解法3:等P可逆相变,373K,P°,△G=0 H20(g) 298K,P9,A 2→H20(1) △G △G9 H0(g) 373K,p9,AG=0→ H0(1) 等P,T1=373K,△G=0 T2-298K,△G2=? a) 识.=?成1= 14 解法 3:等 P θ可逆相变,373K,P θ,ΔG1 =0 H2 O(g)⎯298 ⎯K ⎯,P , ⎯G2 → H2 O(l) H2 O(g)⎯373 ⎯K ⎯,P ⎯,G ⎯1 =0 → H2 O(l) ΔGˊ ΔG” 等 P θ ,T1 =373K,ΔG1 =0 T2 =298K,ΔG2 =? P T G ( ) =? 或 P T T G ] ( ) [ =?