化工热力学答案(第三版)

(完整)化工热力学答案(第三版)(完整)化工热力学答案(第三版)化工热力学课后答案（第三版）陈钟秀编著2-11kmol0.12465（）（）R-K（）普遍化关系式。=0。1246m3/1kmol=124.6c/mol查附录二得甲烷的临界参数：Tc

=190。6K

=4.600MPa c

=99cm3/molcω=0。008理想气体方程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普遍化关系式TTTr

323.15190.61.695 Vr

VVc

124.6991.259＜2∴利用普压法计算，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，查附录三得：=0.8938 Z1=。46230 r0ZZ0Z1=0。8938+0.008×0。4623=0。8975此时,P=Pc

P4。6×4.687=21.56MParZ=0。8975Z1ZP∴P=19。22MPa2-2.Pitzer510K、2。5MPa1480.7cm/mol.解:查附录二得正丁烷的临界参数：Tc

=425.2K

=3。800MPa c

=99cm3/molcω=0。193理想气体方程V=RT/P=。×61.6961.4807100%14.54%1.4807Pitzer对比参数:Tr

TTc

510425.21.199 Pr

PPc

2.53.80.6579—普维法∴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%(摩尔分数)的碳生成二氧化碳，其余的生成一氧化碳。试计算（）含碳量为81.38100kg焦炭能生成1.1013MPa、303K的吹风气若干立方米?（2)所得吹风气的组成和各气体分压。解：查附录二得混合气中各组分的临界参数：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-普维法rm cm rm cm利用真实气体混合物的第二维里系数法进行计算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总（2)

PyPZc11 1 ZmZ

0.240.10130.2950.025MPa0.28450.274PyP c2 2 Zm

0.760.10130.28450.074MPa2-4。03MP477K。83mNH3

0.142，若压缩后温度448.6K,则其压力为若干？分别用下述方法计算：(1)VanderWaals方程；（2)Redlich-Kwang(3）Peng-Robinson（4解：查附录二得NH的临界参数：T=405.6K P=11。28MPa

=7。5c3/mol3 c c cω=0。250(1)

求取气体的摩尔体积对于状态Ⅰ：P=。03MP、T=447、V=。83m3TTTr

477405.61.176 Pr

PPc

2.0311.280.18—普维法∴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对于状态Ⅱ：摩尔体积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普遍化关系式∵ V r

V 9.4581057.251051.305＜2适用普压法，迭代进行计算，方法同c1—1(3）2-6.30％（摩尔分数）氮气70％(摩尔分数）正丁烷气体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(摩尔体积）=。24×10—m/molm RT RT假设气体混合物总的摩尔数为n，则0.3n×28+0.7n×58=7→n=0.1429mol∴V=n×V(摩尔体积)=0.1429×4.24×10-4=60.57c32—8。试用R-KSRK273K1013MPa2。0685解：适用EOS的普遍化形式查附录二得NH的临界参数：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、②两式联立，迭代求解压缩因子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、②两式联立，迭代求解压缩因子Z第三章3-1.物质的体积膨胀系数和等温压缩系数k的定义分别为：

1VVT

， 。 k1V V PVanderWaals和k的表达式.

P

T解：Vanderwaals方程P RT aVb V2Z=f(x,y）的性质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.试求算1kmol氮气在压力为10.13MPa、温度为773K下的内能、焓、熵、CV、Cp和自由焓之值。假设氮气服从理想气体定律。已知：（1)在0。1013MPa时氮的C与温度的关系为C 27.22/molK；p p（2）假定在0℃及0.1013MPa时氮的焓为零；（3298K0.1013MPa191.76J（mol·K（）熵值的计算dSCp

VdT dpT p

dSCpdTRdp对于理想气体： 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）焓值的计算dHCdTpHH0

773273

0.004187T)dT127.22(773273) (77322732)214704.9(Jmol1)HH014704.914704.9(Jmol1)14704.9KJKmol1)其他热力学性质计算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。已知氯在理想气体状态下的定压摩尔热容为Cig31.69610.144103T4.038106T2J/molKp解：分析热力学过程300K0.1MPa真实气体H=0，S=0-HR1

H

500K10MPa真实气体H2-SR13000.1MPa

H、S

S250010MPa理想气体

1 1理想气体查附录二得氯的临界参数为:T=417K、P=7。701MPa、ω=0.073c c∴(1）300K、0.1MPa的真实气体转换为理想气体的剩余焓和剩余熵T=Tr

/T=300/417=0。719 1 c

=P/r 1

=0。1/7。701=0。013—利c用普维法计算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代入数据计算得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—利用普维法计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代入数据计算得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.试用普遍化方法计算二氧化碳在473。2K、30MPa下的焓与熵.已知在相同条件下，二氧化碳处于理想状态的焓为8377J/mol,熵为-25.86J/（mol·K)。解：查附录二得二氧化碳的临界参数为：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压法计算查表，由线性内插法计算得出：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

