(完整)化工熱力學(xué)答案(第三版)(完整)化工熱力學(xué)答案(第三版)化工熱力學(xué)課后答案（第三版）陳鐘秀編著2-11kmol0.12465（）（）R-K（）普遍化關(guān)系式。=0。1246m3/1kmol=124.6c/mol查附錄二得甲烷的臨界參數(shù)：Tc

=190。6K

=4.600MPa c

=99cm3/molcω=0。008理想氣體方程P=RT/V=8.314×323。15/124。6×1—=21.56MPaR—Ka0.42748

R2TcPc

0.42748

8.3142190.62.54.6106

3.222Pam6K0.5mol2b0.08664

RTPc0.08664c

8.314190.64.6106

2.985105m3mol1∴P RTVb

aT0.VVb 8.314323.15 3.22212.462.985105 323.150.512.4610512.462.985105=19.04MPa普遍化關(guān)系式TTTr

323.15190.61.695 Vr

VVc

124.6991.259＜2∴利用普壓法計(jì)算，ZZ0Z1∵ PZRT

PPV crc∴ ZPVPcRT rPV 4.610612.46105Z c PRT r

8.314

P0.2133Pr r迭代：令Z=→P=4.687又Tr=1.695，查附錄三得：=0.8938 Z1=。46230 r0ZZ0Z1=0。8938+0.008×0。4623=0。8975此時(shí),P=Pc

P4。6×4.687=21.56MParZ=0。8975Z1ZP∴P=19。22MPa2-2.Pitzer510K、2。5MPa1480.7cm/mol.解:查附錄二得正丁烷的臨界參數(shù)：Tc

=425.2K

=3。800MPa c

=99cm3/molcω=0。193理想氣體方程V=RT/P=?！?1.6961.4807100%14.54%1.4807Pitzer對(duì)比參數(shù):Tr

TTc

510425.21.199 Pr

PPc

2.53.80.6579—普維法∴B00.0830.422

0.083

0.422

0.2326B10.139

T1.6 1.1991.6r0.1720.139 0.172 0.05874T4.2r

1.1994.2BPcB0RTc

1=—0。2326+0。193×0.05874=-0.2213BP BP

=1—0.2213×0.6579/1。199=0。8786Z1

RT1 c rRT Tc r∴P=ZRT→V=ZRT/P=0。8786×8。61.491.4807100%0.63%1.4807

m/mol2—376%(摩爾分?jǐn)?shù))的碳生成二氧化碳，其余的生成一氧化碳。試計(jì)算（）含碳量為81.38100kg焦炭能生成1.1013MPa、303K的吹風(fēng)氣若干立方米?（2)所得吹風(fēng)氣的組成和各氣體分壓。解：查附錄二得混合氣中各組分的臨界參數(shù)：一氧化碳（1

=132。9K c

=3。496MPa c

=93.1cm3/molω=0049cZ=0.295c

=304.2K

=7376MPa

=94.0c3/molω=0225

=0。274y1

=0.24，y2

c c c c=0.76∴(1）KayT cmiP i

TiciP

0.24132.90.76304.2263.1K0.243.4960.767.3766.445MPacm iT TT

ci303263.11.15

P

0.1011.4450.0157-普維法rm cm rm cm利用真實(shí)氣體混合物的第二維里系數(shù)法進(jìn)行計(jì)算B00.0831

0.4220.083 303132.303132.91.6r1

0.02989B10.1391

0.172T4.2r1

0.139

0.1723039

0.1336RT B c1 B0

B18.314132.90.029890.0490.1336

7.37810611 P c1

1 1 3.496106B00.0832B10.1392

0.422T1.6r20.172T4.22r22

0.0830.139

0.422303304.21.60.1723032

0.34170.03588RT 0

18.314304.2

6B c2 B 22 P 2

2B 7.376106

0.34170.225

119.9310又Tcij

c2TTcicj

0.5

132.9304.20.5201.068KV13V133 333V c1

c2

2

93.55cm3/molcij

Z cij

Z Zc1 c2

0.2950.2742

0.2845

0.2950.225 1cij 2

2 2 0.137Pcij

Zcij

RTcij

/Vcij

0.28458.314201.068/1065.0838MPa∴Trij

TTcij

201.0681.507 Prij

PPcij

0.10135.08380.01990.422 0.422B00.083 0.083 0.13612 T1.6r12

1.5071.6B10.13912

0.1720.139 0.172 0.1083T4.2 1.5074.2r12∴B RT 0 c12 B

B1

8.314201.0680.1360.137

39.8410612 P 12c12

12

5.0838106B y2

2yy

y2Bm 1

1 2

2 220.2427.37810620.240.7639.841060.762119.9310684.27106cm3/mol ∴Z 1

BP PVm

→V=。02486m/molm RT RT∴V=nV=100×1×81.38%/12×0.02486=168.58m3總（2)

