△H≈△,H°=[831-(1351+22673) kJ. mol -1787kJ·mol AU=0 AH=O AS=RIn P1=8.3145xIn-JK-1=19.14J.K-l △A=△G=-TAS=(-300×19.14)J=-5742kJ △U=W, n CymAT=→P外(2-1 C 1.5(72-7)=-72+571 12=788.19K △U=W=1×2×8314×(78819-30315)J=604J AH=1×2÷×8314×(78819-30315)J=10081kJ AS=nC, Inl2+nRInY2anC. Inf2+nRInI2P1=64791K-l P (1)ln{p”} R △Hm=1(314×4979555J=414k △H (2)由hn22 TT P2 =101.2 kPaA 卷 第 5 页 1 o 1 r m 178.7kJ mol [183.1 (135.1 226.73)] kJ mol − − = − ⋅ ΔH ≈ Δ H = − + ⋅ 四、 ( ) 300 19.14 J 5.742 kJ J K 19.14 J K 0.1 1 ln 8.3145 ln 0 0 1 1 2 1 Δ = Δ = − Δ = − × = − ⎟ ⋅ = ⋅ ⎠ ⎞ ⎜ ⎝ ⎛ Δ = = × Δ = Δ = − − A G T S p p S R U H 五、 , ( ) ΔU =W nCV ,m ΔT = − p外 V2 −V 1 2 1 2 1 ,m 1 1 2 2 ,m 1.5( ) 5 2 3 , T T T T C R p RT p RT C T p V V − = − + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ Δ = − 外 − 788.19KJ T2 = 8.314 (788.19 303.15) J 6.049kJ 2 3 1 ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ ΔU =W = × × × − 8.314 (788.19 303.15) J 10.081kJ 2 5 1 ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ ΔH = × × × − 1 2 1 2 1 1 2 ,m 1 2 1 2 ,m ln ln ln ln 6.479 J K− Δ = + = + = ⋅ p T T p nR T T nC V V nR T T S nCV V 六、 ( ) 8.314 4979.5525 J 41.4 kJ 1 (1) ln{ } vap m vap m ∴Δ = × = ⋅ + Δ = − ∗ H C R T H p (2) 由 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − Δ = − 2 1 vap m * 1 * 2 1 1 R ln T T H p p 得 101.2 kPa * p2 = 七