878 
(AVM, (AV), and (AV)g>0 ; (AV)z and (AV)g< 0 
ADm (AH ,(AM¢ and (AH); >0 ; (AMjz <0 
The isovolumetrical reaction becomes now: 
ZJ ZH A + ee 0 = (DIGS 250 
(L) | (4) (Zn) A 
Towards lower T | | Ted: hee of he 
. 
The isentropical reaction becomes: 
ZAA+A+GSL (Avg =< 0 i) 
L) | (ZZ) (A) (@) 
Towards lower P | Towards higher P 
From both these reactions it 
follows that the curves must be 
situated as in fig. 5 with respect 
to their direction of temperature 
and pressure. 
Now we have still to show 
that curve (7) is situated above 
(G), curve (G) above (A) and curve 
(A) above (Z,); this latter appears 
Fig. 5. again in the same way as in J. 
In order to show that curve (7) is situated above curve (G) we take: 
dP dP\ _ ((AH)z (AB) 
Guha Galan eae 
As (AV)z is negative, the second part has the same sign as: 
AV )z.(4H)g—(4V )G.(44)z. 
When we substitute in this: 
(AV)g=yOV)y—aAV)z and (4H)g= y(OA)y—a 4); 
then we find: 
AV) ADA Vt )Z\=y4h)jvy>0 
Hence it is apparent that in ee 5 curve (Z) must be situated 
above curve ((). 
In order to show that curve (() is situated above curve (A) we take: 
BEN (APN AG SAE 
(58), [ae (AV)g (AV) 
In the same way as above we find that the second part must 
have the same sign as x(AH)y, so that this is positive. In fig. 5 
curve (G) must be situated, therefore, above curve (A). 
d. Now we have: 
