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Results of Midterm 1 0 102030405060708090 points # of students GradePoints A> 85 B+B+ 80-85 B60-79 C+C+ 55-59 C30-54 D< 30 100.

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Presentation on theme: "Results of Midterm 1 0 102030405060708090 points # of students GradePoints A> 85 B+B+ 80-85 B60-79 C+C+ 55-59 C30-54 D< 30 100."— Presentation transcript:

1 Results of Midterm 1 0 102030405060708090 points # of students GradePoints A> 85 B+B+ 80-85 B60-79 C+C+ 55-59 C30-54 D< 30 100

2 Problem 1 One mole of a monatomic ideal gas goes through a quasistatic three-stage cycle (1-2, 2-3, 3-1) shown in the Figure. Process 3-1 is adiabatic; P 1, V 1, and V 2 are given. (a) (10) For each stage and for the whole cycle, express the work  W done on the gas in terms of P 1, V 1, and V 2. Comment on the sign of  W. (b) (5) What is the heat capacity (in units R) for each stage? (c) (15) Calculate  Q transferred to the gas in the cycle; the same for the reverse cycle; what would be the result if  Q were an exact differential? (d) (15) Using the Sackur-Tetrode equation, calculate the entropy change for each stage and for the whole cycle,  S total. Did you get the expected result for  S total ? Explain. 1 – 2 2 – 3 P 1 2 3 V V1V1 V2V2 P1P1 P = const (isobaric process) V = const (isochoric process) (a) 3 – 1 adiabatic process

3 Problem 1 (cont.) 1 – 2 P = const (isobaric process) 2 – 3 V = const (isochoric process) 3 – 1  Q = 0 (adiabatic process),  Q = 0 while dT  0 P 1 2 3 V V1V1 V2V2 P1P1 (b)

4 Problem 1 (cont.) 3 – 1 adiabatic process (c) 1 – 2 2 – 3 P = const (isobaric process) V = const (isochoric process) For the reverse cycle: If  Q were an exact differential, for a cycle  Q should be zero. P 1 2 3 V V1V1 V2V2 P1P1

5 Problem 1 (cont.) 1 – 2 V  T  P = const (isobaric process) 2 – 3 V = const (isochoric process) 3 – 1 as it should be for a quasistatic cyclic process (quasistatic – reversible), because S is a state function. Sackur-Tetrode equation: P 1 2 3 V V1V1 V2V2 P1P1  Q = 0 (quasistatic adiabatic = isentropic process) (d)

6 Problem 2 Consider a system whose multiplicity is described by the equation: (a) (10) Find the system’s entropy and temperature as functions of U. Are these results in agreement with the equipartition theorem? Does the expression for the entropy makes sense when T  0? (b) (5) Find the heat capacity of the system at fixed volume. (c) (15) Assume that the system is divided into two sub-systems, A and B; sub-system A holds energy U A and volume V A, while the sub-system B holds U B =U-U A and V B =V-V A. Show that for an equilibrium macropartition, the energy per molecule is the same for both sub-systems. - in agreement with the equipartition theorem When T  0, U  0, and S  -  - doesn’t make sense. This means that the expression for  holds in the “classical” limit of high temperatures, it should be modified at low T. where U is the internal energy, V is the volume, N is the number of particles in the system, Nf is the total number of degrees of freedom, f(N) is some function of N. (a)

7 Problem 2 (cont.) (b) (c)

8 Problem 3 Consider two Einstein “solids”: solid A consists of N A =2 oscillators, solid B consists of N B =4 oscillators. Being in thermal contact, these solids share 6 units of energy. (U A +U B = 6  ). (a)(20) Construct a table showing U A, U B,  A,  B, and  AB for all possible macropartitions of the system and compute the probabilities for each possible macropartition (P AB =  AB /total # of microstates). (b)(5) What is the most probable macropartition of this system? How does the energy stored per oscillator in each solid compare in this macropartition? How does this compare with the energies stored per oscillator for two solids with large N A and N B in thermal equilibrium with each other? qAqA qBqB AA BB  AB P AB 06184 0.182 152561120.242 243351050.227 33420800.173 42510500.108 5164240.052 607170.015 In equilibrium: q A / N A =0.5; q B / N B =1.25. For large N: q A / N A = q B / N B in equilibrium.


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