PHYS-4420 THERMODYNAMICS & STATISTICAL MECHANICS
Quiz 2
SPRING 2005
Friday, April 22, 2005
NAME: ___________ANSWERS__________________________
To receive credit for a problem, you must show your work, or explain how you arrived at your
answer.
1. a) (10%) Consider a system consisting of two identical harmonic oscillators, each with
energy levels 0, , 2, 3, etc. Three units of energy, each equal to , a total of 3, are to
be distributed between these two oscillators. How many ways can the energy be
distributed between the oscillators? Actually list each possibility, and count them up.
Oscillator 1
3

2





Oscillator 2



3
w= 4
b) (5%) The system of two oscillators could be treated by Bose-Einstein statistics where the
number of microstates in a given macrostate can be calculated from the formula
( N j  g j  1)!
wj 
. In this formula, Nj is the number of particles, and gj is the number of
N j !( g j  1)!
states corresponding to the macrostate having a particular energy, and designated by the
subscript j. Each oscillator can be considered to be a “state”, and each unit of energy a
particle (phonon). The macrostate having energy 3 is under consideration. Calculate the
number of microstates for Nj = 3 and gj = 2, using this formula. If it is not the same as
your answer to part a), one of them is wrong.
( N j  g j  1)! (3  2  1)! 4!
wj 


N j !( g j  1)!
3!(2  1)! 3!1!
wj = 4
1
c) (12%) Now consider two identical systems, A and B, like that in part a), each consisting
of two oscillators. The systems are in contact, and there is a total of 3 to be divided
between them. Below is a table that you are to fill in. It lists the four macrostates, j,
possible. For each macrostate, the energies of system A and system B are given. You are
to calculate the number of microstates for system A and system B in each case, as well as
the number of microstates for the combined system consisting of A and B. To get credit
for this, you must indicate how you arrived at the numbers that you put in the table. You
need not show all the calculations, but enough to show the methods being used. I realize
that many of you can do this in your heads, but if you do, and you make any typos, I will
have no idea what you did.
j
UA
wA
UB
wB
wA+B
1
3
4
0
1
4
2
2
3

2
6
3

2
2
3
6
4
0
1
3
4
4
For wA and wB, the formula w j 
Most
probable


( N j  g j  1)!
N j !( g j  1)!
can be used, where gi = 2, and Ni is as
given for each case.
Then, wA+B = wA × wB
d) (5%) Indicate the most probable macrostate (or states) of the combined system by putting a
check in the box for that (or those) state(s) in the column marked “most probable”.
e) (8%) What is the entropy of the combined system when it is in the most probable state(s)?
Leave your answer in terms of k, the Boltzmann constant.
S = k ln wA+B = k ln wA + k ln wB = k ln 6 = k ln 3 + k ln 2 = k ln(3×2)
S = k ln 6
2
2. (25%) In a classroom activity, you analyzed the frictional force on a disk that floated on a
layer of air. In this problem, you will examine it again. The diameter of the disc is 10 cm
(0.10 m) and its mass is 0.30 kg. It is gliding on an air table with a speed of 2.0 m/s. You can
assume that air is mainly nitrogen. The diameter of a nitrogen molecule is 4.0 ×10-10 m. and
its mass is 4.65 ×10-26 kg Assume the air temperature is 300K.
a) (10%) Calculate , the coefficient of viscosity for the air.

