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# 4501 Homework03sol

```University of Minnesota
Statistical Molecular Thermodynamics
Homework Week 3
1. You are given the following partition function for a fictitious gas “Vikonium”. What
is the correct equation of state for Vikonium? In the equation A and B are constants,
the other variables have their usual meanings.
3N/2
A 2m2 β
2
(V − 2N B)N eβN /V
Q(N, V, T ) =
2
N ! 3πh
(a) P V = N kB T
A2 N 2
N
(b) P + V 2
= N kB T
V −2N B
2 2
(c) P − mV N2 (V − 2N B) = NA! kB T
2 2
(d) P + AVN2 (V − 2N B) = kB T
N2
(e) P − V (V − 2N B) = N kB βT
2
(f) P + N
V N = N BkB T
V2
2
1 N
(g) P − N
= N kB T
2
V
V
2
1
(h) 2B P − N
= 3N kB T
2
V
V −2N
N2
(i)
P + 2 (V − 2N B) = N kB T
V
Since we want the equation of state,
calculate the pressure in terms of volume and
∂ ln Q
temperature using P = kB T ∂V N,T . Start by separating ln Q into pieces that
include V and those that do not,
ln Q = ln [terms no V ] + N ln (V − 2N B) +
P = kB T
∂ ln Q
∂V
P =
N
βN 2
= kB T 0 +
− 2
V − 2N B
V
N2
P + 2 (V − 2N B) = N kB T
V
N,T
2
N kB T
N
− 2,
V − 2N B
V
βN 2
V
2. In an Einstein crystal, one principal assumption is that each of the N atoms of the
crystal vibrate independently about their lattice positions. The crystal is thus pictured
in three dimensions as N independent three dimensional harmonic oscillators. Using
the partition function for a harmonic oscillator,
qho (T ) =
∞
X
e
−β(j+ 12 )hν
−βhν/2
=e
j=0
∞
X
e−βjhν ,
j=0
determine the partition function for an Einstein crystal. It will be helpful to make use
of the geometric series,
∞
X
xj =
j=0
1
1−x
to derive this result.
−βhν/2 3
e
(a) Q = 1−e
−βhν
(b) Q =
(c) Q =
e−βhν/2
1−e−βhν
3N
e−βhν/2
1−e−βhν
(d) Q =
(e) Q =
e−βhν/2
1+e−βhν
(f) Q =
eβhν/2
1+e−βhν
1
3N
1−e−βhν
3N
3
The partition function for a one-dimensional harmonic oscillator is given as
qho (T ) =
∞
X
e
−β(j+ 21 )hν
−βhν/2
=e
j=0
∞
X
e−βjhν
j=0
If we let x = e−βhν we can simplify the partition function to
qho (T ) = e−βhν/2
∞
X
e−βjhν = e−βhν/2
j=0
j=0
Note that this summation is the geometric series,
∞
X
j=0
xj =
∞
X
1
.
1−x
xj
We can then write the partition function as
qho (T ) = e
−βhν/2
∞
X
xj = e−βhν/2
j=0
qho (T ) =
1
1−x
e−βhν/2
1 − e−βhν
To find the partition function for a three-dimensional harmonic oscillator, think of the
three-dimensional harmonic oscillator as three independent one-dimensional harmonic
oscillators. The partition function for a three-dimensional harmonic oscillator can then
be written as
−βhν/2 3
e
3
q3D (T ) = [qho (T )] =
1 − e−βhν
Each particle in the system is distinguishable. Therefore, the partition function for a
system of N particles can be written as
Q = [q3D (T )]N
Substituting the partition function for the three-dimensional oscillator gives:
Q=
e−βhν/2
1 − e−βhν
3N
N
)
3. One of the requirements for the statistical result that Q(N, V, T ) = q(V,T
is that
N!
the system have many more available states than particles. Usually, the number of
translational states alone are enough to satisfy this condition. Use the translational
partition function to determine which of the following conditions leads to an increased
number of accessible translational states:
(a) small molar volume
(b) low temperature
(c) small mass
(d) all of the above
(e) none of the above
In lecture video 3.6 it was discussed that the condition under which the number of
available states exceeds the number of particles is
N
V
h2
8mkB T
3/2
1
Large molar volumes, high temperatures, and large masses favor this condition, so the
answer is none of the above.
