12.
Which, if either, contains a greater number of molecules, one mole of hydrogen or
one mole of oxygen? Which, if either, has more mass?
Answer: One mole of any substance contains Avagadro's number (6.02 x 1023) of
particles. Consequently, one mole of hydrogen and one mole of oxygen contain
the same number of molecules.
Since they both contain the same number of molecules and the mass of an oxygen
molecule is about 16 times that of a hydrogen molecule, one mole of oxygen has
about 16 times the mass of one mole of hydrogen.
13.
Calculate the kinetic energy of 1 mole of nitrogen molecules at 300 K.
Answer:
Model Nitrogen as an ideal gas.
We know KE for one particle = 3/2 k T
So for a mole = 3/2 N k T which is 3/2 R T (k = R / NA)
= (3/2) 8.3145 J/(mol K) * 300 K
= 3742 J / mol (or 3.74 kJ/mol)
*Note that at 300 K, gases that behave ideally have the same energy per mol.
14.
The molar masses are 349 and 352 for 235UF6 and 238UF6 respectively. The
separation of these two isotopes of uranium from a gaseous mixture containing
both of them was critically important to US scientists during WWII. How might
they have used the theories we have discussed here (think rms) to achieve this
task. Draw a sketch of the type of apparatus they may have used.
Answer:
Their root-mean-square speeds are 149 and 148 m/s respectively. This allowed
for enrichment via diffusion separation through a small opening. Due to the small
difference in rms speeds, many enrichment stages were necessary and alternative
methods were developed.
15.
If the speed of each molecule in an ideal gas were tripled, would the temperature
also triple?
Answer: No, the temperature will not triple. When the speed of each molecule is
tripled then the root-mean-square velocity will also triple. Consequently, the
kinetic energy of each molecule will increase by a factor of 9. Therefore, the
Kelvin temperature of the gas will also increase by a factor of 9.
16.
If the temperature of an ideal gas is doubled from 50°C to 100°C, does the
average kinetic energy of the molecules double?
Answer: No, the kinetic energy will not double. In fact, the kinetic energy per
particle is related to the absolute temperature through the relationship:
1/2mv2 = (3/2)kT
So, if the Celsius temperature increases from 50°C to 100°C, the absolute
temperature (or Kelvin temperature) increases from 323K to 373K. Thus, the
fractional increase in kinetic energy is (373/323) ~ 1.15.
17.
Can you explain if it is possible for both the pressure and volume of an ideal gas
to change without causing the internal energy of the gas to change?
Answer: According to the kinetic theory, the energy of N molecules of an ideal
gas at a temperature T (Kelvin!) is (3/2)NkT. Thus, the energy depends only on
the temperature. So, if the pressure and volume are changed isothermally there
will be no change in internal energy. This means that the changes in pressure (P)
and volume (V) have to be such that the product PV remains constant, i.e., if you
double the pressure you must halve the volume.
18.
If the atoms in a container of helium gas have the same rms speed as the atoms of
a container of argon, which one has the higher temperature?
Answer: If the rms speed are the same then the argon atoms have the greater
kinetic energy, because the mass of argon atoms is about 10 times that of helium
atoms. Since the temperature of an ideal gas is directly proportional to the kinetic
energy of the atoms then the temperature of the container of argon atoms (in
Kelvins) is about 10 times that of the container of helium atoms.
19.
If a slippery cork is pushed into an almost full bottle of wine and released it will
slowly slide back out. However, if your pour some wine out and re-insert the
cork, the cork does not slide out. Why?
Answer: When the bottle is very full, the volume of air in the bottle above the
wine is relatively small. Consequently, when the cork is inserted the reduction in
volume that occurs is an appreciable fraction of the original volume. Thus, there
is a correspondingly large increase in pressure, which may be sufficient to push
the cork out. If some of the wine is removed, the volume of air above the wine is
now much larger and so the relative reduction in volume when the cork is pushed
in is now much smaller. As a result, the increase in pressure is much smaller and
not sufficient to push the cork out.
20.
When bubbles rise in a glass of beer, ummmm I mean soda, should they change in
size? Should their speed increase, decrease or remain the same as they rise?
