化工熱力學(xué)版第四章習(xí)題答案

本文格式為Word版，下載可任意編輯——化工熱力學(xué)版第四章習(xí)題答案4-11解：（1）已知H?300x1?450x2?x1x2(25x1?10x2)（A）由于x1?1?x2故H?300x1?450x2?x1x2(25x1?10x2)

?300x1?450(1?x1)?x1(1?x1)[25x1?10(1?x1)]

23?450?140x1?5x1（B）?15x1根據(jù)H?H?(1?x1)(?H)T?P?x1H2?H?x1(?H)T?P?x1其中

(?H)T.P??140?10x1?45x12?x123則：H1?450?140x1?5x1?15x1?(1?x1)(?140?10x1?45x12)23?310?10x1?50x1（C）?30x1H2?450?140x1?5x1?15x1?x1(?140?10x1?45x1)?450?5x1?30x1(D)(2)將x1?1及x1?0分別代入式（B），得純組元的焓H，和H2H1?300J?molH2?450J?mol

（3）H1和H2是指在x1?0及x1?1時的H1和H2的極限值。將x1?0代入式（C）中得H1?310J?mol將x1?1代入式（D）中得H2?475J?mol4-13解：根據(jù)摩爾性質(zhì)與偏摩爾性質(zhì)間的關(guān)系得M1?M?(1?x1)????23232dMdx1M2?M?x1當(dāng)M?V時

dMdx1V1?V?(1?x1)dMdx1V2?V?x1dMdx12已知V?109.4?16.8x1?2.64x1

得

dVdV代入V1和V2的表達式中??16.8?5.28x1將V及

dx1dx12得V1?92.6?5.28x1?2.64x1（A）2V2?109.4?2.64x1（B）

由式（A）由式（B）由于?V?當(dāng)x1?1時，得V1?89.96當(dāng)x1?0時，得V2?109.4

?x(V?V)

iii22所以?V?x1(V1?V1)?x2(V2?V2)

?x1(92.6?5.28x1?2.64x1?89.96)?(1?x1)(109.4?2.64x1?109.4)?2.64x1?5.28x1?2.64x1?2.64x1?2.64x1?2.64x1?2.64x1?2.64x1(1?x1)?2.64x1x24-14解：根據(jù)Gibbs-Duhem方程

22323?(xidMi)T,P?0

得恒溫恒壓下x1dM1?x2dM2?0

或x1dM1dM2dM2??x2?x2dx1dx1x2dH1dH2??x2dx1dx1當(dāng)Mi?Hi時，得x12已知H1?a1?b1x22H2?a2?b2x1

則

dH1??2b1?2b1x1dx1dH2?2b2x1dx1

x1dH1??2b1x1?2b1x12?2b1x1(x1?1)??2b1x1x2dx1dH2??2b2x2x1dx1dH1dH2??x2dx1dx1?x2只有當(dāng)b1?b2時x1結(jié)論得證。

4-15，試計算在25℃下，由22.5kgH2SO4與90kg50%(百分數(shù))H2SO4水溶液進行混合時的

熱效應(yīng)。

解：90kg50%的H2SO4水溶液中有：H2SO445kg,H2O45kg混合后溶液中有：H2SO422.5?45?67.5kg

H2O

45kg共計112.5kg混合后H2SO4的濃度為

67.5?100%?60%112.5由此利用有關(guān)的H2SO4—H2O的焓濃圖可進行計算。根據(jù)直線規(guī)則，將代表25℃50%H2SO4的直線交點可讀出t=60℃,該溫度即為絕熱混合之終溫，其相應(yīng)的焓值為—195kJ/kg查得25℃60%溶液的焓值為—275kJ/kg

kJ則Q?m??H?112.5?(?275?195)??90004-17，試用適合的狀態(tài)方程求正丁烷在460K,1.5?10Pa時的逸度與逸度系數(shù)。解：查附錄三得：Tc?425.12KPc?3.796MPaw?0.199

64601.5?106?1.082Pr??0.395Tr?6428.153.796?10查圖2-11，Tr、Pr點落在圖2-11分界限上方，故適用于普遍化其次維里系數(shù)關(guān)聯(lián)式。由式（2-30）得B(0)?0.083?B據(jù)式（4-86）ln?i?則ln?i?