4. The average energy for a monatomic van der Waals gas is
3
aNA2
hĒi = NA kB T −
2
V
Use the definition of the constant volume molar heat capacity discussed in lecture
video 3.4 to determine a formula for the constant volume molar heat capacity of a
monatomic van der Waals gas. Here, NA is Avogadro’s number and R is the universal
gas constant.
(a) C̄V = 23 NA kB
aN 2
(b) C̄V = 23 NA kB T − V A
(c) C̄V = 23 NA kB − aNA2
aN 2
3
(d) C̄V = 2R
− VA
(e) C̄V = 23 R
(f) C̄V = 43 RT 2
(g) C̄V = 34 NA kB T 2 −
2T
aNA
V
(h) a) and e)
(i) b) and g)
(j) none of the above
The constant-volume molar heat capacity is defined in lecture video 3.4 as
∂hĒi
∂ Ū
=
C̄V =
∂T N,V
∂T N,V
Insert the average energy that is given
3
aNA2
hĒi = NA kB T −
2
V
into the expression for the heat capacity:
∂hĒi
∂
3
aNA2
C̄V =
=
NA kB T −
∂T N,V
∂T 2
V
Differentiating with respect to temperature gives the heat capacity for a monatomic
van der Waals gas as
3
C̄V = NA kB
2
Substituting R = NA kB , we can also see that
3
C̄V = R
2
This heat capacity is the same as that for a monatomic ideal gas.
5. In the lecture videos, a general formula for the average energy,
∂ ln Q
2
hEi = kB T
,
∂T
N,V
was derived.
Determine the average energy, hEi, for a monatomic ideal gas given the partition
function for this gas,
1
Q(N, V, T ) =
N!
(a) hEi = (3/4)N kB T 2 +
(b)
2πmkB T
h2
3N/2
V N.
N
V
hEi = (3/2)N kB T
(c) hEi = (3/4)N kB T 2
(d) hEi = (3/2)N kB T +
N
V
(e) hEi = (3/4)N kB T +
3N
2
(f) hEi =
h2
2πmKB
+
N
V
3N
2T
First, rearrange this equation,
1
Q(N, V, T ) =
T 3N/2
N!
2πmkB
h2
3N/2
V N.
Take the natural log of both sides of the partition function and separate the terms to
obtain
3N
2πmkB
3N
ln T +
ln
+ N lnV
ln Q(N, V, T ) = −ln N ! +
2
2
h2
Note that only the second term in the above equation depends on T . Therefore,
differentiating with respect to T produces the following:
∂ ln Q
∂T
=
N,V
3N
2T
Insert this result into the equation for hEi given above:
∂ ln Q
3N
2
2
hEi = kB T
= kB T
∂T
2T
N,V
3
hEi = N kB T
2
6. We found that under certain conditions, the partition function for a collection of particles, Q(N, V, T ), could be written in terms of the partition function for individual
particles, q(V, T ), and the number of particles, N ,
Q(N, V, T ) =
q(V, T )N
.
N!
This equation is correct for a gas described by the van der Waals equation of state.
(a) TRUE
(b) FALSE
We actually accept either answer as correct. The formula given is for a partition function involving particles that are indistinguishable, and all of which have identical and
independent energy levels, because they do not interact with one another. In a van der
Waals gas, the particles are indistinguishable, but the particles do interact with one
another (as measured by the a and b parameters), so each particle’s energy depends
upon its interactions with all other particles, making the given partition function equation inappropriate. However, note that the van der Waals gas partition function, Q,
given in problem 1 of this homework can be rearranged so as to be written in the form
Q(V, T ) =
q(V, T )N
.
