物理化学总习题解答

物理化学习题解答（一）

习题p60〜62

10解：

(1)平均自由程：1=若手，未知数〃怎么求?其物理意义是什么？

ndn

由公式pV=NkBT，fn=N/V=p/(kRT)

n=100xl03/{p(1.381xl0-23x298)}=2.43xl025/n3

I=0.707/{3,14x(03X10'9)2x2.43x1025}=1.03x10%

(2)z!=/2V^d2n,未知数％怎么求?其物理意义是什么？

%=J翳=15.01m-s"

Z!=f2x15.01x3.14x(0.3x10-9)2x2.43xl025=1.47x108/s

(3)Z=inZ/=1x2.43xl025x1.47xl08=1.77x1033/s

13解：

(1)理想气体:pV=〃R7,f(詈).=nR/p

1dv

a=v(dT)/f，giMVFR/nRTTIT

⑵范德华气体:伽+”力正乂展开方程式得,pV-//?p+〃2a/V-〃3ab/v2=〃RT

dV2,,dVa3,3dV

p(防)?+an/V\—)p+2nab/V(^)p=nR

dV,、,

232

(—)p=nR/(p-an/V+2abn/V)

1dV,,,,

a=-(—)„=nR/(pV-an2/V2+2abn3/V3)

=(7?V3-^/?V2)/(7?TV3-2anV2+4a/7n2V-2(z/72/z3)

21解：2C+O2=2COC+O2=CO2

(1)VO2=1x0.21=0.21单位体积；

Vbo=2Vo2x0.92=0.3864单位体积；

Vbo2=VO2x0.08=0.0168单位体积；

V总=V空+Vco+匕o”Vo2=1+0.3864+0.0168-0.21=1.1932单位体积.

(2)xm=k/K总=1x0.78/1.1932=0.654；

XAFVAW总=1x0.0094/1.1932=0.00788

xco=WV,&=0.3864/l.l932=0.324；

XCO2=Vco2/V£l=(lX0.0003+0.0168)/1.1932=0.0143

(3)2C+O2=

2C0C+O2=C02

24g\mol2mol12/g\mol\mol

xgx/24molx/12moly/gy/\2moly/l2mol

x+y=1000x+y=1000x=958.33/g

x/24:y/12=92:8x=23yy=41.67/g

g=x/24+y/12=958.33/24+41.67/12=43.40m。/；

〃co=x/l2=958.33/12=79.86,〃。/；

〃co2=y/l2=41.67/12=3.47机。/；

n%=noJ0.21=206.68〃?。/

n总=〃空+〃co+〃co2-〃O2=206.68+79.86+3.47-43.40=246.61zn。/

V&=n总R77p=246.61x8.314x293/105=6.00/n3

25解：

(1)co=27rx3000/60=1007r/s",V=(wr=407r/〃z7,"

EH2=*WH2V2=±x2/6.023xl()23x(407r)2=262i8xl(y2°j

2232l9

£o2=1wO2V=ix32/6.023x10x(40^)=4.195x10J

(2)〃/〃o(H2)=e卬(-EH247B)=exp[-2.6218x10-20/(l.381x1023x293)]=1.5347x1O'3

192346

n/n(i(O2)=exP(-EO2/kBT)=exP[-4.195x10-/(1.381xl0x293)]=9.477x10

(3)tt(H2)/n(O2)=1.5347x10-3/9477x10-46=16194x1(/2

物理化学习题解答（二）

习题pl29-133

3解：

(1)VV2=V!，:.w=o,

△U=^nG.mdT=1.5/?(T2-7'I)=1.5X8.314X(546-273)=3404.587

'：AU=Q+W,/.e=Z\[7=3404.58J

3

p2=nRT2/V2=\X8.314X546/(22.4X10-)=202.65jtP«

(2)'：T2=T1,：.AU=0

w=j一pdV=，nRT/VdV=-nRTln(V2IV\)

=-lX8.314X546Xln(44.8/22.4)=-3146.50J

"：AU=Q+W,：.Q=-W=3146.50J

P3=〃RTs/V3=lX8.314X546/(44.8Xl(y3)=ioi33攵&

(3)U=pnCvmdT=1.5R/(T1-T3)=1.5X8.314X(273-546)=-3404.58J

Q=「〃CdT=2.57?(T1-T3)=2.5X8.314X(273-546)=-5674.31J

pm

JT}'

W=AU-Q=-3404.58-(-5674.31)=2269.737

3

pi=nRT3/V3=1X8.314X273/(22.4X10-)=101.33da

8解：

(1)^=«7?7|//?1=1X8.314X423/(100X1O3)=35.17XlO'V

W=^-pdVnRT/VdV=-z?/?nn(V2/Vi)

