of Carnot's Theorem. 513 



will be formed at this constant temperature. The system is in 

 mechanical and thermal equilibrium throughout the process ; 

 but the process is not reversible, for if heat is slowly abstracted 

 from the liquid cryohydrate, it is not dissociated — it is simply 

 frozen. 



We define an " equilibrium path " to be a path at every 

 point of which the system under consideration is in equili- 

 brium, which may be either stable or unstable, provided, of 

 course, that the equilibrium is not broken at any point of the 

 path. 



A reversible path is always an " equilibrium path," but an 

 "equilibrium path" is not necessarily reversible, as we have 

 already seen. All the irreversible equilibrium paths I have 

 noticed correspond to an absorption of heat — that is, an in- 

 crease of entropy. 



Now it is well known that in a complete cycle the energy 



dissipated =— 1 \ ~, where dq denotes a quantity of heat 



absorbed, and t is the temperature of the refrigerator. 

 Clausius has shown that this quantity cannot be negative, and 

 that when the cycle is reversible it is equal to zero. It is 

 then generally assumed that in all irreversible cycles the 

 energy dissipated has a positive value. 



The study of Solubilities has compelled me to adopt a dif- 

 ferent view. I have been led to conclude that in any irrever- 

 sible equilibrium- cycle the energy dissipated is zero, but that 

 in all other irreversible cycles the energy dissipated has a 

 positive value. Any irreversible equilibrium-engine is, then, 

 as efficient as Ca7'not's perfectly reversible engine, and all 

 other irreversible engines are less efficient than Carnot's. 



We have already seen that 1— ^ =0 for any equilibrium- 

 cycle. Suppose, then, that a system travels from one position 

 A to another position B by an irreversible equilibrium path, 

 and let the cycle be completed by the reversible path BCA. 

 It has been shown by Clausius that, for the reversible path 

 BCA, l 



s A -s B _r A ^ = o ; 



where S A is the entropy of the system in the position A, and 

 S B the entropy in the position B. But I — ± =0 for the com- 

 plete cycle : hence, for the irreversible equilibrium path AB, 