PyPZc11 1 ZmZ

0.240.10130.2950.025MPa0.28450.274PyP c2 2 Zm

0.760.10130.28450.074MPa2-4。03MP477K。83mNH3

0.142，若壓縮后溫度448.6K,則其壓力為若干？分別用下述方法計(jì)算：(1)VanderWaals方程；（2)Redlich-Kwang(3）Peng-Robinson（4解：查附錄二得NH的臨界參數(shù)：T=405.6K P=11。28MPa

=7。5c3/mol3 c c cω=0。250(1)

求取氣體的摩爾體積對(duì)于狀態(tài)Ⅰ：P=。03MP、T=447、V=。83m3TTTr

477405.61.176 Pr

PPc

2.0311.280.18—普維法∴B00.0830.422T1.6r0.172

0.0830.4221.1761.60.172

0.2426B10.139BP

T4.2r

0.139

1.1764.2

0.05194cB0B10.24260.250.051940.2296RTcBPZ1

PV BP1

P→V=。885×1—m/molrRT RT RT Trc r∴n=2.83/1。885×10—3/mol=1501mol對(duì)于狀態(tài)Ⅱ：摩爾體積V=0.1423/1501mol=。45×10-5/mol T=448。6KVanderWaals27R2Ta c64Pc

278.31426411.28106

0.4253Pam6mol2RTb c8Pc

8.314405.6811.28106

3.737105m3mol1P RT a

8.314

17.65MPaVb V2

9.45870

3.7371052Redlich—Kwanga0.42748

R2TcPc

0.42748

8.314211.28106

8.679Pam6K0.5mol2b0.08664

RTPc0.08664c

8.31411.28106

2.59105m3mol1P RTVb

aTb

8.314448.69.45890

8.679448.60.59.4581059.45890

18.34MPaPeng-Robinson∵TTTr

448.6405.61.106∴k0.37460.2699220.37461.542260.250.269920.2520.7433r T1k1T0.5210.743311.1060.5r

0.9247a

ac

0.45724

R2TcPc

0.45724 0.92470.4262Pam6mol28.3142405.628.3142405.62cb0.07780RTcPc

8.314405.60.07780 2.326105m3mol11.28106∴P RT

aTVb

Vbbb 8.314448.6

0.4262

19.00MPa9.4582.326105 9.4589.4582.32610102.3269.4582.3261010普遍化關(guān)系式∵ V r

V 9.4581057.251051.305＜2適用普壓法，迭代進(jìn)行計(jì)算，方法同c1—1(3）2-6.30％（摩爾分?jǐn)?shù)）氮?dú)?0％(摩爾分?jǐn)?shù)）正丁烷氣體718888MPa

=14cm/molB

=-265c3/mol，B=—9.5c3/mo。12

11 22

y2

2yy

y2Bm 1

1 2

2 221420.30.79.50.72265132.58cm3/molmZ 1Bm

PV→V(摩爾體積）=。24×10—m/molm RT RT假設(shè)氣體混合物總的摩爾數(shù)為n，則0.3n×28+0.7n×58=7→n=0.1429mol∴V=n×V(摩爾體積)=0.1429×4.24×10-4=60.57c32—8。試用R-KSRK273K1013MPa2。0685解：適用EOS的普遍化形式查附錄二得NH的臨界參數(shù)：T=126。2K

=3。394MPa ω=0.043 c c（1)R-K方程的普遍化R2T2.5 8.3142126.22.5a0.42748

0.42748 1.5577Pam6K0.5mol23.394106cRT 8.314126.2b0.08664

c0.08664 2.678105m3mol1P 3.394106cA

BbP A a 1.5577

1.551R2T2.5

RT B

bRT1.5 2.6781058.314∴hB

b bP

2.678105101.3106

1.1952 ①Z V ZRT Z8.314273 ZZ 1

A h

1 1.551 h ②1h B1h 1h 1h ①、②兩式聯(lián)立，迭代求解壓縮因子Z（2）SRK方程的普遍化TTTr

273126.22.163m0.4801.5740.17620.4801.5740.040.1760.0420.5427

11m1T0.5

1 10.542712.160.5

0.2563TrR2T2

r 2.163 8.3142126.22.5a0.42748

Pc c

0.42748 0.25630.3992Pam6K0.5mol23.394106RTb0.08664 Pc

8.314126.20.08664 2.678105m3mol3.394106A a 0.3992

0.3975B bRT1.5 2.6781058.314∴hB

b bP

2.678105101.3106

1.1952 ①Z V ZRT Z8.314273 ZZ 1

A h

1 0.3975 h ②1h B1h

1h

1h ①、②兩式聯(lián)立，迭代求解壓縮因子Z第三章3-1.物質(zhì)的體積膨脹系數(shù)和等溫壓縮系數(shù)k的定義分別為：

1VVT

， 。 k1V V PVanderWaals和k的表達(dá)式.

