Thermodynamics

662 | ThermodynamicsNow we choose the entropy to be a function of T and v; that is, s s(T, v)and take its total differential,ds a 0s0T b dT a 0sv 0v b dvTSubstituting this into the Tdsrelation du Tds Pdv yieldsdu T a 0s0T b dT c T a 0sv 0v b P d dvTEquating the coefficients of dT and dv in Eqs. 12–25 and 12–27 givesUsing the third Maxwell relation (Eq. 12–18), we getSubstituting this into Eq. 12–25, we obtain the desired relation for du:(12–26)(12–27)(12–28)(12–29)The change in internal energy of a simple compressible system associatedwith a change of state from (T 1 , v 1 ) to (T 2 , v 2 ) is determined by integration:u 2 u 1 T 2a 0s0T b c vv Ta 0u0v b T a 0sT 0v b PTa 0u0v b T a 0PT 0T b Pvdu c v dT c T a 0P0T b P d dvvT 1c v dT v 2v 1 c T a 0P0T b v P d dv(12–30)Enthalpy ChangesThe general relation for dh is determined in exactly the same manner. Thistime we choose the enthalpy to be a function of T and P, that is, h h(T, P),and take its total differential,Using the definition of c p , we havedh a 0h0T b dT a 0hP 0P b dPTdh c p dT a 0h0P b dPT(12–31)Now we choose the entropy to be a function of T and P; that is, we takes s(T, P) and take its total differential,ds a 0s0T b dT a 0sP 0P b dPTSubstituting this into the T ds relation dh T ds v dP givesdh T a 0s0T b dT c v T a 0sP0P b d dPT(12–32)(12–33)