Mr. H. Hilton on van der Waals' Equation. 587 



86 in (i.) we have (writing x for x{) 



\{2x*y + 9x-3-3\/l-6x + yx 2 -4:yx*y{^l—6x + yx*-4:yxS-l + 3x-2yx z ) 



■) 



a if 2i ov9 n i (ZSl-6x + 9x*-±yx*-3±9x-2y^ 

 = 4#o,* 8 (y# 2 + 3) (3x — 1) x log I 



2ty.£ 2 (3#— 1) 



as the equation of the border curve between ,r = Jand x=l. 



Similar reasoning shows that the equation of the border 

 curve between #=land^=oo is found by changing the 

 sign of the radical in (8). 



To trace the curve (8) directly is not easy, but we can 

 readily determine a number of points on it with fair accuracy 

 when once a number of curves of the family (a) are carefully 

 drawn. We have in fact to choose the points A, E (fig. 3) 

 so that the areas ABC, EDC are equal, and this may be done 

 by help of a planimeter ; or, more simply, by ensuring the 

 equality of the number of millimetre squares in these two 

 areas when the curves (a) are plotted out on paper ruled in 

 square millimetres. 



The curve (&) evidently passes through the points (1; 1) 

 and (J ; 0) [as may be verified also from its equation]. The 

 tangent at each point is parallel to the axis of x : this is 

 evident in the case of the former point, we may prove it in 

 the case of the latter, thus : — A tangent to (a) parallel to the 



on q 



axis of x is'y=— - — - — - v where (3#— l)=4&c 3 ; if 6 is 



{jiH ^~* J. X 



small, x must be large [supposing 3.?— 1=£0], and we have 

 approximately 40#=9. Then the tangent becomes (sub- 

 stituting —q for x, and neglecting higher powers of Q than the 

 second) y = ^# 2 . Hence the ordinate of the point on the 

 border curve corresponding to this value of 6 is < —= O 2 . 

 Now the curve (a) cuts y = 0, 



where 9-9. /ns7 



-*L-=lU±\i. 



-l+fty 



(if 6 is small), and makes an angle 



tan { (^T^ + 3 } or tan- { W + jj 



or - when 6 is small, with the axis of x\ hence the abscissa 



2Q2 



