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Unformatted text preview: Quiz 4 Solutions, Physics 176 March 28, 2011 The following solutions are far more detailed than what was needed to get full credit. But make the time to read through the details, it will help you understand physics and improve your problem solving abilities. Problems That Require Writing 1. (15 points) Consider a small system whose volume and energy can not change. This system is in thermodynamic equilibrium with a reservoir in such a way that the system can exchange molecules of a certain type with the reservoir. The number N s of molecules in the system (of the given type) therefore varies with the state s of the system. The reservoir has a fixed temperature T and a fixed chemical potential μ , and the system and reservoir together form a closed system so that the total number N of molecules of the given type is conserved. Stating clearly your assumptions and approximations, derive an expression for the probability p s for the small system to be in a particular state s that corresponds to the system having N s particles. Your expression will look like an “exponential of something” over a quantity Z that is similar to a partition function. Answer: You should have obtained the answer p s = e βμN s ∑ s e βμN s , (1) for the probability p s to observe the small system in a state s with N s particles. To see this, observe that, given that the reservoir and small system form an isolated system, all microstates are equally likely and so the probability p s will be proportional to the number of microstates available to the entire isolated system when the system has N s particles and the reservoir has N N s particles where N s N . But since we are asking for the probability for the system to be in a particular state s , all the available microstates come from the reservoir, i.e., we can assume p s = c Ω R ( N N s ) , (2) where c is a constant of proportionality that will be determined by the condition that all the proba bilities have to sum to one. We can then follow step by step the discussion I gave in lecture (and in my lecture notes of March 15). The mathematics looks like this p s = c Ω R ( N N s ) (3) = c exp 1 k S R ( N N s ) (4) ≈ c exp 1 k S R ( N ) + S R ( N ) × ( N s ) + h . o . t . s (5) ≈ h ce S R ( N ) /k i exp 1 k S R ( N ) N s (6) ≈ ˜ c exp 1 k μ T N s (7) ≈ ˜ ce βμN s . (8) with the following justifications: 1 (a) I expressed the multiplicity Ω R in terms of the reservoir’s entropy S R via S R = k ln Ω R or Ω R = e S R /k . The reason is that the entropy is what is more important physically since S , but not Ω, appears in the thermodynamic identity and it is S that can be computed from a knowledge of the heat capacity C ( T )....
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This note was uploaded on 10/20/2011 for the course PHYSICS 176 taught by Professor Behringer during the Spring '08 term at Duke.
 Spring '08
 Behringer
 Physics

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