Thermodynamics and Statistical Mechanics Exam: Spring 2001, Exams of Physics

Download Thermodynamics and Statistical Mechanics Exam: Spring 2001 and more Exams Physics in PDF only on Docsity! 1 PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS SPRING 2001 FINAL EXAMINATION Tuesday, May 1, 2001 Your grade will be sent to you by e-mail by midnight, Wednesday, May 2, 2001 NAME: __________ANSWERS__________________ e-mail address:___________________ There are four pages to this examination. Check to see that you have them all. To receive credit for a problem, you must show your work, or explain how you arrived at your answer. 1. (20%) Consider a collection of N identical, distinguishable harmonic oscillators, all of frequency . The energies that one these oscillators can take on (measured relative to the ground state) are n = nh. Where n is an integer that can take on values from 0 to . a) (5%) Find the partition function , for one of these oscillators. (Hint: x x n n    0 1 1 for x < 1)                    e e e nh n h n n h 0 0 1 1 b) (5%) Find the partition function Z, for the collection of oscillators. (Since the oscillators are distinguishable, no N! is needed in the denominator.) N h N e Z           1 1 c) (5%) Find the Helmholtz function F, for this collection of oscillators. (Hint: F = – kT ln Z)      hh N h eNkTeNkTe kTZkTF         1ln1ln1ln 1 1 lnln   heNkTF  1ln 2 d) (5%) Find the entropy S, for this collection of oscillators. Hint: NVT F S ,       . This will not give a particularly neat expression. However, you can check your result as follows: as T goes to zero, S goes to zero, and as T gets very large, S goes to             h kT NkS ln1 .                                                                                       kT h kT h kT h kT h kT h NV kT h NV h NV e e kT h NkS e e kT h e TNkS eNkT T eNkT TT F S          1ln 1 1 1ln 1 1ln1ln 2 ,,, I said it was not particularly neat. As T goes to zero, kT h gets very large. The exponential in the denominator of the left term gets very large, and that term goes to zero. The exponential in the ln gets very small, and the ln 1 = 0, so S goes to zero. As T gets very large, kT h gets very small, and both exponentials can be expanded.                                                                                h kT Nk kT h NkS kT h kT hkT h Nke e kT h NkS kT h kT h ln1ln1 11ln 11 1 1ln 1 1 5 g) (6%) Calculate the work done by the gas and the heat added to the gas as the system goes from d to a. Express your answers in terms of p0 and V0. W pdVda d a  . Since T is constant, pV p V 0 0 , along the path, so p p V V  0 0 . Then, W p V dV V p V V V p V V V p Vda d a a d                  0 0 0 0 0 0 0 0 0 02 2ln ln ln Since T is constant, E = 0, so Qda = Wda = – p0V0 ln2 Wda = – p0V0 ln2 Qda = – p0V0 ln2 h) (3%) Calculate the net work done by the system as it goes around the complete cycle. Express your answer in terms of p0 and V0. W = Wab + Wbc + Wcd + Wda = p0V0 + 2p0V0 ln2 + [– p0V0] + [– p0V0 ln2] W = p0V0 ln2 i) (3%) Calculate the net heat that goes into the system as it goes around the complete cycle. Express your answer in terms of p0 and V0. Q = Qab + Qbc + Qcd + Qda = (5/2)p0V0 + 2p0V0 ln2 + [– (5/2)p0V0] + [– p0V0 ln2] Q = p0V0 ln2 j) (3%) Calculate the efficiency of this engine. (Hint: This should be a pure number. If you can not get rid of constants, be sure you explain what you are trying to calculate.)   W Qin . W is the net work done, and Qin is what the textbook calls Q1. It is the heat from the high temperature reservoir(s). In this case, Qin = Qab + Qbc = (5/2)p0V0 + 2p0V0 ln2 Then,       W Q p V p V p Vin 0 0 5 2 0 0 0 5 2 2 2 2 2 2 2 ln ln ln ln  = 0.178 = 17.8% k) (3%) What would be the efficiency of a Carnot engine that could be operated between the highest and the lowest temperatures that this system reaches. (Hint: This should be a pure number. If you can not get rid of constants, be sure you explain what you are trying to calculate.) Carnot       1 1 2 1 1 2 1 2 2 1 0 0 T T T T Carnot = 0.50 = 50% 6 l) (2%) Based on your answers to (j) and (k), which would be the better engine? Which would be the better heat pump? (Circle your choices.) Better engine: Carnot cycle Cycle of this problem Better heat pump: Carnot cycle Cycle of this problem 3. (30%) N molecules of an ideal diatomic gas are confined to a vessel of volume V that is maintained at a constant temperature, T. At the temperature T the vibrational degrees of freedom of the molecules are not active. (Each molecule has five degrees of freedom.) a) (4%) What is the internal energy of this gas? Express your answer in terms of N, T, k, and pure numbers. NkTE 2 5  b) (9%) For this gas, find the heat capacities, CV and Cp, and  = Cp/CV. Express your answers in terms of N, k, and pure numbers. (For an ideal gas Cp – CV = Nk) NkNkT dT d dT dE CV 2 5 2 5       NkNkNkNkCC Vp 2 7 2 5  5 7 2/5 2/7  V p C C  , or 5 7 5 2 1 2 1    c) (4%) The gas is irradiated with ultra-violet light, and one third (1/3) of the N molecules dissociate, each into two atoms. What is the internal energy of this mixed gas? Remember, T is held constant. (The vessel is in contact with a heat reservoir.) Express your answer in terms of N, T, k, and pure numbers. There are now N 3 2 diatomic molecules with energy NkTkTNEmol 3 5 3 2 2 5       There are also N 3 2 monatomic atoms with energy NkTkTNEat       3 2 2 3 The total energy is the sum of the two, NkTNkTE  3 5 NkTE 3 8  7 d) (9%) For this mixed gas, find the heat capacities, CV and Cp, and  = Cp/CV. Express your answers in terms of N, k, and pure numbers. As before, NkNkT dT d dT dE CV 3 8 3 8       Since there are now N 3 4 particles, NkNkNkNkNkCC Vp 43 12 3 4 3 8 3 4  2 3 8 12 3/8 3/12  V p C C  , or average  21 , and 4 2 35   average , so 2 3 2 1 1 4 2 1 2 1  average  e) (4%) Determine the ratio of the pressure of the gas after the irradiation to that before the irradiation. Your answer should be a pure number. pV = NkT. Since V and T do not change, pressure is proportional to N. Before the irradiation there are N particles, and after, there are N 3 4 particles. Therefore, 3 43 4  N N p p before after