( 129 ) 
in which for the sake of simplicity the expression --- is 
represented by /, and - by /'. 
We see at once that we must assume 0 = 0,48 to find rp = 2,6, 
in correspondence with the assumed value. Then T c would be =174,4. 
But according to the table for 160° and that of 200 on p. 130, 
a value 1 ) for 0 = 0,59 corresponds to this with y=2,5 which does 
not agree at all with 0 = 0,48. Hence the assumed value rp = 2,5 
is not correct. 
Let us now assume rp = 2,6. 
2,912 | 
So to find rp = 2,6, 0 must be = 0,404, to which corresponds 
T c = 150,7. But as according to the tables 0 = 0,34 corresponds 
with this, also (p — 2,6 is not correct. 
As for (p = 2,5 the first found value of 0 is 0,11 units lower than 
the second, whereas for *p = 2,6 the first is 0,064 higher, the accurate 
value of rp will lie in the neighbourhood of 2,55. 
i 1 m j " 1 / 
Tc II / 
<pz= 2,6 >0=0,4 1,807 0,64810,4214 
150,5 5,911 
|0=O,5| 1,32 0,6680)0.4332 
154,0 I 5,823 
ff— 2,55 < 
[ 0=0,43 
1,294 
0,6643 
/ 
0,4492 
T c I 
160,5 
1 ! 
5,567 
2,535 | 
1 
0=0,44 
1,296 
0,6654 
0,4491 
159,9 1 
1 5,573 
2,572 J 
So if we take 0 = 0,434, we find rp = 2,55. Then T c becomes 
= 160,3. To this corresponds 0 = 0,439, so that <p = 2,55 is still 
slightly too small. 
m|«|/ 
TC fl f 
* 1 
0=0,40 
1.292 0,6600 0,4462 
159,4 5,586 
2,444 . 
0=0,45 
1,303 |0,6643 |o,4437 
157,9 fl 5,645 
2,640 j 
p=2,J 
From this follows 0 = 0,430, T c 
however, gives 0 = 0,421, so that rp : 
The accurate value of 
to which belongs 0 C = 0,433, T c == 159,6. 
This latter value, 
slightly too high. 
For v c we find then further the value 1,064 from 
l ) It appears viz. from the two tables, in connection with that for 144°, that 3 
in the neighbourhood of ^ = 2 or 3 and T= 160 increases by about 0,011 with 
every degree the temperature rises. 
