Statistical Mechanics

Transcription

1 Statistical Mechanics

2 Newton's laws in principle tell us how anything works But in a system with many particles, the actual computations can become complicated. We will therefore be happy to get some 'average' or approximate behavior of the system that will be useful for practical purposes Describing this average dynamics is the goal of thermodynamics A microscopic level fundamental understanding of thermodynamics was found later, and this field is called statistical mechanics

3 Average energy Suppose we have probability p i to have energy E i Then the average energy is hei = P i P i E i P i P i Continuous variables: Probability for any given value is zero, but we have a probability for a range P (v) dv v v + dv v

4 v P (v) Area under curve Z P (v)dv =1 v Sometimes we may give the number of particles per unit velocity range N(v)dv particles have momentum between and v v + dv N(v) Area under curve Z N(v)dv = N total

5 GREPracticeBook 56. A sample of N molecules has the distribution of speeds shown in the figure above. P u du is an estimate of the number of molecules with speeds between u and u + du, and this number is nonzero only up to 3u 0, where u 0 is constant. Which of the following gives the value of a? N (A) a 3 u 0 N (B) a 2 u (C) a (D) a N u (E) a N N 2 u 0

6 GREPracticeBook 56. A sample of N molecules has the distribution of speeds shown in the figure above. P u du is an estimate of the number of molecules with speeds between u and u + du, and this number is nonzero only up to 3u 0, where u 0 is constant. Which of the following gives the value of a? N (A) a 3 u 0 N (B) a 2 u (C) a (D) a N u (E) a N N 2 u 0

7 Hot bath, temperature T The particle can get energy from the hot walls when it touches them Basic law of Statistical Mechanics: Probability for the particle to have energy E P / e E kt Here k = JK 1 is the Boltzmann constant

8 77. An ensemble of systems is in thermal equilibrium with a reservoir for which kt = ev. State A has an energy that is 0.1 ev above that of state B. If it is assumed the systems obey Maxwell-Boltzmann statistics and that the degeneracies of the two states are the same, then the ratio of the number of systems in state A to the number in state B is (A) e +4 (B) e (C) 1 (D) e (E) e -4 GRE0177

9 77. An ensemble of systems is in thermal equilibrium with a reservoir for which kt = ev. State A has an energy that is 0.1 ev above that of state B. If it is assumed the systems obey Maxwell-Boltzmann statistics and that the degeneracies of the two states are the same, then the ratio of the number of systems in state A to the number in state B is (A) e +4 (B) e (C) 1 (D) e (E) e -4 GRE0177

10 e E kt E = kt E For each degree of freedom, hei = 1 2 kt A state with lower energy is more likely...

11 First consider the 1-dimensional problem v P / e E kt E = 1 2 mv2 P (v) / e mv2 2kT The most likely speed is v =0

12 Hot bath, temperature T (v 2 x +v2 y ) P / e 2kT v y Many velocities ~v to the same speed v correspond v =0 v x Higher speeds are suppressed because they have more energy Higher speeds are enhanced because there are more possible velocities for higher speeds

13 57. Which of the following statements is (are) true for a Maxwell-Boltzmann description of an ideal gas of atoms in equilibrium at temperature T? I. The average velocity of the atoms is zero. II. The distribution of the speeds of the atoms has a maximum at u = 0. III. The probability of finding an atom with zero kinetic energy is zero. (A) I only (B) II only (C) I and II (D) I and III (E) II and III GREPracticeBook

14 57. Which of the following statements is (are) true for a Maxwell-Boltzmann description of an ideal gas of atoms in equilibrium at temperature T? I. The average velocity of the atoms is zero. II. The distribution of the speeds of the atoms has a maximum at u = 0. III. The probability of finding an atom with zero kinetic energy is zero. (A) I only (B) II only (C) I and II (D) I and III (E) II and III GREPracticeBook

