iof;2 



jniiclioii with tluil in 1, [). Hlll, can he for a fir.sl orientation con- 

 cerning- I lie critical data. 



Now tlic eqnation ol' state (37) passes into (assuming' the value 

 () also for /') 



from whicii may be solved. 



_ 3,424w 



^ "~ " ~ 7 + 5 ; n' ' 



The following values have all been derived from Comm. 1J8'>, 

 p. 19 et seq. (Kamert.ingh Onn'KS and (^rommklin) and from these 

 Proceedings of Oct. 1913 p. 477 et seq. (Crommelin). (See also 

 Comm. 138). 



The value i^k follows from r = vk ■ l>k = 1 : th-, hence 3k = 1 : r. 

 Now r={l-{-y) : 7 — see I, formula (14) — so that iik = 7 ■ (l+r)=0,427. 

 With r = l + 8:/ (see I, (7)) follows with / — 6 for r the Vcilue 

 2,33, hence for ^k the value '/. ~ 0,429. As further 2y = hk : />„ = 

 = ^k : i^,, we have ,i, = ih : 2/ = 0,429~1 ,5 = 0,286. 



We are not jet ready to carry out the calculations, as the given 

 volumes must all be changed into "reduced" volumes. Now q = 0,1073 

 corresponds with (/^= 60,21, (Comm. 118, p. 8), so (> = 0,5308 (the 

 critical density, corresponding to/z=:l) would correspond tod a =297,84, 

 i. e. to Fj = 1 : 297,84. This is therefore the value of Va at the 

 critical point. To reduce this to 1 (// = 1), this and all the other 

 volumes must be multiplied by the factor 297,84. 



a. The isotherm of 20°,39 C, i. e. (with~~273,09) 7'= 293,48 

 absolute. From this folloNvs ni = 293,48 : 1 50,65 — 1,948, so 3,424 m = 

 = 6,670, so that 



n — ^=z G,670 : (f -f 5 : n-). 

 Now e. g. cLi = 20,499 has been given (for /> = 21,783). So the value 

 ]/,i.is 1:20,499, hence according to the above yi = 297,84 : 20,499. 

 We must therefore divide 297,84 by all the given values of (Ia- 

 Thus we calculate the following survey. 



Mean 0.551 



