復(fù)旦 物理化學(xué) 第二章 習(xí)題答案

1、第二章習(xí)題解答1.U=H=02.等溫可逆膨脹：向真空膨脹：由于始態(tài)和終態(tài)同上，體系的熵變也為19.14 JK13先求冷熱水混合后溫度：500Cp(T343)+100 Cp (T300) = 0T=336.3 K (63.3C)再計(jì)算熵變：=41.27 + 43.63 = 2.34 JK14Sn摩爾數(shù)nCp,m,Sn(T473)+1000Cp,H2O(T283)=02.10624.14(T473)+10004.184(T283)=0T=285.3 K5體系熵變按可逆相變計(jì)算：真空蒸發(fā)熱：Q=HnRT=106708.314373.2=37567 J環(huán)境熵變：S總= S體 + S環(huán)= 8.28 JK

2、1 0自發(fā)進(jìn)行6.體系：設(shè)計(jì)可逆過(guò)程：T=263KlT=263KsSS3S2S1T=273KlT=273Ksp恒定JK1 JK1JK1S體=S1+S2+S3=20.66 JK1環(huán)境：J JK1總熵：S總=S體+S環(huán)=0.81 JK10過(guò)程自發(fā)。7等量氣體混合，熱容相同，平衡溫度T=15C=288 K=0.006+11.526=11.53 JK1 0自發(fā)過(guò)程8W=30.65 kJ，即環(huán)境做功30.65 kJ。W=Q1+Q2Q2=WQ1=30.650334.7103=365.35 kJ即向環(huán)境放熱365.35 可kJ9S總=S體+S環(huán)=2.6 JK10不可逆過(guò)程。10Q = H = 92.1436

3、2 = 33355 JW= p(VgVl) pVg = nRT = 8.314383.2 = 3186 JU = QW = 30169 JG = 0S總= 011（1）等溫可逆U = H = 0Q = W = JS隔S體 +S環(huán)0（2）等溫恒外壓U = H = 0 S體14.9 JK1G = 4442 JQ = W = p外（V2V1）=S隔S體 +S環(huán)26.67 JK1設(shè)計(jì)恒溫變壓可逆過(guò)程：12計(jì)算G：H2O(g)pH2O(l)pGG3G2G1H2O(g)3167PaH2O(l)3167PaT=298KSS2S1S3G2=0G30G總=8591 J計(jì)算S： JK1S30S體S1 +S2+S3

4、105 JK1 JK1S總S體 +S環(huán)28.81 JK113UHSGF（1）理想氣體卡若循環(huán)00000（2）H2和O2絕熱鋼瓶中反應(yīng)0（3）非理想氣體絕熱節(jié)流膨脹0（4）水在373K，p壓力蒸發(fā)0（5）理想氣體節(jié)流膨脹00（6）理想氣體向真空自由膨脹00（7）理想氣體絕熱可逆膨脹0（8）理想氣體等溫可逆膨脹0014.(1) 可逆過(guò)程 JK1(2) 自發(fā)進(jìn)行時(shí)JK1S總S體 +S環(huán)147.6 JK1(3) 根據(jù)F物理意義，等溫條件下，Wmax=F=(UTS)=TSH (凝聚相，U=H)=QrQp(恒壓)=4000(40000) = 44 kJ設(shè)計(jì)恒溫變壓可逆過(guò)程：G1+G50GT=268KG2G

5、1苯(l) p苯(l) 2.64kPa苯(g) 2.64kPa苯(s) p苯(s) 2.28kPa苯(g) 2.28kPaG4G5G315G1=Vl(plp)G5=Vs(pps)G2=G4=0G=G3 JK116.2C(石墨)+ 3H2 (g) C2H6 (g)Sm/JK1mol15.74130.6229.5=229.5(25.74+3130.6)=173.8 JK117C6H6(g)+ C2H2(g) C6H5C2H3(g)fHm/kJmol182.93226.25147.36Sm/JK1mol1269.2200.82345.1rHm= (fHm)產(chǎn)物(fSm)反應(yīng)物=147.6(82.93

6、 + 226.75) = 162.32 kJmol1rSm= (Sm)產(chǎn)物(Sm)反應(yīng)物=345.1(269.2 + 200.82) = 124.92 JK1rGm=rHmTrSm=162.32298.2(124.92)103 =125.07 kJ18C(石墨) C (金剛石)cHm/kJmol1393.5395.4Sm/JK1mol15.742.38rHm=(cSm)反應(yīng)物(cHm)產(chǎn)物=393.5(395.4)=1.9 kJmol1rSm= (Sm)產(chǎn)物(Sm)反應(yīng)物=2.385.74=3.36 JK1rGm=rHmTrSm=1900 298(3.36)=2901 JrGm0，常溫下，反應(yīng)

7、不會(huì)自發(fā)進(jìn)行，石墨穩(wěn)定。石墨p金剛石pGG3G2G1石墨p金剛石p19設(shè)壓力為p時(shí)，石墨金剛石，此時(shí)G=0由18題，G2=2901 JG=G1+G2+G3=0忽略p對(duì)V影響積分：p=1.532109 Pa=15120pH2O(l)pT=373.2KG1H2O(g)pT=373.2KG2H2O(g)pT=473.2KG3H2O(g)0.5pT=473.2K20變化過(guò)程：G1=0G2=H(T2S2T1S1)，分別計(jì)算H和始、終態(tài)的熵298K373K積分：298K473K積分：G2=H(T2S2T1S1)=3490(473.2204.63373.2196.34)=20067 JG終=G1+G2+G3

8、=22793 J*G2的另一解法：a=30.54，b=10.29103=81223+61161=20062 J21C2H2(g)+ 2H2(g) C2H6(g)Sm/JK1mol1200.82130.59229.49Cp,m/JK1mol143.9328.8452.65rSm= (Sm)產(chǎn)物(Sm)反應(yīng)物=229.49(200.82+2130.59) = 232.51 JK1Cp = 52.65(43.93+228.84) = 48.96 JK1 JK122.偏摩爾量偏摩爾量偏摩爾量化學(xué)勢(shì)化學(xué)勢(shì)化學(xué)勢(shì)23證明：dU=TdSpdV，恒溫對(duì)V求導(dǎo)：對(duì)于范氏氣體，代入上式：24證明：dU=TdSpd

9、VdH=TdS+Vdp25.溫度為T(mén)時(shí)，F(xiàn)=UTSdF=SdTpdV合并F項(xiàng)：兩邊除以T2：即26由集合公式：V=n水Vm,水+n乙醇Vm,乙醇，設(shè)總物質(zhì)量為1molV=0.4Vm,水+0.657.5解得V=40.97 mLVm,水=16.175 mLmol127V=1001.38+16.625n+17738m2/3+0.1194n2當(dāng)n=1時(shí)，V=1018.2 mL，代入集合公式：V=n水Vm,水+n鹽Vm,鹽1018.2=55.49Vm,水+119.5245Vm,水17.99 mLmol1101，p，l100，p，l100，2p，lSldTace101，p，g100，p，g100，2p，gSgdTbdfG=0VldpVgdp28純物質(zhì)，化學(xué)勢(shì) =Gm(1) 相平衡 a = b(2) 液相壓力增加 ca 因?yàn)閏a=Gm(l)=