( 130 ) 
» = 1 — 7.^ - ^). whereas p t = 1630 - 2385 = — 755 i8 
found for p e from p = 638,4 </ — 
So the values found theoretically agree perfectly with the values 
of the point of inflection derived from the table for 160°. 
12. Now the tables for T=200 and T=400 may follow. 
T = 200 . 
With 6 = = — the formula for ft becomes : 
T~ = 32000 X C/.)’/. 
L ~ P <P 
i.e. 
ft* 
^ l# YZf = “ °t 3238 + 0,4343 <p — log 10 y, 
the pressure being given by p = 800 ( P — 2700 
4P 
Uf-T- 
9 1-F 
i—^ 
' 1 ’ 
V 
- 
-■ 
00 
1 
0,50 
00 
9 
2,631 
427 
1 800 
0,61 
— 16 
( 6 
1,504 
31,9 
/ 0,98 
0,67 
—1150i 
5 
1,149 
14,1 
0,97 
0,71 
—1300 
14 
0,811 
6,48 
i0,93 
0,78 
-1280/ 
3 
0,502 
3,18 
0,87 
0,88 
-1120 
2 
0,244 
1,75 
0,80 
1,05 
- 850 
. *V 8 
0,152 
1,42 
0,77 
1,21 
— 660 
1 
0,110 
1,29 
0,75 (min) 
1,5 
— 400 
0,5 
0,194 
1,56 
0,78 
2,4 
— 72 
/0,25 
0,387 
2,44 
/0,84 
4,3 
+ 54. 
0,15 
0,565 
3,67 
0,89 
6,8 
H 
(o ,10 
0,720 
5,24 
10,92 
10 
+ «! 
0,01 
1,681 
47,9 
0,990 
100 
+ V 
0,001 
2,677 
475 
1 ~950 
1000 
0,80 
0 
00 
00 
CO 
0 