P

T解：Vanderwaals方程P RT aVb V2Z=f(x,y）的性質(zhì)zx

xy

yz

得 P1 V

VT

TP

1 y

z

T

P V又 P

2a RT

P R VT

V3 Vb

TV

Vb所以 2a

RT V Vb

1V

V

Vb2T RPRV3bPTP

RTV32aVb2V 故 1V P k1V

RV2bRTV32aVb2V2Vb2VPT

RTV32aVb23-2.3445MPa93℃，3.45MPa之U、H、、、G、TdS、pdV、QW。U=0、H=0∴ —=pdV2pdV1RTdVRTln2=2109.2J/mol1V V1∴ J/mol又 dSC dT

V

dP

、V RPT TP

T PP ∴ dSR 2∴ SS2dSR2dlnPRlnP2

R

=5。763J/（mol·K)1AUTS=—366×5。763=—2109.26J/（mol·K)GHTSA=—2109。26J/（mol·K）TdSTSA=—2109。26J/（mol·K)pdV

pdV1RTdVRTln2=2109.2J/molV3—3.試求算1kmol氮?dú)庠趬毫?0.13MPa、溫度為773K下的內(nèi)能、焓、熵、CV、Cp和自由焓之值。假設(shè)氮?dú)夥睦硐霘怏w定律。已知：（1)在0。1013MPa時(shí)氮的C與溫度的關(guān)系為C 27.22/molK；p p（2）假定在0℃及0.1013MPa時(shí)氮的焓為零；（3298K0.1013MPa191.76J（mol·K（）熵值的計(jì)算dSCp

VdT dpT p

dSCpdTRdp對(duì)于理想氣體： T pdS73 73CdST

dT

113

Rdpp298

298

0.1013SS

73(2220418T)1

dT

113

Rdp0 T p298 0.10130.004187(773298)27.22ln7738.314ln10.13 298 0.101310.354Jmol1K1SS0

10.354191.7610.354181.4(Jmol1K1)(2）焓值的計(jì)算dHCdTpHH0

773273

0.004187T)dT127.22(773273) (77322732)214704.9(Jmol1)HH014704.914704.9(Jmol1)14704.9KJKmol1)其他熱力學(xué)性質(zhì)計(jì)算UHpVHRT14704.98.3147738278.178(KJKmol1)AUTS7278.178773181.4132944.022(KJKmol1)GHTS14704.9773181.4125517.3(KJKmol1)C 773mol1K)pC CV

R30.458.31422.14(KJKmol1)3-4.27℃、0.1MPa227℃、10MPa熵值。已知氯在理想氣體狀態(tài)下的定壓摩爾熱容為Cig31.69610.144103T4.038106T2J/molKp解：分析熱力學(xué)過(guò)程300K0.1MPa真實(shí)氣體H=0，S=0-HR1

H

500K10MPa真實(shí)氣體H2-SR13000.1MPa

H、S

S250010MPa理想氣體

1 1理想氣體查附錄二得氯的臨界參數(shù)為:T=417K、P=7。701MPa、ω=0.073c c∴(1）300K、0.1MPa的真實(shí)氣體轉(zhuǎn)換為理想氣體的剩余焓和剩余熵T=Tr

