Contents of Today S.J.T.U. Phase Transformation and Applications Review previous /Quiz Equlibrium 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 Spring 2007 ©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2007 © X. J. Jin Lecture 8 Chemical equilibrium Contents of Today Review previous / Quiz Equlibrium 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 Spring2007©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2007 © 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 Spring 2007 ©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2007 © 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 Spring2007©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2007 © 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=CvX(T2-T+R(T2T)尸Cp×(T2T1) 热：Q1=AU-W=Cv×(T2T1) 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 Spring 2007 ©X.J.Jin Lecture 8 Chemical equilibrium
Phase Transformation and Applications S. J. T. U. SJTU Thermodynamics of Materials Spring 2007 © 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