天大物化五版上册习题答案(7)

物理化学上册习题解（天津大学第五版）

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

l4mo l n B?6mo lCV,m,mix?Cp,m,mix?R?2.1R nA?yAn?0.4?10mo??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-1 或

?S(B)?nBCp,m(B)lnpB,2T27483.87120?nBRln ?{6?R?ln?6?Rln}J?K?1 T1pB,1230030 ?8.924J?K?13-18 解：先求混合物的摩尔定压热容

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 解：Qp = 0，△H = 0，△H1 +△H2 = 0

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物理化学上册习题解（天津大学第五版）

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 ℃

330.15330.15???1K-1 ?S??100?4.184?ln?200?4.184?ln?J?K= 2.68 J·300.15345.15??100?103?50?10?3?3-20 解：nCH????mol?16.667mol ??4300??nH2?100?103?100?10?3???300?? ??mol?33.333mol?V2V?nCH4Rln2V1,H2V1,CH4?S??SH2??SCH4?nH2Rln ?33.333?8.3145?ln

150150?16.667?8.3145?ln10050 = （13.516 +18.310）J·K-1= 31.83 J·K-1

3-21解：QV = 0，W = 0， △U = 0，则有 △U（单）+△U（双） = 0 2?35R?(T2?200K)?3?R?(T2?400K)?0 22解得 T2 = 342.86K ?S(A)?nACV,m(B)lnT2V342.86150??3?1?nARln2 ??2?R?ln?2?R?ln?J?K T1,AV1,A?220050? = 31.714 J·K-1 TV342.86150??5?1?S(B)?nBCV,m(B)ln2?nBRln2??3?R?ln?3?R?ln?J?K

T1,BV1,B?2400100? = 0.502 J·K-1

△S = △S（A）+△S（B）= （31.714 + 0.502）J·K-1= 32.216 J·K-1 =32.22 J·K-1

3-22 解：设左侧的N2（g）用A代表，左侧的N2（g）用B代表。混合过程示意如下： A ，2 molA B，4 mol 2 mol B ， 4 mol ΔSA 3350 dmVA， VB， ， 75 dm ， 200K,pT ;p T;p A 500K,p B ΔSB

QV = 0，W = 0， △U = 0，则有 △U（A）+△U（B） = 0

2?CV,m?(T2?200K)?4?CV,m?(T2?500K)?0 解得 T2 = 400 K

方法一：若用分体积计算熵变：

VA?(2/6)?125dm3?41.67dm3,VB?(4/6)?125dm3?83.33dm3, ?S(A)?nACV,m(B)lnV2,A?5T240041.67??1?nARln??2?R?ln?2?R?ln?J?KT1,AV1,A?220050? ?( 28.816-3.03 0) J?K?1?25.786J?K?1V2,B?5T40083.33??1?S(B)?nBCV,m(B)ln2?nBRln ??4?R?ln?4?R?ln?J?KT1,BV1,B50075??2 ?(-18.553?3.503)?-15.05J?K?132

物理化学上册习题解（天津大学第五版）

△S = △S（A）+△S（B）= （24.786– 15.05）J·K-1= 10.736 J·K-1 方法二：先计算A和B各自初始压力及终态压力 pA?p?2?8.315?2004?8.315?500；?66.52kPap??221.73kPa B?3?350?1075?106?8.315?400?159.65kPa

125?10?3T2p40066.52??7?1?nARlnA ??2?R?ln?2?R?ln?J?KT1,Ap?2200159.65??S(A)?nACp,m(B)ln ?( 40.345-14.559) J?K?1?25.786J?K?1?S(B)?nBCp,m(B)lnT2p400221.73??7?1?nBRlnB ??4?R?ln?4?R?ln?J?KT1,Ap?2500159.65?

