practice problem 1
Despite the lengthy introduction, this is a comparatively simple question. Temperature and pressure are directly proportional (if we assume that volume remains constant).
|P1 =||26 kPa|
|T1 =||−60 °C = 213 K|
|P2 =||81 kPa|
|26 kPa||=||81 kPa|
|T2 = 663 K = 390 °C|
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
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.
Harold McGee, 1984 (paid link)
Verify Mr. McGee's claim.
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|
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
- 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)
- the density of air at room temperature (25 °C) and one atmosphere of pressure
Use the complete ideal gas law to determine this somewhat famous number.
PV = nRT V = nRT P V = (1 mol)(8.314 462 618 J/K mol)(273.15 K) 101 325 Pa V = 0.022 414 m3 = 22.414 liter
The volume of a cube is the cube of one side. Conversely the side of a cube is the cube root of its volume.
V = s3 s = ∛V = ∛(0.022 414 m3) s = 0.281 95… m = 28.2 cm
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.
ρ = m V ρ = 0.288 71 kg/mol 0.022 414 m3/mol ρ = 1.29 kg/m3
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.
ρ ∝ 1 (m constant) ⎫
⇒ V ρ ∝ 1 (m & P constant) V ∝ T (P constant) T ρ2 ρ1 = T1 T2 ρ2 1.288 kg/m3 = 273 K 298 K ρ2 = 1.18 kg/m3
practice problem 4
- its volume
- its radius
- its density
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 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 of events prior to 6,000 years before present is nonsense, 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 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 by definition.)
V2 = V1 T2 T1 V1 = T2 V2 T1 V1 = 2.75 K V2 3,000 K V1 = 1 V2 1,100
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.
T2 = V1 = ⎛
3 T1 V2 r2 r1 = ⎛
⅓ r2 T1 r1 = ⎛
⅓ r2 1,100 r1 = 1 r2 10.3
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.
ρ = m V ρ1 = V2 ρ2 V1 ρ1 = 1,100 ρ2 1
The universe had one thousand times greater density at the moment of recombination than it does now.