/T=300/417=0。719 1 c

=P/r 1

=0。1/7。701=0。013—利c用普維法計(jì)算B00.083

0.422 dB0 0.6324 0.675T B10.139

T1.6r0.172T4.2r

0.5485

dTrdTr

r0.722T5.24.014rHR PB0

dB0

B1

dB1

SRPdB0

dB1又RT r

rdT

rdT

R rdT dTc r r r r代入數(shù)據(jù)計(jì)算得1

=-91。、SR=—0。20371（2）300K、0.1MPa500K、10MPaTH 2CigdTT

500

31.69610.144103T4.038106T2dT1 T 1

300=7。02kJ/molS

Cigp

dTRlnP

50031.696T10.1441034.038106TdTRln1021 T T P21

300

0.1=—20。39J/(mol·K)500K、10MPaT=T/T=500/417=1。199

=P/

=10/7.701=1.299—利用普維法計(jì)r 2 c算

r 2 cB00.083

0.422dB00.2326 0.675T2.60.4211dB0Tr

dT rrB10.139

0.172dB10.05874 0.722T5.20.281dB1HR

T4.2rdB0

dB1

dT rrSRPdB0

dB1PB0T B1T 又RT r rdT

rdT

R rdT dTc r r r r代入數(shù)據(jù)計(jì)算得HR=-3.41KJ/mo、SR=-4。768J（mo·)2 2∴H=H—H=H=-HR+H+HR=91.41+7020—3410=3.701KJ/mol2 1 2 1 2S=S-SSSRS1SR=0.2037-20.39-4.768=-24。952 1 2 1 23—5.試用普遍化方法計(jì)算二氧化碳在473。2K、30MPa下的焓與熵.已知在相同條件下，二氧化碳處于理想狀態(tài)的焓為8377J/mol,熵為-25.86J/（mol·K)。解：查附錄二得二氧化碳的臨界參數(shù)為：T

=304.2K、P

=7。376MPa、ω=0。225∴ T=T/r

c=473。2/304.2=1。556 c

c=P/Pr

=30/7.376=4.067-利用普c壓法計(jì)算查表，由線性內(nèi)插法計(jì)算得出：H 0HRTc

1.741

1HH

0.04662

0SRSR

0.8517

SRR

0.296HR

0 1HR HR SR SR SRRR∴由RT RTRRc c

RT 、c

計(jì)算得：KJ/mol 。635(∴H=HR+Hig=-4。377+8。377=4KJ/molS=SR+Sig=-7.635-25.86=—33。5J/(mol·K)3—621℃時(shí),1molUVHS乙炔在01013MPa0℃的理想氣體狀態(tài)的HS84℃,21℃4。459MPa.3-7.10kg373.15K、0.1013MPaU、H、、和G之值。解法一：查表U,kJ/kg；H,kJ/kg；S，kJ/kg/K飽和液體U飽和蒸汽飽和液體飽和蒸汽飽和液體飽和液體U飽和蒸汽飽和液體飽和蒸汽飽和液體飽和蒸fUHHSSg f g f g418.942506.5419.042676.11。30697。354g f△H=m(HH)=26570。6kJg f△S=m(SS）=60。48kJ/kg f解法二思路:查出水的汽化潛熱H,根據(jù)熱力學(xué)基本關(guān)系式依次求出△H，△S，△A，△U，△G熱力學(xué)基本關(guān)系式：dH=TdS+VdpdA=－SdT－pdVdU=TdS－pdVdG=－SdT+Vdp

fgT,p不變,V變dH=TdS+Vdp=TdSdA=－SdT－pdV=－pdVdU=TdS－pdV(完整)化工熱力學(xué)答案(第三版)=·m1 f=1.0435×10-3×10=0.010435m3終態(tài)水為蒸汽，V2

=Ｖ·mg=1673.0×10-3×10=16.730m3△V=V－V2 1將△V代入△U=T△S－P△V，得3 6△U=373.15×60.485×10－0.10113×10×16.720=20879084J≈20879kJ3—80。1013MPa、801.013MPa、180℃的飽和蒸汽時(shí)該過(guò)程的VH和3733J/mol；957c3/mol;Cigp

16.036/molK

1 1 。B=-78T103 cm3/mol 解:1.查苯的物性參數(shù)：T=562.1K、P=4。894MPa、ω=0。271c c求ΔV由兩項(xiàng)維里方程

(完整)化工熱力學(xué)答案(第三版)PV BP P

2.4Z 2 RT

1

1

78T103 1.013106

1 2.41

78

103 0.85978.314106453ZRT 0.85978.314453

453

V 2 P

1.013

3196.16cm3molVVV1 2VVV2 1

3196.1695.73100.5cm3mol HHV

R)HidPHidT

HR2SSV

(S1

R)SidPSidT

SR2計(jì)算每一過(guò)程焓變和熵變（1)飽和液體（T、P）→飽和蒸汽ΔH=30733KJ/KmolV(完整)化工熱力學(xué)答案(第三版)(完整)化工熱力學(xué)答案(第三版)ΔS=ΔH/T=30733/353=87。1KJ/Kmol·KV V(2）飽和蒸汽（353K、0。1013MPa）→理想氣體∵TT r TC

353562.1

P0.628 P r PC

0.10130.02074.894點(diǎn)（TP2—8r r由式(—6、(3-6）計(jì)算HR dB0 B0 dB1 B1rr1 -Prr

RT c

rdT T

dT Tr r-0.02070.6282.26261.28240.2718.11241.7112=-0.0807∴HR0.08078.314562.1∴1-377.13KJ KmolSR dB0 dB11

