Exam2SolnME501F2011 - 1 Name_ ME 501 Exam #2 November 16,...

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1 Name___________________________ November 16, 2011 Prof. Lucht ME 256 1. POINT DISTRIBUTION Problem #1 40 points _________________ Problem #2 20 points _________________ Problem #3 40 points _________________ 2. EXAM INSTRUCTIONS Write your name on each sheet. This exam is closed book and closed notes. Several equation sheets are attached. When working the problems, list all assumptions, and begin with the basic equations. If you do not have time to complete evaluation of integrals or of terms numerically, remember that the significant credit on each problem will be given for setting up the problem correctly and/or obtaining the correct analytical solution. ME 501 Exam #2
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2 ME 501 Exam #2 11/16/2011 Name___________________________ 1. (40 points) One (1.0) kmol of gaseous CO is contained in a rigid tank at pressure of 10 4 Pa (0.1 bars) at a temperature of 200 K (state 1). The temperature of the tank is raised to 2000 K (state 2) by heat transfer. Assume that CO exhibits ideal gas behavior at both states 1 and 2, and that the CO does not dissociate. Recall that CO has a 1 + ground electronic level. (a) Assuming that CO is a rigid-rotator and harmonic oscillator, calculate the internal energy change U 2 U 1 (in J ) for the constant volume heating process. (b) Calculate the entropy change S 2 S 1 (in J/K) for the constant volume heating process. (c) Calculate the number of CO molecules in the v1 , 1 0 J level at state 2 (T 2 =2000 K) in the rigid tank. Note: we are after the number of molecules in this vibration-rotation level regardless of translational state. Thus you do not need to consider the translational mode or calculate the translational partition function for part (c). For CO: rot (K) vib (K) m (amu) 2.78 3120 28.01 Solution: (a)      2121 21 2 1 2 1 1 2 01 33 2000 200 2700 2 22 2000 200 1800 2 3120 exp 1 3120 exp 3120 20 tr rot vib elec elec BB B tr B rot Bv i b vib vib vib vib B UU EE U U Nk T T Nk Nk K U U Nk T T Nk Nk K Nk UK T K UN k    4 2 26 23 7 5.238 10 2 3120 830.0 2 exp 3120 2000 1 830.0 5330 6.022 10 1.381 10 / 5330 4.433 10 3 B vib B B B vib B Nk K K k N k K N k K N k K J K K J 
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3 (b)     21 3/2 2 22 1 2 12 1 1 2 01 ln 1 ln ln 1 2 3 ln ln tr rot vib elec elec tr tr tr tr B B tr B tr tr tr tr BB tr tr SS ZE E SN k N k Z N NT T mk T Z h E T N k N k ZT T T                 2 1 2 1 1 3 33 ln ln 10 3.454 4 ln ln ln 10 2.303 3 1 3120 1e x p 1 1.000 1 exp 3120 200 B tr int int B int rot rot rot B B rot B rot vib vib vib vib Nk Nk T S S Nk Nk Nk T E k Z T E T N k TN k T T S S Nk Nk Nk T ZK T Z   1 4 11 4 1 121 1 1.266 1 exp 3120 2000 5.238 10 830.0 830.0 5.238 10 ln ln 1.266 2000 200 0.6509 4 3.454 2.30 vib vib vib B vib vib B vib vib vib vib vib B vib B Z EU N k K N k K E S S Nk Nk T N k SSN k 26 23 4 3 0.6509 6.022 10 1.381 10 / 6.408 5.329 10 / 3 JK J K 
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4 (c)  exp / v, 1 1 v= 10 11 1 121 121 2.78
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This note was uploaded on 12/26/2011 for the course ME 501 taught by Professor Na during the Fall '10 term at Purdue University-West Lafayette.

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Exam2SolnME501F2011 - 1 Name_ ME 501 Exam #2 November 16,...

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