(1)0.422??0.2891.61.0820.172?0.139??0.015

1.0821.6Pr(0)[B?wB(1)]Tr0.395?(?0.289?0.199?0.015)??0.10441.082?i?0.9009

fi?P?i?0.9009?1.5?106?1.351?106Pa4-18，試估算1-丁烯蒸氣在478K、6.88?10Pa時的逸度。

解：查附錄三得1-丁烯的臨界參數(shù)Tc?419w?0.187.5KPc?4.02MPa則對比溫度對比壓力為Tr?6T478P6.88MPaPr???1.139??1.711

Tc41.95Pc4.02MPa參照圖2-11普遍化關(guān)系適用范圍圖，Tr、Pr點落在分界限下方，適用于普遍化圖查圖4-3~4-6得：據(jù)ln?i?ln?i(0)?i(0)?0.700?i(1)?1.091

?wln?i(1)

ln?i?ln0.700?0.187ln1.091??0.3405

?i?0.7114

66fi?P?i?6.88?10?0.7114?4.894?10Pa

4-19，在25℃、20atm條件下，由組元1和組元2組成的二元液體混合物中，組元1的逸

?度f1由下式給出

?23f1?50x1?80x1?40x1式中，x1是組元1的摩爾分數(shù)，f1的單位為atm。在上述T和P下，試計算：

（1）純組元1的逸度f1；（2）純組元1的逸度系數(shù)；（3）組元1的亨利常數(shù)H1；

（4）作為x1函數(shù)的活度系數(shù)r1的表達式（組元1以Lewis—Randall規(guī)則為標準

態(tài)）。

??23解：在25℃、20atm，f1?50x1?80x1?40x1(1)在給定的溫度壓力下，當(dāng)x1?1時f1?10atm（2）根據(jù)定義?1?f110??0.5P20??ff50x1?80x12?40x13ii（3）根據(jù)lim?Hi得Hi?lim?lim?50atm

x1?0xx1?0xi?0xx1i23?50x?80x?40xf111（4）?r1?1?r1??5?80x1?40x12

x1(10)x1f14-20某類氣體的容量性質(zhì)由下式表示P?RT。式中，b只是組元的函數(shù)。對于V?b混合物b??yibi，式中，bi是純組元i的常數(shù)。試導(dǎo)出這類氣體下述性質(zhì)的表達式：

?（1）ln?i；（2）lnfi；（3）ln?i解：P??。；（4）lnfiRTRTbP或V?b?或Z?1?V?bPRTdPP（1）混合物的逸度系數(shù)公式可寫為ln???0(Z?1)

PdPbPPbP???ln???0RTPRT

對純組元i，ln?i?biPRT（2）?fi??iP，lnfi?ln?i?lnP?lnfi?biP?lnPRTP???(Z?1)（3）根據(jù)ln?i0?Zi?[dPPP?(yibi)P?(nibi)?(nZi)，nZ?n?]T,P,ni?j而Z?1?RTRT?nibiPRTbiPRT?Zi?1???因而ln?i??y??（4）因fiiiP

??biP?lny(iP)?lnfiRT4-21假使?1?G1?RTlnx1系在T、P不變時，二元溶液系統(tǒng)中組元1的化學(xué)位表達式，試證明?2?G2?RTlnx2是組元2的化學(xué)位表達式。G1和G2是在T和P的純液體組元1和組元2的自由焓，而x1和x2是摩爾分數(shù)。解：根據(jù)Gibbs-Dubem方程，

x1d?1?x2d?2?0（T、P恒定）

即x1d?1d?d?d??x22?0或x11?x22?0dx1dx1dx1dx2x1d?1d?1dx2?dlnx2（T、P恒定）

x2dx1dln?1d?1?RT（T、P恒定）

dlnx1?d?2???1?G1?RTlnx1故

由x2?1（此時?2?G2）積分到任意組成x2，得

即4-22

?2?G2?RT(lnx2?ln1)

?2?G2?RTlnx2

P2??P(B?y212)(B11?y212)ln?2221RTRT??解：（1）?ln?1其中12?2B12?B11?B22??9.5?2?14?265?232cm3?mol?1

??N2：ln?17?106(14?0.72?232)?10?6?0.2332

8.3145?461??1.2626?167?10??(?265?0.32?232)?10?6??0.4458n?C4H10：ln?28.3145?461??0.6403?222B?y1B11?2y1y2B12?y2B22

B?0.32?14?2?0.3?0.7?9.5?0.72?265??132.58(cm3?mol?1)V?（2）ij112212

RT8.3145?461?6?63?1?B??132.58?10?414.99?10(m?mol)6P7?10Tcij(K)

126.10425.12231.53

Pcij(MPa)

3.3943.7963.438

Vcij(cm3?mol?1)Zcij

90.1255158.44

1/31/3Vci?Vcjbi?106

26.773

80.670

aij

1.553429.00957.0113

0.2920.2740.283

根據(jù)混合規(guī)則Tcij?TciTcj，Vcij?(2)3，Zcij?Zci?Zcj2，Pcij?ZcijRTcijVcij，

2.50.42748R2Tcij0.08664RTci及bi?，aij?