R rdT dT-0.02072.26260.2718.1124 -0.02072.26260.2718.1124-0.09234∴SR-0.092348.314∴10.7677KJKmolK（3）理想氣體（353K、0。1013MPa）→理想氣體(453K、1.013MPa)Hid

2CiddTTP T PT145316.0360.23TdT 16.0363530.235745323532211102.31KJ KmolSid

CidPTPPT

dTRln 2353

T T P16.03616.0360.2357 8.314ln1.013dTT0.101316.036ln4530.23573533538.47KJKmolK（4）理想氣體（453K、1。013MPa）→真實(shí)氣體(453K、1。013MPa)T 453r 562.1

P1.0130.2070r 4.894點(diǎn)（TP）2-8r r由式（3-61（3-62）計(jì)算rrHR-TPrr

dB0

B0

dB1

B1RT c

rdT T

dT Tr r-0.8060.20700.51290.2712.2161-0.3961SR-P

dB0

dB1R r dT dTr r-0.20700.2712.2161-0.3691∴HR1850.73KJ Kmol SR3.0687KJ KmolK2 24.求HHV

(H1

R)HidPHidT

H

40367KJ KmolV

(S1

R)SidPSidSRT 293.269KJ KmolKH /kg H H /kg H /kgl gV1.1273cm3/g V 194.4cm3/gl gxV xV g lx194.4x1.1273解之得：x0.577%HHxH xH0.005772778.10.00577672.81774.44kJ/kggl3－11260℃1.0336MPa0.2067MPa衡,試問(wèn)蒸汽在噴嘴出口的狀態(tài)如何？解:查1.03MPa過(guò)熱水蒸汽表TS6.8817kJkg1K1TS7.0463kJkg1K1TS1

6.9641kJkg1K10S2

S kg1K1p0.2067MPa時(shí)：S 1.5301kJkg1K1，Hl

504.7kJkg1S kgK，Hg

2706.7kJkg1比較S和S、S可知，出口處體系處于氣液平衡狀態(tài)。2 l gxS2

(1x)Sl

S S代入已知,解得干度為：x Sg

Sl0.971lHxHg

(1x)Hl

2642.8(kJkg)3-12.試求算366K、2。026MPa下1mol乙烷的體積、焓、熵與內(nèi)能。設(shè)255K、0.1013MPa時(shí)乙烷的焓、熵為零。已知乙烷在理想氣體狀態(tài)下的摩爾恒壓熱容Cigp

10.038239.304103T73.358106T2J/molK：初態(tài)的溫度T 273.1518255.15K，末態(tài)溫度為：T

273.1593366.15K1 2計(jì)算從初態(tài)到末態(tài)的熱力學(xué)性質(zhì)變化，計(jì)算路徑為：255.15K,0.1013MPaH1SR1理想氣體

H,S 366.15K,2.026MPaHR2SR2理想氣體55.15K,0.1013MPa

366.15K,2.026MPaig,ig）計(jì)算剩余性質(zhì)烷的臨界參數(shù)為:T＝305.32K,p＝4.872MPa，＝0。099c c態(tài)壓力為常壓，HR0, SR01 1366.15 2.026態(tài)：T 1.1992，p 0.4158r2 305.32 r2 4.872據(jù)圖2—11，應(yīng)該使用普遍化的第二維里系數(shù)計(jì)算。0.422 0.422 0.172 0.172)0.083 0.083 0.2326 B0.139 0.139 0.058T1.6r

1.1992

1.6

T4.2r

1.1992

4.2(0) 0.675

0.675

0.4209T2.6r r

199260.722

0.722

0.2807T5.2r r

19922式（3－78）得：HR B(0) dB(0) BdBp RT r Tr

dT Tr r

rdT r0.41580.21260.42090.0990.05880.28070.2652

1.1992

1.1992 HR0.26528.314366.15807.30Jmol-1式（3－79)得：SRp

dB(0)

dB(1)

0.4158(0.42090.0990.2807)0.1866R r

dT dT r rSR0.18668.3141.5511Jmol-1K-1）計(jì)算理想氣體的焓變和熵變ig

2CigdTT3615 T3615 3 6 1 10.83239.30410 T73.35810 d255.15

239.304103

73.358106 10.83T T T2T2 T3T32 1 2 2 1 3 2 18576.77Jmol-1igSS

Rln 1

Cig2366.15 pdT2p T p p

255.15T0.1013

8.314

2.026

3661510.083239.304103T73358106T2255.15 T2.7386Jmol-1K-1）計(jì)算末態(tài)的體積式（2－30)和(2－31)得：Bp

p

0.41581RT

1

B(0)B(1)

