Practice
practice problem 1
The graphs to the right show the pressure and temperature inside the cabin of a commercial jet airliner on a two hour flight. As can be seen in the pressure graph, the plane spent half an hour on the ground waiting to take off, fifteen minutes ascending, an hour and a quarter cruising, fifteen minutes descending, and fifteen minutes on the ground approaching the terminal. The interesting segment from a gas laws perspective occurred when the plane was cruising.
Jet aircraft of the type from which this data was collected typically fly at altitudes greater than 10,000 m; well above the vertical limit of human survivability. Pressure and temperature outside the cabin on this flight are about 26 kPa (one-quarter atmosphere) and −60 ℃, respectively. That's low enough that most humans would suffocate in under thirty seconds and freeze solid in a few hours.
Although the environmental conditions outside are harsh, life inside a jet isn't all that bad. While passengers would find it most comfortable if the cabin was kept at one atmosphere there are engineering and economic reasons to maintain the pressure at a slightly lower value. During this flight the cabin pressure was kept at a compromise value of 81 kPa — much higher than the pressure outside, but still lower than what most people live under. (80 kPa is the average atmospheric pressure at an elevation of 2000 m.) Temperature hardly varied at all, staying nearly constant at 23 ℃. (People are far more sensitive to temperature changes than to pressure changes.)When air is drawn into the cabin from outside, to what temperature does it rise after it has been compressed?
solution
Despite the lengthy introduction, this is a comparatively
simple question. Temperature and pressure are directly proportional (if we assume
that volume remains constant).
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| P1 |
= |
26 kPa |
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P1 |
= |
P2 |
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| T1 |
= |
−60 ℃ = 213 K |
T1 |
T2 |
| P2 |
= |
81 kPa |
26 kPa |
= |
81 kPa |
| T2 |
= |
?? |
213 K |
T2 |
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T2 |
= |
663 K |
= 390 ℃ |
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This temperature is about the same as one would find inside a pizza oven.
Since the volume of the air does not remain constant but is somewhat reduced
during pressurization, the actual temperature of the air drawn into the cabin
is even higher than we've calculated. To prevent roasting the passengers
to golden brown crispness this air must be refrigerated — a comparatively
expensive procedure given the size and weight limitations imposed by flight.
Thus, a significant portion of the air breathed in a typical commercial airliner
is recirculated. That is, the air exhaled by the passengers is stirred up
by the plane's ventilation system with a small amount of fresh, refrigerated
air is continuously added to the mix. This recirculation is what makes airplane
air so particularly nasty.
practice problem 2
There is a truly excellent book on food science written by Harold McGee called
On Food and Cooking: The Science and Lore of the Kitchen
. Mr. McGee's
book is vast in scope and interesting on every page. There is one peculiar essay in
the chapter on legumes called "The Problems of Legumes and Flatulence" that
lends itself particularly well to the gas laws.
We are indebted to high-altitude aircraft flight and the space program
for the recent spate of interest in flatulence. After World War II, it
appeared that intestinal gas might prove a serious problem for test pilots.
The volume of a given amount of gas increases as the pressure surrounding
it decreases. This means that a pilot's intestinal gas will expand as
he flies higher into the atmosphere in an unpressurized cockpit. At 35,000
feet, for example, the volume will be 5.4 times what it would be at sea
level. The resulting distention could cause substantial pain ….
So the word went out across the land: study flatulence.
Verify Mr. McGee's claim.
solution
For those of you who are still a bit unclear, legumes are the third largest
family of flowering plants (which includes beans, peas, and peanuts) and
flatulence is the medical term for intestinal wind (which is the polite term
for farts — yes, the gas laws apply even to "gas"). The typical
atmospheric pressure at sea level is 101 kPa. According to the standard
atmospheric tables, at 35,000 feet (11,000 m) typical atmospheric
pressure is more like 22.7 kPa. If we assume that a person's body temperature
doesn't change much while flying, its volume would be inversely proportional
to the external pressure acting on it.
| P1V1 = P2V2 |
⇒ |
V2 |
= |
P1 |
= |
101 kPa |
= 4.4 |
| V1 |
P2 |
22.7 kPa |
At 35,000 feet the volume would be 4.4 times what it would be at sea level,
not 5.4 times as Mr. McGee claims.
practice problem 3
Determine …
- the volume of one mole of an ideal gas at standard temperature and pressure,
- the dimensions of a cube that could hold one mole of an ideal gas at STP,
- the density of air at standard temperature and pressure (air has an average
molecular mass of 28.871 u), and
- the density of air at room temperature (25 ℃) and one atmosphere
of pressure.
solution
- Use the complete ideal gas law to determine this somewhat famous number.
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| PV = |
nRT |
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| V = |
nRT |
= |
(1 mol)(8.31450 J/K mol)(273.15 K) |
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| P |
(101,325 Pa) |
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| V = |
0.022414 m3 = 22.414 liter |
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- The volume of a cube is the cube of one side. Conversely the side of
a cube is the cube root of its volume.
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| V = |
s3 |
| s = |
∛V = ∛(0.022414 m3) |
| s = |
0.28195 … m = 28.2 cm |
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For those familiar with the English system of units, this is roughly
eleven inches.
