Rate of Explosion in Gases. 97 



If 



v=v , Y 2 =oo, 



Hence one minimum lies between these values*. 



C 

 If H is large compared to p v pp ? V will be a minimum or 



maximum when 



/ \ rOp— C„ Cp + 0„ "1 



is a maximum or minimum. 



Writing this F: F will be zero for 



G p — G v 



v = oTTc^ 



For v = + go it will be negative ; hence between the above 

 values of v there will be a maximum of F or a minimum of V. 

 Also 



2C v ^ = (C p -G v )v -(G p +G)v-(v-v )(C p + C v ) 



=2G P v -2(G p + G v )v. 



d 2 ¥ 

 And yt * s a ^ wa y s negative ; hence F must be a maximum 



when 



— u+o (5) 



By eliminating v from (4) and (5) we obtain the value 

 of V 2 . This elimination leads to the result 



2R 

 = -j-Qi [{{m-n)G p +mG v }G p t + (C p + C )/i], 



since p Q VQ=nRt . 



It is assumed throughout that the exploded gas behind the 

 wave remains at constant temperature and pressure, and has 



dV 

 * The other value of v obtained from the quadratic equation -j~ =0 is 



much larger than v , and gives to V a very small value. It has therefore 

 no connexion with the wave we are considering. 



Phil. Mag. 8. 5. Vol. 47. No 284. Jan. 1899. H 



