(2)(aH/a nr=(a F/a nr+TaS/a nr+p+v(ap/a nT =p+7(p01)y+p+Wap1r=0 17.5分(2111 211l [答]Gbs- Duhem公式: nB duB+S dT-Vdp=0 (1分) 恒压下 nB duB=-SdT' 边除以∑n1代入灿=+RTh(pmp)得 ∑xadn(pp)=Sa( I)dT/RT (在∑中下标=B) xadIn(pyAp)(1-xADdInp(l-yAp=Sm()dInT/R (1分) 恒压dn(pp2)=0 (1分) xA(dyMyA)=(1-xA)[dl-yA(1-yA)=Smdnnp/R (xA-VAVLYA(-yAJXR/Sm(=(a InT/ayA)P (1分) (aInT/ayA)P=(aInT/alxa)P=0 (1分)(2) ( ∂ H/ ∂ V)T = ( ∂ F/ ∂ V)T +T( ∂ S/ ∂ V)T +p+V( ∂ p/ ∂ V)T = -p + T( ∂ p/ ∂ T)V +p+V( ∂ p/ ∂ V)T= 0 (5 分) 17. 5 分 (2111) 2111 [答] Gibbs-Duhem 公式： ∑ dµ = n i 1 nB B+S dT-V dp=0 (1 分) 恒压下 ∑ dµ = n i 1 nB B = -S dT 二边除以 ∑ 代入 µ = n i 1 nB B=µB * +RT ln (pB/p ∃ ) 得 ∑ dln(p = n i 1 Bx B/p θ )= Sm(l)dT/ RT (在 中下标 i= B ) i n = ∑ 1 xAdln(pyA/p ∃ )+(1-xA)dln[p(1-yA)/p ∃ ]=Sm(l)dlnT/R (1 分) 恒压 dln(p/p ∃ )=0 (1 分) xA (dyA/ yA)P=(1-xA)[d(1-yA)/(1-yA)]P =Sm(l)(dlnT)P/R (xA-yA)/[ yA(1-yA)]×R/Sm(l)=( ∂ lnT/ ∂ yA)P (1 分) 当xA=yA时 ( ∂ lnT/ ∂ yA)P=( ∂ lnT/ ∂ lnxA)P=0 (1 分)