Phys-4420 Thermodynamics & Statistical Mechanics Spring 2006

PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS SPRING 2006

Preview Quiz 2 Tuesday, April 18, 2006

NAME: ______ANSWERS______

To receive credit for a problem, you must show your work, or explain how you arrived at your answer.

1. (25%) A small time gambler is running a game on a street corner. He tosses five coins at once, and bettors can bet on how many heads will appear. Anyone who bets on zero or five, and wins, is paid $30 for each dollar bet. Those who win with one or four are paid $6 for each dollar bet, and they get $3 for each dollar bet if they win with two or three heads. a) (10%) Find the probabilities that 0, 1, 2, 3, 4, and 5 heads will appear when the coins are tossed.

N! The probability of getting n heads out of N coins is given by: P(n)  p n q N n . n!(N  n)! 0 50 5 5!  1   1  5!  1  1 Then, P(0)           0.03125 0!(5  0)! 2   2  0!(5)! 2  32 1 51 5 5!  1   1  5!  1  5 P(1)           0.15625 1!(5 1)! 2   2  1!(4)! 2  32 2 52 5 5!  1   1  5!  1  10 P(2)           0.3125 2!(5  2)! 2   2  2!(3)! 2  32 3 53 5 5!  1   1  5!  1  10 P(3)           0.3125 3!(5  3)! 2   2  3!(2)! 2  32 4 54 5 5!  1   1  5!  1  5 P(4)           0.15625 4!(5  4)! 2   2  4!(1)! 2  32 5 55 5 5!  1   1  5!  1  1 P(5)           0.03125 5!(5  5)! 2   2  5!(0)! 2  32

b) (5%) Suppose someone played the game, and always bet that three heads would show up. Based on the probability that you calculated, and the payoff offered for winning, how would this person do financially, after playing for a long time, and making many bets? Circle the correct answer, and support your choice with some numerical evidence. WIN LOSE BREAK EVEN

If the player wins 0.3125 of the time, and gets back $3 each time, then the average return is $3 × 0.3125 = $0.9375 for each $1 bet. After 100 plays, and $100 bet, the player would have on average $93.75 left. In this game, that really means $93 or $94.

1 c) (5%) A street-smart character plans to improve his odds by switching the coins used in the game. He has a set of five coins, each of which comes up heads 60% (0.60) of the time. What is the probability that three heads will come up when these five loaded coins are tossed?

3 53 3 2 5!  3   2  5!  3   2  216 P(3)             0.3456 3!(5  3)! 5   5  3!(2)! 5   5  625

d) (5%) Based on the probability that you just calculated, and the payoff offered for winning, how would this character do after playing for a long time with the loaded coins, and always betting on three heads? Circle the correct answer, and support your choice with some numerical evidence. WIN LOSE BREAK EVEN

If the player wins 0.3456 of the time, and gets back $3 each time, then the average return is $3 × 0.3456 = $1.0368 for each $1 bet. After 100 plays, and $100 bet, the player would have on average $103.68 left. In this game, that really means $103 or $104.

Here are the overall odds:

For p = 0.5 n prob "payoff" "gain" 0 0.03125 30 0.9375 1 0.15625 6 0.9375 2 0.3125 3 0.9375 3 0.3125 3 0.9375 4 0.15625 6 0.9375 5 0.03125 30 0.9375

For p = 0.6 n prob "payoff" "gain" 0 0.01024 30 0.3072 1 0.0768 6 0.4608 2 0.2304 3 0.6912 3 0.3456 3 1.0368 4 0.2592 6 1.5552 5 0.07776 30 2.3328

3. (30%) Consider a system consisting of 4 magnetic dipoles lined up in a row. The dipoles are labeled 1, 2, 3, or 4 depending on their position in the row. Each dipole either points up or down (nothing else is allowed) and has an energy –  if the ith dipole is pointing up (parallel to the magnetic field) and an energy  if the ith dipole is pointing down (antiparallel to the magnetic field). Suppose the spin system is contact with a heat reservoir which is at a constant temperature of T.

2 a.) (10%) What is the partition function Z for a single dipole? (Note that it has two energy levels.)

 n  ( )    Z  e  e  e  e  e n It is not necessary, but this could be expressed as: Z = e  + e -  = 2 cosh 

b.) (5%) What is the partition function for the entire system of four dipoles? You can give your answer in terms of Z from part (a).

Skip this part.

c.) (10%) What is the average energy of the 4-dipole system at temperature T?