?( -25.976?10.925) J?K?1?15.05J?K?1△S = △S（A）+△S（B）= （24.786– 15.05）J·K-1= 10.736 J·K-1

3-23 解：n = （1000÷32）mol = 31.25 mol Q = Qp = △H = n△vapHm = （31.25×35.32）kJ = 1103.75 kJ W = - pamb（Vg – Vl ）≈ - pambVg = -ng RT = {- 31.25×8.3145×337.80} = - 87770 J= - 87.77 kJ

△U = Q – W = （1103.75 - 87.77）kJ = 1015.98 kJ

△S = n△vapHm / Tvap = （1103750÷337.80） = 3267 J·K-1 = 3.267 k J·K-1

3-24解：常压绝热混合，Qp = 0，设末态温度为T2（T2＞273.15K），于是有 500×333.3 + 500×4.184×（T2 – 273.15K）+ 1000×4.184×（T2 – 298315K）=0 解得 T2= 263 K 显然，- 10℃这个结果不合理。因此，只是高温水放出热量使部分冰熔化，温度仍是0℃。设0℃冰量为 m，则0℃水量为（500 – m）g，其状态示意如下

500g,H2O(s), 273.15KQp?0(500?m)gH2O(l), mH2O(s), 273.15K ????1000g, H2O(l), 298.15K1000g, H2O(l), 273.15K（500-m）g×333.3 J·g-1+ 1000g×4.184 J·g-1·K-1×（273.15K– 298.15K）=0

333.3 m = 62050 g

m = 186.17 g 0℃熔化的水量 = （500 – 186.17）g = 313.83 g ?S??fusS(H2O,s)??S(H2O,l)

273.15??313.83?333.3?1?1 ???1000?4.184?ln?J?K?16.52J?K273.15298.15??3-25 解：常压绝热混合，Qp = 0， 500g×333.3 J·g-1+ 500×4.184 J·g-1·K-1×（T2-273.15K）

+1000g×4.184 J·g-1·K-1×（T2– 353.15K）=0

12.552 T2 = 3764.7188 K T2 = 299.93 K 冰的熵变：

?S1??fusS(H2O,s)??S(H2O,l)

299.93??500?333.3?1?1 ???500?4.184?ln?J?K?805.765J?K273.15273.15??299.93??1?1水的熵变：?S2???1000?4.184?ln?J?K??683.430J?K 353.15??△S = △S1 + △S2 = 122.33 J·K-1

3-26 解：和3-24题类似，高温水放出热量使部分冰熔化，温度仍是0℃。设0℃冰量为 m，

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物理化学上册习题解（天津大学第五版）

则0℃水量为（500 – m）g，其状态示意如下

500g,H2O(s), 263.15KQp?0(500?m)gH2O(l), mH2O(s), 273.15K ????1000g, H2O(l), 298.15K1000g, H2O(l), 273.15K500×2.00 J·g-1·K-1×（273.15K– 263.15K）+（500-m）g×333.3 J·g-1

+ 1000g×4.184 J·g-1·K-1×（273.15K– 298.15K）=0 333.3 m = 72050 g

m = 216.17g 熔化的水量 = （500 – 216.17）g = 283.83 g 冰的熵变：

?S1??S(H2O,S)??fusS(H2O,s)

273.15283.83?333.3???1?1 ??500?2?ln???J?K?383.63J?K263.15273.15??273.15??1?1水的熵变：?S2???1000?4.184?ln?J?K??366.42J?K 298.15??△S = △S1 + △S2 = 17.21 J·K-1

3-27 解：设液态苯全部凝固，冰全部融化，于是示意如下

8molH2O(l), 2molH2O(l), t8molH2O(s), 2molH2O(l), 0oCQ?0 ????o5molC6H6(s), 5molC6H6(s),t5molC6H6(s), 5molC6H6(l), 5.51C 8mol×6004 J·mol-1+10mol×75.37 J·mol-1·K-1（T2 - 273.15K）

+5mol×（-9832）J·mol-1 +10mol×122.59 J·mol-1·K-1×（T2-278.66K）=0 1979.6 T2 =548610.395K T2 =277.13K 所以，t=3.98℃，0℃＜3.98℃＜5.51℃，假设合理。