Answer: Yes, as bubbles rise from the bottom of a glass their volume increases!
Why? Because, as a bubble rises it experiences decreasing pressure. Although
we likely haven’t covered this yet, this is due to the fact that there is less ‘weight’
of liquid above it. The ideal gas law tells us that, assuming the beer (I mean
Mountain Dew) is at constant temperature, the product of the pressure and volume
must be constant also. As a result, the reduction in pressure on the bubble will
lead to an increase in volume.
However, the effect of the change in pressure on volume is actually very small in
the case of beer (red bull) in a glass. A much larger effect is the increase in
volume due to an increase in the number of molecules of gas in the bubble! Why
is that? Because a bubble acts as a "nucleation" (or "birth") point for other
bubbles; they quickly merge to form a larger bubble. Kinda like raindrops on a
window….
But what about the speed of a bubble as it rises? The speed will increase also
because the increase in volume leads to an increase in the ‘buoyant force’ on the
bubble. The ‘buoyant force’ is equal to the weight of beer (jolt) displaced, which
increases as the volume of the bubble increases.
Mini-Project
You have recently been promoted to Captain of an Academy Craft for Exploration (ACE)
due to excellent performance on earlier scientific inquiry projects. You are on a mission
to study the unusual properties of a gas reportedly existing on planet Irona in a far off
galaxy in the Leonid cluster.
When you land on the planet, everything seems normal, except the atmosphere seems to
be much ‘thicker’ from what you are used to, and you can barely see your hand when you
hold it in front of your face.
Quickly you decide to collect a sample and return to the ACE to analyze it. In the lab
you subject a sample of the gas to an isothermal pressure test to watch volume response,
and you also took a sample and heated it up under constant pressure conditions.
Question 0. Explain in detail the actual apparatus you could have
used to accomplish these tests. What I am asking for is the
description of the equipment you might use to perform an
isothermal pressure test, and also the equipment you could use to
heat a gas under constant pressure conditions. Sketch the
apparatus also.
On the following page are the test results:
LAB TEST 1 Results
Volume
1000
900
815
770
708
680
617
593
550
520
500
485
466
449
437
426
Volume vs. Temperature
(under constant pressure)
960
volume (mL)
Temperature
200
220
240
260
280
300
320
340
360
380
400
420
440
460
480
500
860
760
660
560
460
360
150
200
250
300
350
400
450
500
550
temperature (degrees K)
LAB TEST 2 Results
Volume
103
122
155
169
200
224
252
280
300
321
352
380
400
425
449
477
500
Volume vs. Pressure (constant T)
600
500
volume (ml)
Pressure
1
1.25
1.5
1.75
2
2.25
2.5
2.75
3
3.25
3.5
3.75
4
4.25
4.5
4.75
5
400
300
200
100
0
0
1
2
3
pressure (atm)
Answer all the questions on the following page.
4
5
6
1. Using the data seen above, determine a mathematical
relationship between Volume and Pressure. Explain if there
are any significant differences between how the atmospheric
gas on Irona behaves vs. those on earth.
2. Repeat the previous task, this time focusing on the
relationship between Volume and Temperature.
3. Are there any limitations on Pressure? What are they? Why
do you think so?
4. Can you develop a mathematical expression for the Ideal Gas
Law on Irona?
5. When we derived the ideal gas law, we stated that the
Kinetic Energy of gas particles is directly proportional to the
temperature. Is this still true on Irona? If not, can you
develop a new one?
Answers:
1. There is a direct relationship as seen from the graph. V = kP. This can be thought
of as V1 / P1 = V2 / P2. The gas is a lot weirder than on earth as it seems to increase in
pressure even as its volume expands.
2. There is an inverse relationship, as VT = Constant (hyperbola). This can be
thought of as V1 T1 = V2T2. The gas is still weird, as volume drops with increasing
temperature.
3. The only pressure limitation is that it could never be zero, or the volume would be
zero.
4. PT = nRV (remember R would now have different units, what would they be?)
5. Well I guess it is true that KE depends on T, it’s just that KE = k / T now. So I’d
say that KE is INVERSELY proportional to T.