Trr

10.23260.0990.058801.1992

0.9214ZRT 0.92148.314366.15 2 1.384103m3mol1p 2.0261062此：H HH HRHigHR1 1 1 2 008576.77 7769.47Jmol1S SS SRSigSR1 1 1 2 002.7686 1.2175Jmol1K1U H p

7769.472.0261061.3841032 2 2

jie4965.5Jmol13-13.試采用RK方程求算在227℃、5MPa下氣相正丁烷的剩余焓和剩余熵。解:查附錄得正丁烷的臨界參數(shù)：T=425.2K、

=3。800MPa、ω=0。193c c又R—K方程：P RT aVb T0.VVb∴ a

R2TcPc

0.42748

8.3142425.22.53.8106

29.04Pam6K0.5mol2cb0.08664RTcPc

8.314425.20.08664 8.06105m3mol13.8106∴51068.314500.15V8.06105

500.150.5V

29.04V8.06105試差求得：=5。61×10—3/mol∴ h

b8.061050.1438V 56.1105A a 29.04

3.874B bRT1.5 8.061058.314∴Z

A h

1 3.874 0.1438 0.6811h B1h

1

10.1438∴HR

Z1

1.5a

ln1

bZ11.5Alnh1.0997RT bRT1.5 V BHR1.09978.314500.154573J/molSR Pbln

a ln1

b0.809 R RT 2bRT1.5 V SR0.8098.3146.726J/K3—14.RK50℃、10.13MPa度。解：查附錄得二氧化碳的臨界參數(shù):Tc

=304。2。2K、P

=7。376MPac∴ R2T2.5 8.3142304.22.5a0.42748

0.42748 6.4661Pam6K0.5mol27.376106cRT 8.314304.2b0.08664

c0.08664 29.71106m3mol1P 7.376106c又P RT aVb Tb∴10.131068.314323.15V29.71106

323.150.5V

6.4661V29.71106迭代求得：=294.9cm/mol∴hb29.710.1007V 294.9A a 6.466

4.506B bRT1.5 29.711068.314∴Z

A h

1 4.506 0.1007 0.69971h

h

1

10.1007B1 ∴l(xiāng)nf

PbZ1ln

a ln1

b0.7326P RT bRT∴f=4。869MPa

V3-15.3℃下,壓力分別為(飽和蒸汽壓（b)100×1Pa（30℃=5100×10Pa范圍內(nèi)將液態(tài)水的摩爾體積視為常數(shù)，其值為0.01809m3/kmol;（3)1×1Pa（a)30℃，=0.0424×15PafLi

fVfi i1×15Pa=0。0424×105P＜1×105Pa∴30℃、。0424×15Paf=Pi∴fSPS0.0424105Pai iSfi

S PS1i(b）30℃，10×10Pa∵f

PSSexpP

V dP S

S PSLii i Li

PSRTi

i i ifL VL V

PPS

0.018091031000.0424105ln

P i dP i

i 0.07174fSi∴f

PSRTi

RT 8.314303.15i 1.074fSifL1.074fi

S1.0740.04241054.554103Pa3—16.AB05MPaA981kg/sB473的過(guò)熱蒸汽，試求B股過(guò)熱蒸汽的流量該為多少？解：A股：查按壓力排列的飽和水蒸汽表，0.5MPa（151。9℃）時(shí),HH640.23kJ/kglH 2748.7kJ/kggH H 0.982748.70.02640.232706.53kJ/kgAH H 2855.4kJ/kgBHHQp忽略混合過(guò)程中的散熱損失，絕熱混合后焓值不變

H=所以 混合前Bxkg/s12706.532706.5312855.4xx解得：x2748.72706.530.3952kg/s2855.42748.7該混合過(guò)程為不可逆絕熱混合，所以S0只有可逆絕熱過(guò)程,S0因?yàn)槭堑葔哼^(guò)程，該題也不應(yīng)該用U0第四章

混合前后的熵值不相等.進(jìn)行計(jì)算。4-120℃0.1013MPaH2

O（2）所形成的溶液其體積可用下式

58.3632.46

42.98x2

58.77x323.45x4VV表示為濃度x2

2 2 2 2 1 2的函數(shù)。解：由二元溶液的偏摩爾性質(zhì)與摩爾性質(zhì)間的關(guān)系：M M

M

M M

M1 2x2

T

2 2 x2

T,P得： V

V

V

VV

V1又 V

2x2

T,P

2 2 x2

T,Px

32.4685.96x

176.31x293.8x32 T,P

2 2 2所以V58.3632.46x42.98x258.77x323.45x4x

32.4685.96x176.3x293.8x31 2 2

2 2 2

2 2 258.3642.98x2117.54x370.35x4J/mol2 2 2V58.3632.46x42.98x258.77x323.45x41x32.4685.96x176.31x293.8x32 2 2 2 2 2 2 2 225.985.96

