Advanced Calculus fi..

346 Advanced Calculus, Fifth EditionSince this is impossible (when U has continuous second derivatives), theheat introduced is dependent on the path. For a simple closed path C,, as inFig. 5.38, the heat introduced is*Since U is a given function of p and V, $ d = ~ 0; thus the heat introducedreduces toJ pdv.This integral is precisely the area integral l y dx considered in Section 5.5,with p replacing y and V replacing x. Hence for such a counterclockwisecycle the heat introduced is negative; there is a heat loss, equal to the areaenclosed (a unit of area corresponding to a unit of energy). The integrallp d V can also be interpreted as the mechanical work done by the gas on thesurrounding medium or as the negative of the work done on the gas by thesurrounding medium. For the process of the curve C1 considered above,the heat loss equals the work done on the gas; the total energy remains .:unchanged, in agreement with the law of conservation of energy expressedby the first thermodynamic law (5.150).While the integral SdU + p d V is not independent of path, it is anexperimental law that the integralis independent of path. One can accordingly introduce a scalar S whose 2ddifferential is the expression being integrated:aS is termed the entropy. In the first equation here, one can consider U and V asindependent variables; in the second, p and V can be considered independent.Thus the first equation gives 4and hence 41a2s -2-- a2sAccordingly, one finds 3ap ar aTT- - p- + - = 0 (U, V indep.).au au av