277.31??8?6004?1?S(H2O)???10?75.37ln?J?K

273.15??273.15=（175.845+11.392）J·K-1= 187.24 J·K-1

)277.31??5?(?9832?1?S(C6H6)???10?122.59ln?J?K

278.66??278.66=（-176.416 - 5.953 ）J·K-1= -182.37 J·K-1

△S = △S1 + △S2 = 187.24 J·K-1 - 182.37 J·K-1 = 4.87 J·K-1

3-28解：（1）p乙醚?nRT/V??0.1?8.314?308.66/10?kPa?25.664kPa （2） 画出如下框图：

0.1mol 乙醚（g）?H ?S ?????0035.51C, p?0kPa35.51C, 25.664kPa △H0 △S0

0.1mol 乙醚（l）0.1mol 乙醚（l） △H1 △S1

35.510C, 101.325kPa △H2 △S2

0.1mol 乙醚（g）035.51C, 101.325kPa

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物理化学上册习题解（天津大学第五版）

?H??H1??H2?2.5104kJ; W?0; Q??U??H-?(pV)??H-ngRT?2.2538kJ ?S1??H1/T?(2.5104?103/308.66)J?K?1?8.133J?K?1 ?H0?0; ?S0?0: ?H1?0.1?25.104kJ?2.5104kJ ?H2?0?S2?nRln(p1/p2)??0.1?8.314ln(101.325/25.664)?J?K?1?1.142J?K?1

△S= △S1+ △S2=9.275J·K-1

3-29 解：把苯蒸气看作是理想气体，恒温可逆压缩时，△U1=0，△H1=0，于是有 W1?nRTln(p2/p1) ?{1?8.3145?353.25?ln(101.325/40.53)}J?2691J W2 = -pamb（Vl – Vg） ≈pambVg = ng RT= （1×8.3145×353.25）J =2937.1 J W3 ≈ 0; W= W1 + W2 + W3=（2691+2937.1+0）J= 5628 J = 5.628 kJ △U1 = 0，Q1 = W1 = 2937 J； Q2 = -30878 J Q3??333.15K353.25KnCp,mdT?{1?14.27?（333.15?353.25）}J??2868J

Q = Q1 + Q2 + Q3 = {（-2691）+（ -30878）+（ – 2868）}= - 36437J = -36.437 kJ

△U = Q + W = - 36.437 kJ + 5.628 kJ = - 30.809 kJ

△H = △H1 + △H2 + △H3 ={ 0 +（-30.868）+（-2.868）} kJ = - 33.746 kJ

3-30 解：因液态水占的体积小，可以认为20 dm3 的密闭容器体积是气体的体积，于是，与液态成平衡的水气的物质的量为

47.343?103?20?10?3?始态：ng,353.15K?p1V1????mol?0.3225mol ??RT1?8.314?353.15?末态：ng,373.15Kp1V1?101.325?103?20?10?3?????mol?0.6532mol ??RT1?8.314?373.15?始态液态水的物质的量 = 2mol – 0.325 mol =1.6675 mol

末态液态水的物质的量 = 2mol – 0.6532 mol =1.3468 mol 为求过程的Q，△U，△H及△S，设计如下途径：

1.6775 mol H2O(l)1.3468 mol H2O(l)?H0.3225 mol H2O(g)???0.6532 mol H2O(g) 353.15K, 20 dm3373.15K, 20 dm3

△H1 △H4

1.6775 mol H2O(l)1.3468 mol H2O(l)H20.3225 mol H2O(g)?????0.6532 mol H2O(g)

298.15K298.15K△H1 = {1.6775×75.75×（298.15 – 353.15）+ 0.3225×33.76×（298.15 – 353.15）}J

= （- 6988.88 – 598.82 ）= - 7587.7 J △H2 = {（1.6775 – 1.3468 ）×44106}J = 14556.1 J △H3 = {1.3468×75.75×（373.15.15 – 298.15）+ 0.6532×33.76×（373.15 – 298.15）}J

= （7651.5 + 1653.9 ）= 9305.4 J

△H = △H1 + △H2 + △H3 =16274 J =16.274 kJ

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