=-lX8.314X423X/〃(10/35.17)=4422.78J

2

(^)P\Vm,\=RT\+bpi-a/Vm,i+ab/VmA

5352

100X103vm,i=8.314X423+3.71XIO-X100X10-0.417/%1+0.417X3.71X10-/VmJ

5325

10Vm.i=3520.532Vm.i-0.417Vm.1+1.54707X10-,Vm.]=35.087X10V

W=^-pdV^^{RT/(Vm-b)-a/V^}dVm

=-RTln{(%2』)/(Km.iM}-a(1/%,2-1/Vm,l)

=-8.314X4231n{(10-0.0371)/(35.087-0.0371)}-0.417X103(1/10-1/35.087)

=4423.826-29.815=4394.01J

9解:

36

(1)W=-/?e(V2-Vi)=-100X10X(1677-1.043)X18X10'=-3016.72J

36

⑵W=-pe(V2-V^-PeV2=-100X10X1677X18X10-=-3018.60J：

AW%=(3018.6-3016.72)/3016.72X100%=0.063%

33

(3)V2=HWP2=1X8.314X373/(100X10)=31.0112i//n

33

W=-pe(V2-Vi)^-peV2=-100X10X31.011X10'=-3101.12J

11

(4)AvapH,n=40.69V-fnor；AvapU,n=AvapHm+1V=40.69-3.02=37.61kJmol

(5)'：AvapUm>0(实际上是T、P的函数)，.••△，“收"〉-W

由于体积膨胀，分子间的平均距离增大，必须克服分子间引力做功，热力学能

也增大，故蒸发的焰变大于系统所做的功。

23解：

(1)T2=TI,:.AU=O

w=P_pdV=-PnRT/VdV=-nRTln(V2/V1)=-1X8.314X298Xln2=-l717.32J

'：AU=Q+W,：.Q=-W=1717.32J

(2)绝热，。=0,T2=TI(VI/V2)'"=298X(1/2严"=187.73K

Z\U=〃G”"(T2-TI)=L5X8.314(187.73-298)=-1375.2J

'：AU=Q+W,：.W=AU=-\315.2J

(3)X8.314X298/(200X103)=12.38786Jm3

pi=l()4%+b,/,=200X103-104X12.38786=76121.4

443

p2=\0Vx+Pi=\0X12.38786+200X10=323.8786JtP«

T2=P2V2/nR=323.8786X2X12.38786/8.314=965.16K

7622

卬=^-pdV^-^(10V+b)dV=-5X10(V2-V1)-b(V2-Vi)=-15X1()6匕24修

=-15X106(12.38786X103)2-76121.4X12.38786X10-3=-3244.87J

-W3>-Wi>-W2;

AU3>AU\>AU2

28解:

Ag(s)+l/2Cl2(g)=AgCl(s)AfHm(AgCl,s,298.15K)=?

由1/2⑴+1/2(2)得:(5)Ag(s)+HCl(g)+l/4O2(g)=AgCl(s)+l/2H2O(l)

由⑸+(3)得：⑹Ag(s)+l/4O2(g)+l/2H2(g)=AgCl(s)+l/2H2O(l)

由⑹-1/2(4)得：⑺Ag(s)+l/2Cl2(g)=AgCl(s)

即由1/2(1)+1/2(2)+(3)-1/2(4)可得：Ag(s)+l/2Cl2(g)=AgCl(s)

故：AfH,n(AgCl,S,298.15K)=l/2(4"m；+)+

=1/2(-324.9-30.57+285.84)-92.31=-127.125kJ-mol-

31解：

C(g)+4H(g)=CH4(g)4凡:(298.15幻=?

(1)C(石墨)=C(g)4汕”肛「(298.15K)=711.1kJmoP

⑵C(石墨)+2H2(g)==CH£g)4",/(298.15K)=-74.78kJ-moC1

(3)H2(g)=2H(g)15K)=431.7kJ-mol'

由(2)-(1)-2(3)可得:C(g)+4H(g)=CH4(g)

故:4H,；(298.15K)=AfHm^-AsubHm,\-2AdiYHm3

=-74.78-711.1-2X431.7

=-1649.28kJ-morl

35解：可近似看作绝热反应AHHJ^O

(298K,100kPa)2C2H2(g)+5O2(g)+20N2(g)-4CO2(g)+2H2O(g)+20N2(g)(T=?,100kPa)

1AHf

'Ar2H；

(298K,100kPa)2C2H2(g)+5O2(g)+20N2(g)-4CO2(g)+2H2O(g)+20N2(g)(298K,lOOkPa)

ZA”i=O；«(N2)=80/20X5=20moZ

42H；=42/:(CO2)+24H；(H2O)-2△/「(C2H2)

=4X(-393.51)+2X(-241.82)-2X226.7=-2511.08kJ-mol'

