Download Test 1 solutions

Chem 3411 test #1 brief answers (not fully worked out) – out of 22 marks,
given on Feb. 8, 2006. Class average 13.6/22
You must attempt questions 1, 2, and 3. You can choose to do either question 4 or
question 5. This makes four questions in total. You can use the back of the test pages for
extra space, if needed.
Ideal gas equation of state is PV = nRT, R=8.31451 J mol−1 K−1 . Using the molar volume
V̄ gives the equation of state as P V̄ = RT because V̄ = V /n.
Boltzmann’s constant kB =1.38066×10−23 J K−1
β = (kB T )−1
Boltzmann factor: e−βEi for a state of energy Ei .
P
Partition function: Q = i e−βEi
(fractional) population is the Boltzmann factor divided by the partition function.
Problem 1 – 4 marks
(a – 2 marks) The coefficient of thermal expansion α is defined as
1 ∂ V̄
α=
V̄ ∂T P
(1)
Show that α = 1/T for an ideal gas.
Answer:
P V̄ = RT ⇒ V̄ =
RT
P
⇒
Hence α =
1 R
V̄ P
=
P R
RT P
=
∂ V̄
∂T
=
P
R
P
(2)
1
T
(b – 2 marks) The isothermal compressibility κ is defined as
1 ∂ V̄
κ=−
V̄ ∂P T
show that κ = 1/P for an ideal gas.
1
(3)
Answer:
V̄ =
RT
P
⇒
So κ =
1 RT
V̄ P 2
=
∂ V̄
∂P
=−
T
RT
P2
(4)
1
P
Problem 2 – 4 marks
Why do we get an exponentially decreasing probability of populating energy levels (eg.
homework #1 and examples in class) when we are following the fundamental postulate
of statistical mechanics, which says something about assigning equal probabilities to each
microstate? In other words, why don’t the populations come out equal?
Answer:
Because the total energy serves as a constraint.
Problem 3 – 8 marks
For the statistics which were used in counting microstates (see question 2, also recall
homework #1 and examples in class) we had a weight factor which entered into the expressions of the form
W (a) =
N!
a0 !a1 ! . . . an !
(5)
where a0 +a1 +· · ·+an = N where ai counted the number of particles with energy i . Later on,
we learned that the fundamental quantum mechanical nature of all known particles (fermions
or bosons) was not compatible with this “classical” counting scheme. What changes if we
have a system of N identical non-interacting fermions? (2 marks)
Answer:
indistinguishability and also, can’t have two or more particles occupying the same state.
What changes if we have a system of N identical non-interacting bosons? (2 marks)
Answer:
indistinguishability
2
Even though fermions and bosons have different properties from each other, we unified
our treatment of them in what is called Boltzmann statistics. How was this possible? How
did we manage to combine them into a single treatment? (3 marks)
Answer:
We assumed there were enough available states that the probability of finding doubly (or
more) occupied bosonic states was negligible.
What conditions might be expected to lead to a breakdown of this unified description?
(1 mark)
Answer:
Conditions where double occupied bosonic states are likely: extremely low temperature
can give a Bose-Einstein condensate (a new form of matter).
3
Problem 4 – 6 marks
Suppose we have a two-level system. This is a system with only two allowed states. Take
the energy of the two states to be zero and E, respectively. Write down the partition function
for this system. (2 marks)
Answer:
Q = 1 + e−βE
(6)
Write down explicit expressions for the populations of the two states. (2 marks)
Answer:
P(1) =
1
1 + e−βE
(7)
P(2) =
e−βE
1 + e−βE
(8)
What is the population of the excited state in the limit T → 0? What is the population
of the excited state in the limit T → ∞? (2 marks)
Answer:
If T → 0 then β → ∞. So e−βE → 0. Hence the population of the excited state is zero
at zero temperature. On the other hand, if T → ∞ then β → 0. So e−βE → 1. This gives
an excited state population of one half (50%).
Problem 5 – 6 marks
In the harmonic approximation, the vibrational energy levels for a bond between two
atoms are
1
(n + )hν
2
n = 0, 1, 2, . . .
where h = 6.626 × 10−34 Js is Planck’s constant, and where
s
1
k
ν=
2π µ
4
(9)
(10)
where k is the bond force constant and
µ=
m1 m2
m1 + m2
(11)
is the reduced mass, where m1 and m2 are the masses of the two atoms, respectively. Consider replacing hydrogen with deuterium in a C − H bond with a force constant of k = 500
N/m. How would this change the relative populations of the first two levels at room temperature? Note that, because I have only asked for a ratio of populations, Q cancels and
you don’t need to compute it. Assume that k is unchanged under isotopic substitution
(this is actually a reasonable approximation). One atomic mass unit is 1.66×10−27 kg and
Avogadro’s number is NA = 6.02 × 1023 mol−1 .
Answer:
e−βE1
P(1)
= −βE0 = e−βhν
P(0)
e
(12)
Need to compute this for C-H and C-D. Also need to sort out the units. I’m not going
to do this here, but you are responsible for being able to work out the numbers properly,
taking care of all the conversion factors.
Essential part of the argument:
µCH =
mC
≈1
1 + mC
(13)
µCD =
2mC
≈2
2 + mC
(14)
So the CH versus CD ratio of populations are roughly
e−β~
√
k
versus
√
− √1 β~ k
e
2
which can be quite large (depending on the values of β and k).
5
(15)