N!
While not derived here, this comes from assuming a uniform distribution of particles
which generates a ”mean-field” potential energy with which each particle interacts
”independently”, in which case use of the non-interacting ensemble partition function
can be viewed as justified. Clearly, the assumptions are not really internally consistent,
since you can’t both interact and not interact simultaneously! But, that is how the
van der Waals equation of state is derived.
7. For a system that has equally spaced non-degenerate energy levels, at a fixed temperature what will happen to the value of the partition function, Q(N, V, T ), if the spacing
between the energy levels INCREASES?
(a) Q will increase
(b) Q will decrease
(c) Q will remain the same
(d) either (a) or (b) depending on the temperature
(e) none of the above
See the plots in lecture video 3.2. If the temperature is fixed, and the spacing between
the non-degenerate energy levels increases, then there are fewer states available to the
system at that given temperature. Therefore, the partition function will decrease.
8. The following expresses the partition function of the system in terms of the partition
functions of the individual particles (such as atoms or molecules),
Q=
qN
.
N!
For this to be true, it must be the case that there are many more particles than available
states.
(a) TRUE
(b) FALSE
We had to assume that the number of available states was greater than the number of
particles. Therefore the answer is false.
9. Derive that the fraction of harmonic oscillators in the ground vibrational state is given
by
f0 = 1 − e−hν/kB T .
Then, given that ν̃ for N2 is 2330 cm−1 , calculate f0 for N2 (g) at 300 K and 1000 K.
(a) 1.0000 and 0.7457
(b) 0.9000 and 1.0000
(c) 1.0000 and 0.9650
(d) 1.0000 and 0.0000
(e) 0.9578 and 0.8762
(f) 0.9200 and 0.4578
To find an expression for the probability that a harmonic oscillator will be found in
the jth state, start with the general equation for the probability
pj =
e−βEj
qvib
where Ej = hν(j + 12 ) for a harmonic oscillator. Substituting into the equation for the
probability
1
e−βEj
e−βhν(j+ 2 )
e−βhνj e−βhν/2
pj =
=
=
qvib
qvib
qvib
We have shown that the partition function for the harmonic oscillator is
qvib =
e−βhν/2
1 − e−βhν
Insert this into the equation for the probability
pj =
e−βhνj e−βhν/2
e−βhν/2
1−e−βhν
=
e−βhνj e−βhν/2 (1 − e−βhν )
= e−βhνj (1 − e−βhν )
e−βhν/2
From the preceding equation we can see that for the ground state (j=0),
p0 = e−βhνj (1 − e−βhν ) = 1 − e−βhν
Therefore, the fraction in the ground state is equal to this probability.
f0 = p0 = 1 − e−βhν = 1 − e−hν/kB T
To find the fraction of harmonic oscillators in the ground state at different temperatures, first convert the fundamental vibrational frequency for N2 to s−1 . For
ν̃ = 2330 cm−1 .
ν=
c
= cν̃ = (3 &times; 1010 cm &middot; s−1 )(2330 cm−1 ) = 6.99 &times; 1013 s−1
λ
We can then obtain a value for hν/kB :
hν
(6.626 &times; 10−34 J &middot; s)(6.99 &times; 1013 s−1 )
=
= 3352 K
kB
1.38 &times; 10−23 J &middot; K−1
This simplifies the equation for the fraction of harmonic oscillators to:
f0 = 1 − e−hν/kB T = 1 − e(−3552
K)/T
Insert the different temperatures for T to obtain the fraction in the ground state:
Temperature (K)
f0
300
1.000
1000
0.9650
10. Imagine you are given one mole of a homonuclear diatomic molecule, X2 , at a fixed
volume and temperature. If you increase the mass of the molecules, for example you
go from F2 to I2 , the partition function will,
(a) increase
(b) decrease
(c) not change