The Physics
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Opus in profectus

Gas Laws

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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.

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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 °C, respectively. That's low enough that most humans would suffocate in under thirty seconds and freeze solid in a few hours.

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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 °C. (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).

P1 =  26 kPa
T1 =  −60 °C = 213 K
P2 =  81 kPa
T2 =  ?
   
P1  =  P2  
T1 T2  
26 kPa  =  81 kPa  
213 K T2  
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

There is an 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.

Harold McGee, 1984 (paid link)

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
V1P2
V2  =  101 kPa
V1 22.7 kPa
V2  = 4.4
V1

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…
  1. the volume of one mole of an ideal gas at standard temperature and pressure
  2. the dimensions of a cube that could hold one mole of an ideal gas at STP
  3. the density of air at standard temperature and pressure (air has an average molecular mass of 28.871 u)
  4. the density of air at room temperature (25 °C) and one atmosphere of pressure

solution

  1. Use the complete ideal gas law to determine this somewhat famous number.

    PV =  nRT  
     
    V = 
    nRT
    P
    		 
     
    V = 
    (1 mol)(8.31450 J/K mol)(273.15 K)
    (101,325 Pa)
    		 
     
    V = 0.022414 m3 = 22.414 liter  
     

  2. 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 = ∛(0.022414 m3)
    s = 0.28195… m = 28.2 cm

    For those familiar with the English system of units, this is roughly eleven inches.

  3. 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 =  0.28871 kg/mol
    0.022414 m3/mol
    ρ = 1.29 kg/m3  
     

  4. 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
    1.2881 kg/m3
    		 = 
    273 K
    298 K
    		 
     
    ρ2  =  1.18 kg/m3  
     

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 speck inflated, starting 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 the 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 roughly 380,000 years, temperatures reduced to a relatively cool 3,000 K and the universe finally became diffuse enough for light and matter to live independent lives. When we look out at the universe around us now, all 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 time when nearly every free electron joined up with a hydrogen or helium nucleus — the period of recombination. In the intervening 13,799,620,000 years since recombination the oldest radiation has been stretched by the expansion of space-time 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 currently observable universe…
  1. its volume
  2. its radius
  3. its 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 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…

  1. 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.

  2. 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  =  ⎛
    ⎜
    ⎝    		 r1 ⎞
    ⎟
    ⎠    		 3
    T1 V2 r2
    r1  =  ⎛
    ⎜
    ⎝    		 T2 ⎞
    ⎟
    ⎠    		 ⅓
    r2 T1
    r1  =  ⎛
    ⎜
    ⎝    		 1 ⎞
    ⎟
    ⎠    		 ⅓
    r2 1,100
    r1  =  1
    r2 10.3

    The universe was roughly one tenth its current diameter at the moment of recombination.

  3. 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.