Homework #8 Solutions - UMD

Homework #8 Solutions

Question 1) K+K, Chapter 5, Problem 6 Gibbs sum for a 2-level system has the following list of states: 1) Un-occupied N = 0; = 0 2) Occupied with energy 0; N = 1, = 0 3) Occupied with energy ; N = 1, =

a) The Gibbs sum can be written explicitely using the above list of all states.

Zg(T, ?) =

e(N ?-s)/

(1)

N =0 S(N )

= e(0?-0)/ + e(1?-0) + e(1?-)/

(2)

= 1 + + e-/

(3)

where we have defined e?/ and Zg denotes the grand canonical partition function.

b) Find the thermal average occupancy.

d

<

N

>

=

d

log Zg

(4)

1d

= Zg d Zg

(5)

+ e-/

=

(6)

Zg(T, ?)

c) Average thermal occupancy of state of energy .

e-/ P (1, ) =

Zg

This is the probability of occupancy of the state with N = 1 and energy .

d)The thermal average of the system is given by a weighted sum of energies

and corresponding probabilities of occupation. In this case there is only one

term

e-/ U=

Zg

e) Now suppose that there can be an N = 2 state in which energy levels 0 and can be occupied simultaneously. This will add an N = 2 term to the sum:

e(2?-(0+))/ = 2e-/

And the total Gibbs function becomes Zg = 1 + + e-/ + 2e-/

1

Question 2) K+K, Chapter 5, Problem 7 As usual we start by forming the partition function.

2

Zg =

e(N ?-s)/

(7)

N =0 s(N )

= e(0?+/2) + e(?+/2)/ + e(?-/2)/ + e(2?-/2)/

(8)

= e/2 + 2e?/ cosh + e(2?-/2)/

(9)

2

We can use eq(5) for < N >.

Zg = e/2 + 2 cosh

2

+ 2e-/2

(10)

Zg = 2 cosh + 2e-/2

(11)

2

1 =

2 cosh

+ 2e-/2

(12)

Zg

2

Now enforce < N >= 1. Using the explicit form of Zg we get

Zg = 2 cosh 2 2e-/2 = e/2

2 = e/

+ 22e-/2 = e/2 + 2 cosh 2

+ 2e-/2

(13) (14) (15)

Since = e?/ we get ? = /2 .

Question 3a) K+K, Chapter 5, Problem 8 First consider the case with O2 only. In this case this is a two state system. The heme site can be occupied or unoccupied. The Gibbs sum has only two terms:

Zg = 1 + e(?-A)/ = 1 + e-A/

(16)

We want to find A such that < N >= 0.9N . Again using eq(5):

<

N

>

=

log Zg

=

e-A/ Zg

e-A/

1

=

=

= 0.9

1 + e-A/ 1 + -1eA/

(17) (18)

Note that in the last equality we used 0.9 instead of 0.9N . The reason for this is the following. Zg we used above is that for a single site (for N sites there would be more than 2 states). We are assuming that the heme sites are distinguishable and independent. Hence the average occupancy of N sites is just N times that of one site.

2

Now solve for A to get:

A = log 9

(19)

= 1.38 ? 10-23 J (273.16 + 37)K log 10-5

(20)

K

9

= -5.87 ? 10-20J = -0.37eV

(21)

This is the bonding energy of O2 on Hb. b) Now allow CO as well. The Gibbs sum now has 3 terms. The heme site

can be un-occupied, occupied by O2 or occupied by CO.

Zg = 1 + O2 e-A/ + COe-B/

(22)

We are looking for the condition on B such that only 10% of the heme sites are occupied by O2.

< NO2

>=

O2 e-A/ Zg

=

O2 e-A/ 1 + O2 e-A/ + COe-B/

= 0.1

(23)

Again it is to be understood that this calculation is done for one heme site. Solving this for B we get:

B = - log

9O2 e-A/ - 1 CO

= -0.548 eV = B

(24)

Discussion: An alternative method: N sites rather than one site If we study the system with N sites for part a), the partition function will be:

l=N

Z=

l e-lA /

l=0 s

l=N

=

N ! (e-A/ )l

l!(N - l)!

l=0

= (e-A/ + 1)N

(25) (26) (27)

Then, we can use the formula for average < N >:

1 Z

< N >=

= 0.9N

(28)

Z

The same method for part (b), but the math will be more complicated.

3

Question 4a) K+K, Chapter 5, Problem 10

< N > = Zg Zg ? ,V

Zg =

e(?N -s)/

N s(N )

Zg =

N e(?N -s)/

?

N s(N )

2 2Zg =

N 2e(?N -s)/

?2

N s(N )

< N2 > =

N

s(N ) N 2e(?N -s)/ Zg

=

2 2Zg Zg ?2

b) Consider

(29) (30) (31) (32) (33)

=

ASN N e(?N -s)/

?

?

Zg

=

-

Zg ?

ASN N e(?N -s)/ Zg2

+

1 Zg

N 2e(?N -s)/

ASN

=-

?

Zg ?

2 2 Zg2

+

2 Zg

2Zg ?2

=< (N )2

>

(from above)

(34) (35) (36)

Question 5) K+K, Chapter 5, Problem 12 Ascent of sap in trees. Assume the water is in diffusive equilibrium in the tree. The vapor at the base has a chemical potential given by that of the ideal gas

? = log(n/nQ)

n = n0 at the pool.

?pool = log(no/nQ) (internal chem. pot. of water at the base)

The external chemical potential due to gravity is M gh, where M is the mass of the water molecule, and h is the height of the tree.

At the top of the tree, the total chemical potential of the water molecule is

?top = log(rn0/nQ) + M gh

where r is the relative humidity at the top of the tree. In diffusive equilibrium, ? is uniform in the tree, so:

?pool = ?top

(37)

log(no/nQ) = log(rno/nQ) + M gh

(38)

4

Solve for the maximum height of the tree

1 h = log

n0/nQ

= - log(r) = h

(39)

Mg

rn0/nQ

Mg

For the mass of the water molecule we will use M = 18 amu = 18 ? 1, 66 ? 10-27kg (Back inside cover of K+K). r = 0.9 is given in the problem. If you

plug in everything in eq(39) you get

h = 1, 480 meters

That is one tall tree!

In practice trees rarely get over 100 meters in height, 110 meters is about

the maximum.

This is an over-estimate because the tree does not grow in water, and there

is often a temperature gradient between the roots and top. Other mechanical

limitations may come into play as trees grow taller.

Question

6)

K+K

Chapter

6,

Problem

1

FD

Distribution

f ()

=

. 1

e(-?)/ +1

f -=

1

1 e(-?)/

(e(-?)/ + 1)2

Evaluate the derivative at the chemical potential:

f

-1e0

1

- (

=

?)

=

(e0

+ 1)2

=

4

Question 7a) K+K Chapter 6, Problem 3 Suppose such a system is in thermal and diffusive contact with a reservoir at and ?. There are 3 distinct states: 1) Un-occupied with energy 0 2) Single occupancy with energy . 3) Double occupancy with energy 2. The Gibbs sum is calculated as follows:

2

Zg =

e(?N -s(N))/

N =0 s(N )

= e(0?-0)/ + e(1?-)/ + e(2?-2)/

= 1 + e-/ + 2e-2/

(40)

(41) (42)

We can calculate < N > "by hand"

0 ? 1 + 1 ? e-/ + 2 ? 2e-2/ =

Zg e-/ (1 + 2e-/ ) = 1 + e-/ + 2e-2/

(43) (44)

5

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