mkT
mkT

3
3 d 2

(28 amu)(1.66  10 27 kg/amu) (1.38  10 23 J/K)(300 K)
3 (4.0  10 10 m) 2
 = 9.20 ×10-6 poise
units
b) (10%) The magnitude of the frictional force on the disk is measured to be 1.0 ×10-3 N.
How thick is the layer of air on which the disk is gliding?
 v A (9.20 10 6 poise)(2.0 m/s)  (0.05 m) 2
F
dv
v
  x   x . Then, y  x 
A
dy
y
F
1.0 10 3 N
y = 1.4 ×10-4 m = 0.14 mm
units
c) (5%) A careful examination of the equations reveals that the frictional force is not
constant. It depends on the speed of the disk. With that in mind, find the time needed for
the disk to slow to one half of its original speed. (Remember, the frictional force is
negative, which means it is directed opposite to the velocity.)
F
dv
v
A
dv
A
  x   x , so F  
vx .
vx , or m x  
A
dy
y
dt
y
y
Integrating gives, ln vx  
A
my
t = 0. If vx = ½ vx0 we have,
t
1
2
t  Constant , or vx  vx 0e
e

A
t
my
, and ln 2 
A
my

A
t
my
t , or
Then,
dvx
A

dt
vx
m y
, where vx0 is the speed at
t
(0.30 kg)(1.4  10 4 m)
ln 2
(9.20 10 6 poise)  (0.05 m) 2
my
ln 2
A
t = 420 s = 7 min
The calculation is easier if you note that the numerical value of
3
A
y

vx
1

.
F 2000
3. (20%) The tungsten filament of an incandescent light bulb operates at a temperature of about
3000 K. Assume the filament is an ideal blackbody and answer the following questions.
a) (10%) If the bulb gives off 100 watts, what is the surface area of the filament?
P   T 4 A , so A 
P
100 W

4
-8
T
(5.67  10 Wm - 2K - 4 )(3000 K)4
A = 2.18 ×10-5 m2 = 21.8 mm2
units
b) (5%) At what wavelength does the peak of the bulb’s spectrum occur?
maxT = 2.90 ×10-3m·K, so max 
2.90  103 mK 2.90  103 mK

T
3000 K
 = 9.67 ×10-7 m = 967 nm
units
c) (5%) To increase the fraction of the bulb’s output that is in the visible range (400 nm to
700 nm), what should be done to the temperature at which the bulb operates? (Circle the
correct answer.)
INCREASE IT
DECREASE IT
TEMPERATURE CHANGE WILL NOT IMPROVE THE BULB’S SPECTRUM
4
4. (15%) The measured heat capacity of liquid 4He below 0.6 K is proportional to T 3. This
suggests that long wavelength phonons are responsible for the heat capacity, as in the Debye
model of a solid. The only important difference is that the liquid can support only
longitudinal waves, while the solids support both longitudinal and transverse waves.
a) (10%) Calculate the Debye temperature for liquid 4He. The formula for the Debye
1/3
c 
 N 
temperature of a solid is:  D  6 2   , where c is the speed of sound. Make any
k 
 V 
changes that you might think necessary in this formula, and indicate why they are made.
The speed of sound in liquid 4He is 238 m/s, and the number of atoms per volume is
2.18 ×1028 m-3.
The density of states will be 1/3 that of the solid, but still proportional to 2. Therefore
when the density of states is integrated up to the cutoff frequency the result will be
1
3 N  const He D3 . Since const He  const Solid , D3 , He  3D3 , Solid and D, He  31 3 D, Solid .
3
1/3
c 
 N 
Then,  D 
18 2  

k 
 V 

(1.054 10 34 J  s)( 238 m/s )
[18 2 (2.18 1028 m  3 )]1/3
 23
1.38 10 J/K
Alternatively: Speed of sound used in solid was,
3
1 2
1
 3 3  3.
3
c
cl ct
cl
Replace c 3 with 3cl3 , or c  31 3 cl
D = 28.5 K
units
b) (5%) The Debye heat capacity, at low temperatures, can be written:
3
CV 12 4  T 

  KT 3 . Based on your answer to a), calculate a value for K for liquid

Nk
5  D 
4
He. Again, make any changes that you might think necessary in the formula that is
given, and indicate why they are made.
No changes are necessary. The Debye temperature took care of that.
12 4
12 4
K

53D 5(28.5 K)3
K = 1.01 ×10-2 K-3
units
5