1. Lecture 9 Overview (Ch. 1-3) Format of the first midterm: four problems with multiple questions. The Ideal Gas Law, calculation of  W,  Q and dS for various ideal gas processes. Einstein solid and two-state paramagnet, multiplicity and entropy, the stat. phys. definition of T, how to get from the multiplicity to the equation of state. The test is an open textbook exam (but no open HW solutions!). I recommend to list all essential equations – you won’t have time to read the textbook!

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. T 1 and T 2 are given. (a) (10) Calculate the work done by the gas. Is it positive or negative? (b) (20) Using two methods (Sackur-Tetrode eq. and dQ/T), calculate the entropy change for each stage and for the whole cycle,  S total. Did you get the expected result for  S total ? Explain. (c) (5) What is the heat capacity (in units R) for each stage? T V V1V1 V2V2 T1T1 T2T2 1 2 3 1 – 2 V  T  P = const (isobaric process) 2 – 3 V = const (isochoric process) 3 – 1 T = const (isothermal process) (a)

3. Problem 1 (cont.) T V V1V1 V2V2 T1T1 T2T2 1 2 3 1 – 2 V  T  P = const (isobaric process) 2 – 3 V = const (isochoric process) 3 – 1 T = const (isothermal process) as it should be for a quasistatic cyclic process (quasistatic – reversible), because S is a state function. (b) Sackur-Tetrode equation:

4. Problem 1 (cont.) T V V1V1 V2V2 T1T1 T2T2 1 2 3 1 – 2 V  T  P = const (isobaric process) 2 – 3 V = const (isochoric process) 3 – 1 T = const (isothermal process) (b) - for quasi-static processes

5. Problem 1 (cont) (c) Let’s express both  Q and dT in terms of dV : T V V1V1 V2V2 T1T1 T2T2 1 2 3 1 – 2 V  T  P = const (isobaric process) 2 – 3 V = const (isochoric process) 3 – 1 T = const (isothermal process), dT = 0 while  Q  0 At home: recall how these results would be modified for diatomic and polyatomic gases.

6. Problem 2 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

7. Problem 2 (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

8. Problem 2 (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)

9. Problem 3 Calculate the heat capacity of one mole of an ideal monatomic gas C(V) in the quasistatic process shown in the Figure. P 0 and V 0 are given. 3030 2020 we need to find the equation of this process V=V(T) 4040 P V P0P0 V0V0 0 1010 Start with the definition:

10. Problem 3 (cont.) P V P0P0 V0V0 0 T=const isotherm S=const adiabat V 0 /25V 0 /8 C/R V/ V 0 2.5 1 0 1/25/8 1.5 the line touches an isotherm the line touches an adiabat 5050 Does it make sense?

11. Problem 4 You are in possession of an Einstein solid with three oscillators and a two-state paramagnet with four spins. The magnetic field in the region of the paramagnet points “up” and is carefully tuned so that µB = , where µB is the energy of a spin pointing “down”, -µB is the energy of a spin pointing “up”, and  is the energy level separation of the oscillators. At the beginning of the experiment the energy in the Einstein solid U S is 4  and the energy in the paramagnet U P is -4 . (a) (4) Using a schematic drawing of the Einstein solid, give an example of a microstate which corresponds to the macrostate U S = 4 . (b) (4) Using a schematic drawing of the paramagnet, give an example of a microstate which corresponds to the macrostate U P = -4 . (c) (8) Considering that the “system” comprises the solid and the paramagnet, calculate the multiplicity of the system assuming that the solid and paramagnet cannot exchange energy. (d) (14) Now let the solid and paramagnet exchange energy until they come to thermal equilibrium. Note that because this system is small, there will be large fluctuations around thermal equilibrium, but let’s assume that the system is not fluctuating at the moment. What is the value of U S now? Draw an example of a microstate in which you might find the solid. What is the value of U P now? Draw an example of a microstate in which you might find the paramagnet.

12. Problem 4 (cont.) 22 E 1 = -  B B E 2 = +  B B Einstein solid Two-state paramagnet  (a) U S = 4  (b) U P = -4  E 2 = +  B B E 1 = -  B B (c) Most of the confusion came from the fact that we usually measure the energy of an oscillator in the Einstein solid from its ground state (which is 1/2  above the bottom of the potential well), whereas for the two-state paramagnet we’ve chosen the zero energy in the middle of the energy gap between “spin- up” and “spin-down” levels. The avoid confusion, consider the number of energy quanta  available for the system.

13. Problem 4 (cont.) 22 E 1 = -  B B E 2 = +  B B Example of one of the equilibrium microstates:  (d) In equilibrium, the multiplicity is maximum. The two-state paramagnet can absorb only multiples of 2 . Two options: 2  is transferred from S to P, and 4  is transferred from S to P. 2  transfer S – N = 3, q = 2, P – N = 4, N  = 1 4  transfer S – N = 3, q = 0, P – N = 4, N  = 2 Thus, the equilibrium situation corresponds to the transfer of 2  from the Einstein solid to the two-state paramagnet Note that the equilibrium condition U A /N A = U B /N B holds if both systems have only “quadratic” degrees of freedom.

14. Problem 5 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)

15. Problem 5 (cont.) (b) (c)

16. Problem 6 (10) The ESR (electron spin resonance) set-up can detect the minimum difference in the number of “spin-up” and “spin-down” electrons in a two-state paramagnet N  -N  =10 10. The paramagnetic sample is placed at 300K in an external magnetic field B = 1T. The component of the electron’s magnetic moment along B is   B =  9.3x10 -24 J/T. Find the minimum total number of electrons in the sample that is required to make this detection possible. - the high-T limit