物理化学（天大第五版全册）课后习题答案(13)

解：过程为

2mol 双原子气体2mol 双原子气体2mol 双原子气体恒容加热恒压加热 T1?300K?????T0?400K?????T2?？50dm3,p150dm3,p0100dm3,p0p1?2RT/V1?{2?8.3145?300/(50?10?3)}Pa?99774Pa p0?p1T0/T1?{99774?400/300}Pa?133032Pa T2?p0V2/(nR)1?{133032?100?10?3/(2?8.3145)}K?800.05K

W1=0； W2= -pamb（V2-V0）= {-133032×（100-50）×10} J= - 6651.6 J 所以，W = W2 = - 6.652 kJ ?H?nCp,m(T2?T1)?{2??U?nCV,m(T2?T1)?{2?7R?(800.05?300)}J?29104J?29.10kJ 25R?(800.05?300)}J?20788J?20.79kJ 2-3

Q = △U – W = （27.79 + 6.65）kJ≈ 27.44 kJ

TT?S??SV??Sp?nCV,mln0?nCp,mln2

T1T0-1 -1

= {2?5Rln400?2?7Rln800.05} J·K= 52.30 J·K

230024003-13 4 mol 单原子理想气体从始态750 K，150 kPa，先恒容冷却使压力降至 50 kPa，再

恒温可逆压缩至 100 kPa。求整个过程的Q，W，△U，△H，△S。

解：过程为

4mol 单原子气体4mol 单原子气体4mol 单原子气体恒容冷却 T1?750K?????T0?？?可逆压缩????T2?T0V1,p1?150kPaV1,p0?50kPaV2,100kPaT0?T1p0/p1?{50?750/150}K?250K W1?0，

W?W2?nRT0ln(p2/p0)?{4?8.3145?250ln(100/50)}J?5763J?5.763kJ 3?U2?0，?U??U1?{4?R?(250?750)}J??24944J??24.944kJ

2 ?H2?0，?H??H1?{4?5R?(250?750)}J??41570J??41.57kJ

2Q = △U – W = （-24.944 – 5.763）kJ = - 30.707 kJ ≈ 30.71 kJ

Tp?S??SV??ST?nCV,mln0?nRln2

T1p0= {4?3Rln250?4?Rln100} J·K= - 77.86 J·K

275050-1

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3-14 3 mol 双原子理想气体从始态100 kPa ，75 dm，先恒温可逆压缩使体积缩小至 50

33

dm，再恒压加热至100 dm。求整个过程的Q，W，△U，△H，△S。

解：过程为

3mol 双原子气体3mol 双原子气体3mol 双原子气体恒温可逆压缩恒压加热 V1?75dm3??????V0?50dm3?????V2?100dm3T1,p1?100kPaT1,p0?？T2,p0?p2T1?p1V1/(nR)?{100?103?75?10?3/(3?8.3145)}K?300.68K

3

p0?nRT?300.68/(50?10?3)}K?150000Pa?150kPa 1/V0?{3?8.3145T2?p2V2/(nR)?{150?103?100?10?3/(3?8.3145)}K?601.36K W?W1?W2??nRTV0/V1)?p0(V2?V0) 1ln(?{?3?8.3145?300.68ln(50/75)?150?103?(100?50)?10?3}J

= - 4459 J = - 4.46 kJ

5?U1?0，?U??U2?{3?R?(601.36?300.68)}J?18750J?18.75kJ

2 ?H1?0，?H??H2?{3?7R?(601.36?300.68)}J?26250J?26.25kJ

2Q = △U – W = （18.75 + 4.46 ）kJ = 23.21 kJ

pT?S??ST,r??Sp??nRln0?nCp,mln2

p1T0-1 -1

= {?3?R?ln150?3?7Rln601.36} J·K= 50.40 J·K

1002300.683-15 5 mol 单原子理想气体从始态 300 K，50kPa，先绝热可逆压缩至 100 kPa，再恒

3

压冷却使体积缩小至 85 dm，求整个过程的Q，W，△U，△H，△S。

解：过程示意如下： 5mol 单原子气体5mol 单原子气体5mol 单原子气体绝热可逆压缩恒压冷却热 , T0,V1?？,T1?300K,??????V0?？?????V2?85dm3,T2,p2p1?50kPa p0?100kPa

T0?(p0/p1)R/Cp,mT1?{(100/50)2/5?300}K?395.85K

V0?nRT0/p0?{5?8.3145?395.85/(100?103)}m3?0.16456m3 3Q1?0，W1??U1?{5?R?(395.85?300)}J?5977J?5.977kJ

2T2?p2V2100000?0.085?{}K?204.47K nR5?8.3143

-3

W2 = - pamb ( V2 – V1 ) = {- 100×10×（85 – 164.56）×10} J = 7956 J W = W1 + W2 = 13933 J = 13.933 kJ ?U2?{5?3R?(204.47?395.85)}J??11934J 2△U = △U1 + △U2 = -5957 J = - 5.957 kJ 5R?(204.47?300)}J??9929J??9.930kJ 2Q?Q2??U?W2?(?11.934?7.956)kJ??19.89kJ ?H?{5??S??S绝热，r??SP?0?nCp,mln(T2/T0) ?{5?5204.47R?ln}J?K?1??68.66J?K?1 2395.85 3-16 始态 300 K，1Mpa 的单原子理想气体 2 mol，反抗 0.2 Mpa的恒定外压绝热不