PcijPci得到表上數(shù)據(jù)。

對二元物系，a?y1a11?2y1y2a12?y2a22

22?0.3?1.5534?2?0.3?0.7?7.0113?0.7?29.0095?17.2992(Pa?K0.5?m6)

22b?y1b1?y2b2?(0.3?26.763?0.7?80.670)?10?6?64.4979?10?6(m3)

求Z，V

1ah?()1?hbRT1.51?hbPh?

ZRT據(jù)Z?代入數(shù)據(jù)，得：Z?117.299h1h?()??3.259()1?h64.4979?10?6?8.3145?4611.51?h1?h1?h64.4979?10?6?7?1060.118?h?

Z?8.314?5461Z迭代求解：設(shè)Z0?0.8?h0?0.1475?Z1?0.7541?h1?0.1565?

Z2?0.7439?h2?0.1586?Z3?0.7423?h3?0.1589?Z4?0.7421?Z?0.7421V?ZRT0.7421?8.3145?46163?1??406.35?10(m?mol)6P7?10

N2:??lnV?b1?2(y1a11?y2a12)lnV?b?ab1(lnV?b?b)?lnZ?0.2679ln?1V?bV?bVVV?bRT1.5bRT1.5b2??1.3073??14-23解：（1）根據(jù)

?i?yiPPisxiy1P0.634?24.4??2.2361s23.06?0.3P1x1y2P(1?y1)P(1?0.634)?24.4???1.2694P2sx2P2s(1?x1)10.05?(1?0.3)得：

?1?

?2?E（2）根據(jù)G?RT?xiln?i

得：G?8.3145?318(0.3ln2.2361?0.7ln1.2694)?1079.8(J?mol)

E?1?i根據(jù)?G?RT?xilna得：?G?RT?xiln(?ixi)?RT?x1ln(?1x1)?x2ln(?2x2)?

[o.3ln(2.2361?0.3)?0.7ln(1.2694?0.7)]?8.3145?318??535.3(J?mol?1)

（3）已知

?H?0.437RTHE?H?據(jù)RTRTHE?0.437R得T?(GE/T)HE0.437R]P.x??2???[?TTTGE0.437R)??dT（恒P，x）?d(TT將T1?318K，T2?333K，G1E?1079.8代入上式得

EG2G1ET1079.8333??0.437Rln2??0.437?8.3145lnT2T1T1318318?G2?1075.0(J?mol)

4-24解：兩個公式在熱力學(xué)上若正確，須滿足恒T，P的G?D方程，即x1E?1dln?1dln?2?x2?0dx1dx1x1dln?1dln?2?x2?x1(b?a?2bx1)?x2(?b?a?2bx2)dx1dx122?a(x2?x1)?b(x2?x1)?2b(x2?x1)?(a?b)(x2?x1)?(a?b)(1?2x1)?0（a?b）

?這兩個公式在熱力學(xué)上不正確。

GE?Ax1x2，對組元14-25已知RT?(nGE/RT)又?ln?1?[]T,P,n2

?n1由于x1?n1n和x2?2

nnnGEAn1n2??RTn則ln?1?An2[?(n1/n)nn1n]n2?An2(?1)?A2(1?1)?n1nn2nn2或ln?1?Ax2(1?x1)?Ax22同理，對組元2，ln?2?Ax1

E圖4-1所示為ln?1，ln?2和G/RT作為x1函數(shù)的關(guān)系圖線，設(shè)圖取A?1。標準

態(tài)的選擇對兩個組元都以Lewis-Randall規(guī)則為基準，這是一種很普通的選擇法。超額Gibbs自由能在x1?0和x1?1兩處都為零。兩個活度系數(shù)符合下述必要條件lim?i?1。

xi?1?id?fx，而f?f(T,P)?常數(shù)，結(jié)論正確。4-26解：（1）理想溶液fiii（2）錯。?vid?0，?uid?0，?Hid?0而?Gid?RT?xilnxi?0?Sid??R?xilnxi?0

