the Molecular Velocities in Gases. 63 



and its edges of the length Jit parallel to the velocities: its 

 volume being lit cos 9, the number of the molecules included 

 in it is l$ht cos 8; and this is also the number of the collisions. 

 Moreover v = h cos 6, whence results 



p = 2mNh 2 cos 2 6. 



Case II. The molecules have one and the same velocity h, 

 in all directions, indifferently. 



They will be divided into groups, of which the velocities 

 correspond to the various elements a> of a typical sphere ; 



and we shall have to replace N by Nj-, taking for a a thin 



zone parallel to the side. We shall have, as we have seen, 

 co = 27rsin6dd; and on integration from to \tr, reckoning 

 only the molecules moving towards the side, we get 



>=mB 



2 Tcos 2 

 Jo 



a known formula, often deduced from other considerations, 

 and from which is derived the numerical value of h. 



Case III. By taking account of the inequalities of velocity, 

 the molecules will be divided into groups. The number of 

 those of one and the same group in the unit of volume is 

 l^(p(x)dx, which will have to be substituted for N; and on 

 adding the result for all the groups, we get 



p = ^ mN 1 x 2 <p(x)dx. 

 Jo 

 Hence results 



■ h 2 (16) 



I x 2 ${x)dx- 



Solution of the Equation yty( K x) = ty'(x). 



According to formula (9), the function f(u, v) having two 

 distinct forms, the integral (lo) is to be divided into two 

 others, which gives 



rt *) = *(*) £*(« + £) <t>(v)dv + *(*)£" (v + ~) *( «) dv. 



In the same way, in the value (14) of ty'(z), dudv can be 

 regarded as the surface-element of the plane, and the integral 

 can be divided into several others corresponding to the differ- 

 ent regions of the figure. According to the values (12) of 



