf= =  =737.6=   ==ft=− lbf=  =12=  in= = =




 in=2==  ==   == kJ= 

(1) = kgmol (8.314) (1000) K  === 1===  =kgmolC6=H8=N2O9=== (10) g =




 == 1==  =kg=C6=H8=N2O9= =
2=

= = = = = = =
kgmol=  kgmol=-=K==  252= kg=  C6=H8=N2O9= =
1,000  g
= = 
=


N=

 ==
N2=

, = uytrewuytrew= =




 =C=H=N=O=== = =(= )=

 = =  = =  =C=H==N=O= = =  =C=H==N==O===== =p=== =




 
6====8=====2====9=== = = =
6====8=====2====9= =


6====8=====2====9===== =

pN2===292 =lbf= = 10=in=3=  ==737.61==      ===ft=−kJ=lbf==      ==121==      ==inft==   ===


  in=2==  =

2= 2=




Then=the=total=pressure=is=

pp======1,168pH==O=+ ==lbfpCO= ==+==1,460pN  =lbf= =+=29 =




2 =lbf= ===2,920 =lbf= =in=2=  +=  in=2==  = =




  in=2==     in=2== 
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2.2 Other Gas Laws
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Problem=2=-=Perform=the=same=calculation=as=in=problem=1=but=use=the=Noble-
Abel=equation=of=state=and=assume=the=covolume=to=be=32.0=in3/lbm=
=




Answer:=p===314.2 =lbf= =


  in=2==  =
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Solution:=
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This=problem=is=again=straight-forward=except=for=those=pesky=units=–
=but=we’ve=done=this=before.=We=start=with=equation=(VW-2)=
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p(V=−=cb)===mg RT= = (VW-2)=
=


Rearranging,=we=have=
=

mg RT=
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p===
V=−=cb=
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Here=we=go=