Z\"2=({4G，M(CO2)+2C/),„,(H2O)+20Cp,ni(N2)}dT

=’(4X37.1+2X33.58+20X29.12)"+

X8.79XI022X10.29X10-3+20X3.76XIO3)TdT

=797.96X(7-298)+65.47X10-3(T2-2982)=(817.47+0.065477^(7-298)

AriHn^AH\+Ar2H,n+AH2,：.AH2=-&H；

(817.47+0.065477)(r-298)=2511.08XIO3

T=2806.1K

或者：AH^{4Cp.m(CO2)+2Cp.m(H2O)+20Cp.m(N2)}JT

=(4X37.1+2X33.58+20X29.12)(72-「)=797.96X(T-298)

ArlHm'=AH{+Ar2Hl,^AH2,/•AH2=-Ar2H^

797.96X(7-298)=2511.08X103

T=3444.87K

物理化学习题解答（三）

习题p200〜203

4解：

⑴p总x(VO2+VN2)=〃总AT，P02X%2="02夫7'PN2*VN2="N2RT

P,e=n怠RT/(〃o2RTi/po2+〃N2RT/pN2)="总/("02/P02+〃N2/pN2)

=l/(0.2/20+0.8/80)=50kPa

(2)，.,等温，：.AU=0,AH=Q,故：。=-卬

"：AV=0,：.Q=-W=0

为了计算状态函数，设计如下可逆途径：

(a)O2(298K>V、20kPa)-*O2(298K>2V、lOkPa),等温可逆

Wo2,=

r-nmRTln2=-0.2X8.314X298=-343.46J

2O2,F-WO2=343.46J

ASO2=<2O2/T=1.1526J/K

/GO2=-〃O2R77〃2=-343.46J

(b)(N2(298K.V、80kPa)->O2(298K、2V、40kPa)

Wm.r=-nmRTln2=-0.8X8.314X298=-1373.85J

如2,尸-%2=1373.85J

/SN2=如2,〃=4.6103J/K

ZAGN2=-〃N2RT/"2=-1373.85J

AS=ASm+Z\SN2=5.763J/K

WG=Z\GO2+/GN2=-1717.3J

(2),等温可逆，：.AU=0,AH=Q,AS=-5.7627J/K,Z\G=1717.3J

故：-W=Q=TAS=-5.7627X298=-1717.3J

6解：

(1)..•等温可逆，，ZiU=0,AH=Q,故：Q=-W

W=-nRTlnV2/Vi=2X8.314X300/n50/20=-4570.82J

Z\S=2/7=4570.82/300=15.324J/K

(2)：真空膨胀，：.AU=Q,AH=Q,Q=-W=0,AS=15.324J/K

(3),等温，：.AU=O,AH=G

W=-Pe(%--)=-100X103X(50-20)X103=-3000J

Q=-W=3000J,Z1S=15.324J/K

8解：

(a)(0.15kg,273K)H2。⑸fH2O(l)(0.15kg,273K)等温等压可逆相变

。尸尸m//"sH=0.15X103gX333.4Jg」=50.01kJ

■Si=0/T=5OO10/273=183.19J-K"

(b)miG，(T-273)=〃?2Cp(298-7),0.15(7-273)=0.1(298-7),T=294.74K

(0.15kg,273K)H2O(l)^H2O(l)(0.15kg,TK)等压可逆

23

ZiS2=mCp/dT=mCpInT^T^O.15X10X4.184/rt(294.74/273)

=48.09J/K

(c)(1.0kg,298K)H2O(l)-*H2O(l)(1.0kg,7K)等压可逆

3

/S3=^mCP/dT=mCplnT2/T]=1.OX10X4.184/n(294.74/298)

=-46.02JK"

1

故：AS=AS\+AS^AS3=183.19+48.09-46.02=185.26J-K-

U解:

⑴,等温可逆，/.Z\t/=O,AH=0

W=-nRTlnV2!V\=-nRTlnPi/P2=\X8.314X298//?(100/600)

-4439.21J

Q=-W=4439.21J

/SS»=0/T=4439.21/298=14.90JK」

△A=/U-T/S=0-4439.21=-4439.21J

AG=AH^TAS=0-4439.21=-4439.21J

△Ss“b=-Q"=-4439.21/298=14.90JK-'

WSiso=CSsys+Z\S*"/,=0

⑵..'等温，'.AU=0,AH=0

•・,等外压,W=-Pe(v2-V!)=-Pe(nRT/p2-nRT/pi)

=-600X103XlX8.314X298X(l/600-l/100)X10-3=12.39kJ

Q=-W=-12.39kJ

Z\S=14.90JK-1

/A=-4439.21J

△G=-4439.21J

1

Z\5sub=-2/T=41.57J-K,/Siso=.S»s+/Ss迎=41.57+14.90=56.47J-K"