15 For each degree of freedom, hei = 1 2 kt v hei = h 1 2 mv2 i = 1 2 kt Hot bath, temperature T hei = h 1 2 mv2 i + h 1 2 Kx2 i K = 1 2 kt + 1 kt = kt 2 m

16 Hot bath, temperature T Particle in 3-d hei = h 1 2 mv2 x mv2 y mv2 zi = 3 2 kt

17 5. A three-dimensional harmonic oscillator is in thermal equilibrium with a temperature reservoir at temperature T. The average total energy of the oscillator is GRE0177 (A) 1 2 kt (B) kt (C) 3 2 kt (D) 3kT (E) 6kT

18 5. A three-dimensional harmonic oscillator is in thermal equilibrium with a temperature reservoir at temperature T. The average total energy of the oscillator is GRE0177 (A) 1 2 kt (B) kt (C) 3 2 kt (D) 3kT (E) 6kT

19 48. A gaseous mixture of O 2 (molecular mass 32 u) and N 2 (molecular mass 28 u) is maintained at constant temperature. What is the ratio u u rms ( N ) rms( O 2 2) of the root-mean-square speeds of the molecules? GRE0177 (A) 7 8 (B) (C) (D) FH 8 I K 7 2 FH (E) ln 8I K 7

20 48. A gaseous mixture of O 2 (molecular mass 32 u) and N 2 (molecular mass 28 u) is maintained at constant temperature. What is the ratio u u rms ( N ) rms( O 2 2) of the root-mean-square speeds of the molecules? GRE0177 (A) 7 8 (B) (C) (D) FH 8 I K 7 2 FH (E) ln 8I K 7

21 Avagadro number N = (1 mole) Avagadro number of H atoms weigh 1 gram Avagadro number of He atoms weigh 4 grams Molar mass of H = 1gm = mass of H atom times N Molar mass of He = 4 gm = mass of He atom times N We define kn = R R =8.31 JK 1 /mole One degree of freedom hei = 1 2 kt N degrees of freedom hei = 1 2 NkT = 1 2 RT

22 9. The root-mean-square speed of molecules in an ideal gas of molar mass M at temperature T is (A) 0 GREPracticeBook (B) (C) (D) (E) RT M RT M 3RT M 3RT M

23 9. The root-mean-square speed of molecules in an ideal gas of molar mass M at temperature T is (A) 0 GREPracticeBook (B) (C) (D) (E) RT M RT M 3RT M 3RT M

24 Blackbody radiation is made of photons p y P e E kt p x p =0 Wien displacement constant W = mk W peak = T peak Wien displacement law: Double T double peak E halve wavelength

25 GRE The distribution of relative intensity I ( l ) of blackbody radiation from a solid object versus the wavelength l is shown in the figure above. If the Wien displacement law constant is m K, what is the approximate temperature of the object? (A) 10 K (B) 50 K (C) 250 K (D) 1,500 K (E) 6,250 K

26 GRE The distribution of relative intensity I ( l ) of blackbody radiation from a solid object versus the wavelength l is shown in the figure above. If the Wien displacement law constant is m K, what is the approximate temperature of the object? (A) 10 K (B) 50 K (C) 250 K (D) 1,500 K (E) 6,250 K

27 Approximations e E kt E = kt E For x 1 e x 1 x +... Low energy e E kt 1 E kt

28 GRE0177 C 3kN hv A kt hv / kt F I = H K hv / kt - 2 (e e 1) Einstein s formula for the molar heat capacity C of solids is given above. At high temperatures, C approaches which of the following? (A) 0 (B) 3kN A F H (C) 3kN hv A (D) 3kN A (E) Nhv A hv kt I K

29 GRE0177 C 3kN hv A kt hv / kt F I = H K hv / kt - 2 (e e 1) Einstein s formula for the molar heat capacity C of solids is given above. At high temperatures, C approaches which of the following? (A) 0 (B) 3kN A F H (C) 3kN hv A (D) 3kN A (E) Nhv A hv kt I K

