PES 3950/PHYS 6950: Homework Assignment 5
Handed out:
Due in:
Wednesday April 8
Wednesday April 22, at the start of class at 3:05 pm sharp
Show all working and reasoning to receive full points.
Question 1 [25 points]
a)[10 points] Find the partition function for an ideal gas in two dimensions. You should
get
N
1
2mπ
A
,
Z=
N!
βh2
where A is the Area that the gas is confined to.
b)[5 points] Find the average energy < E > in two dimensions.
c)[5 points] Find the average pressure < p > in two dimensions.
d)[5 points] Find the chemical potential in two dimensions. Using Stirling’s approximation
ln(N !) = N ln(N ) − N , you should get
A2mπ
.
µ = −kT ln
N βh2
Question 2 [5 points]
a)[2.5 points] You are making strawberry shortcake. You cut up the strawberries, then
sprinkle on some powdered sugar. A few moments later the strawberries look juicy. What
happened? Where did this water come from?
b)[2.5 points] You are trying to make artificial blood cells. You have managed to get pure
lipid bilayers to form spherical bags of radius 10 µm, filled with hemoglobin. The first time
you did this, you transferred the cells into pure water and they promptly burst, spilling the
contents. Eventually you found that transferring them to a 1 millimolar salt solution prevents bursting, leaving the cells spherical and full of hemoglobin and water. If 1 millimolar
is good then would 2 millimolar would be twice as good? What happens when you try this?
Question 3 [10 points]
Consider a 3 state system problem where O2 can be in one of three states:
- In solution floating around
- On receptor of type 1 - binding energy Eb1
- On receptor of type 2 - binding energy Eb2
a) Find an expression for the probability for the O2 to be attached to receptor 1.
b) Find an expression for the probability for the O2 to be attached to receptor 2.
c) Find the ratio of these two probabilities. Does it depend on the chemical potential?
Question 4 [10 points] GRADUATE STUDENTS ONLY
Laser light can be used to trap micron-sized beads. The dynamics of such beads can be
thought of as the Brownian motion of a particle in a quadratic energy well. Compute the
mean-squared excursion < x2 > of such a bead in a one-dimensional quadratic well with a
potential-energy profile U (x) = 12 kx2 and show that we can then determine the trap stiffness
as
kB T
k=
.
< x2 >
Homework 5 SOLUTIONS
Question 1) a) Partition function for an ideal gas in two dimensions
We need to sum over all states – but what is a state? A state for one particle
is know when we know the position and momentum of the particle. So to sum
over all states we sum (integrate) over all positions and all momenta.
Our energy is the kinetic energy only (no interactions)
E=
p 2x  p 2y
2m
The partition function is then given by
Z   e  E dp x dp y dxdy
The integration over coordinates gives a Area A. Break up the integral on momenta

ZA

2
e   p x / 2m dp x


e

 p 2y / 2m
dp y

But each integral is identical, so
    p 2 / 2m
Z  A  e
dp

 




2
We can look up the integral
    Bp 2 
 e
dp  
 

 


B
So the partition function is
 2m  

Z  A
  
If we had 2 particles instead of just 1, we would have to do our "sum" as
1
Z   e   E dp1x dp1y dx1dy1
sum over
particle 1
states
  2m  
Z  A
 
   
dp 2 x dp 2 y dx 2 dy 2
sum over
particle 2
states
2
For N particles we would expect
  2m   
Z  A

   
N
We have made two mistakes
1) We have treated the particles as if they are each distinguishable – we can tell particle 1
is at (x1, y1) and particle 2 is at (x2, y2). But this isn't right. We really can't tell which
particle is where. So we have to reduce the number of states to account for this. (Gibb's
paradox). To get rid of this problem we have to divide by N!
2) We didn't count the states correctly. From our original definition we need to sum over
states! Instead we did integrals over position and momentum. But how many states does
this correspond to? From quantum mechanics we know that there is a fundamental limit.
Essentially 1 state (in phase space) is given by the uncertainty principle
dx dp x  h and dx dy dpx dpy = h2
So we need to divide the answer by h2 to get the right number of states!
The correct answer – for N particles is
1   2m 
Z
A

N!    h 2 
N
2
b) Find average energy

E  
ln Z

0
N

  1   2m  

  ln  A
    ln(1 / N!)
  N!    h 2  




  2m   

 )
N ln(  A
2 


  h 
Or, getting rid of the other parts which give no contribution
E  N


ln( 1 )  N ln() 


E  NkT
Note: This answer is consistent with the equipartition theorem where there are two
degrees of freedom (two dimensions) and each degree of freedom is associated with ½ kT
c) Find the average pressure
Normally one has
p
1  ln( Z)
 V
But in two dimensions, we should take the derivative with respect to area, not volume.
So
p
1  ln( Z)
 A
Here we only care about the A dependence of the partition function. The ln(Z) allows us
to isolate this
N

1  ln( Z) 1   1   2m   
1  ln A
p 

ln  A

 N
 A
 A  N!    h 2   
 A


p 
NkT
A
This is the ideal gas law in two-dimensions
3
d) Find the chemical potential
N

1  ln( Z)
1 l  1   2m  

 
ln 
A
 
 N
 N  N!    h 2  



 
1    1 
 2m 

ln    N ln  A 2 
 N   N! 
  h 












Look at the first term
ln(1/N!) = -ln(N!)
Now use Stirling's approximation ln(N!) = N ln(N) – N
We get
  2 m  
1  

 N ln( N )  N  N ln A 2  
 N 
   h  

Or
  2m   
1 

    ln( N )  1  1  ln A
2  


   h   

Or

  2m  

   kT  ln( N)  ln A
2  



h

 
 
Now, combining the ln terms
  A  2m  

   kT ln  
2 
  N   h  
The minus sign indicates that the chemical potential is usually negative for typical values.
4
Question 2 answer
a) The water did not come from the atmosphere. Some of the sugar dissolved in the
surface water layer, creating a layer of concentrated sugar solution. Osmotic flow then
pulled water out of the cells of the fruit to dilute this exterior solution.
b) From the text of the problem we can conclude that the osmotic pressure due to the
various solutes on the inside of the bag is balanced by osmotic pressure due to the 1mM
solution of salt on the outside. Thus 2mM solution of the salt would exert larger osmotic
pressure inward and collapse the bags.
Question 3
There are 3 important energies in this problem, E1, E2 and the chemical potential.
Take the relative energy for being in solution as zero.
Eo = 0
The energy for the oxygen to be attached to receptor 1 involves both the binding energy
and the chemical potential
E1 = Eb1 - 
Similarly
E2 = Eb2 - 
The partition function is
 e  E
Z
s
 e 0  e E1  e E 2 = 1  e ( E b1 )  e ( E b 2 )
states
s
The probability to be in state 1 is
P(state 1) =
P(state 2) =
e ( E b1 )
1  e ( E b1 )  e ( E b 2 )
e ( E b 2  )
1  e ( E b1 )  e ( E b 2 )
The ratio
e ( E b1 )
e ( E b 2 )

e E b1
e E b 2
It is independent of the chemical potential
5
Question 4 answer
6