Contents of Today S.J.T.U. Phase Transformation and Applications Review previous /Quiz TA:黄宝旭 Equlibrium Huangbx3@sjtu.edu.cn Thermodynamic activity Chemical equilibrium Gaseous equilibrium Solid-vapor equilibrium Sources of information on Chemical equilibrium and adiabatic flame temperature etc. Science research is an adventure,is an interest-driving learning process. It takes more time to think than to do.-Ke Lu SJTU Thermodynamics of Materials Spring2006©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 8 Chemical equilibrium Contents of Today Review previous / Quiz Equlibrium TA:黄宝旭 Huangbx3@sjtu.edu.cn Science research is an adventure, is an interest-driving learning process. It takes more time to think than to do. – Ke Lu Thermodynamic activity Chemical equilibrium Gaseous equilibrium Solid-vapor equilibrium Sources of information on Chemical equilibrium and adiabatic flame temperature etc
Quiz Question 1 S.J.T.U. Phase Transformation and Applications 1)Consider an isolated system consisting of a kilogram of lead and a kilogram of water illustrated below. T=20C funiform throughout system) Tfinal=???(uniform throughou system) I Kg 1 Kg Liquid Water meter I Kg Liquid Water Figure:Isolated system illustrated before and after. The heat capacity of 1 kilogram of Pb is given by Cp;the heat capacity of 1 kilogram of water is given by Coall other heat capacities in the isolated system can be neglected.Cb and CH2o may be considered independent of any constraints(e.g.,constant pressure or constant volume)and to be independent of temperature. i Derive an expression for the final temperature after a process leading to the figure on the right of the illustration. ii Would the temperature be larger or smaller if the block of lead had fallen to the left(i.e., into the wateTermodynamics of Matera Spring2006©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 8 Chemical equilibrium Quiz Question 1 1) Consider an isolated system consisting of a kilogram of lead and a kilogram of water illustrated below. Figure : Isolated system illustrated before and after. The heat capacity of 1 kilogram of Pb is given by CPb ; the heat capacity of 1 kilogram of water is given by CH2O; all other heat capacities in the isolated system can be neglected. CPb and CH2O may be considered independent of any constraints (e.g., constant pressure or constant volume) and to be independent of temperature. i Derive an expression for the final temperature after a process leading to the figure on the right of the illustration. ii Would the temperature be larger or smaller if the block of lead had fallen to the left (i.e., into the water)?
Quiz Q1 Answer S.J.T.U. Phase Transformation and Applications 1-3-i Derive an expression for the final temperature after a process leading to the figure on the right of the illustration. Change in internal energy of system =Heat flow into Pb and water Imghl =Cpb(Tyinal-20)+CHO(Tyinat-20) mgh] Trmal=20+Cpb+CHaO 1-3-ii Would the temperature be larger or smaller if the block of lead had fallen to the left (i.e., into the water)? Tricky.The answer is that the temperature will be lower. The center of mass of the water will raise and so part of the potential energy of the lead weight will be converted to potential energy of the water.mg would be decreased by an amount corresponding to the raise of center of mass of the water. SJTU Thermodynamics of Materials Spring2006©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 8 Chemical equilibrium Quiz Q1 Answer
Quiz Question 2 S.J.T.U. Phase Transformation and Applications 2)1mol理想气体等压膨胀到状态2，求Q,W,△U,△H。若将理想 气体先等容加热到状态3，然后再等温（可逆）膨胀到状态2， 求Q,W,△U,△H,并与直接从1到2的途径相比较。 P↑3(V1,T2) 1(V,T) 2(V2,T2) SJTU Thermodynamics of Materials Spring 2006 ©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 8 Chemical equilibrium Quiz Question 2 2) 1mol 理想气体等压膨胀到状态2，求Q,W,ΔU,ΔH。若将理想 气体先等容加热到状态3，然后再等温（可逆）膨胀到状态2， 求Q,W,ΔU,ΔH，并与直接从1到2的途径相比较。 2 (V2,T2 1 (V ) 1,T1) 3 (V1,T2 P ) V