30 The partition function

31 The average energy is hei = P i P i E i P i P i with P i / e E i kt To compute this, we define the Partition Function Z = X i e E i kt We write 1 kt = This gives Z = X i e = X i e E i E i = P i e E i E i Pi e E i = hei

32 98. Suppose that a system in quantum state i has energy E i. In thermal equilibrium, the expression GRE0177  i  i Ee i e - E - E i i / kt / kt represents which of the following? (A) The average energy of the system (B) The partition function (C) Unity (D) The probability to find the system with energy E i (E) The entropy of the system

33 98. Suppose that a system in quantum state i has energy E i. In thermal equilibrium, the expression GRE0177  i  i Ee i e - E - E i i / kt / kt represents which of the following? (A) The average energy of the system (B) The partition function (C) Unity (D) The probability to find the system with energy E i (E) The entropy of the system

34 49. In a Maxwell-Boltzmann system with two states of energies and 2, respectively, and a degeneracy of 2 for each state, the partition function is GRE0177 (A) e- /kt (B) 2e-2 /kt (C) 2e-3 /kt (D) e - /kt + e-2 /kt (E) 2[e - /kt + e -2 /kt ]

35 49. In a Maxwell-Boltzmann system with two states of energies and 2, respectively, and a degeneracy of 2 for each state, the partition function is GRE0177 (A) e- /kt (B) 2e-2 /kt (C) 2e-3 /kt (D) e - /kt + e-2 /kt (E) 2[e - /kt + e -2 /kt ]

36 An unusual situation

37 76. The mean kinetic energy of the conduction electrons in metals is ordinarily much higher than kt because (A) electrons have many more degrees of freedom than atoms do (B) the electrons and the lattice are not in thermal equilibrium (C) the electrons form a degenerate Fermi gas (D) electrons in metals are highly relativistic (E) electrons interact strongly with phonons GRE0177

38 76. The mean kinetic energy of the conduction electrons in metals is ordinarily much higher than kt because (A) electrons have many more degrees of freedom than atoms do (B) the electrons and the lattice are not in thermal equilibrium (C) the electrons form a degenerate Fermi gas (D) electrons in metals are highly relativistic (E) electrons interact strongly with phonons p y GRE0177 E kt p x p =0

39 Deriving hei = 1 2 kt

40 First consider the 1-dimensional problem v P e E kt E = 1 2 mv2 P (v) / e mv2 2kT Adding over all possibilities : Z 1 1 dv

41 First consider the 1-dimensional problem v hei = P i p i E i P i p i p / e E kt E = 1 2 mv2 p(v) / e mv2 2kT hei = R 1 1 dv( 1 2 mv2 ) e mv2 2kT R 1 1 mv2 dv e 2kT = 1 2 kt

42 Hot bath, temperature T K p / e E kt m E = 1 2 mv Kx2 p / e mv2 2kT e Kx 2 2kT Adding over all possibilities of velocity : Adding over all possibilities of position : Z 1 1 Z 1 1 dv dx

43 Hot bath, temperature T K p / e E kt m E = 1 2 mv Kx2 p / e mv2 2kT e Kx 2 2kT Adding over all probabilities Z Z dvdx e mv2 2kT e Kx 2 2kT = Z dv e mv2 2kT Z dx e Kx2 2kT

44 Hot bath, temperature T K m p / e mv2 2kT e Kx 2 2kT E = 1 2 mv Kx2 h 1 2 mv2 i = R dv dx ( 1 2 mv2 ) e mv R dv dx e mv 2 2kT 2 2kT e Kx2 2kT e Kx2 2kT = R dv ( 1 2 mv2 ) e mv R dv e mv 2 2kT 2 2kT R dx e Kx 2 2kT R dx e Kx 2 2kT = 1 2 kt

45 Hot bath, temperature T K m p / e mv2 2kT e Kx 2 2kT E = 1 2 mv Kx2 h 1 2 mv2 i + h 1 2 Kx2 i = 1 2 kt kt = kt