12解：

•.•绝热可逆，，。=0，z\s=o

A'=Ti'(pi/p2)'YInT2rYlnT\+(\-Y)lnpi/p2>/〃7W〃Ti+(lT/Y)/〃6

277

T2/TI=6,T2=455.5K

Z\(/=nCl,m(7'2-7'i)=1X2.5X8.314X(455.5-298)=3237.64J

AH=nCp,m(T2-T\)=1X3.5X8.314X(455.5-298)=4583.09J

W=/U=3237.64J

S2=S|+Z\S=Si

AG=AH-A(TS)=AH-S(T2-T\)

=4583.09-205.14X(455.5-298)=-27726.46J

AA=Z\U-A(TS)=AU-S(.T2-T])

=3237.64-205.14X(455.5-298)=-29071.91J

ASSUb=0jASiso=^Ssys+ASsub=0

16解：

(1)••,等温可逆，.,./〃=(),Z\H=0

W=-n/?nM(V2/V|)=n/?nn(p2/pi)=lX8.314X273/«2=1573.25J

Q=-W=-1573.25J

AS=Q/T=-5.7628J/K

AA=AU-TAS=W=1513.25J

AG=AH-TAS=W=1573.25J

(2).恒压可逆,W=-pe(V2-V])=-pVi=-nRT=-1X8.314X273=-2269.72J

<2=nCp,m(T2-Tl)=nCp,m(pV2/nR-pVl/nR)=2.5pb

=2.5X1X8.314X273=5674.305J

Z\H=Q=5674.305J

AU=Q+W=5614.305-2269.72=3404.585J

PiV2/piVi=nRTMRTi，T2=2T]

1

AS=nCp.mlnT2IT\=\X2.5X8.314/«2=14.41JK-

1

S2=S!+Z\S=100+14.407=1U4lJ-K-'-mor

AG=AH-A(.TS}=AH-(2T\S2-T\S\)

=5674.305-273(2X114.41-100)=-29493.555J

AA=AU-A(TS)=AU-(2T\S2-T\S\)

=3404.585-273(2X114.41-100)=-31763.275J

(3),恒容可逆，：.AV=0,W=0

Qv=nCv,m(T2-Ti)=nCv,m(p2V\lnR-p\V\/nR)=\.5p\V,

=1.5X1X8.314X273=3404.583J

/U=Q"=3404.583J

△S=pnCv,m/TdT="GmlMTHi)=1.5nRln(p2V]/Pi%)=1.5nRln2

=1.5X8.3141n2=8.644J/K

AH=AU+A(pV)=AU+^1y\-p\V\')=AU+p\V\=AU+nRT\

=3404.583+1X8.314X273=5674.3051

PW\/p\V\=nRTVnRT\,T2=2T}

ll

S2=S1+AS=100+8.644=108.644J-K--mor

AG=AH-Z\(TS)=AH-(T2S2—T5)=/4-(2T|S2-TIS。

=5674.305-273X(2X108.644-100)=-26345.32J

AA=AV-A(TS)=AU-(T2S2-T\S\)=AU-(1T\S2-T\S\)

=3404.583-273X(2X108.644-100)=-28615.04J

(4)•.•绝热可逆,.*.Q=0,AS=0

T2、=TI'SI/P2)…，YInT尸Y/nTi+(l-Y)ln2,lnT^lnT\+(\/YT)/〃2

I2/5

T2/T=2-"5,T2=2'T|=206.9K

AU=nCvjn(T2-T\)=\X1.5X8.314X(206.9-273)=-824.33J

AH=nCp,m(T2-Ti)=lX2.5X8.314X(206.9-273)=-1373.89J

W=/U=-824.33J

11

S2=AS+Si=Si=100J-K-mol

AG=AH-△(TS)=4H-(T2SI-TIS\)

=-1373.89-100(206.9-273)=5236.11J

AA=AU-Z\(TS)=/U-(T2s2-T5)

=-824.33-100(206.9-273)=5785.67J

(5)•.•绝热，：.Q=Q

,恒外压，'.AU=W=-pe(V2-Vi)=-pe{nRT2/p2-nRT\lp\}

=nR(peT2/p2-peTi//?i)

W=AU=nCv,m(T2-T\)=\.5nR(T2-T\)

-(PeW-PeTi/p1)=1.5(72-7!),死+0.5八=1.541.5/,T2=O.8T1

△U=nCv,m(T2-T|)=-0.2nCv,mTi=-0.2X1X1.5X8.314X273=-680.92J

△H=nCp,m(T2-Ti)=-0.2nCv,mTi=-0.2X1X2.5X8.314X273=-1134.86J

W=/U=-680.92J

设计如下可逆途径：

(273K,lOOkP.绝热不可逆’50kPa)

