Written 4-hour exam in Thermodynamics and Statistical ... - dirac


Written 4-hour exam in Thermodynamics and Statistical ... - dirac

1Written 4-hour exam inThermodynamics and Statistical MechanicsThursday 10/01-12, 10.00-14.00The exam consists of three problems on two pages altogether (+ this front page). The weights of theproblems are given, and within a problem sub-questions have the same weight.Writing aids and a simple calculator (that is one which can not do graphs or symbolic manipulations)are allowed. The “Golden sheet” (3 pages) are included with this exam. No further helping aids areallowed.

3Problem 1 (40 %)In the following we consider a system with 6 places and 2 particles (each place can maximal beoccupied by one particle). The energy of the system depends on which places are occupied, in thefollowing way: 1 place exists which contributes 0ǫ if occupied, 2 places exists which contributes 1ǫ ifoccupied, and 3 places exists which contributes 2ǫ if occupied (ǫ > 0). Figure 1 shows three examplesof configurations and the corresponding total energy.The system is assumed to be in equilibrium, and in good thermal contact with a heat bath at a giventemperature, T.2ǫ1ǫ 0ǫ 2ǫ 1ǫ 0ǫ 2ǫ 1ǫ 0ǫE = 4ǫ E = 1ǫ E = 1ǫFigure 1: Three examples of configurations with corresponding energy. The 6 ringsillustrates the possible places and the filled circles the two particles.a) State the partition function for the system, furthermore show that there are 15 configurations intotal.b) Calculate the mean energy for the system as function of temperature. Furthermore, calculate thehigh and low temperature limits of the mean energy.c) Calculate the entropy of the system as function of temperature. Furthermore, calculate the highand low temperature limits of the entropy.The system is now changed in such a way that 2 places exists which each contributes with 0ǫ ifoccupied, 3 places exists which each contributes 1ǫ if occupied, and 4 places exists which eachcontributes 2ǫ if occupied (total of 9 places).d) For the changed system, state (without explicit calculation) the high and low temperature limitof the entropy, and low temperature limit of the mean energy.Problem 2 (40 %)a) Show that the following general relation holdsIn the following a mono atomic ideal gas is considered.b) Calculate κ S and κ T for a mono-atomic ideal gas.κ S = κ T − TVα2 PC P. (1)Consider a system as sketched on figure 2. The system consists of an outer isolating box, with anisolating piston. The box is partitioned by an inner wall which is a good heat conductor. The boxis filled with mono-atomic ideal gas. The system is considered to be in thermal equilibrium.

4V 2 ,N 2 V 1 ,N 1P 2 ,T 2 P 1 ,T 1S 2 S 1Figure 2: Illustration of the system for problem 2.We now define the “externally seen” (that is in volume 1) adiabatic compressibility as (where thesecond equality sign follows from the “fake chain rule”):( )κ S,1 = − 1 ( ) ∂V1= 1 ∂S total∂P 1 V( ) 1(2)V 1 ∂P 1 S totalV 1 ∂S total∂V 1 P 1where S total = S 1 +S 2 .In the following it is used that the entropy of an mono-atomic ideal gas can be written as:[ ]S = Nk B ln V (k BT) 3/2+5/2 . (3)Nc) Calculate κ S,1 .Hint: First show that the total entropy of the system can be written as:(S total = N 1 k B ln V ( ) (3/21 P1 V 1+5/2)+N 2 k B ln V ( ) ) 3/22 P1 V 1+5/2N 1 N 1 N 2 N 1. (4)d) Calculate κ S,1 in the limits N 1 ≫ N 2 and N 1 ≪ N 2 . Compare the results with the results fromsub-question b).Opgave 3 (20 %)Consider a system consisting of an isolated container which is separated into two compartments(each with volume V 1 and V 2 ) by a heat conducting wall. The two compartments are filled withmono-atomic ideal gas (N 1 and N 2 particles).For an mono-atomic ideal gas, entropy can be written as:[ ]S = Nk B ln V (k BT) 3/2+5/2 . (5)Na) Discuss the fundamental criterion for equilibrium, and show explicitly thatT 1 = T 2 in equilibrium.Volume 2 is now considered as a heat bath, and the two compartments are assumed to have the sametempe-rature.b) Show that the change in total entropy, for the total system, coming from a change in an innerdegree of freedom in sub-volume 1, can be found as:dS total = − 1 T dF 1 (6)where F 1 is Helmholts free energy of sub-volume 1.End of the exam.

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