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Number
Chemistry Name
Section 1314 Honors
Midterm Exam #2
October 6, 2010
This exam is closed book, notes, papers, etc. You may use a calculator. If you need a physical
constant or unit conversion factor that is not provided, please ask. This exam must represent your
own work.
Constants and Conversion Factors:
= 6.022 x 1023 particles/mol
NA
R
= 8.20578 x 10-2 (atm L)/(mol K)
h
= 6.62607 x 10-34 J s
c
= 2.99792 x 108 m/s
me
= 9.10939 x 10-31 kg
1J
= 1 kg m2 s-2
n2a
van der Waals equation: p + 2 (V nb ) = nRT
V
Rydberg equation:
Scores:
1
1
= 2 2 , =1.096776 x 107 m-1
n
1 n2
1
/ 20 (25 possible)
1.
/ 10
/ 10
a.
/9
b.
/6
c.
3.
a.
b.
2.
/5
4.
/ 20 (25 possible)
5.
/ 20 (25 possible)
Total:
/ 100 (115 possible)
CHEM 1314 H
Midterm Exam #2, 2010
Page 2
1. In terms of the postulates and conclusions of the kinetic molecular theory of gases, give a
brief molecular interpretation of each of the following experimental observations:
a. The volume of a gas at fixed pressure increases proportionally with the temperature.
Increasing T increases average speed, resulting in greater force of collisions of particles
with walls. If resisting pressure is fixed, the pressure due to collisions will push back
container walls, thus expanding volume.
b. For a fixed volume and temperature, the total pressure of a mixture of one mole of
oxygen and one mole of nitrogen is equal to the total pressure of two moles of nitrogen.
Force generated by a particle colliding with a wall depends on kinetic energy, not mass or
chemical character. Thus, pressure of two moles is independent of what molecules those
particles are, because there are no interactions.
c. The temperature of a gas can be raised by heating it with a flame or by stirring it with a
blender.
Temperature is a measure of kinetic energy. Molecules can be accelerated thermally or
mechanically.
d. For a fixed volume and temperature, the pressure exerted by a mole of gas with strong
intermolecular attractions is significantly less than the pressure exerted by a mole of gas
with weaker intermolecular attractions.
When molecules attract each other, they experience a decelerating force as they approach
the walls. This lowers the force of impact, lowering the observed pressure.
e. One mole of gas A occupies the same volume as one mole of gas B (at the same
temperature and pressure), even if the molecules of compounds A contain many more
atoms than do the molecules of compound B.
In an ideal gas, the particles are at great distances from one another. At these distances, the
sizes of the molecules are insignificant, so differences in sizes are irrelevant.
5 points each
CHEM 1314 H
Midterm Exam #2, 2010
Page 3
2. Assuming ideal gas behavior, answer the following questions by performing appropriate
calculations.
a. On an old television commercial, the distinguished Shell Answer Man used to tell us
(with a suitably grim expression on his face) that a car requires 10,000 gallons of air to
burn 1 gallon of gasoline. Determine the validity of the Answer Mans claim by
calculating the volume of air (at 25 C and 1 atm pressure) required to burn 1 gallon of
gasoline. You need to assume that gasoline is pure octane, and that a clean-burning car
produces only carbon dioxide and water in the combustion of octane. You also need to
know that the density of octane is 2.65 kg/gallon and that there are 3.7854 liters in one
gallon. Finally, you need to take into account that air is, on average, 20.95% oxygen by
volume.
2.65 kg C8 H 18 1000 g C8 H 18 1 mol C8 H 18
1 gal C8 H
1 gal C H 1 kg C H
114.229 g C H = 23.2 mol C8 H 18
8
18
8
18
8
18
25
C8 H 18 + O2 8 CO2 + 9 H 2 O
Balanced Equation: 1 point
2
Determining Moles of Oxygen: 3 points
25
mol O2
Determining Volume of Oxygen: 3 points
2
= 290.0 mol O2
nO2 = 23.2 mol C8 H 18
Determining Volume of Air: 2 points
1 mol C8 H 18
Evaluating Answer Mans Claim: 1 point
(290.0 mol ) 8.20578 10 2 L atm (298.15 K )
mol K
100 L air
nRT
= 7095 L O2
=
V=
20.95 L O = 33,860 L air
p
1 atm
2
1 gal air
33,860 L air
3.7854 L air = 8,946 gal air
The Answer Mans claim is pretty close to right.
b. In preparation for a combustion demonstration, a professor fills a balloon with equal
molar amounts of H2 and O2, but the demonstration has to be postponed until the next
day. During the night, both gases leak through pores in the balloon. If 35% of the H2
leaks, what is the O2/H2 ratio in the balloon the next day?
