50 



As 



dx dn dm 



(10) passes into : 



dt dt dt 



dn I dm ) 



m . r ^M 1" ^^^ • ^1 = « . ^' • «~^*' . . . (12) 



dt {dt ] 



Let us now take ??< so that 



dm, 



j- 2pk . m ^ 0. 



dt 



On integration of rhis last equation we then find for m : 



m = e-'2M-< (13)^) 



Introducing this value into (12), we get: 



g-2j>kt . — z=a . k .e .— ^*-'', 

 (if 



from which by integration : 



n = — . g(2/>-3jA:< \ C (14) 



2p—3 ^ ^ 



Now follows from equations (11), (13), and (14): 



.^■ = m.n= — «-3^« + C . e-2y^^' . ... (15) 



2p-3 ^ ^ ^ 



Bearing in mind that for i =z also .i' = 0, we find for the 

 integration constant C: 



C= " 



2p— 3' 

 through which (15) passes into: 



d) = —^.{e-^^(-e-^p^t^ ...... (16) 



The change of the number of molecules E (hence also of F and 

 G) is represented by : 



dy 



f^=2pk.x-qk.y (17) 



If the value of x from (16) is substituted in (17), then follows: 



— ^ qk.y = a. ^ . ie^^^t —e-^P^i\ . . . . (18) 

 dt ^ ^ ^ 2p— 3 ' ' ^ ' 



This differential equation can be solved in the same way as (10). 

 We then find : 



1) A constant of integration can of course be omitted here 



