6i6 Albert P. Mathews 



held at low temperatures and that T^ /3 was equal to 2.19- 

 (T c 6)/N 2/3 dynes. This, however, gives values uniformly 

 lower than method i and I have accordingly reverted to Eotvos 

 original formula as already stated according to which 



(14) TV 2 / 3 = C(T C T). 



The theoretical surface-tension energy at absolute zero 

 should be, then, CT C ergs for a gram mol, or CT C /N 2/3 ergs for 

 a single molecule. Substituting in (13) we have 



(15) M 2 KS/3^ C = CT C /N 2/3 . 



Changing to the total volume V, in place of the volume of a 

 single molecule, we have : 



(16) a = N 2 M 2 K =3CN 1/3 VcT c /S. 



Since S = RT C /P C V C by substitution in the foregoing we 

 have: 



(17) ' a =3CN V3 P C V*/R ergs. 



The value of C has been given in (i i). It can be found also 

 by comparing (17) with (29) which follows. Substituting 

 its value we have : 



(18) a = (S 2 S + 2)P c Vc/(S 2) 



which is identical with the formula derived from van der 

 Waals' equation on the basis that b c = 2V C /S. Young's 

 formula, then, at absolute zero yields the same result as the 

 others when combined with Eotvos, if, however, we make 

 no assumption as to C, but take it, as Eotvos thought it should 

 be as constant for all substances, we obtain the approximately 

 correct value of "a" for all except very simple substances: 



(19) a = 3 X 2.2 7 N V3 VcT c /S. 



Computations of "a" by formula 18 and 19 are given in Table 9. 

 We may check the foregoing reasoning in the following way 

 avoiding any assumption of what fraction of V c b c is. Taking 

 Young's formula: T ^= rK/3, r being equal to v l/3 at absolute 

 zero, multiplying both sides by v 2 o /3 and then by N we have : 