计算得：KJ/mol 。635(∴H=HR+Hig=-4。377+8。377=4KJ/molS=SR+Sig=-7.635-25.86=—33。5J/(mol·K)3—621℃时,1molUVHS乙炔在01013MPa0℃的理想气体状态的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饱和液体U饱和蒸汽饱和液体饱和蒸汽饱和液体饱和液体U饱和蒸汽饱和液体饱和蒸汽饱和液体饱和蒸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:查出水的汽化潜热H,根据热力学基本关系式依次求出△H，△S，△A，△U，△G热力学基本关系式：dH=TdS+VdpdA=－SdT－pdVdU=TdS－pdVdG=－SdT+Vdp

fgT,p不变,V变dH=TdS+Vdp=TdSdA=－SdT－pdV=－pdVdU=TdS－pdV(完整)化工热力学答案(第三版)=·m1 f=1.0435×10-3×10=0.010435m3终态水为蒸汽，V2

=Ｖ·mg=1673.0×10-3×10=16.730m3△V=V－V2 1将△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℃的饱和蒸汽时该过程的VH和3733J/mol；957c3/mol;Cigp

16.036/molK

1 1 。B=-78T103 cm3/mol 解:1.查苯的物性参数：T=562.1K、P=4。894MPa、ω=0。271c c求ΔV由两项维里方程

(完整)化工热力学答案(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计算每一过程焓变和熵变（1)饱和液体（T、P）→饱和蒸汽ΔH=30733KJ/KmolV(完整)化工热力学答案(第三版)(完整)化工热力学答案(第三版)ΔS=ΔH/T=30733/353=87。1KJ/Kmol·KV V(2）饱和蒸汽（353K、0。1013MPa）→理想气体∵TT r TC

353562.1

P0.628 P r PC

0.10130.02074.894点（TP2—8r r由式(—6、(3-6）计算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）理想气体（353K、0。1013MPa）→理想气体(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）理想气体（453K、1。013MPa）→真实气体(453K、1。013MPa)T 453r 562.1

P1.0130.2070r 4.894点（TP）2-8r r由式（3-61（3-62）计算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,试问蒸汽在喷嘴出口的状态如何？解:查1.03MPa过热水蒸汽表TS6.8817kJkg1K1TS7.0463kJkg1K1TS1

6.9641kJkg1K10S2

S kg1K1p0.2067MPa时：S 1.5301kJkg1K1，Hl

504.7kJkg1S kgK，Hg

2706.7kJkg1比较S和S、S可知，出口处体系处于气液平衡状态。2 l gxS2

(1x)Sl

S S代入已知,解得干度为：x Sg

Sl0.971lHxHg

(1x)Hl

2642.8(kJkg)3-12.试求算366K、2。026MPa下1mol乙烷的体积、焓、熵与内能。设255K、0.1013MPa时乙烷的焓、熵为零。已知乙烷在理想气体状态下的摩尔恒压热容Cigp

10.038239.304103T73.358106T2J/molK：初态的温度T 273.1518255.15K，末态温度为：T

273.1593366.15K1 2计算从初态到末态的热力学性质变化，计算路径为：255.15K,0.1013MPaH1SR1理想气体

H,S 366.15K,2.026MPaHR2SR2理想气体55.15K,0.1013MPa

366.15K,2.026MPaig,ig）计算剩余性质烷的临界参数为:T＝305.32K,p＝4.872MPa，＝0。099c c态压力为常压，HR0, SR01 1366.15 2.026态：T 1.1992，p 0.4158r2 305.32 r2 4.872据图2—11，应该使用普遍化的第二维里系数计算。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）计算理想气体的焓变和熵变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）计算末态的体积式（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.试采用RK方程求算在227℃、5MPa下气相正丁烷的剩余焓和剩余熵。解:查附录得正丁烷的临界参数：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试差求得：=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：查附录得二氧化碳的临界参数: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 ∴lnf

PbZ1ln

a ln1

b0.7326P RT bRT∴f=4。869MPa

V3-15.3℃下,压力分别为(饱和蒸汽压（b)100×1Pa（30℃=5100×10Pa范围内将液态水的摩尔体积视为常数，其值为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的过热蒸汽，试求B股过热蒸汽的流量该为多少？解：A股：查按压力排列的饱和水蒸汽表，0.5MPa（151。9℃）时,HH640.23kJ/kglH 2748.7kJ/kggH H 0.982748.70.02640.232706.53kJ/kgAH H 2855.4kJ/kgBHHQp忽略混合过程中的散热损失，绝热混合后焓值不变

H=所以 混合前Bxkg/s12706.532706.5312855.4xx解得：x2748.72706.530.3952kg/s2855.42748.7该混合过程为不可逆绝热混合，所以S0只有可逆绝热过程,S0因为是等压过程，该题也不应该用U0第四章

混合前后的熵值不相等.进行计算。4-120℃0.1013MPaH2

O（2）所形成的溶液其体积可用下式

58.3632.46

42.98x2

58.77x323.45x4VV表示为浓度x2

2 2 2 2 1 2的函数。解：由二元溶液的偏摩尔性质与摩尔性质间的关系：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。试确定在该温度、压力状态下(1）用x1