The amount of each gas that leaks from the balloon is
proportional to its effusion rate.
M W ,H 2
Rate O2
2.016 g / mol Rate O2 Recognizing this is an effusion problem: 2 points
=
=
=
Rate H 2
M W ,O2
32.00 g / mol
35%
Calculating the O2 lost: 3 points
Determining the amount of each remaining: 3 points
Rate O2 = 8.78%
Finding the requested ratio: 2 points
Amount H2 remaining = 100% - 8.78% = 91.22%
Amount O2 remaining = 100% - 35% = 65%
O2 91.22%
=
= 1 .4
H2
65%
CHEM 1314 H
Midterm Exam #2, 2010
Page 4
3. For this problem we will develop the ideal gas law and explore its limits.
a. State Boyles Law, Charless Law, and Avogadros Hypothesis in words and in
mathematical notation, being sure to identify what quantities must be held constant in
each case.
Boyles Law: Volume is inversely proportional to pressure for an ideal gas when
1
1
temperature and amount of gas are held constant. V or V = k1
p
p
Charless Law: Volume is directly proportional to temperature (in Kelvin) for an ideal gas
when pressure and amount of gas are held constant. V T or V = k 2T
Avogadros Hypothesis: Equal volumes of gases contain equal numbers of particles for an
ideal gas, or volume is directly proportional to number of particles for an ideal gas when
pressure and temperature are held constant. V n or V = k 3 n
Verbal statement: 1 point each law
What is held constant: 1 point each law
Mathematical statement: 1 point each law
b. Explain how Boyles Law, Charless Law, and Avogadros Hypothesis can be combined
to form the Ideal Gas Law (IGL). As part of your answer, be sure to include (i) the
mathematical form of the IGL, and (ii) an explanation of why the constant in the IGL is a
true universal constant.
1
(1), with n & T held constant. From Charless law we know that V
p
is directly proportional to T when p is held constant, and the only part of expression (1) that
T
(2). From
could be proportional to T is k1, so k1 T , or k1 = k 4T . Substituting in, V = k 4
p
Avogadros Hypothesis we know that V is directly proportional to n when p and T are held
constant, and the only part of expression (2) that could be proportional to n is k4, so k 4 n , or
nT
k 4 k = 5 n . Substituting in, V = k 5
(3). In this expression, we have accounted for the
p
variation of V with n, T, and p, so there is nothing left for k5 to depend on; it is therefore a
universal constant, which we call R, the gas constant. We can then rewrite expression (3) into
its more familiar form, pV = nRT .
Finding variation of k1 from another law: 1 point
Substituting in for k1: 1 point
Finding variation of k4 from remaining law: 1 point
Substituting in for k4: 1 point
Universality of gas constant: 1 point
Mathematical form of IGL: 1 point
From Boyles law, V = k1
CHEM 1314 H
Midterm Exam #2, 2010
Page 5
c. The van der Waals equation is sometimes used to model deviations from the IGL. Sketch
a plot that illustrates how deviations from the IGL can be observed (being sure to label
the axes clearly), and explain how the van der Waals equation accounts for each of the
two major types of deviations.
The IGL states that pV=nRT, or
cases where
pV
= 1 . Any deviations from the IGL should be observable as
nRT
pV
1 . These deviations would look like this:
nRT
(The y-axis should be density or pressure, and the y=axis should be pV/nRT. A curve such as the
one for ammonia should be shown, thereby demonstrating both the attractive deviations (the plot
going below 1) and repulsive deviations (the plot going above 1).)
The van der Waals equation accounts for these deviations by applying two correction factors to the
n2a
IGL. The first, 2 , is a correction added to the pressure, and accounts for attractive deviations.
V
The second, nb , is a correction subtracted from the volume, and accounts for repulsive deviations.