忘Jr

、Q73K,50kPa)/

g,=-W1=nRTlnV^V^nRTlnpi/p2=-1X8.314X273//?2=683.25J

AS\=Q\/T=683.25/273=2.5028J/K

Z\S2=fnCpmJTdT=nCp,mlnT2/T}

=lX2.5X8.3141n(0.8Ti/Ti)=-4.638J/K

Z\S=Z\SI+Z\S2=2.5028-4.6383=-2.135J/K

1

S2-SI=AS,S2=AS+SI=-4.638+100=95.362J-K4-moF

△A=At/-A(rS)=Z\U-(T2s2-T1S1)

=-680.92-273(0.8X95.362-100)=-7153.86J

Z\G=AH-A(rS)=AH-(T2S2-T\S\)

=-1134.86-273(0.8X95.362-100)=-7607.8J

19解：

e

⑴C(石墨)一C(金刚石)Z\trsG,„=?

£八乩「=八阳"「(石墨)-4”“「(金刚石)=-393.51-(-395.40)=1.89/，〃。尸

金刚石)-(石墨)=2.45-5.71=-326/K小。尸

e3

ArsG,,,°=/\lrsffni-TZ\trsSm°=1.89-298X(-3.26X10-)=2861.48J-moL'

(2)4rsG「＞0,；・石墨比金刚石稳定。

(3)4、G,"=Z^G/+J；Vdp=Q,112Xl(y3(i/p*1/p石)dp=-2861.48

12X1O-3(1/3513-l/2260)(p-pe)=-2861.48

p-p°=1510932234.8Pa,p=1511032234.8P«=1.511X1OgPa

故：增加压力可使不稳定晶体向稳定晶体转化。

20解：

pVm=RT+ap

VmA=RT/p]+a,Vm,2=RT/p2+a

W=-pdV=一jRT/(y,„+a)dVm=RTln{(Vm,2+a)/(Vm,i+a)}

=-RTln(p[/p2)

Z\G=fVdp=/(RT/P+a)dp=RTln(pdpi)+a(p2-pi)

Z\A=-fpdV=W=-RTln(px/p2)

,dS._

%)T=T需p=RP

AS=r一R/pdp=Rl〃(P2/pi)

Jpl

Q=TAS=-RTln(pypi)

AU=Q+W=-RTln(p2/p\)-RTln(p1/p2)=0

AH=AU+A(pV)=O+nRZ\T=O

23证明:

i,。八i,。八八八VTa2

a=M（而=M（而）T，证明：Cp-c、=

K

rr《UHdu_(d{U+Pv)du

,eu、,ev\,eu、月u、1ZQU、

=(君)。+H乔)"一(寿*(方)。*&'V-(而)v

=(―)+(—)(—)+apV-(-)=«V(—)+apV

STv5VTrSTpdTvdVTr

S(A+TS)dAdS

=aV(-----dV-----)'+apV=aV^'dv^r+aVT^'dV^r+中丫

_VTa2

z:—apV+aVTaP^=—ccVT(T-----

K

24证明：

范氏方程：p=RT/(Vm-b)-a^

dU=TdS-pdV=(祭),=T鲁)T-P=T(3-p=RT

dVeVdT

25证明：

—)s

ss

dVdp-pV-nRTn

-atT-=-=—i-=-nR

26解:

(600K,lOOkPa)CaSO4-2H2O(s)-CaSO4(S)+2H2O(g)(600K,lOOkPa)

HDt(3)

(2)

(298K,lOOkPa)CaSO4-2H2O(s)-CaSO£S)+2H2O(g)(298K,lOOkPa)

e

ArHm°(2)=zAf//„,°(CaSO4)+2AHm°(H2O)-AfH,n(CaSO4-2H2O)

=-1432.68+2(-241.82)-(-2021.12)=104.8kJ/mol

ArS,,,°(2)=2X188.83+106.70-193.97=290.39JK-1

4U：(2)=zAM「(2)-A(pV)=ArHin\2)-pV=AH；Q)-nRT

=104.8-2X8.314X298X10-3=99.84kJ-mor1

Z\rG「(2)=Z\r”,「(2)-T/⑸」(2)=104.8-298X290.39X10-3=18.26kJ/mol

4A,J(2)=ZiG:(2)-A(pV)=4G/(2)-pV=4G「(2)-〃RT

=18.26-2X8.314X298X10-3=13.30kJ/mol

Z\//,„(l)=/?Cp,m(298-600)=186.20X(298-600)=-56.23kJmo「

ZA1)=/",”(1)-pZW〜1)=-56.23kJ-mor'

1

Z\S(1)=1nCP.m/TdT=nCp,Jn(T2/ri)=lX186.20/n(298/600)=-130.31JK-

zAG(l)=Z\A(l)=

AHm(3)=E呜,m(600-298)=(99.6+2X33.58)X(600-298)

