34 PROFESSOR KELLAND, ON MOLECULAR EQUILIBRIUM. 



doubtless be zero. This difficulty will be surmounted by taking U and 

 V, as I have done, not for the absolute forces, but for the sum of all the 

 forces if the particles were distributed symmetrically: the pressure, in 

 this case, which arises from the action of the particles contiguous to that 

 under consideration, will obviously vary as the square of the density. 



12. I proceed in the next place, to determine the value of V, by 

 two different methods, the comparison of which, will prove that no 

 inaccuracy arises from the adoption of direct variation in deducing the 

 equation (2), provided such variation is known to be uninterrupted. 



To integrate equation (2) we must observe that 



ft 2 ?.— r/ 2 S— rf 2 ?.— 



a2j B a *R R _ n 



~inr + dy* + d%* 



and must consequently include 2-^ in the value of V in the place and 

 with the interpretation of an arbitrary constant. 



_ . d*V d 2 V d*V \ ^Ae-"* 



We have -d^ + -dy + ^ = ~ ^^-R-- 



The complete value of V will therefore be 



C being given by the equation 



Co 2 = 4ttP; 



Ae~ aR \ 4ttP 



tr-v\ B Ae-° R \ 47T. 



13. In order to calculate the value of V directly, it is most con- 

 venient to employ polar co-ordinates; the particle M being the pole. 

 Let A be the place of the particle acted on, P that of any other 

 particle, AP = r, AM = R, angle PMA = 9, MP = P ; radius of a 

 particle = /, then 2 7rp 2 sin 0d6dp is the volume of an elementary 

 annulus ; 