可逆膨胀平衡态。求整个过程的W，△U，△H，△S。

解：Q = 0，W = △U

3?pamb(V2?V1)?n?R(T2?T1)2

?nET2nRT1?3??pamb???n?R(T2?T1)?p?p21??amb代入数据整理得 5T2 = 3.4 T1 = 3.4×300K；故 T2 = 204 K W??U2?{2??H?{2?3R?(204?300)}J??2395J??2.395kJ25R?(204?300)}J??3991J??3.991kJ 2?S?nCp,mlnT2p?nRln2T1p152040.2 ?{2?R?ln?2?Rln}J?K?123001?1 ?{?16.033?26.762}J?K?10.729J?K?1?10.73J?K?1 3-17 组成为y（B）= 0.6的单原子气体A与双原子气体B的理想化合物共 10 mol，从始态T1 =300K，p1 = 50kPa，绝热可逆压缩至p2= 200 kPa的平衡态。求过程的W，△U，△H，△S（A），△S（B）。

解：先求混合物的摩尔定压热容

75Cp,m,mix??yBCp,m(B)?0.6?R?0.4?R?3.1R

22BT2?(p2/p1)R/Cp,mT1?{(200/50)1/3.1?300}K?469.17K

?H?{10?3.1R?(469.17?300)}J?43603J?43.60kJ

nB?6mol CV,m,mix?Cp,m,mix?R?2.1R nA?yAn?0.4?10mol?4mol?U?{10?2.1R?(469.17?300)}J?29538J?29.54kJ

pB,1?ybp1?0.6?50kPa?30kPa, pA,1?20kPa pB,2?ybp2?0.6?200kPa?120kPa, pA,2?80kPa

pA,2T25483.8780?nARln?{4?R?ln?4?Rln}J?K?1 T1pA,1230020?S(A)?nACp,m(A)ln ?{37.18?46.105}J?K?1??8.924J?K?1

因是绝热可逆过程，△S=△SA+△SB=0，故有△SB = - △SA = 8.924J·K或

?S(B)?nBCp,m(B)lnpB,2T27483.87120?nBRln ?{6?R?ln?6?Rln}J?K?1 T1pB,1230030-1

?8.924J?K?13-18 单原子气体A与双原子气体B的理想气体化合物共8 mol，组成为 y（B）= 0.25，

始态 T1 = 400 K，V1 = 50 dm。今绝热反抗某恒定外压不可逆膨胀至末态体积V2 = 250 dm3的平衡态。。求过程的W，△U，△H，△S。

解：先求混合物的摩尔定压热容

3

53CV,m,mix??yBCp,m(B)?0.25?R?0.75?R?1.75R

22BQ = 0，W = △U

?pamb(V2?V1)?nCV,m,mix(T2?T1) nET2?V2?V1??n?1.75R(T2?T1)?V2将数据代入，得 2.55 T2 = 1.75 T1= 1.75×400K，故 T2 = 274.51 K W??U?{8?1.75R?(274.51?400)}J??14610J??14.61kJ Cp,m,mix?CV,m,mix?R?1.75R?R?2.75R

?H?{8?2.75R?(274.51?400)}J??22954J??22.95kJ

nB?yBn?0.25?8mol?2mol, nA?6mol

?S(A)?nACp,m(A)lnT2V?nARln2T1V1

3274.51250 ?{6?R?ln?6?Rln}J?K?1 ?{?28.172?80.29}J?K?1?52.118J?K?1240050?S(B)?nBCp,m(B)lnT2V?nBRln2T1V15274.51250 ?{2?R?ln?2?Rln}J?K?1 ?{?15.651?26.763}J?K?1?11.112J?K?1

240050?S??S(A)??S(B)?(52.118?11.112)J?K?1?63.23J?K?13-19 常压下将 100 g，27 ℃的水与 200g，72℃的水在绝热容器中混合，求最终温

-1-1

度t 及过程的△S。已知水的比定压热容 cp = 4.184 J·g·K。

解：Qp = 0，△H = 0，△H1 +△H2 = 0

100×4.184×（T2 – 300.15K）+200×4.184×（T2 – 345.15K）=0 T2 – 300.15K + 2×（T2 – 345.15K）=0 T2 = 330.15 K 即 t = 57 ℃

-1330.15330.15???1= 2.68 J·K ?S??100?4.184?ln?200?4.184?lnJ?K?300.15345.15??33

3-20 将温度均为300K，压力为100 kPa的 100 dm的H2（g）与 50 dm 的CH4（g）恒温恒压混合，求过程的△S。 …… 此处隐藏：4510字，全部文档内容请下载后查看。喜欢就下载吧 ……