and vaporization, we can obtain expe enthalpy variation with temperature ist (Equation 6.5). For each phase, dH=CpdT Table 6.1 Enthalpy and entropy values for the three major phases of water at the melting and boiling points at 1 Note: Tm and Ty are melting and vaporization temperatures at 1atm (237.15 and 373.15K, respectively). Enthalpies and entropies are in kJmol1 and kJmol1K1, respectively. The molar enthalpy of ice at the melting point, Hrice(Tm), is arbitrarily set to zero. The molar entropy of ice at the melting point, Sc(Tm), is set to the standard molar entropy of ice, S ice,, which assumes that the entropy of ice is zero at absolute zero. The molar enthalpies of fusion and vaporization are Hm=6.01 and Hv=40.68kJmol1, respectively. Combining these enthalpies with Tm and Ty (Equations 6.9 and 6.10 ) gives molar entropies of melting and vaporization of Sm=0.022 and Sv=0.109kJmol1K1, respectively. Values are from the CRC Handbook of Chemistry and Physics. Unlike Chapter 3, we are using partial molar quantities rather than total extensive analogs but ti: same relations apply. Basic Equations: At T=Tv:(Tv,p)=(Tv,1atm)+1pVdp, and liq(Tv,p)=stm(Tv,p) with (Tv,1atm)=H(Tv,1atm)TvS(Tv,1atm) Assume Hm,Hv,Sm, and Sv=f(T,p). Use values for these properties that are given in the caption of Table 6.1. Assume Vliq is a constant, and water is 18g/ mole. Problem \#2 Leave these expressions in integral form (i.e., don't solve Cp integrals). (A) Show the expression for Hstm(Tv,1atm) beginning with Hice(Tm,1atm). (B) Show the expression for Sstm(Tv,1atm) beginning with Sice(0K,1atm)