Mary Anne Rooney
Hmwk 3 - Page 1
MISE - Physical Basis of Chemistry
Third Set of Problems - Due November 26, 2006
Submit electronically (digital drop box) by Sunday, November 26 – by 6 pm. Note: When
submitting to digital Drop Box label your files with your name first and then the label HmwkThree. Please PUT YOUR NAME in the ‘HEADER’ along with the alreadyinserted page #.
The following information may be useful when solving the problems:
Ideal gas law: PV = nRT ;
where R = 0.0821 L•atm•mole-1•K-1.
1 mole = 6.022 x 1023 “things”.
The atomic weight of an element listed in the periodic table is the mass of one mole of
atoms in grams. This is termed the molar mass (i.e., mass per mole) of the element.
The molecular weight - determined from molecular formula as the sum of the atomic
weights of all of the constituent atoms - is the mass of one mole of molecules in grams. This
is termed the molar mass (i.e., mass per mole) of the molecule.
1 atmosphere (atm) = 760 mm Hg = 76 cm Hg = 101.3 kilopascals.
1000 cm3 = 1 L ; K = ºC + 273 .
1. Given: 3 identical containers (same volume), at 1 atm pressure and 0 ºC. Each flask is
filled with a different gas.
Flask A: CO
Flask B: NO2
Flask C: He
Answer each of the following and briefly explain your answer using the concepts covered in
class:
(a)
Which flask contains the greatest number of moles of gas?
Each flask contains one mole of gas because the volume is the same at standard
temperature and pressure.
(b)
Which flask contains the largest total mass of gas?
The B flask containing NO2 has the greatest total mass because the molecular
weight of the particles of gas per mole is the greatest for NO2. Flask A has a total
mass per mole of 28 g. Flask B has a total mass per mole of 46 g. Flask C has a
total mass per mole of 4 g.
(c)
In which flask will the gas molecules have the greatest average kinetic energy?
All three flasks of gases will have the same average kinetic energy because each
flask contains one mole of gas at the same volume, temperature, and pressure.
The formula used to calculate average kinetic energy is (3RT)/2; therefore, the
average kinetic energy would be (3 x 8.314 J/mol*K x 273 K)/2 in all three flasks.
(d)
In which flask will the molecules have the greatest average velocity?
The C flask containing He has the greatest average velocity because the average
velocity in inversely proportional to the mass of the gas. The formula used to
calculate average velocity is √(8RT)/πM; therefore,
uavgCO = √((8 x 8.314 x 273)/π x 0.028 = 454.3 m/s
uavgNO2 = √((8 x 8.314 x 273)/π x 0.046 = 354.5 m/s
uavgHe = √((8 x 8.314 x 273)/π x 0.004 = 1202.1 m/s
Mary Anne Rooney
Hmwk 3 - Page 2
2. A mixture of gases effusing …
A chamber is filled with a gaseous mixture consisting of 1.00 mole of He gas and 4.00 moles
of SO2 gas at a certain temperature. [The atomic weight of He is 4.00 g/mole and the molecular
weight of SO2 is 64.0 g/mole] The mixture is allowed to escape into a series of two evacuated
chambers (“1” & “2”) connected by pinholes. Please refer to the diagram below.
1
2
Initial Chamber
You may assume that all three chambers have the same volume and are rigid. You may also
assume that both pinholes are exactly the same area. Finally, temperature is constant throughout
the entire process. [Recall the kinetic theory lecture slides (especially #15) depicting the process
and relevant relationship to determine how the composition of a mixture of gases changes when
the mixture effuses (leaks) through a pinhole into an evacuated chamber.] A useful proportion
when working in moles is the mole percent. If we have a mixture of substances, “A” and “B”,
the mole % is defined as:
moles A
moles A
mole % A =
• 100 % =
• 100 %
moles A + moles B
total moles
mole % B =
moles B
moles B
• 100 % =
• 100 %
moles A + moles B
total moles
Of course, mole % A + mole % B = 100 %.
Now we are ready to go. Please show your work.
(a) Determine the mole percent of each gas in the initial chamber before effusion occurs.
To determine the mole percent of each gas we need to use the mole ratio
which was presented in the problem. He = (1 mole/5 total moles) * 100 = 20%
and SO2 = (4 moles/5 total moles) * 100 = 80%
(b) Determine the ratio of moles He: moles SO2 in chamber 1 a few moments after the gaseous
mixture begins to effuse through the pinhole into evacuated chamber 1. Now, determine the
mole percent of each gas in this effused mixture.
Zeffusion = (#moleculesHe/ΔtimeHe)
= (A/4 x NHe/V x √(8RT)/(π x MHe)
(#moleculesSO2/ΔtimeSO2)
(A/4 x NSO2/V x √(8RT)/(π x MSO2)
By canceling before we substitute numbers, we have simplified the equation
for anything that is the same for both gases; therefore:
Zeffusion = # molecules He = N He x √(M SO2/M He)
# molecules SO2
N SO2
(Based on 100 molecules of mixture):
20He / 80SO2
= (1 molHe / 4 molSO2) x √(64.0 g/molSO2)/(4.00 g/molHe)
= (1/4) x √16
=¼x4
= 1 which is written as the ratio 1:1
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or 1 particle He to 1 particle SO2; therefore,
The mole percent of Helium is 1:2 x 100 = 50% Helium and the mole percent
of SO2 is also 1:2 x 100 = 50% SO2 after a few minutes of effusion.
