PHYSICALCHEMISTRY物理化学-用物理学的原理和实验手段来研究

1、物理学校长1制作：李乃军，物理学校长？ thephysicalchemistryisabranchofsciencethatsolvedtheproblemofchemistrybymeansofphysicalprincipleandexperiments.(用物理学原理和实验手段解决化学问题的科学1 .热动态(firstlawandsecondlawofthermodynamics )问题解决(1)反应方向判断(2)反应限度2 .动态()反应基理分析和反应过程的推定3.Electrochemistry使用电化学手段4界面和colloid chemistry，chapter1thefirst

2、lawofthermodynamicsandapplication本章的目的和要求是： 1.1 theconceptionsofthermodynamics1. theintroductionofconceptions1. system (我们将讨论的对象)孤立系统*封闭系统* 在Open System 2.Surroundings (与体系密切相关的周边部分)3. thepropertiesofsystemandstatefunctions、state functions * * importantpportant c.1周期中du=04 equilibriumstateofthermody

3、namicsthermalequilibriummechanicalequilibriumphaseequilibriumchemicalequilibrium，热(q )与heattheenergyexchangedbetweensystemandsurroundings，becauseofthetemperancedifferences. wfworkexceptexpansionworkthedefinitiono *，1.2 First law of Thermodynamics一The First Law of Thermodynamics能量转化和守恒定律自然段的所有物质都有能量，

4、能量有多种形式，可以从一种形式转换为另一种形式，转换过程中，能量的订正energykineticenergyofwholesystempotentialenergyofsysteminternalenergy (studyonlyinternalenergyinfirstlaw )，二等式du=q-w qandwistheprocessfunction.theuisstatefunction .1.3quasi p外Pi，Pi P外，膨胀，dV0，对外工作，W0 Pi P外，压缩，dV0，内工作，W0 W=P外dV*，reset uisastatefunction.2.workandproce

5、ssconclusion 3360 theworkisapathfunction.3.uasi -静态processandreversiblepple the解决方案330 p外部dv=p2(v2- v1)=(nrt/v2)=n rt-t2=373 k w=w1w2=p外部1 dV P外部2 dv=p2(v2- v1)=p2v2- p2v2- p2=nrt p (v2/v2)=2nrt nrt (0. 50.5 )=nrt=rt (4) w=p外dv=nrt/vdv=nrt ln (v2/v1)=RRT简易处理cessesneveroccurunderequilibriumco whybeco

6、ncernedwithreversibleprocessesiftheycannotoccur？ oneplausibleansweristhatcalculationofthereversibleworkprovideduswithausefulupperlimittothemagnitudeofworkthatcanbeobtainedin the ghthisiscertainlytrue，itisnotthemajorreasonweconsiderthereversibleprocess .itismucheasiertocomputesimilarchangesinareversi

7、ble spontaneous paths.ifthequantityweareattemptingtocomputehasanexactdifferential。 wecanalwaysreplacetheactualirreversiblepathwithareversibleonewithoutchangingthevalueoftheintegral .1.4焓散热箱、散热器状态函数、Let us prove it: U=Q W When Wf=0、恒定压力3360 u2u1=qpp (v2- v (u2p2v2)、qp=(u2p2v2) (u1p1v1)定义3360 hupvprop

8、erties 3360.u.p.varestatefunctions sohisastatefunctiontoo.2.absolutevalueofhcannotbeascertained。 andhhasnodefinitephysicalsignificance.3.hhasaunitofenergy.h=qpusingconditions:wf=0， 常数继续注意事项： h=u (PV )=upvvpconstant-pressure:h=u (PV )=upv常数卷： h=u () under常数继续预测到100 计算3360 q、w、h、u .解决方案3360 wf=0、恒定压力q

9、=h=22.02 kjw=p外dv=p (v2- v1)=p2v2- p1v1=NR (t2- t1)=8. 3=22020-5820 j=16.2 kjuandhmayuseinanypathsbetweenthestates.answer 3360 (omit )、summit2. hupvproperties 3360 a.hisastatefunction andhhasnodefinitephysicalsignificance.c.hhasaunitofenergy.h=qpusingconditions:wf=0， 恒定无阻系统4.h=u (PV )=u PV VP、家庭3360

10、 p 893360、5(1)、6、9和1.5热容量c=热容量t2.特定热容(比热)3. molarheatcapacity5.恒定-恒定-恒定-卷1.6 theapplicationoffirst-lawofthermodynamicstoperfectgas1. theinternalenergyandenthalpyofperfectgas (joulesexperimenns ) 结论：选择： theinternalenergyandenthalpyofidealgasdependonlyupont.u=f (t ) h=f (t )条件：理想瓦斯气体，例如：摩尔理气。(1)向真空的膨胀

11、(2)P外=1atm .膨胀(3)等温可逆膨胀解： (1)向真空的膨胀W=P外dV=0； Q=0(吸热=散热或不赶趟热交换) U=0 H=U PV=U nRT=U=0 (2)P外=1atm .膨胀W=P外dV=P外(v2-v1) Q=W=0.9RT (3)等温可逆膨胀W=P外dv=pdv=nrtlnv2/v1=n 另一方面，cpandcvofidealgasderivation 3360 h=upv，对其进行等压下偏置t 3360 h/TP=u/tppv/TP=CP (1) u=f (t ) du=u/tpdt u/pt DP两侧同乘(dt ) v 3330 结果：理想堆瓦斯气体TP=CV nR、4. Carnot cycle of perfect gas*# carnotcycleinclud