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1、1mol水蒸气(HO,g)在100,101.325kPa下全部凝结成液态水。求过程的功。假设:相对于水蒸气的体积,液态水的体积可以忽略不计。2.2H2O(l)100℃101.325kPaH2O(g)100℃101.325kPaW=-p环(V2-V1)=-p(-V1)=nRT=18.314373.15=3102.4J7/22/20211始态为25℃,200kPa的5mol理想气体,经a,b两不同途径到达相同的末态。途径a先经绝热膨胀到-28.57℃,100kPa,步骤的功Wa=-5.57kJ;再恒容加热到压力200kPa的末态,步骤的热Qa=25
2、.42kJ。途径b为恒压加热过程。求途径b的Wb及Qb。5molPgp1=200kPaT1=25℃5molPgp2=100kPaT2=-28.57℃5molPgp3=200kPaT3(1)绝热(2)恒容2.5途径b恒压7/22/202125molPgp1=200kPaT1=25℃5molPgp2=100kPaT2=-28.57℃5molPgp3=200kPaT3(1)绝热(2)恒容T3=2T2=2(273.15-28.57)=489.16KU=Qa+Wa=Qb+WbWb=-p环(V3-V1)=-nRT=-58.314191.01=-794
3、0.3JQb=(Qa+Wa)-Wb=(25.42-5.57)+7.94=27.79kJ途径b恒压7/22/20213某理想气体CVm=3/2R。今有该气体5mol在恒容下温度升高50℃。求过程的W,Q,U及H.W=02.85mol理想气体T1V15mol理想气体T2=T1+50℃V2=V1恒容Q=U=nCVmT=53/2R50=3.118103JH=nCpmT=55/2R50=5.196103J7/22/202144mol某理想气体,Cp,m=7/2R。由始态100kPa,100dm3先恒压加热使体积增大到150dm3,再恒
4、容加热使压力增大到150kPa。求过程的W,Q,U,和H.U=nCV,m(T3-T1)=CV,m(p3V3-p1V1)/R=18750JH=nCp,m(T3-T1)=Cp,m(p3V3-p1V1)/R=31250JW=W1+W2=W1=-p2(V2-V1)=-100103Pa(150-100)10-3m3=-5000JQ=U-W=18750+5000=23750J2.11第四版4molPgp1=100kPaV1=100dm34molPgp2=100kPaV2=150dm34molPgp3=150kPaV2=150dm33(1)恒压(
5、2)恒容7/22/20215解法27/22/202167/22/202171mol某理想气体于27℃,101.325kPa的始态下,先受某恒定外压恒温压缩至平衡态,再恒容升温至97℃,250.00kPa。求过程的W,Q,U,H。已知气体的CV,m=20.92J·mol·K-1。2.11第五版1molPgT1=300Kp1=101.325kPa1molPgT2=T1=300Kp2=1molPgT3=370Kp3=250.00kPa(1)恒温(2)恒容7/22/20218U=nCV,m(T3-T1)=20.92×(370-300)=1464JH
6、=nCp,m(T3-T1)=(20.92+8.314)×(370-300)=2046Jp2=p3×T2/T3=250×300/370=202.70kPaW=W1+W2=W1=-p2(V2-V1)=-nRT(1-p2/p1)=-8.314300(1-202.70/101.325)=2496JQ=U-W=1464-2496=-1032J1molPgT1=300Kp1=101.325kPa1molPgT2=T1=300Kp2=1molPgT3=370Kp3=250.00kPa(1)恒温(2)恒容7/22/20219容积为0.1m3的恒容密闭容器中有
7、一绝热隔板,其两侧分别为0℃,4mol的Ar(g)及150℃,2mol的Cu(s)。现将隔板撤掉,整个系统达到热平衡,求末态温度及过程的H.已知:Ar(g)和Cu(s)的摩尔定压热容Cp,m分别为20.786J·mol·Kˉ1及24.435J·mol·Kˉ1,且假设均不随温度而变。Q=0,W=0,∴U=0U=U{Ar(g)}+U{Cu(s)}=0U=nCV,m{Ar(g)}(T-T1)+nCp,m{Cu(s)}(T-T2)=0T=347.38KH=nCp,m{Ar(g)}(T-T1)+nCp,m{Cu(s)}(T-T2)=2469J2.1
8、57/22/202110已知水在100℃,101.325kPa下的摩尔蒸发焓vapHm=40.668kJ·mol-1,试