74 Mr. J. Kam on Molecular Attraction and 



in which u (the volume reduced to v =l,0°andll== 1 = 760 mm.) 



T 



is always = ^-<f). 



Solving the cubic expression for the critical state, we find 

 (equation D : ) : 



dp _ n 4c 1 . , 



dv 2 U ' v c *~ (v c -pf [1 ' } 



dv 2 ~ U ' i V 4 ~ (tv-/3) 3 ' W 



whence ( >. V ) 



Hence (wd* equations 1-4) 



T 



iv = 2/> =2 — - fo 



Solving the quadratic equation (Dx), we find for the critical 

 state, 



( JP = ±c 0= 2(r e + b) 1. 



dl? ' ?' c 3 l'c 3 ^r 2 



whence 



fy = 2 Vc= -o&.R-Te, . . • • (17) 



but in respect of </>o=l, 



T i rr i * f273^ 2 1 . 273 n « 



l c and n = 1, c* c = ;>M ""p - r = 2^ , ""f"' * * ' '' 



Substitution of c — ^v c m the original equation leads at 

 once to 



*-*.«3>*H ^ 19 > 



2p e . v c = 1 (constant) , (20) 



Zp c = 3p, c =p + 2p lc = $.j=n e . . . (21> 

 For temperatures <273 we have the equations 



p T273 "I 2- v '273 ~ v-P ' - ' K > 



IT"'; t ■•" 



?*_£+», 1_ 1 



