Download Physics 406 Final Exam: Thermodynamics and Statistical Mechanics and more Exams Physics in PDF only on Docsity! Ethermal 0.191 eV= d) Three translational d.o.f. Etrans 3 ε⋅:= Etrans 7.043 10 21− × J= Etrans 0.044 eV= e) Two rotational d.o.f. Erot 2 ε⋅:= Erot 4.695 10 21− × J= Erot 0.029 eV= f) One molecule: Einternal Ethermal µ 1⋅+:= Einternal 6.56− 10 20− × J= Einternal 0.409− eV= Physics 406 Final Exam 12/16/05 1. Natoms 3:= T 340 K⋅:= µ 0.6− eV⋅:= a) Equipartition Theorem: ε 1 2 k⋅ T⋅:= ε 2.348 10 21−× J= ε 0.015 eV= b) For linear molecules: Nν 6 Natoms⋅ 5−:= Nν 13= c) Equipartition Theorem: Ethermal Nν ε⋅:= Ethermal 3.052 10 20− × J= 2 7 55 3.125 103×= 27 128= 55 27⋅ 400000= 55 27⋅ Total 0.044= Total 9120800= Ptotal 1:= b) The most probable energy macrostate is E1 = 3. c) A total of 9120800. d) S k ln 35 47⋅( )⋅:= S 2.099 10 22−× J K = S 1.31 10 3−× eV K = S 15.197 k= e) Sf k ln Total( )⋅:= Sf 2.213 10 22− × J K = Sf 1.382 10 3− × eV K = Sf 16.026 k= f) Sinitial k ln 2 5 57⋅( )⋅:= Sinitial 2.034 10 22−× JK= ∆S Sf Sinitial−:= ∆S 1.787 10 23− × J K = ∆S 1.116 10 4−× eV K = ∆S 1.294 k= Physics 406 Final Exam 12/16/05 2. a) The poossible energy macrostates are listed below. ν1 10:= ν2 14:= E1 E2 Etotal Ω1 Ω2 Ωc P(E1) ___________________________________________________________________ 2 5 7 25 32= 57 7.813 104×= 25 57⋅ 2500000= 25 57⋅ Total 0.274= 3 4 7 35 243= 47 1.638 104×= 35 47⋅ 3981312= 35 47⋅ Total 0.437= 4 3 7 45 1.024 103×= 37 2.187 103×= 45 37⋅ 2239488= 45 37⋅ Total 0.246= 5 dQldt 373.235 watts=dQldt dQhdt dWdt−:=e) dWdt 226.765 watts=dWdt ηactual dQhdt⋅:=d) ηactual 0.378=ηactual COU ηcarnot⋅:= COU 60 %⋅:= dQldt 222.058 watts=dQldt dQhdt dWdt−:=b) dWdt 377.942 watts=dWdt ηcarnot dQhdt⋅:=c) ηcarnot 0.63=ηcarnot 1 Tlow Thigh −:=a) dQhdt 600 watts⋅:=Thigh 600 oC⋅ 273.15 K⋅+:=Tlow 50 oC⋅ 273.15 K⋅+:=5. Physics 406 Final Exam 12/16/05 Physics 406 Final Exam 12/16/05 6. a) First: dE = δQ - δW + µ*dN, where dE is the energy change of the system, δQ is the heat added, δW is the work done by the system, µ is the chemical potential (per particle) and dN is the number of particles added. Second: dS > 0 for systems approaching equilibrium and dS = 0 for systems at equilibrium where S is the entropy. Third: S --> 0 at T--> 0, where S is the entropy. b) The First and Third are rigorous while the Second is only statistically true [It is possible for dS < 0 in small systems]. c) Measureable properties of a system include E, N and µ. The process dependent quantites are Q and W. d) 1) The heat capacity of all systems must appoach 0 as T --> 0. 2) The volume of all systems is finite as T --> 0. S k ln 1 1 e β− εo 2 µ⋅−( )⋅ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ β εo 2 µ⋅−( )⋅ e β− εo 2 µ⋅−( )⋅ 1 e β− εo 2 µ⋅−( )⋅ − ⋅+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ⋅:= S k ln z( ) β Eavg µ Navg⋅−( )⋅+⎡⎣ ⎤⎦⋅:=d) pavvg 0:= Since z does not depend on the volume this quantity must be zerto. pavg 1 β d dV ⋅ ln z( )⋅:=c) Eavg εo e β− εo 2 µ⋅−( )⋅ ⋅ 1 e β− εo 2 µ⋅−( )⋅ − :=Eavg µ Navg⋅ e β− εo 2 µ⋅−( )⋅ − εo 2 µ⋅−( )−⎡⎣ ⎤⎦⋅ 1 e β− εo 2 µ⋅−( )⋅ − +:= Eavg µ Navg⋅ d dβ ln 1 e β− εo 2 µ⋅−( )⋅ − ⎡ ⎣ ⎤ ⎦⋅+:=Eavg µ Navg⋅ d dβ ln 1 1 e β− εo 2 µ⋅−( )⋅ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅−:= Eavg µ Navg⋅ d dβ ln z( )⋅−:=b) Navg 2 e β− εo 2 µ⋅−( )⋅ ⋅ 1 e β− εo 2 µ⋅−( )⋅ − :=Navg 1−( ) β e β− εo 2 µ⋅−( )⋅ − 2⋅ β⋅ 1 e β− εo 2 µ⋅−( )⋅ − ⋅:= Navg 1−( ) β d dµ ⋅ 1 e β− εo 2 µ⋅−( )⋅ − ⎡ ⎣ ⎤ ⎦⋅:= Navg 1 β d dµ ⋅ ln 1 1 e β− εo 2 µ⋅−( )⋅ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅:=Navg 1 β d dµ ⋅ ln z( )⋅:=a) z 1 1 e β− εo 2 µ⋅−( )⋅ − :=7. Physics 406 Final Exam 12/16/05
