第二章 热力学第一定律 习题物理化学 天津大学第五版上课专用课件！

习题2.1

W=?p(V2?V1)=?nR￠T=?8:314J习题2.2

W=?===

?[p2V2(l)?p1V1(g)]

p1V1(g)

RT 8:314￡(273:15+100) 3:102kJ

习题2.3

W=?[p2V2(H2;O2;g)?p1V1(H2O;l)]

?p2V2(g)

X =nBRT=(1+0:5)￡8:314￡(273:15+25)

=3:718kJ

习题2.4

￠U=Qb+W)Wbb=Qa+Wa=2:078?4:157=?2:079kJ

=￠U?Qb=?2:079?(?0:692)=?1:387kJ

习题2.5 5mol T0=25oC P=200kPa V0

绝热膨胀 5mol T1=-28.57oC P1=100kPa V 途径a

恒容加热 5mol T2 p=200kPa V 5mol T0=25oC P=200kPa V0 恒压加热 5mol T2 P=200kPa V 途径b

Wb=====?p￠V3nRTnRT0′1?p?p1p03TT0′1?nRp?p1p0

3244:58298:25′?5￡8:314￡200￡10￡?100￡103200￡103?7:940kJ3b￠U=Qb+W)Wb=Qa+Wa=25:42?5:57=19:85kJ=￠U?Qb=19:85+7:940=27:79kJ习题 2.6

H=U+pV)￠H?￠U=￠(pV)

)￠H?￠U=====￠(pV)￠(nRT)nR￠T4￡8:314￡20665:1J

习题2.7

￠H=￠U+￠(pV)?￠(pV)=V￠p (1)

18:02￡10?3￠H?V￠p=￡100￡103=1:8J997:04

(2)

18:02￡10?3￠H?V￠p=￡(106?100￡103)=16:3J 997:04

习题2.8

￠U=Q+W=nCV;m￠TW=0)

￠H=nCp;m

习题2.9

￠U=Q=nCV;m￠T=5￡1:5￡8:314￡50=3:118kJ

￠T=n(CV;m+R)=5￡2:5￡8:314￡50=5:196kJ

W=?p￠V=?nR￠T=?5￡8:314￡(?50)=2:079kJ

￠U=Q+W=nCV;m￠T=5￡2:5￡8:314￡(?50)=?5:196kJ ￠H=Q=￠U?W=?5:196?2:079=?7:275kJ

习题2.10

p0V0=p2V2))T0=T2

￠U=0;￠H=0

W=?pe￠V=?200￡103￡(25?50)￡10?3=5:00kJ

￠U=Q+W=0)Q=?W=?5:00kJ

习题2.11

￠U=nCV;m￠T=1￡20:92￡(97?27)=1:464kJ

T300:15p1=p20=250:00￡103￡=202:7kPa T2370:15 3RT′RT20W=?p1￠V=?p1?=2:497kJ

p2p0

￠U=Q+W)Q=￠U?W=?1:033kJ

￠H=nCp;m￠T=n(CV;m+R)￠T=1￡29:23￡(97?27)=2:046kJ

习题2.12

RT2 Cp;mdTT1 Cp;m=T2?T1

R800=300(26:75+42:258￡10?3T?14:25￡10?6T2)dT800?300==ˉ800?3T2?114:?6T3)ˉ(26:75T+142:258￡1025￡10ˉ23300500?145:38J￠mol￠K?1￠H=nCp;m

103￠T=￡45:38￡500=515:6kJ

44:01习题2.15 恒容绝热过程

￠U=QV=n(Ar;g)CV;m(Ar;g)￠T(Ar;g)+n(Cu;s)CV;m(Cu;s)￠T(Cu;s)=0

￠T(Ar;g)=T;￠T(Cu;s)=T?150

CV;m(Ar;g)=Cp;m(Ar;g)?R);CV;m(Cu;s)?Cp;m(Cu;s)

n(Cu;s)CV;m(Cu;s)T(Cu;s)T==74:23oCn(Ar;g)CV;m(Ar;g)+n(Cu;s)CV;m(Cu;s)￠H=n(Ar;g)Cp;m(Ar;g)￠T(Ar;g)+n(Cu;s)Cp;m(Cu;s)￠T(Cu;s)=2:47kJ

习题2.16

3300￡10n(CO)=n(H2)==9990mol

M(CO)+M(H2)

把水煤气和水看作一个系统，则此系统与环境的热交换Q＝0

ZT2ZT2Q=n(CO;g)Cp;m(CO;g)dT+n(H2;g)Cp;m(H2;g)dT

T1T1

+m(H2O;l)Cp(H2O;l)￠T(H2O;l)=0

n(CO;g)m(H2O;l)=?

习题2.17 等压绝热过程

RT2T1Cp;m(CO;g)dT+n(H2;g)RT2T1Cp;m(H2;g)dTCp(H2O;l)￠T(H2O;l)=2863:8kg

3519￠U=￠UB=yBnCV;m(B)￠T=3￡R￠T+2￡R￠T=R(T2?T1)

222 p5W=?p2￠V=?p2(V2?V1)=?nRT2+2nRT1=RT1?5RT2

p12

￠U=Q+W=W

24)T2=T=331:0K

291XX￠H=

X￠HB=XyBnCp;m(B)￠T=￠U+￠(pV)=￠U+nR￠T= 19R￠T=?5:448kJ2

29R￠T=?8:315kJ2￠U=Q+W=W=