表示的H和H1

;（2）纯组分焓H1

H无限稀释下液体2的偏摩尔焓H和H的数值。1 2解:（1）已知H400x1

600x2

xx12

40x1

20x2

（A）x=1-x带入(A）,并化简2 1H400x6001xx1x40x201x600180x20x3

(B）1 1 1 1 1 1 1 1由二元溶液的偏摩尔性质与摩尔性质间的关系: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)（摩尔比)组成。试求需要多少体积的25℃的甲醇与水混合。已知甲醇和水在25℃、摩尔分数1

38.632cm3/mol2

/mol25℃下纯物质的体积：V1

40.727cm3/mol,V2

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

49.95moln1

0.349.9514.985moln 0.749.9534.965mol2则所需甲醇、水的体积为：V1t

14.98540.727610.29molV 34.96518.068631.75mol2t将两种组分的体积简单加和：V V1t 2t

610.29631.751242.04mol1242.0412003.503%12004-4。有人提出用下列方程组表示恒温、恒压下简单二元体系的偏摩尔体积：VVaaxbx2 VV aax bx21 1 1 1 2 2 2 2式中，V1

和V是纯组分的摩尔体积,a、b只是T、P的函数.试从热力学角度分析这2些方程是否合理?解：根据Gibbs-Duhem方程 dMi i

T

0得恒温、恒压下 xdV1 1或 dV

xdV 02 222dV dV22x 11dx

x2

dx2x dx由题给方程得

1 1 2dV

（A）x 11dx

b

x2bx21 112x dV22dx2

ax2

2bx22

（B)（A（方程，故不合理。4—5.试计算甲乙酮(1)和甲苯(2）323K2。、1解,解,B1,B, 数据同例4。ij 12lnˆ Py220501RT112 128314323（13870.521.054ˆ 3481lnˆ Py220502RT221 128314323（18600.521.415ˆ 243ln2xlnˆ 5ln348.5ln243527870732350.2908iif0.290820510459.61104Pa24-6.vanderwaals方程的气体的逸度表达式。

vl为气液两相平衡的一个基本限制,试问平衡时下式是否成立？i ifvfvNylniiyiNylnvNi iylnyi ii1i1i1lnflNxlniixNxlnlNi ixlnxi i根据平衡常数Ky/x即y Kx和lnvln则i1ii1i1iiiilnfvNylnvNi(Kx)lni iylnyiNiiiKN xlnˆlNi1iKxlnKxi ii1i1iiNi1xlnxi ii1K（lnfi1N xlnK）ii1lNxlnK）ii1若K则lnfllnfv；即flfvK即x＝y共沸点时，才有flfvi i133701272026MPa。求容器内混合物的摩尔数、焓和熵。假设混合物为12720。26MPaV、HS表中焓值和熵值的基准是在绝对零度时完整晶体的值为零.V(cmo)mo-)mol-）氮 6 18090 0乙烷

31390

190。2i MxMi 理想溶液中各组份的偏摩尔性质与他们纯物质之间的关系为:V V H H S S Rlnxi i i i i i i混合物的摩尔体积:VxVxVi i i

0.3179.6mol1tV 110n 7504.1molt混合物的摩尔数： V 133.26混合物的摩尔焓：HxH xH 0.31809031390mol1i i i iHtnH7504.1327400205613162J混合物的摩尔熵的计算N的偏摩尔熵：2S SN N2

RlnxN2

1548.314ln164.0Jmol1KCH28S S RlnxCH CH CH28 28 2

8.314ln193.17Jmol1K混合物的摩尔熵：SiSi3164719171842Jmol1K1混合物的熵：S nS7504.13184.421383911.65Jt4—9.344。75K时，由氢和丙烷组成的二元气体混合物，其中丙烷的摩尔分数为0。792，混合物的压力为3.7974MPa。试用RK方程和相应的混合规则计算混合物中氢

=0.07,ij

的实验值为1。439.H2解：已知混合气体的T=344.75K P=3.7974MPa，查附录二得两组分的临界参数氢（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 联立①、②两式,迭代求解得:Z=0。7375 h=0.09615所以，混合气体的摩尔体积为：ZRT 0.73758.314344.75V 5.567104m3mol1P 3.7974106∴ˆ

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 mˆ 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分别代入数据计算得：4—10.某二元液体混合物在固定T和P下其超额焓可用下列方程来表示：HE=xx12（40x+20x）J/mol。试求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试计算：（1）纯物质摩尔体积V1

V2(2）x=0.2、0.50.8V和V；2 1 2（3）x2

=0.2、0.50.8；(4）无限稀释混合物中偏摩尔体积V和V的数值1 2（）V＝113.64（cm3/mol）和V＝101.46（cm3/mol）;1 2V V 1

Vx2x2

Vx2x1

Vx2x12当x 时；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当x 时；2VV(xVxV)1 1 2 2111.310(0.8113.640.2101.46)0.106cm3/mol同理：当x2

0.5;V0.