=50.36kJmor'

e

Z\t/„I(3)=zAH„,(3)-Zi(pV)^Z\rHffl(3)-n/?(7'2-rl)

=50.36-2X8.314X(600-298)=45.34kJ-mol”

Z^3)=,£nCP.m/TdT^nCp,nln(T2/Ti)=(99.6+2X33.58)/n(600/298)

=116.70JK-'

Z\G(3)=ZU(3)=

AH,n=AHm^+AHm(3)+Z\r/7„(°(2)=-56.23+50.36+104.8=98.93kJ-mo「

W=-pe(V2-Vi)^-pV=-2X8.314X600=-9.98kJ

1

ASm=AS,„(\}+AS,„(3)+/B/(2)=-130.31+116.7+290.39=276.78JK-

Z\U,"=Z\U“,(l)+Z\U,"(3)+Z\rU:(2)=-56.23+45.34+99.84=88.95kJmol1

Q=ZM「W=88.95+9.98=98.93kJ

AG,n=AGm{\}+AGm(3)+CrG,「(2)

AAm=ZU,“(1)+ZU,“(3)+/rA,「(2)

27解：

设计如下可逆途径：

(298K)I2(s)-----＞匕(0(457K)

[⑴f(4)

(387K)L⑤@12(1)(387K)为l2(D(457K)

△SQ)=15.66X103/387=40.47J-K-I

△S@=tnCp.MTdTX79.59加457/387=13.23J-K"

3

Z\S(4)=AvapH„/T=25.52X10/457=55.84J-K-'

△S=/S(1)+Z\S(2)+Z\S(3)+/S(4)=14.29+40.47+13.23+55.84=123.83JK

eO,,

S,„(457K,I2,g)=S„,(298K,I2,s)+Z\S=116.14+123.83=239.97JK--mor

物理化学习题解答(四)

习题p266〜270

1、在298K时，有0.10kg质量分数为0.947的硫酸H2s。4水溶液，试分别用(1)

