Problem #1 30 points Problem #2 30 points Problem #3 30 points Problem #4 30 points Problem #5 30 points

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Name ME 5 Exam # November 5, 7 Prof. Lucht ME 55. POINT DISTRIBUTION Problem # 3 points Problem # 3 points Problem #3 3 points Problem #4 3 points Problem #5 3 points. EXAM INSTRUCTIONS You must do four (4) out of the five (5) problems and indicate clearly which problem you decided not to have graded. Please write DO NOT GRADE in the space next to the problem that you decide not to do. 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 5 Final Exam //7 Name. (3 points) The time-independent Schrödinger wave equation for the harmonic oscillator is d V( x) ( x) ( x) dx where V( x) k x and x r re. The internuclear spacing is r, and the equilibrium internuclear spacing is r e. The normalized eigenfunction and energy eigenvalue for the ground state (v = ) are given by a ( x) exp ax, a, v h h, k Calculate the expectation values in Joules for the potential energy V( x ) and for the kinetic energy p x, for N in its ground vibrational state. For nitrogen, the mass of each nucleus is 4 amu and the vibrational frequency c = 33 cm -. Hint: you may find it useful to express the spring constant k in terms of the parameter a. Also note: /4 Answers: bx dx for b b exp 35... (n ) exp,,..., b b n x bx dx for n b n n n! exp,,,..., b n x bx dx for n b n V x p x.6 J

3 ME 5 Final Exam //7 Name c h K, 5 K, K, 5K,. (3 points) The system shown below has energy levels of j k B and K. The degeneracy of each of three lower levels is 8 and the two upper levels is. The system has only these energy levels; there are no translational, rotational, or vibrational modes. The number of particles in the assembly is 5. (a) Calculate the energy of the assembly at temperatures of 5, 5, and 5 K, assuming corrected Maxwell-Boltzmann statistics. Why is CMB statistics a good assumption? (b) Estimate the heat capacity of the system at 5 K and at 37.5 K by considering the results of part (a). Comment on the results. (c) Calculate the assembly entropy S (in J/K) at temperatures of 5, 5, and 5 K. Assume corrected-maxwell-boltzmann statistics. Comment on the results. j j /k B (K) g j 4 3 5 8 5 8 8 6 7 Answers (a) E5 K.64 J (b) S 5 K.73 J / K

4 ME 5 Final Exam //7 Name 3. (3 points) (a) Given the translational energy distribution for a Maxwell-Boltzmann gas / f d exp / k T d show that the speed distribution is given by tr tr tr / 3/ tr B tr kt B (b) 3/ m f VdV V exp mv / k B TdV kt B Calculate the value of the mean reciprocal speed for an assembly of nitrogen (N ) V molecules at a temperature of K and a pressure of bars. For N, the mass of each nucleus is 4 amu. You will need to use the definite integral formulae from Problem #. Answer: (b) V.46 3 s m

5 ME 5 Final Exam //7 Name 4. (3 points) A rigid pressure vessel with a volume of m 3 contains. kmols of diatomic oxygen (O ) and. kmols of diatomic nitrogen (N ) at 3K. The gas is then heated to NN NNO 3 K. At 3 K, find the equilibrium ratio for the reaction. Consider the N N reaction O N O N NO N Assume that the molecules are rigid rotators and harmonic oscillators, and use the following data: Species rot (K) vib (K) (D /hc) (cm - ) N.998 358 78,7 NO.74 94 5,3 For the calculation of the electronic partition functions, consider the following electronic levels: O: Level [g =5, /hc = cm - ], Level [g =3, ( /hc) = 58 cm - ], Level [g =, ( /hc) = 6 cm - ] N : Level [g =, ( /hc) = cm - ] NO: Level [g =, ( /hc) = cm - ], Level [g =, ( /hc) = cm - ] N: Level [g =, ( /hc) = cm - ] Answer: NN N N N O NO N 7. 6

6 ME 5 Final Exam //7 Name 5. (3 points) The temperature of a. cm 3 copper metal crystal is raised from T = 4 K to T = 5 K. Calculate the internal energy change of the crystal U U due to lattice vibrations U U and due to the free electron gas U U. Assume that the vib elec temperature is high enough that the Einstein model is accurate for determining the internal energy due to lattice vibrations. The Einstein temperature of the copper crystal is 65 K. The density of the copper is 9 kg/m 3, the atomic weight of the metal is copper is 63.5 kg/kmol, and there is one free electron per lattice site. Answers: U U 347 J vib U U 3. J elec E