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Physics 4230 Final Exam: Statistical Mechanics and Thermodynamics, Exams of Physics

Material Type: Exam; Class: Thermodynamics and Statistical Mechanics; Subject: Physics; University: University of Colorado - Boulder; Term: Fall 2008;

Typology: Exams

2019/2020

Uploaded on 11/25/2020

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Physics 4230, Fall 2008, Final Exam
Saturday 12/13/08
You have 2 1/2 hours to finish this exam. You may use calculators, and two 8 1
2x 11 sheets of paper, with your
own notes written on both sides.
Express all answers in terms of given quantities.
Explain your reasoning clearly and concisely for full credit.
Good luck!
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Physics 4230, Fall 2008, Final Exam Saturday 12/13/

You have 2 1/2 hours to finish this exam. You may use calculators, and two 8 21 x 11 sheets of paper, with your own notes written on both sides.

Express all answers in terms of given quantities.

Explain your reasoning clearly and concisely for full credit.

Good luck!

  1. (a) (7 pts) What are macrostates and microstates, and how are they related to one another?

(b) (6 pts) Clearly state the fundamental assumption of statistical mechanics. (c) (6 pts) Consider a system in equilibrium with a reservoir at temperature T. The system and reservoir can exchange energy. In this case the probability for the system to be in a state s is given by the Boltzmann distribution, Ps = (1/Z)e−Es/kT^ , where Es is the energy of the state. This tells us that the system has a higher probability to be in states with lower energy. Explain how this is consistent with the fundamental assumption of statistical mechanics. (d) (6 pts) Now, let’s think about an example that illustrates how the second law of thermodynamics follows from the fundamental assumption of statistical mechanics. Consider two 2-state paramagnets each contain- ing the same number of dipoles – call these systems 1 and 2. Initially system 1 has all its dipoles pointing up (maximum positive M ), and system 2 has all its dipoles pointing down (maximum negative M ). Then, the two systems are brought into contact, so that they can exchange magnetization M. Describe what happens to the entropy of system 1 (S 1 ) and of system 2 (S 2 ) when the systems are brought into contact. Explain how this behavior follows from the fundamental assumption of statistical mechanics.

  1. Hint: In each of the following two questions, rather than starting from scratch, think about whether we have already done the same counting problem, but in a different setting. Remember, you know how to count states in the two-state paramagnet, in the Einstein solid, for a deck of cards, ....

(a) (13 pts) Suppose we have N energy levels and k identical fermions. These fermions don’t have spin, so only one of them can go into each energy level. How many distinct ways are there to fill the N energy levels with the k fermions? Explain the reasoning behind your answer. (b) (12 pts) Now suppose we have N energy levels and k identical bosons. How many ways are there to fill the N energy levels with the k bosons? Explain the reasoning behind your answer.

Thermally insulated container

Gas

Piston

Constant pressure environment

FIG. 1: Illustration for part (b). The gas is maintained at the same pressure as the constant pressure environment.

  1. (a) (10 pts) Consider two containers of an ideal gas with f degrees of freedom. Both containers hold the same kind of gas (for example, if one holds nitrogen, then so does the other one). Both have the same pressure P and number of molecules N. Container 1 has temperature T 1 and volume V 1 = N kT 1 /P , and container 2 has temperature T 2 and volume V 2 = N kT 2 /P. We are interested in knowing the difference in entropy between these two gases, ∆S = S 2 − S 1. In part (b), we will take the gas in container 1 and consider a quasistatic process. The process of course starts with the gas having volume V 1 , pressure P , temperature T 1 , and number of molecules N. At the end of the process, the gas has volume V 2 , pressure P , temperature T 2 , and number of molecules N – just like container 2. Explain why the entropy change during this process is equal to ∆S = S 2 − S 1. (b) (15 pts) Consider an ideal gas with f degrees of freedom that is thermally isolated from its surroundings. The gas is held at constant pressure P by a piston, which separates the gas from a constant-P environment, as shown in the figure. The gas has a constant number of molecules N. Thermal isolation means that no heat can flow through the piston or through the walls of the container. Initially the gas has temperature T 1 and volume V 1 = N kT 1 /P. Suppose that heat is quasistatically added to the gas (maybe by turning on a heater inside the container), so that the temperature is raised to T 2 (and thus the volume changes to V 2 = N kT 2 /P ). The pressure of the gas is always P , throughout the process. What is the change in entropy during this process? Do not use the Sackur-Tetrode equation, since this is not a monatomic ideal gas. Hint: Think about the change in enthalpy. Hint 2 : You will have to do an integral (not a hard one) to get the final answer.

T

P

Phase 1

Phase 2

FIG. 2: Phase diagram for problem 4. There is a phase transformation between phase 1 and phase 2 at the curved line in the P - T plane. Please note that the particular shape of the line is not important for either part of the problem – the questions asked are very general and apply to any phase transformation.

  1. (a) (10 pts) Consider a homogeneous substance that undergoes a phase transformation along some line in the pressure-temperature plane, as sketched in the figure. This could be, for example, a phase transition between liquid and solid phases of this substance. As we move across the phase transformation (from phase 1 into phase 2), the Gibbs free energy G is continuous. However, the volume V and entropy S do not have to be continuous. Explain why this is so. That is, explain both why G must be continuous, and why V and S don’t have to be. (b) (15 pts) The Clausius-Clayperon relation tells us about the slope of the phase transformation line. It says

dP dT

∆S

∆V

where ∆S = S 2 − S 1 and ∆V = V 2 − V 1 are the discontinuous changes in entropy and volume as one crosses the line. The line is the graph of a function P (T ), which simply tells us, for a given temperature, what pressure we need to have in order to be at the phase boundary. dP/dT is the derivative of this function. Derive the Clausius-Clayperon relation.

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