质量摩尔浓度mB;(2)物质的量浓度CB和(3)摩尔分数期来表示硫酸的含量。已

知在该条件下，硫酸溶液的密度为1.0603Xl()3kg.m-3,纯水的密度为997.1kgm-3o

解：

加(B)=WBXX%=0.947X0.10kg=0.0947kg=94.7g

A

/iB=m(B)/MB=94.7/98.079=0.9655mol

-，〃(B)=0.10-0.0947=0.0053kg=5.3g

A

“A=m(A)/MA=5.3/l8.015=0.2942mol

⑴机B="B/""A)=0.9655/(5.3XW3)=182.17mol-kgI

(2)V=2%/P=0.10/(1.0603X103)=0.0943X10-3m3=0.0943dm3

A

1

cB=nB/V=0.9655/0.0943=10.24mol-L-

(3)XB=«B/Z%=0.9655/(0.9655+0.2942)=0.7664

A

2、在298K和大气压力下，含甲醇(B)的摩尔分数冲为0458的水溶液的密度为

0.8946kg-dm-3,甲醇的偏摩尔体积VB=39.80cm3.mo「，试求该水溶液中水的偏摩

尔体积VA«

解：

设〃B=L0mol,贝ijnii=〃B/xB=2.183mol,〃A=L183mol

〃?(B)=〃BMB=1.0X32.042=32.042g,6(A)=nAMA=1.183X18.015=21.312g

V={m(A)+机(B)}/=(21.312+32.042)/0.8946=59.64cm3

1

V=〃AVA+〃BVB，VA=(V-MBVB)/^A=(59.64-1.OX39.80)/1.183=16.77cm^mol-

3、在298K和大气压下，某酒窑中存有酒lO.On?,其中含乙醇的质量分数为0.96,

今欲加水调制含乙醇的质量分数为0.56的酒，已知该条件下,纯水的密度为999.1

kg.m0,水和乙醇的偏摩尔体积为：

63

w(C2H5OH)MH2O)/106m3.mo「V(C2H5OH)/10'mmor

0.9614.6158.01

0.5617.1156.58

试计算：

(1)应加入水的体积；

(2)加水后，能得到含乙醇的质量分数为0.56的酒的体积。

解：

(1)〃BMB/{几AMA+〃BMB}=0.96,46.068〃B/(18.015〃A+46.068即)=0.96,

18.015/7A+46.068〃B=46.068〃B/0.96=47.988几B，〃B=9.38"A

33

V=nAVA+nBVB=10.0m,(14.61nA+58.01nB)XlO^10.0m,

63

(14.61nA+58.01X9.38nA)X10-=10.0m,nA=17897.3mol,nB=167876.6mol,

n^M^I{〃/AMA+〃BMB}=0.56,

167876.6X46.068/(18.015〃么+167876.6X46.068)=0.56

18.015〃A+167876.6X46.068=167876.6X46.068/0.56=13810248.6

/

MA=337302.8mol,4=337302.8-17897.3=319405.5mol

pV*/l8.015=319405.5

V^=18.015X319405.5/999.1kg-m3=5.76m3

(2)V=/AVA+〃BVB=(337302.8X17.11+167876.6X56.58)X10-6=15.27m3

4、在298K和大气压下，甲醇(B)的摩尔分数&为0.30的水溶液中，水(A)和甲

3

醇(B)的偏摩尔体积分别为VA=17.765cm3.mo「，VB=38.632cm-mor',已知该条

件下，甲醇(B)和水(A)的摩尔体积为V„,.B=40.722cm3.mo「，匕”,A=18.068cm3.mo「,

现在需要配制上述水溶液1000cn?,试求：

(I)需要纯水和纯甲醇的体积；

(2)混合前后体积的变化值。

解：

3

(1)V=nAVA+wBVB=1000cm,17.765wA+38.632/jB=1000

n^M^I{〃AMA+〃BMB}=0.96,32.042/7B/(18.015HA+32.042HB)=0.30

18.015〃A+32.042〃B=32.042〃B/0.30=106.807〃B，;?A=4.15"B

17.765X4.15Z?B+38.632/?B=1000,/iB=8.900mol,HA=4.15HB=36.935mol

3

nAVm,A=36.935X18.068=667.34cm

3

Vw=Vm,B=8.900X40.722=362.43cm

(2)V水+V甲龄=667.34+362.43=1029.77cm3

Z\V=V混合前-V混合国=1029.77-1000=29.77cm3

5、在298K和大气压下，溶质NaCl(s)(B)溶于LOkgHzO⑴(A)中，所得溶液的体

积V与溶入NaCl(s)(B)的物质的量“B之间的关系式为：

3/223

V=[1001.38+1.625(nB/mol)+1.774(/7B/mol)+0.119(nB/mol)]cm

试求：(1)H2O(1)和NaCl的偏摩尔体积与溶入NaCl(s)的物质的量“B之间的关系；

(2)nB=0.5mol时-，氏0⑴和NaCl的偏摩尔体积；

(3)在无限稀释时，氏0⑴和NaCl的偏摩尔体积。

解：

3/223

(1)V=[1001.38+1.625(HB/mol)+1.774(rtB/mol)+0.119(/?B/mol)]cm

1/23

氏=(段)后[16.625+3/2X1.774(«B/mol)+2X0.119(nB/mol)]cm

(2)V=(1001.38+16.625X0.5+1.774X0.53/2+0.119X0.52)=1010.349cm3

1/231

VB=(16.625+3/2X1.774X0.5+2X0.119X0.5)=18.6256cm.mof

〃A=加(A)/MA=1.0X103/18.015=55.509mol

VA=(V-HBVB)/«A=(1010.349-0.5X18.6256)/55.509=18.0334cm3-mor'

33

(3)nB-*0,V=1001.38cm,VB=16.625cm,

3

VA=(V-/7BVB)//?A=1001.38/55.509=18.04cmmor'

6、在293K时、氨的水溶液A中NH3与H2O的量之比为1:8.5,溶液A上方NH3

的分压为10.64kPa,氨的水溶液B中NH3与H2O的量之比为1:21,溶液B上方

NH3的分压为3.579kPa,试求在相同的温度下：

(1)从大量的溶液A中转移lmolNH3(g)到大量的溶液B中的ZiG；

(2)将处于标准压力下的ImolNHKg)溶于大量的溶液B中的/G。

解：

⑴PNH3=Z3A^NH3，h,A=PNH3/-^NH3=10.64/(1/9.5)=101.18

PNH3=七即NH3，4X,B=PNH3/XNH3=3.579/(1/22)=78.738

〃NH3(A,aq)+nNH3(B,aq)-*(n-l)NH3(A,aq)+(/?+1)NH3(B,aq)

Z\G=(”+1)"2,B(aq)+(n-l)"2,A(aq)-nMi,B(aq)-”i,A(aq)