U     ln Z   ln(e  e ) N  

 e  e e e U  N  4  4 tanh  e  e e  e The last step is not necessary.

d) (5%)What is the energy of this 4-dipole system in the limit of T? There is no need to do a calculation in this case. Simply present a well-reasoned argument if you prefer.

As T,   0, so   0 also. Since tanh 0 = 0, U  0.

Without the mathematics, you should realize that as T, the + and – states will become equally populated. The average energy of one dipole will then be zero, and therefore, the average energy of the entire system will be zero.

3. (30%) A container is divided into two equal chambers, each of the same volume V. One chamber contains N0 molecules of an ideal gas at temperature T, and the other chamber is completely empty, a perfect vacuum. A small hole, of area A, is punched in the wall separating the two chambers, and gas begins to leak into the empty chamber. The temperature of the gas is kept constant. a) (10%) Find an expression for the rate at which molecules leave the filled chamber at the instant the hole is punched. Express your answer in terms of A, V, N0, and v , the average speed of an air molecule. (Hint: the flux of molecules moving in the + x direction can be 1 written,   nv , where n is the number of molecules per volume.) 4

3 dN  N   A  1 n vA  1  0 vA dt 4 4  V 

dN N vA  0 dt 4V b) (10%) The number of molecules in the chamber that was initially filled will decrease until the two chambers are equally populated (each with N0/2 molecules). Then, the rate at which molecules leave the chamber will be equal to the rate at which they enter it from the other side. Derive an expression for N, the number of molecules in the initially filled chamber, as a function of time after the hole is punched. Again, the result can be in terms of A, V, N0, and v .

If there are N molecules in the first chamber, there are N0 – N molecules in the other. NvA Then based on part a), the rate at which molecules are leaving is , and the rate at 4V (N  N)vA which they are entering is 0 , Then 4V dN NvA (N  N)vA vA    0   (2N  N ) dt 4V 4V 4V 0 dN vA  N     N  0  . This differential equation can be separated and solved. dt 2V  2  dN vA   dt  N  vA N 2V , and after integrating, ln N  0    t  const . Then, N  0  2  2V 2 vA  t N N0 0 N   Ce 2V . At t = 0, N = N0, so C  . Then, 2 2 vA vA N N  t N   t  N  0  0 e 2V  0 1 e 2V    2 2 2   c) (10%) As the gas redistributes itself between the two chambers, the entropy of the system increases. Calculate the difference between the entropy of the final state with both chambers holding N0/2 molecules, and the initial state when all the molecules were in one chamber.

4  w   f  S  S f  Si  k ln wf  k ln wi  k ln   wi 

N0! N0! N0! wi   1 wf   2 N0!0! (N0/ 2)!(N0/ 2)! (N0/ 2)!  w   N !  S  k ln f   k ln 0   k[ln N !2ln(N / 2)!]    2  0 0  wi   (N0/ 2)! 

S  k[N ln N0  N0  2{(N0/ 2)ln(N0/ 2)  N0/ 2}]  k[N0 ln N0  N0ln(N0/ 2)]  N   0  S  N0k ln   N0/ 2 

 S = N0 k ln 2

4. (15%) Assume that Earth and the Sun are perfect blackbodies, and that Earth’s only source of heat is the Sun. What will be the temperature of the surface of Earth when it is radiating energy at the same rate that it is absorbing it from the sun, i.e. when it reaches steady state. The following information may be useful.

Temperature of the surface of the Sun: TS = 5800 K 8 Radius of the Sun: RS = 6.96 ×10 m 6 Radius of Earth: RE = 6.37 ×10 m Distance from Earth to the Sun: r = 1.50 ×1011 m Stephan-Boltzmann constant:  = 5.67 ×10-8 W/m²·K4

Be sure to make your work clear. In case you do not complete this correctly, that will make it possible to assign partial credit.

4 2 The sun radiates energy at the rate: PS  TS 4RS . When this reaches the orbit of Earth, it is spread over a sphere of area 4 r 2 . The fraction of the emitted energy that is absorbed by Earth is equal to the fraction of the  R2 R2 area just calculated that Earth occupies. That fraction is E  E . 4 r 2 4r 2 R2 R2 The rate at which energy is absorbed by Earth is, T 4 4 R2 E  T 4 R2 E S S 4r 2 S S r 2 4 2 The rate at which Earth radiates energy is, TE 4 RE . This must equal the expression above. R2 R2 R2 R T 4 4 R2  T 4 R2 E , so 4T 4  T 4 S , and T 4  T 4 S . Then, T  T S . E E S S r 2 E S r 2 E S 4r 2 E S 2r 6.96108 m T  (5800 K) = 279 K = 6° C = 43° F. E 2(1.51011 m)