- Start with the definition of density and substitute the value just
computed for volume. Air is a mixture of gases, so its molecular weight
is the weighted average of its constituent molecules. Watch the units.
Molecular weights are almost always given in grams per mole, but the
SI unit of mass is the kilogram.
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| ρ = |
m |
= |
0.28871 kg/mol |
= 1.29 kg/m3 |
| V |
0.022414 m3/mol |
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- This is a problem of proportionality. Density and volume are inversely
proportional for a constant mass of gas while volume and temperature
are directly proportional at constant pressure. Thus density and temperature
are inversely proportional when mass and pressure are constant. Be sure
to use absolute temperatures.
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⎫
⎪
⎬
⎪
⎭ |
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| ρ |
∝ |
1 |
(m constant) |
⇒ |
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| V |
ρ ∝ |
1 |
(m & P constant) |
| V |
∝ |
T |
(P constant) |
T |
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| ρ2 |
= |
T1 |
⇒ |
ρ2 |
= |
273 K |
⇒ |
ρ2 = 1.18 kg/m3 |
| ρ1 |
T2 |
1.2881 kg/m3 |
298 K |
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practice problem 4
According to the current interpretation of the big bang theory, the universe
began some
13.8 billion years ago when space, time, matter, and energy arose spontaneously
in an infinitesimally small region of space called a singularity. Luckily
for us, this tiny little speck of something in a vast sea of nothing quickly
blew up (or inflated in more appropriate terms) and started itself on a journey
of cosmic expansion that continues to this day. For the first 380,000 years
of its existence, the space, time, matter, and energy of the universe were
so dense that everything was effectively opaque. Light and other electromagnetic
waves were tightly bound to the matter of the universe much like electrons
in a wire are tightly bound to the metallic network of the metal from which
it was constructed. (Have you ever been shocked while passing an electrical
outlet in your house? No, of course not. And why not? Because the electrons
are bound to the atoms of the solid metal conductor quite tightly.) After
expanding for 380,000 years, when temperatures had reduced to a relatively
cool
3000 K, the universe finally became thin enough for light to separate
from matter and live an independent life. When we look out at the universe
around us now the radiation we see is at least
380,000 years younger than the universe as a whole. Everything before
this moment is lost in time. This is also the moment when the energy density
of the universe dropped low enough to allow nearly every free electron to
join up with a hydrogen or helium nuclei. Since these entities were all created
at the same moment, this event is known as the period of recombination. In
the intervening 13,699,620,000 years since recombination the oldest
radiation has been diluted to the point where it is no longer visible, but
instead
lies wholly
within the microwave part of the spectrum. This cosmic microwave background
radiation (CMB) has been chilled to a mere 2.725 K by the overall expansion
of the universe. Determine the following quantities at the moment of recombination
in comparison
to their current value for the universe as a whole …
- volume,
- radius, and
- density.
solution
If your life in physics is entirely determined by your ability to solve problems,
then you must surely regard the previous paragraph as 10% useful and 90%
wasteful information. If you appreciate physics as an opportunity to be exposed
to and to learn new ways of thinking, then you must surely view the previous
paragraph as wholly incomplete and lacking in interesting detail. If you
believe that
all knowledge prior to 6000 years before present is nonsense for some
imagined religious reason, then you must surely believe that reading every
word is a waste of your precious time. Since one of the fundamental principles
of writing is write to your audience and since my audience has multiple reasons
for using this page, I am certain that no one will be entirely satisfied
with the information presented or the nature of the questions asked. Nonetheless,
let's answer them …
- The basic principle behind this problem, and the underlying assumption
given that it appears in this section, is that the universe is some kind
of ideal gas and that it obeys one of the basic gas laws. My guess would
be that temperature and volume are directly proportional when pressure
is constant. (A more realistic appraisal would be that the transition
is adiabatic — with no heat transfer between the universe and whatever's
outside it; which would be nothing.)
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| V2 |
= |
V1 |
⇒ |
V1 |
= |
T2 |
= |
2.75 K |
= |
1 |
| T2 |
T1 |
V2 |
T1 |
3000 K |
1100 |
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The universe was roughly one thousandth its current volume at the time
of recombination.
- Knowing the way the volume changed, we need only use the fact that volume
is proportional to the cube of radius to determine the change in radius.
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| V1 |
= |
⎛
⎝ |
r1 |
⎞
⎠ |
3 |
⇒ |
r1 |
= |
⎛
⎝ |
T2 |
⎞
⎠ |
⅓ |
= |
⎛
⎝ |
1 |
⎞
⎠ |
⅓ |
= |
1 |
| V2 |
r2 |
|
r2 |
T1 |
|
1100 |
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10.3 |
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The universe was roughly one tenth its current diameter at the moment
of recombination.
- Density is inversely proportional to volume, so this last part is quite
easy to calculate.
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| ρ = |
m |
⇒ |
ρ1 |
= |
V2 |
= |
1 |
= |
1100 |
| V |
ρ2 |
V1 |
1100 |
1 |
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The density was roughly one thousand times greater at the moment of recombination
than it is now.