=(n+1)RTlnkx,BX2,B+(«-1)RTlnkx,AX2,/<-nRTlnkx.Bx\,^-nRTlnkx,AXi,A

*.*n,.,.X2,BQXI,B=XB=1/22,*2,人心为必=工人=1/9.5

AG=RTlnkx,B-RTlnkx,AXA

=8.314X293{加(78.738X1/22)-Zn(101.18X1/9.5)}=-2656.5J

(2)NH3(g)+〃NH3(B,aq)-(〃+l)NH3(B,aq)

o

Z\G=(”+1)“2,B(aq)-M⑺-〃“i,B(aq)

ee

=("+1)R77〃ZX,B/PX2,B-nRTlnk^pxi,B

n,.,.X2,BRSXI,B=XB=1/22

AG=RTlnk^lp°XZB=8.314X293/n(78.738/100X1/22)=-8112.1J

7、300K时，纯A与纯B形成理想混合物，试计算如下两种情况的Gibbs自由

能的变化值。

(1)从大量的等物质量的纯A与纯B形成的理想混合物中，分出Imol纯A的ZiG。

(2)从A与纯B各为2moi所形成的理想混合物中，分出Imol纯A的ZiG。

解：“AB-A⑴+(〃-l)AnB

(1)G2="A"(1)+(〃A-1)"2,A⑴+〃B"2,B(1)，G|=MA»1,A⑴+〃B"1.B0)

△G=G2-G\=〃A*(1)+(〃-1)"2,A⑴+〃"2,B⑴-n«i,A(1)-nwi,B(1)

=(n-1)RTI〃X2,A-“RTl〃x[,A+〃RT(J〃X2,B-l〃xi,B)

Vrt-oo,•,.X2,AQX1A=XA=0-5,X2.BQ%1,B=XB=0.5,

故=-RTlnxA=-8.314X300勿0.5=1728.85J

(2)G2=«A(1)+M2,A(1)+2M2,B(1)»GI=2MI,A(1)+2WI,B(1)

△G=Gz-Gi="A*⑴+〃2,A(1)+2"2.B⑴-2M|.A(1)-2«I,B(1)

=RTlnx2.A+2RTlnxi^-2RTlnx\,\-2RTlnxI.B

=RT(lnx2,A+2lnxi,B-2lnx\,K-2lnxI.B)

，«，X2,A=1/3»X2.B=2/3,XI,A=XI,B=2/4=0.5,

：.AG=8.314X300X(//?1/3+2/〃2/3-2/〃0.5-2/«0.5)=2152.61J

10、在293K时，纯C6H6⑴(A)和C6H5cH3⑴(B)的蒸气压分别为9.96kPa和

2.97kPa,今以等质量的苯和甲苯混合形成理想液态混合物，试求：

(1)与液态混合物对应的气相中，苯和甲苯的分压；

(2)液面上蒸气的总压力。

解：

(1)PA=PA*XA=PA""?(A)/M\/［租(A)/MA+m(B)/MB)=PA如B/(MA+/B)

=9.96X92.138/(78.112+92.138)=5.39kPa

PB=PB*XB=PB*根(B)/MB/［根(A)/MA+〃Z(B)/MB)=PB"MA/(MA+MO

=2.97X78.112/(78.112+92.138)=1.36kPa

(2)p总=PA+〃B=5.39+1.36=6.75kPa

14、在室温下，液体A与液体B能形成理想液态混合物。现有一混合物的蒸气

相，其中A的摩尔分数为0.4,把它放在一个带活塞的汽缸内，在室温下将汽缸

缓慢压缩。已知纯液体A与B的饱和蒸气压分别为40.0kPa和120.0kPa,试求:

(1)当液体开始出现时，汽缸内气体的总压；

(2)当气体全部液化后，再开始汽化时气体的组成。

解：

(1)当PA=p.FyA=40.0kPa时,液体开始出现，故p总=40.0/0.4=lOOkPa

AAX；BB；

(2)/?=P*XA=40.0X0.4=16.0kPa/?=P*XxB=l20.0X0.6=72.0kPa

>A=PA(PA+PB)=16.0/(16.0+72.0)=0.182

)，B=PB(PA+PB)=72.0/(16.0+72.0)=0.818

19、可以用不同的方法计算沸点升高常数。根据下列数据，分别计算CS2⑴的沸

点升高常数。

(1)3.20g的蔡(Go%)溶于50.0g的CS2⑴中，溶液的沸点较纯溶剂升高了1.17K；

(2)1.0g的CS2⑴在沸点319.45K时的汽化焰值为351.91g1；

(3)根据CS2⑴的蒸气压与温度的关系曲线，知道在大气压力101.325kPa及其沸

点。319.45K时,CS2⑴的蒸气压随温度的变化率为3239Pa.K」(见Clapeyron方程)。

解：

(1)kb=AT/mn=Z\T/［m(B)/(m(A)MB)

=1.17/(3.20/128.17/50.0X10-3)]=2.34K-mor'.kg

(2)kb=RM\(Tb*)2/Z\r卬H*m,A=8.314X319.52/351.9X10-3=2.41K-mol'-kg

f2

(3)dlnp/dT=AvapHmA/RT^kb=MA(dT/dlnp)

Ab=MAd77d/"p=MA/?d77dp=76.143X10-3x101325/3