Quiz Q2 Answer S.J.T.U. Phase Transformation and Applications 1ol理想气体等压膨胀到状态2，求Q,W,△U,△H。若将理想气体先等容加热到状态3， 然后再等温（可逆）膨胀到状态2，求Q,W,△U,△H,并与直接从1到2的途径相比较。 1mol理想气体直接从1到2等压膨胀 1mol理想气体从1到3等容加热 功：P外×(V2V)=-R(T2T) 功：W1=0 焓变：CpX(T2T) 焓变：△H1=CpX(T2T) 内能变化：CvX(T2T) 内能变化：△U1=Cv×(T2-T1) 热：Q=AU-W=Cv×(T2-T+R(T2-T尸Cp×(T2T) 热：Q1=AU-W=Cv×(T2T) P↑3(V1,T2) 1mol理想气体从3到2（可逆）等温膨胀 △U2=0,△H2=0 PdV=PdV Q2=△U2-W2 2=-所naw=gaw=-mn 1(VT) 2(V2,T2) SJTU Thermodynamics of Materials Spring2006©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 8 Chemical equilibrium Quiz Q2 Answer 2 (V2,T2 1 (V ) 1,T1) 3 (V1,T2 P ) V 1 mol 理想气体直接从1到2等压膨胀 功：-P外×(V2-V1)＝-R (T2-T1) 焓变：Cp×(T2-T1) 内能变化：CV×(T2-T1) 热：Q=ΔU-W= CV×(T2-T1)+ R (T2-T1)= Cp×(T2-T1) 1mol 理想气体等压膨胀到状态2，求Q,W,ΔU,ΔH。若将理想气体先等容加热到状态3， 然后再等温（可逆）膨胀到状态2，求Q,W,ΔU,ΔH，并与直接从1到2的途径相比较。 1 mol 理想气体从1到3等容加热 功：W1＝0 焓变： ΔH1 = Cp×(T2-T1) 内能变化： ΔU1= CV×(T2-T1) 热：Q1=ΔU-W= CV×(T2-T1) 1 mol 理想气体从3到2 （可逆）等温膨胀 ΔU2=0, ΔH2=0 P外dV=PdV Q2=ΔU2-W2 ∫∫ −=−= −= 21 1 2 2 2 2 1 2 ln V V V V V V RTdV V RT W pdV
15.进行下述过程时，系统的△U、△H、△S和△G何者为零？ (1)非理想气体的卡诺循环； (2)隔离系统中的任意过程： (3)在100℃，01325Pa下1mo水蒸发成水蒸气； (4)绝热可逆过程。 16.改正下列错误： (1)在一可逆过程中熵值不变： (2)在一过程中熵变是 9： (3)亥姆赫兹函数是系统能做非体积功的能量： (4)吉布斯函数G是系统能做非体积功的能量； (5)焙H是系统能以热的方式交换的能量。 SJTU Thermodynamics of Materials Spring2006©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 8 Chemical equilibrium Quiz Q3 and Q4
Quiz Q3 Answer S.J.T.U. Phase Transformation and Applications 15.进行下述过程时，系统的△U、△H、△S和△G何者为零？ (])非理想气体的卡诺循环； (2)隔离系统中的任意过程： (3)在100℃，01325Pa下1mo水蒸发成水蒸气； (4)绝热可逆过程。 15题：1)内能变化为0，焓变为0，熵变为0，自由能变化为0。 2)：内能变化为0，焓变不一定为0，熵变不一定为0，自由能变化不 一定为0。 3)自由能变化为0。错，主要认为相变过程中内能为0，焓变也为0。 4)只有熵变为0。错误很多。 SJTU Thermodynamics of Materials Spring 2006 ©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 8 Chemical equilibrium Quiz Q3 Answer 15题：1）内能变化为0，焓变为0，熵变为0，自由能变化为0。 2）：内能变化为0，焓变不一定为0，熵变不一定为0，自由能变化不 一定为0。 3）自由能变化为0。错，主要认为相变过程中内能为0，焓变也为0。 4）只有熵变为0。错误很多
Review previous lecture (1) S.J.T.U. Phase Transformation and Applications Condition of equilibrium Phase equilibrium相平衡/化学反应的平衡 W,e.1→2=0 42=41G2=G SJTU Thermodynamics of Materials Spring2006©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 8 Chemical equilibrium Review previous lecture (1) Condition of equilibrium Phase equilibrium相平衡/化学反应的平衡 δWrev →21. = 0 1,2, = GG ii 1,2, μ = μii
恒温下(O△GOP)T=△V关系式的应用 S.J.T.U. Phase Transformation and Applications 在298K和1atm下，石墨为稳定态 298K 石墨(G)>金刚石(D) 己知：石墨和金刚石的标准生产热和标准熵为 0,1900J.mo1,5.73J.mol1.K-1,2.43J.mo1.K1 298K下石墨和金刚石密度为：2.22g.cm3,3.515g.cm-3 V,=12/3.515=3.414cm3.mo-1 'c=5.405cm3.mo1 Diamond g △2Gnm(298K,latm)=△Hm-TASm Graphite T=Constant =1900-298(2.43-5.73)=2883J.mo1>0 Po Equilibrium pressure a△G Pressure- ap =AV='o-'o<0 Figure 4.3 Specific Gibbs free energy versus pressure at con- stant temperature for graphite and diamond. SJTU Thermodynamics of Materials Spring 20ub x.J.JIn Lecture 8 Cnemical equllibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 8 Chemical equilibrium 恒温下 (∂ΔG/∂P)T =ΔV 关系式的应用 在298K和1atm下，石墨为稳定态 已知：石墨和金刚石的标准生产热和标准熵为 0，1900 J.mol-1， 5.73 J.mol-1.K-1, 2.43 J.mol-1.K-1 298K下石墨和金刚石密度为：2.22 g.cm-3, 3.515 g.cm-3 金刚石(D) 2883)73.543.2(2981900 0 )1,298( 1 >⋅=−−= Δ Δ−Δ= − molJ m m STHatmKG m DG 3 1 414.3515.3/12 − VD = = ⋅molcm 3 1 405.5 − VG = ⋅molcm
Review previous lecture (2) S.J.T.U. Phase Transformation and Applications Clapeyron equation in vapor equilibria Clapeyron equation pand Talong this line △H Te9△V B Change of the melting point of tin resulting from a pressure change of 500 atm △H 7196 T> Figure 4.4 Pressure-temperature relationship Teg△V 505×4.39×10-7 for equilibrium between phases A and B. △p9=500atm△Te9=+1.58K SJTU Thermodynamics of Materials Spring2006©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2006 © X. J. Jin Lecture 8 Chemical equilibrium Review previous lecture (2) Clapeyron equation in vapor equilibria VT H dT dp eq eq eq Δ Δ = Clapeyron equation Change of the melting point of tin resulting from a pressure change of 500 atm 7 1039.4505 7196 − ×× = Δ Δ = Δ Δ VT H T p eq eq eq p Tatm K eq eq =Δ 500 +=Δ 58.1
