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Thermodynamics and Statistical Mechanics Exam - Spring 2000, Exams of Physics

Final examination document for phys-4420 thermodynamics & statistical mechanics course held in spring 2000. Contains problems related to probability distributions, coffee temperature, paramagnetic salts, and carnot engines.

Typology: Exams

Pre 2010

Uploaded on 08/09/2009

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Download Thermodynamics and Statistical Mechanics Exam - Spring 2000 and more Exams Physics in PDF only on Docsity! PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS SPRING 2000 FINAL EXAMINATION Thursday, May 4, 2000 Your grade will be sent to you by e-mail by 5:00 p.m., Friday, May 5, 2000 NAME: _____SOLUTIONS_____________________ 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. (15%) In the card game Bridge, it is important to know as much as possible about the distribution of cards in your opponents hands. The game is played by four players formed into two teams of two each. An example that is often given in books on the subject is the following. Suppose you know that your opponents team has only four cards of the spade suit in their possession, divided between their two hands. Then the most probable distribution of those four cards between the two hands is the one that has three cards in one hand and only one in the other. The even split, of two cards in each hand, is less probable. Calculate the probabilities of finding the cards split 3 and 1, and also the probability of finding them split 2 and 2, to verify the books’ assertions. Calculate the probability that one opponents hand will contain 0, 1, 2, or 3 spades. 16 1 2 1 2 1 !4)!44( !4 )4( 4 1 16 4 2 1 2 1 !3)!34( !4 )3( 8 3 16 6 2 1 2 1 !2)!24( !4 )2( 4 1 16 4 2 1 2 1 !1)!14( !4 )1( 16 1 2 1 2 1 !0!4 !4 )0( !)!( ! )( 04 13 22 31 40                                                                         P P P P Pqp nnN N nP nNn To find the probability of a 3-1 split, add the probability that a player has 3 spades to the probability that the same player has 1 spade. Either results in a 3-1 split. 2 1 )3()1()1,3(  PPP , while 8 3 )2()2,2( PP , so the books are correct. 1 2. (10%) A student who drinks coffee with cream in it, gets a cup of coffee, which is very hot, and a small container of cream which is much colder. The student wants the coffee to be as hot as possible when it is drunk. The student has to carry the coffee to another room before drinking it, a trip that could take two or three minutes. In order to have the coffee as hot as possible when it is drunk, which of the following strategies should the student follow? A. Carry the hot coffee and cool cream to the other room before mixing, and then add the cream to the coffee. B. Add the cream to the coffee at once, and carry the mixture of coffee and cream to the other room. Circle your choice, and briefly explain why you made that choice. (Hint: Assume that radiation is the primary mechanism of cooling.) By adding the cream at once, the temperature is lowered at the start, and the amount of heat radiated will be less than it would have been if the coffee had been left hot. 3. (20%) For this problem, you will consider a paramagnetic salt with a fixed number of atoms (i.e. N is constant, so dN = 0 throughout). When a magnetic field, B, is applied to the salt, it becomes polarized, and develops a magnetic moment, M. When the field is changed, the magnetic moment changes, and the magnetic energy changes. The work done when the field is changed can be shown to be, dW = MdB. a) The Helmholtz function is defined as F = E – TS. For a paramagnetic salt, where the only work done is magnetic, show that, dF = – MdB – SdT. (Hint: What form does the First Law assume with only magnetic work?) dF = dE – TdS – SdT, and the first law is dE = TdS – dW = TdS – MdB. Then, dF = TdS – MdB – TdS – SdT = – MdB – SdT b) Use the result of part (a) to show that, M B F T      . Fince F is a function of B and T, we can write dTT F dB B F dF BT            . A comparison of the multiplier of dB in this equation with that of part (a) shows that, M B F T      . 2
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