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H 2 O(l) t 1 =80℃,p=101.325kpa

H 2 O(l) t 1 =80℃,p=101.325kpa. H 2 O(g) t 1 =80℃,p=101.325kpa.  S. dT=0. H 2 O(l) t 1 =100℃, p. H 2 O(g) t 1 =100℃,p. dp=0. 3 、( 4 ). 4 、不可逆循环, W=10kJ ,  U=0, 所以 Q = -W= -10kJ  S (系) =0  S (环) = - Q/T= 10kJ/300K=33.3J/K. 5、1 molH 2 或 O 2 , 在恒温下由纯态变为混合态,由于

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H 2 O(l) t 1 =80℃,p=101.325kpa

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  1. H2O(l) t1=80℃,p=101.325kpa H2O(g) t1=80℃,p=101.325kpa S dT=0 H2O(l) t1=100℃, p H2O(g) t1=100℃,p dp=0 3、(4)

  2. 4、不可逆循环,W=10kJ , U=0, 所以Q = -W= -10kJ S (系)=0 S (环)= - Q/T= 10kJ/300K=33.3J/K 5、1molH2或O2,在恒温下由纯态变为混合态,由于 P(H2)=P (H2) ,故S (H2) =n (H2) Rln( P/ P )=0 同理 S ( O2 )=0 S= S (H2) +S ( O2 )=0 因为是理想气体恒温过程,故H= U=0 G= H-T S =0

  3. 7、因为是理想气体恒温不可逆过程,故 U=0,Q=1000J , W = Wr/2= -Q= -1000J Qr = -Wr =2000J ,故 S(环)= -Q/T (环) = -1000J/400K= 2.5J/K S(系)= Q/T = 2000J /400K=5 J/K S(隔)= S(环)+ S(系)= 2.5J/K 8、Sp- SV=nCp,mln(T2/T1)- nCv,mln(T2/T1) =n(Cp,m-Cv,m) ln(T2/T1)=nR ln(T2/T1)=2.392J/K

  4. n=1mol, H2O(g) 100℃101.325kpa n=1mol, H2O(l) 100℃101.325kpa 在真空容器中 恒容 11、 Qv=37.53kJ= U, (PV)=ngRT H = U+ (PV) =37.53kJ+8.314373.15 10-3kJ=40.63kJ 因始末之间处于平衡态,故可视为恒温、恒压可逆相变, S= H /T= (40.63/ 373.15 ) 103J/k=108.88J/k G= H-T S =0

  5. 恒温-5℃ H2O(l) -5℃,Pl  H2O(s) -5℃,Ps =401Pa H2O(g) -5℃,Pl  H2O(g) -5℃,Ps  12 、 G1= G3=0 , G=G2=nRTln(Ps / Pl )=-106J (Ps / Pl )=0.953, Pl = 420.53Pa

  6. 反抗恒外压力 n=1mol T1=300K,p1=10.325kpa p2=p(环) Q=0 二、选择填空题 4、 P(环) =P2=202.65kPa,为绝热不可逆压缩。T升高。 S=nCp,mln(T2/T1)+nRln(p1/p2)>0 若为恒温、恒外压P(环) 压缩至P2= P(环)=202.65kPa S(系)= nRln(p1/p2)<0, U=0系统得功放热,故 S(热源)= -Q/T>0, S(隔)= S(系)+ S(热源)>0

  7. 恒外压P3 Qr=0压缩 P1,V1,T1 Qr=0膨胀 P3=P, V3=V1,T3 P2,V2,T2 要实现上述过程,需P3 > P1 ,因V3=V1 ,P1 /T1 =P3 / T3, 所以T3 > T1 。则W=W1+W2=n Cv,m(T3- T1 ) > 0 H=n Cpm(T3- T1 ) > 0因S 1=0, S 2 > 0, S= S 1+ S 2 > 0 5 、n=1mol, 理想气体(设Cv,m为定值)

  8. 6、A为始态,BC可逆绝热线,TB>TC, (TB-TA)> (TC -TA)>0, ∆UAB> ∆UAC 因∆SBC=0, ∆SAB+ ∆SBC= ∆SAC ∆SAB = ∆SAC 8、液体苯在其沸点下恒压蒸发 H=n vapHm(苯) > 0 ;因 H > >ngRT, U= H- (PV)= H- ngRT >0 S >0, G= H-T S =0(可逆相变)

  9. 9、 苯(s), lsCp,m <0,Vm(s) Vm(l) t=0 ℃, t(熔)=5.9 ℃ 苯(l) t=0 ℃ (过冷态) dT=0,dP=0,不可逆相变, Q= H<0, (PV)=0 所以U=  H=Q<0 S= - fusHm/T (熔)+ 所以在数值上S > H/T, G<0

  10. 12、因Q=0,dV=0,W’=0发生化学反应, T=500K, p=2026.30kPa, 此过程的r U=0, r H= r (PV)=V rP>0 r S >0 (绝热不可逆), r A= r U - r (TS)<0(系统的T,S皆变大) 13、 Q=0 ,W’=0,恒外压化学反应,体积变大,温度升高, 此过程的P1= P2 =P(环) , 则W=- P(环)(V2-V1) <0 r U= Q+ W= W <0 , rH=0 ( Q=0, P1= P2 =P(环) ) r S >0 (绝热不可逆), rG=-r (TS)<0 (系统的T,S皆变大)

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