altitude - Klemmer

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Altitude Problems
the relationship between
altitude and air pressure
“rule of thumb”
air pressure drops about 25 mm Hg
for every 1000 ft (300 m) of elevation
1. What is air pressure on top of Mt. Battie (740 ft)?
740 ft x (25 mm/1000 ft) = 19 mm drop in pressure
760 mm – 19 mm = 741 mm
2. At what temperature would water boil
on top of Mt. Battie?
From our vapor pressure curve, @ 741 mm
the bp of H2O is about 99°C.
“rule of thumb”
air pressure drops about 25 mm Hg
for every 1000 ft (300 m) of elevation
3. What is air pressure on top of Mt.
Katahdin(4800 ft)?
4800 ft x (25 mm/1000 ft) = 120 mm drop
760 mm – 120 mm = 640 mm
2. At what temperature would water boil
on top of Mt. Katahdin?
From our vapor pressure curve, @ 640 mm
the bp of H2O is about 95°C.
“rule of thumb”
air pressure drops about 25 mm Hg
for every 1000 ft (300 m) of elevation
3. What is air pressure on top of Mt.
Everest(29,000 ft)?
29,000 ft x (25 mm/1000 ft) = 725 mm drop in
pressure
760 mm – 725 mm = 35 mm
Hmm … this is not reasonable: the limit for
humans is about 150 mm!
• The rule of thumb breaks down at high
altitudes.
• The relationship is not a proportional or
linear relationship.
rule of thumb is proportional
(linear)
Altitude 
the actual relationship is
curved – real P at high altitude
is higher than predicted by
linear model
Pressure 
exponential relationships
y = 10x or y = ex
• exponential equations can be used to
represent this kind of a curved
relationship.
• in an exponential equation one of the
variables is an exponent
a word about “e”
• Worried about working in something other
than base 10 (10x)?
• “e” is just an irrational number, like the pi
(p) you use in geometry. It’s named for
Euler, a really good mathematician (look
him up some time …).
• We are using base “e” because it gives
us a simple formula your calculators can
handle quickly and easily.
new pressure formula
P=
-h/7
e
• “P” is pressure, which must be in
atmospheres.
 1 atm = 760 mm
• “h” is height (altitude), which must
be in kilometers.
 1 mile = 1.61 km
 1 mile = 5280 feet
5. What is the pressure at the top of Mt.
Everest (29,000 ft)?
P= e-h/7
h = 29,000 ft
P=?
convert to km
29,000 ft x (1 mile/5280 ft) x (1.61 km/1 mile)
= 8.84 km
P = e-8.84/7 plug in & solve
= 0.283 atm
This is a reasonable answer, and in fact matches
measured pressures. The exponential mathemathical
model works!
6. At what temperature would water boil
on top of Mt. Everest?
0.283 atm x (760 mm / 1 atm) = 215 mm
From our vapor pressure curve, @ 215 mm
the bp of H2O is about 68°C.
7. What is the altitude of Mount Hood, if air
pressure there is 0.613 atm?
P= e-h/7
h=?
P = 0.613 atm
plug in 0.613 = e-h/7
To solve for “h”, we need to get it
out of the exponent. Do do that,
take the natural logarithm (ln) of
both sides : “ln” IS the exponent.
ln (0.613) = ln (e-h/7)
-0.489 = -h/7
multiply both sides by -7 to get h
3.43 km = h

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