219.29x2211.34x370.35x4J/mol2 2 2 24-2.T及PH400x1

600x2

xx12

40x1

20x2

H。試確定在該溫度、壓力狀態(tài)下(1）用x1

表示的H和H1

;（2）純組分焓H1

H無(wú)限稀釋下液體2的偏摩爾焓H和H的數(shù)值。1 2解:（1）已知H400x1

600x2

xx12

40x1

20x2

（A）x=1-x帶入(A）,并化簡(jiǎn)2 1H400x6001xx1x40x201x600180x20x3

(B）1 1 1 1 1 1 1 1由二元溶液的偏摩爾性質(zhì)與摩爾性質(zhì)間的關(guān)系:M M

M

M M

M1 1 x1

T

2 1 x1

T,P得：

H

H

, H H

H1由式（BH

1x1

T,P

2 1x1

T,Px

18060x211 T,P所以

600180x20x31x18060x2 42060x240x3J/mol （C）1 1 1 1 1 1 1H 600180x20x3x18060x260040x3J/mol (D）2 1 1 1 1 1（2x=1x=0（B）HH1 1H 400J/mol H1

1 2600J/mol(3H和H是指在x=0x=1H和

x=0（CH420J/mol，1 2 1

1 2 1 1x=1D）H640J/mol。1 24—1200c330％的甲醇）7H2

O（2)（摩爾比)組成。試求需要多少體積的25℃的甲醇與水混合。已知甲醇和水在25℃、摩爾分?jǐn)?shù)1

38.632cm3/mol2

/mol25℃下純物質(zhì)的體積：V1

40.727cm3/mol,V2

18.068cm3/mol。解：由M M得：VxVxVi i 11 2 2代入數(shù)值得：V=0。3×38。632+0。7×17。765=24。03cm3/moln120024.03

49.95moln1

0.349.9514.985moln 0.749.9534.965mol2則所需甲醇、水的體積為：V1t

14.98540.727610.29molV 34.96518.068631.75mol2t將兩種組分的體積簡(jiǎn)單加和：V V1t 2t

610.29631.751242.04mol1242.0412003.503%12004-4。有人提出用下列方程組表示恒溫、恒壓下簡(jiǎn)單二元體系的偏摩爾體積：VVaaxbx2 VV aax bx21 1 1 1 2 2 2 2式中，V1

和V是純組分的摩爾體積,a、b只是T、P的函數(shù).試從熱力學(xué)角度分析這2些方程是否合理?解：根據(jù)Gibbs-Duhem方程 dMi i

T

0得恒溫、恒壓下 xdV1 1或 dV

xdV 02 222dV dV22x 11dx

x2

dx2x dx由題給方程得

1 1 2dV

（A）x 11dx

b

x2bx21 112x dV22dx2

ax2

2bx22

（B)（A（方程，故不合理。4—5.試計(jì)算甲乙酮(1)和甲苯(2）323K2。、1解,解,B1,B, 數(shù)據(jù)同例4。ij 12ln? Py220501RT112 128314323（13870.521.054? 3481ln? Py220502RT221 128314323（18600.521.415? 243ln2xln? 5ln348.5ln243527870732350.2908iif0.290820510459.61104Pa24-6.vanderwaals方程的氣體的逸度表達(dá)式。

vl為氣液兩相平衡的一個(gè)基本限制,試問(wèn)平衡時(shí)下式是否成立？i ifvfvNylniiyiNylnvNi iylnyi ii1i1i1lnflNxlniixNxlnlNi ixlnxi i根據(jù)平衡常數(shù)Ky/x即y Kx和lnvln則i1ii1i1iiiilnfvNylnvNi(Kx)lni iylnyiNiiiKN xln?lNi1iKxlnKxi ii1i1iiNi1xlnxi ii1K（lnfi1N xlnK）ii1lNxlnK）ii1若K則lnfllnfv；即flfvK即x＝y(tǒng)共沸點(diǎn)時(shí)，才有flfvi i133701272026MPa。求容器內(nèi)混合物的摩爾數(shù)、焓和熵。假設(shè)混合物為12720。26MPaV、HS表中焓值和熵值的基準(zhǔn)是在絕對(duì)零度時(shí)完整晶體的值為零.V(cmo)mo-)mol-）氮 6 18090 0乙烷

31390

190。2i MxMi 理想溶液中各組份的偏摩爾性質(zhì)與他們純物質(zhì)之間的關(guān)系為:V V H H S S Rlnxi i i i i i i混合物的摩爾體積:VxVxVi i i