Plotting pV/nRT vs. something reasonable: 2 points
Showing a curve such as the one for ammonia: 1 point
Explaining the first correction factor: 1
Explaining the second correction factor: 1
CHEM 1314 H
Midterm Exam #2, 2010
Page 6
4. For each of the following conclusions, describe an experiment that leads to that conclusion,
the results of that experiment, and any reasoning necessary to reach that conclusion.
a. Light behaves as a waves.
Light passing through multiple slits or around an object, forms interference patterns. You
only see interference patterns from waves, so the light must be wave-like in nature.
5 points each. There are other
possible answers other than the ones
listed.
b. Light behaves as particles.
The photoelectric effect shows that the kinetic energy of ejected electrons varies linearly
with the frequency of the incident light, not the amplitude/intensity of the light. The only
way this can be is if the energy of the light is coming in packets, called photons, whose
individual energy is determined by frequency, and it is the energy of one of those packets
that determines the kinetic energy of the ejected electrons.
c. Electrons behave as particles.
The Millikan oil drop experiment (from Chapter 2) could be used here; this experiment
demonstrated that the charges on oil droplets came in multiples of a small, fixed value,
indicating that there are units of charge. Alternately, it can be observed that any single
electron passing through the double-slit experiment (described in the answer to (d) below)
hits the detector in a single spot, just as a particle would.
d. Electrons behave as waves.
Electrons passed one-at-a-time through a pair of slits form an interference pattern on a
film/detector behind the slits. You only see interference patterns from waves, so the
electrons must move as though they were waves.
e. Atoms have quantized energy levels.
Atomic emission spectra consist of discrete frequencies of light rather than continuous
bands. Since the frequency of light corresponds to the energy of the photon, this must mean
that there are only specific energy losses that are possible for the atom. If only certain
energy losses are possible, then the energies of the atom must exist in discrete, quantized
levels.
CHEM 1314 H
Midterm Exam #2, 2010
Page 7
5. Explain the error(s), if any, in each of the following statements (include both a correction, if
necessary, and a justification for that correction or a justification for why a correction isnt
needed):
5 points each
a. A hydrogen atom has a 3s orbital.
This statement is correct. The 3s orbital may be unoccupied in a particular hydrogen atom,
but the wavefunction describing its probability density will be present.
b. In the quantum mechanical treatment of the hydrogen atom, the probability of finding an
electron at any point is proportional to the wave function .
Replace with 2 and this statement is correct. There is no physical interpretation of the
wavefunction itself.
c. An electron moving from the 1p orbital to the 3s orbital emits longer-wavelength light
than an electron moving from the 2s orbital to the 4p orbital.
There are several problems with this statement. First, there is no 1p orbital, so it would have
to be a 1s orbital. Second, an electron moving from the 1s to the 3s orbital or from the 2s to
the 4p orbital would come from an absorption of light, not cause an emission of light. Third,
the energy difference between the n=1 and n=3 states are greater than the energy difference
between the n=2 and n=4 states, so the light absorbed for the first transition would be higher
in energy (shorter wavelength).
d. The material in parentheses is correct background information. Only evaluate and
respond to the portion outside the parentheses. (The smallest feature that can be
resolved in a microscope is of the wavelength of whatever is being used to form the
images. Electron microscopes use beams of high-speed electrons instead of light to form
images.) Individual atoms of copper that are 128 pm apart can be resolved using a beam
of electrons that are traveling at 1.14 x 104 m/s.
The wavelength of an electron is given by the de Broglie wavelength, which is given by
h
. Upon plugging in Plancks constant, the mass of the electron, and twice the
=
mu
distance between the copper atoms for the wavelength, we find the correct answer speed for
the electron to be traveling is 1.14 x 107 m/s.
e. An electron in the 1s orbital of a H atom has a higher average velocity than an electron in
the 1s orbital of a He+ ion.
The only relevant difference between these two systems is that the He+ ion has a nuclear
charge of +2, whereas the H atom has a nuclear charge of +1. This means that the He+ ions
electron will be held in tighter, thus having a smaller uncertainty in its position than the H
atoms electron will. This smaller uncertainty, due to the Heisenberg Uncertainty Principle,
means that the He+ electron will have a larger uncertainty in its momentum, and thus must be
moving faster.

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