(c) Determine the ratio of moles He : moles SO2 in chamber 2 a few moments after the gaseous
mixture begins to effuse through the pinhole into evacuated chamber 2. Now, determine the
mole percent of each gas in this effused mixture.
Using the same formula from the previous section of this problem, we can
calculate the effused mixture again but with the new mole percents from the
first chamber.
Zeffusion = # molecules He = N He x √(M SO2/M He)
# molecules SO2
N SO2
= (1 molHe/ 1 molSO2) x √(64.0 g/mol)/(4.00 g/mol)
= (1/1) x √16
=1x4
= 4 which is written as the ratio 4:1 (He/SO2) so
The mole percent of He is 4:5 x 100 = 80% He and the mole percent of SO2 is
1:5 x 100 = 20 % SO2 after a few minutes of effusion.
(d) In class, we said that the effusion process creates a gaseous mixture that is enriched in the gas
of lower molar mass. Thus, if more than one effusion occurs, each newly effused mixture is
even more enriched in the gas of lower molar mass. Compare your results to (a), (b), and (c)
and briefly comment on the validity of these assertions.
The validity of these assertions can be seen in the following chart:
mole % He
mole % SO2
initial
20
80
first effusion
50
50
second effusion
80
20
and in addition to this chart, the problem stated that the molar mass of SO2
was 64.0 g/mol and the molar mass of He was 4.00 g/mol.
3. “Real” gases - relaxing some of the postulates from kinetic theory.
Recall the postulates of the kinetic theory of gases (see kinetic theory lecture slide #2).
These postulates rationalize the ideal gas law (PV = nRT) at the molecular level.
The two postulates that may be the most “unrealistic” are the first two:
1. Gas particles have no volume, i.e., they are point particles.
2. There are no inter-particle forces - neither attractive nor repulsive.
A more “realistic” model of gas behavior modifies each of the above postulates.
The two modified postulates are:
1. Gas particles have volume. If we presume that the particles are “hard spheres”- like
“micro” billiard balls - they will occupy some non-negligible volume of the container
volume. Thus, the “empty space”, termed, the available volume, is necessarily less than
the full volume of the container. So, if V is the volume of the container, the volume
available to the gas particles is V minus the unavailable volume due to the spheres. In
brief, due to particle size, there is less space available to the gas molecules than the
full container volume would imply.
2. There are inter-particle (usually termed intermolecular) forces between the particles.
The most important are attractive forces. Thus, gas particles “feel each other’s
presence” and “pull” on each other. We may thus imagine that a molecule in the
“middle” of the sample - if evenly surrounded by its identical neighbors - may not feel
any pull at all due to the balancing forces, as you learned in physics. However, a
Mary Anne Rooney
Hmwk 3 - Page 4
molecule “next to the wall” only has neighbors beside it and behind. Now the forces
will be unbalanced and so there may be some net pull. So, a molecule approaching the
wall may hit with less force (why?) and/or be diverted from hitting the wall at all.
Let’s now consider how the pressure of the gas - as predicted by the ideal gas law - is altered by
the two modified postulates above. Of course, pressure is still force/area - this is its definition.
We will continue to presume that the microscopic origin of the macroscopic gas pressure is due
to the molecules colliding with the container walls.
(a) How would the pressure (P) of the gas be altered from its ideal value (Pideal), due only to
modified postulate 1? [The ideal pressure (= nRT/V) is what we would expect if the particles
followed the original postulates from kinetic theory.] Please select one of the choices below
and then carefully explain your choice. No calculations required or expected.
P < Pideal
P > Pideal
P = Pideal (no change)
If the particles have a larger size than their volume will be larger which means that
the space available to them will be less. Boyles Law states that less volume
means more pressure so therefore P > Pideal
(b) How would the pressure (P) of the gas be altered from its ideal value (Pideal), due only to
modified postulate 2? [The ideal pressure (= nRT/V) is what we would expect if the particles
followed the original postulates from kinetic theory.] Please select one of the choices below
and then carefully explain your choice. No calculations required or expected.
P < Pideal
P > Pideal
P = Pideal (no change)
Attraction would decrease the number of collisions. Since it is the collisions that
are the pressure, when there are attractions there would be fewer and less
frequent collisions; therefore, P < Pideal
(3c) Now imagine two “real” gases “X” and “Y”, where each gas is presumed to fulfill
both of the modified postulates (molecules have volume and experience intermolecular
attractive forces). Gases X and Y are in separate containers of the same volume at the
same temperature. Each container contains the same number of moles of gas. A
measurement of the pressure of gas X at these conditions is 15.0 atm. A measurement of
the pressure of gas Y at these conditions is 11.0 atm.
Please answer the following and explain you answers. No calculations required or
expected.
(i) According to the ideal gas law, Pideal for each gas should be the same. Why?
They should be the same because they would only depend upon how many
particles there are  the # of moles would be inversely proportional to the MW.
(ii) The calculated ideal pressure is 12.0 atm. Considering each gas separately, which
“non-ideal” factor (molecular volume or attractive forces) is contributing most to the
measured pressure? How do you know?
Since the pressure of gas X increased from the ideal, the size of the particles
would be larger based on modified postulate 1. With gas Y more intermolecular
attractions would explain the decrease in pressure based on modified postulate 2