Illustration of the Theory of Gases. 



429 



The form of / is that found by Maxwell. To estimate the 

 mean value of it 2 we must divide 



Now 



I 



+ QO 



>f(u)di 



by £ 



f{u)du. 



so that 



§u 2 e-^ v2 du=-±qtf{ue-^^—§ e - u2 lw 2 du}, 

 f + " u 2 e- u2 'v 2 du = \qv 2 ( + °° e- u2 ^ 2 du. 



J —oo J — oo 



The ratio in question is thus iqv 2 , showing that the mean 

 kinetic energy of a mass is one half that of a projectile, 

 deviating from the law of equal energies first (1845) laid 

 down by Waterston. We must remember, however, that we 

 have thus far supposed the velocities of the projectiles to be 

 all equal. 



The value of A in (14) may be determined as usual. If 

 N be the whole (very great) number of masses to which the 

 statistics relate, 



so that 



H = / (u) du = A\ e- u2 lv 2 du = Av */ (irq) 



J-oo J-oo 



f(u)du = 



N 



e -u2/q?*d Ut 



(15') 



v V{jrq) 



If we were to suppose that the chances of a favourable or 

 unfavourable collision were independent of the actual velocity 

 of a mass, there would still be a stationary state defined by 

 writing -y 1 = co in (15). Under these circumstances the mean 

 energy would be twice as great as that calculated above. 



It is easy to extend our result so as to apply to the case of 

 projectiles whose velocities are distributed according to any 

 given law F(v), of course upon the supposition that the pro- 

 jectiles of different velocities do not interfere with one another. 

 We have merely to multiply by F(v)dv and to integrate be- 

 tween and co . Thus from (13) we obtain 



2uf(u) \ + ™vF(y)dv + qf'(u) | v 3 F{v)dv=0. 

 IfF(v)=e- J ™ 2 , we find 



(17) 



