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第二章习题解答
1.U=H=
02.等温可逆膨胀向真空膨胀由于始态和终态同上,体系的熵变也为
19.14JK–13.先求冷热水混合后温度500CpT–343+100CpT–300=0T=
336.3K
63.3C再计算熵变=–
41.27+
43.63=
2.34JK–14.Sn摩尔数nCpmSnT–473+1000CpH2OT–283=
02.106
24.14T–473+1000
4.184T–283=0T=
285.3K5.体系熵变按可逆相变计算真空蒸发热Q=H–nRT=10670–
8.314
373.2=37567J环境熵变S总=S体+S环=
8.28JK–10自发进行
6.体系设计可逆过程JK–1JK–1JK–1S体=S1+S2+S3=–
20.66JK–1环境JJK–1总熵S总=S体+S环=
0.81JK–10过程自发7.等量气体混合,热容相同,平衡温度T=15C=288K=
0.006+
11.526=
11.53JK–10自发过程8.W=–
30.65kJ,即环境做功
30.65kJW=Q1+Q2Q2=W–Q1=–
30.650–
334.7103=–
365.35kJ即向环境放热
365.35可kJ9.S总=S体+S环=
2.6JK–10不可逆过程10.Q=H=
92.14362=33355JW=pVg–VlpVg=nRT=
8.314
383.2=3186JU=Q–W=30169JG=0S总=011.
(1)等温可逆U=H=0Q=W=JS隔=S体+S环=0
(2)等温恒外压U=H=0S体=
14.9JK–1G=4442JQ=W=p外(V2–V1)=S隔=S体+S环=
26.67JK–112.计算GG2=0G30G总=–8591J计算S JK–1S30S体=S1+S2+S3=–105JK–1JK–1S总=S体+S环=
28.81JK–113.UHSGF
(1)理想气体卡若循环00000
(2)H2和O2绝热钢瓶中反应0
(3)非理想气体绝热节流膨胀0
(4)水在373K,p°压力蒸发0
(5)理想气体节流膨胀00
(6)理想气体向真空自由膨胀00
(7)理想气体绝热可逆膨胀0
(8)理想气体等温可逆膨胀
0014.1可逆过程JK–12自发进行时JK–1S总=S体+S环=
147.6JK–13根据F物理意义,等温条件下,Wmax=–F=–U–TS=TS–H凝聚相,U=H=Qr–Qp恒压=4000––40000=44kJ15.G1=Vlpl–pG5=Vsp–psG2=G4=0G=G3JK–
116.2C石墨+3H2gC2H6gSm/JK–1mol–
15.
74130.
6229.5=
229.5–2
5.74+3
130.6=–
173.8JK–117.C6H6g+C2H2gC6H5C2H3gfHm/kJmol–
182.
93226.
25147.36Sm/JK–1mol–
1269.
2200.
82345.1rHm=fHm产物–fSm反应物=
147.6–
82.93+
226.75=–
162.32kJmol–1rSm=Sm产物–Sm反应物=
345.1–
269.2+
200.82=–
124.92JK–1rGm=rHm–TrSm=–
162.32–
298.2–
124.9210–3=–
125.07kJ18.C石墨C金刚石cHm/kJmol–1–
393.5–
395.4Sm/JK–1mol–
15.
742.38rHm=cSm反应物–cHm产物=–
393.5––
395.4=
1.9kJmol–1rSm=Sm产物–Sm反应物=
2.38–
5.74=–
3.36JK–1rGm=rHm–TrSm=1900–298–
3.36=2901JrGm0,常温下,反应不会自发进行,石墨稳定19.设压力为p时,石墨金刚石,此时G=0由18题,G2=2901JG=G1+G2+G3=0忽略p对V影响积分p=
1.532109Pa=15120p20.变化过程G1=0G2=H–T2S2–T1S1,分别计算H和始、终态的熵298K373K积分298K473K积分G2=H–T2S2–T1S1=3490–
473.2
204.63–
373.2
196.34=–20067JG终=G1+G2+G3=–22793J*G2的另一解法a=
30.54,b=
10.2910–3=–81223+61161=–20062J21.C2H2g+2H2gC2H6gSm/JK–1mol–
1200.
82130.
59229.49Cpm/JK–1mol–
143.
9328.
8452.65rSm=Sm产物–Sm反应物=
229.49–
200.82+2
130.59=–
232.51JK–1Cp=
52.65–
43.93+2
28.84=–
48.96JK–1JK–
122.偏摩尔量偏摩尔量偏摩尔量化学势化学势化学势23.证明dU=TdS–pdV,恒温对V求导对于范氏气体,代入上式24.证明dU=TdS–pdVdH=TdS+Vdp
25.温度为T时,F=U–TSdF=–SdT–pdV合并F项两边除以T2即26.由集合公式V=n水Vm水+n乙醇Vm乙醇,设总物质量为1molV=
0.4Vm水+
0.6
57.5解得V=
40.97mLVm水=
16.175mLmol–127.V=
1001.38+
16.625n+17738m2/3+
0.1194n2当n=1时,,V=
1018.2mL,代入集合公式V=n水Vm水+n盐Vm盐
1018.2=
55.49Vm水+1
19.5245Vm水=
17.99mLmol–128.纯物质,化学势=Gm1相平衡a=b2液相压力增加ca因为c–a=Gml=Vldp=Vm2p–p=1810–6101325=
1.82J3气相压力增加db因为d–b=Gmg=Vgdp4由123得dc5aea=bbf SgSlef29.1液1中NH3的化学势液2中NH3的化学势转移过程,2溶解过程T=263KlT=263KsSS3S2S1T=273KlT=273Ksp恒定设计恒温变压可逆过程H2OgpH2OlpGG3G2G1H2Og3167PaH2Ol3167PaT=298KSS2S1S3设计恒温变压可逆过程G1+G50GT=268KG2G1苯lp苯l
2.64kPa苯g
2.64kPa苯sp苯s
2.28kPa苯g
2.28kPaG4G5G3石墨p金刚石pGG3G2G1石墨p金刚石pH2OlpT=
373.2KG1H2OgpT=
373.2KG2H2OgpT=
473.2KG3H2Og
0.5pT=
473.2K101,p,l100,p,l100,2p,l–SldTace101,p,g100,p,g100,2p,g–SgdTbdfG=0VldpVgdp。