274 Prof. Potter on the Law of the Expansion 



volumes of the gas are represented by the ordinates, as PM = // 

 when C M=ar. Let CA=fl= semi-major axis of the hyperbola, 

 CO = m, and OM = a/ the temperatures t° on the Centigrade 

 scale, so that #=?/*+<*/; then, the equation of the hyperbola 

 being 



b 2 

 we have to find from three conditions the values of -%, m, and a 2 . 



Now, taking the volume of the gas at the freezing-point of 

 water, and (represented in the figure hy pO = y ) as 100 measures, 

 and OM=^' being the Centigrade degrees, we have, if PM = t/, 



b 2 



a: 



Vo i =-2( m ' 2 - a< *)> 



and similarly 



2/' 2 =5((m + ^) 2 -« 2 )> 



which by eliminating give 



a 2 (y ^ 2 -y 2 \ J _ a* l )fn-y* \ * 

 m ~2b 2 \ ~S ) 2~2b 2 \ x" ) 2' 

 b* _a«W*-y *)-W*-y *) 



a 2 

 a*=m*-jp.y *. 



Calculating the volumes y v y%, y 3 . . . y 6 from M. Regnault's 

 values of a at the Centigrade temperatures od l} x 1 ^ x* 3 . . . x ! 6J we 

 have as follows : at 



x\= 98-12 C. then y x -y Q x 1-37532, 



^2= 102-45 „ y 2 =y x 1-39162, 



^3 = 185-42 „ z/ 3 =?/ x 1*70458, 



*' 4 = 257-17 „ y 4 =y x 1-97527, 



a/ 5 =299-90 „ y 5 = y x 2-13701, 



^=310-31 „ y 6 =y x 2-17586. 



b 2 

 Taking x' y , oc ! l y ]) x' 5 y 5 to determine the constants —^ m } and 



o. 



a 2 , we find 