（3）正確。ME??M??Mid對理想溶液ME??Mid??Mid?0

（4）錯。P?0，limf?1P?0P234-27解：ln??y1y2(1?y2)?(1?y2)y2(1?y2)?(1?y2)y2?y2?y2

?是ln?的偏摩爾量，根據(jù)截距發(fā)公式得?ln?i??ln??yln?12dln?323?y2?y2?y2(1?3y2)?2y2dy2??e2y2?13

??yP???4ye1y2??f?f111113??ln??(1?y)同理ln?22dln?3223?y2?y2?(1?y2)(1?3y2)?1?3y2?2y2dy2323)(1?3y2?2y2)???e(1?3y22?2y2??f?4ye222??2.568ΜPαf??3.29M當(dāng)y1?y2?0.5時：f7Pa124-28解：當(dāng)汽—液兩相平衡時，須滿足

?v?f?lfii上標v和l分別指的是汽相和液相。

若汽相可視為理想氣體，則系統(tǒng)的壓力比較低，液相的標準態(tài)逸度fi??Pis，則

有

?Py??fx?Py??Psx?iiiiiiiii假使存在恒沸物，即yi?xi

?1P2s對二元物系來說，則有?s

?2P1?APBs8?104對此題?s??0.675?BPA1.2?10GE?0.5xAxB已知RT?(nGE/RT)2?ln?A?[]T.P.nB?0.5xB?nA

?A?e20.5xB

同理?B?e?0.5x2A?A?e0.5(x?BB?xA)?0.67

解得：xA?0.9056

0?xA?1，說明在353K時該系統(tǒng)有共沸物存在。4-29解：

（1）VanLaar方程ln?1?A12(A21x2A12x1)2，ln?2?A21()2

A12x1?A21x2A12x1?A21x2式中A12和A21由恒沸點的數(shù)據(jù)求得。在恒沸點，yi?xi??i?PyiP?ssPixiPi則?1?P101.3P101.3，??1.0191????1.04142ss99.4097.27P1P2A12?ln?1(1?x2ln?220.475ln1.04142)?ln1.0191(1?)?0.1635

x1ln?10.525ln1.0191x1ln?120.475ln1.01912)?ln1.0414(1?)?0.0932

x2ln?20.525ln1.0414A21?ln?2(1?全濃度范圍內(nèi)，苯和環(huán)己烷的活度系數(shù)為

20.0932x20.1635x22ln?1?0.1635()?20.1635x1?0.0932x2(1.7543x1?x2)0.1635x10.0932x122ln?2?0.0932()?20.1635x1?0.0932x2(x1?0.5700x2)（2）Statchard和Hildebrand方程

V1?89cm?mol，V2?109cm?mol

2V1?22109x20.461x7892222?ln?1?(?1??2)??()(18.82?14.93)?2RT8.3145?350.889x1?109x2(0.816x51?x2)2V2?1289x20.565x51092222?ln?2?(?1??2)??()(18.82?14.93)?2RT8.3145?350.889x1?109x2(x1?1.224x72)3?13?1?1?18.82(J0.5?cm?1.5)，?2?14.93(J0.5?cm?1.5)?1?x1V189x1x2V2109x1，?2???x1V1?x2V289x1?109x2x1V1?x2V289x1?109x2在恒沸點，x1?0.525，x2?0.475

0.4617?0.4752?ln?1??0.1276?1?1.13612(0.8165?0.525?0.475)0.565?50.5225ln?2??1?1.1357?0.12732(0.525?51.224?70.47)5活度系數(shù)比（1）中計算偏大。

（3）當(dāng)x1?0.8時，x2?0.2

0.1635?0.22用VanLaar方程：ln?1??0.002544?1?1.00252(1.7543?0.8?0.2)0.0932?0.82ln?2??0.07140?2?1.0740(0.8?0.5700?0.2)2y1??1P1sx1P?1.0025?99.40?0.8?0.79

101.3用Scatchard和Hildebrand方程：

0.4617?0.22ln?1??0.02537?1?1.02569

(0.8165?0.8?0.2)20.5655?0.82ln?2??0.3315?2?1.39302(0.8?1.2247?0.2)y1??1P1sx1P?1.02569?99.40?0.8?0.81

101.3

4-30解：根據(jù)Wilson方程ln?i?1?ln(??ijxj)??jk?kixk?kjxj將上式應(yīng)用于三元系統(tǒng)，并將已知的各參數(shù)代入，即可求得該三元系統(tǒng)各組分的

活度系數(shù)：ln?1?1?ln(x1?x2?1