0.3179.6mol1tV 110n 7504.1molt混合物的摩爾數(shù)： V 133.26混合物的摩爾焓：HxH xH 0.31809031390mol1i i i iHtnH7504.1327400205613162J混合物的摩爾熵的計(jì)算N的偏摩爾熵：2S SN N2

RlnxN2

1548.314ln164.0Jmol1KCH28S S RlnxCH CH CH28 28 2

8.314ln193.17Jmol1K混合物的摩爾熵：SiSi3164719171842Jmol1K1混合物的熵：S nS7504.13184.421383911.65Jt4—9.344。75K時(shí)，由氫和丙烷組成的二元?dú)怏w混合物，其中丙烷的摩爾分?jǐn)?shù)為0。792，混合物的壓力為3.7974MPa。試用RK方程和相應(yīng)的混合規(guī)則計(jì)算混合物中氫

=0.07,ij

的實(shí)驗(yàn)值為1。439.H2解：已知混合氣體的T=344.75K P=3.7974MPa，查附錄二得兩組分的臨界參數(shù)氫（1)：y1

=0.208 Tc

=332K Pc

=1297MPa Vc

=65.0c3/mol ω=-22丙烷2y=792 T=3698K P=4.246MPa V=203c3/mol ω=01521 c c c∴ R2T2.5 8.314233.22.5a 11

c1 0.42748 0.1447Pam6K0.5mol2P 1.297106c1R2T2.5 8.3142369.82.5a 22

c2 0.42748 18.30Pam6K0.5mol2P 4.246106∵a aij i

c20.51kij∴a a12 1

0.51k12

0.144718.300.510.071.51Pam6K0.5mol2y2am 1

2yya1 2

y2a2 220.20820.144720.2080.7921.5130.792218.3011.98Pam6K0.5mol2cb0.08664RT1c1 Pc1

8.31433.20.08664 1.844105m3mol11.297106cb 0.08664RTc2 Pc2

8.314369.80.08664 6.274105m3mol14.246106 ybm ii

0.2081.8441050.7926.2741055.3526105m3mol1A

11.98 4.206B bRTm

5.35261058.314344.751.5B bP 5.35261053.7974106 0.07091 ①h m Z ZRT Z8.314344.75 ZZ 1

A h

1 4.206 h ②1h B1h

1

1h 聯(lián)立①、②兩式,迭代求解得:Z=0。7375 h=0.09615所以，混合氣體的摩爾體積為：ZRT 0.73758.314344.75V 5.567104m3mol1P 3.7974106∴?

V

2y

y

V

a

V

PVln

ln

1

111

212

ln

m

m1 ln

m

m ln 1 Vb Vb

bRT

V b2RT1.5 V V

RTm m? V b 2y

mya

m mVb ab Vb b PVlnln 2

121 222ln m m2 ln m m ln 2 Vb Vb bRT1.5 V b2RT1.5 V V

RTm m m m m分別代入數(shù)據(jù)計(jì)算得：4—10.某二元液體混合物在固定T和P下其超額焓可用下列方程來(lái)表示：HE=xx12（40x+20x）J/mol。試求HE和HE（

表示）.1 2 1 2 11.4-11333K10Pa（1（2Vcm3/mol）如下表所示.X V X VX V1110.00101.4600。20104。0020。85111。8970。02101。7170.30105.2530.90112。4810。04101。9730.40106.4900.92112.7140。06102。2280。50107.7150。94112.9460.08102。4830。60108.9260。96113。1780.10102.7370。70110。1250。98113。4090.15103.3710.80111.3101.00113。640試計(jì)算：（1）純物質(zhì)摩爾體積V1

V2(2）x=0.2、0.50.8V和V；2 1 2（3）x2

=0.2、0.50.8；(4）無(wú)限稀釋混合物中偏摩爾體積V和V的數(shù)值1 2（）V＝113.64（cm3/mol）和V＝101.46（cm3/mol）;1 2V V 1

Vx2x2

Vx2x1

Vx2x12當(dāng)x 時(shí)；2VVx11111.3100.2

111.310110.125

111.897

}/2（113.669cm3/mol

0.80.7 0.850.8V (VxV)/x2 1 1 2(111.3100.8113.669)/0.2101.87cm3/mol同理：x＝0。5V2

113.805cm

/mol和V2

101.65cm

/mol1x＝0.8V114.054cm12

/mol和V 101.489cm2

/molVV(xVxV)1 1 2 2當(dāng)x 時(shí)；2VV(xVxV)1 1 2 2111.310(0.8113.640.2101.46)0.106cm3/mol同理：當(dāng)x2

0.5;V0.