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ChemMatters February 1983 Page 7
© Copyright 1983, American Chemical Society
Calculating Chemistry
Calculating Chemistry—Gases
Trips to the moon, deep-space probes to other planets in our solar
system, and spacecraft moving beyond our solar system encounter the
nearly perfect vaccum of outer space. In this nearly perfect vacuum
typical petroleum oil used to lubricate moving parts of a spacecraft or
satellite evaporates approximately 250 billion times faster than at
average air pressures in the Earth’s atmosphere. In fact, even soft metals
such as copper, tin, and cadmium evaporate at significantly faster rates.
The physical properties of materials in outer space sometimes
change as a consequence of being outgassed, a process that removes
absorbed gases from the surfaces as well as gases dissolved in the
material. The outgassing of some substances, especially plastics and
metals, makes them more brittle.
These and other effects of the nearly perfect vacuum of outer space
must be considered by the scientists and engineers who design our
spacecraft and satellites. Even the best vacuums in research laboratories
seldom exert less pressure than about 1.0 x 10-11 atmospheres.
Can you determine the pressure exerted by the molecules that are
present in outer space? Measurements have shown that there are about
1.0 x 106 molecules/cm3 in the exosphere (outer space at altitudes above
about 700 km). The temperature in the exosphere is about 1200 K. The
mean free path exceeds 13,000 km. Can you calculate the gas pressure at
this altitude in units of atmosphere?
Hint
Use the ideal gas law, PV = nRT. Remember that P represents the
pressure of the gas (in units such as atmospheres), V is the volume
occupied by the gas (in units such as cubic centimeters), n equals the
amount of gas (in moles), T is the absolute temperature of the gas (in
kelvins), and R is a constant. The value of R depends on the units
chosen for the other variables. For the units mentioned here, the value
of R is:
82.1 atm . cm3/mol . K
(The answer can be found on The Puzzle Page, page 16)
SIDE BAR
The Mass of the Earth’s Atmosphere
At first sight, the task of finding the mass of the air surrounding the
Earth would seem a formidable one. Many of you, however, have either
seen or read about Torricelli’s experiment, from which this seemingly
elusive mass can readily be calculated. In this experiment, a long tube
closed at one end and completely filled with mercury is inverted into a
large dish of mercury. At sea level, the mercury inside the tube
stabilizes at a height approximately 76 cm above that of the mercury in
the dish. The familiar explanation is that the downward pressure of the
mercury is exactly counterbalanced by the downward pressure of all the
air subtended by (contained above) an equivalent area of the Earth’s
surface. Thus, the mass of air contained above 1 cm2 of the Earth’s
surface counterbalances 1 cm2 x 76 cm x 13.6 g cm-3, i.e., 1034 grams of
mercury. The mass of the atmosphere is therefore 4π r2 times 1034
grams, since 4π r2 is the formula for the surface of a sphere, r being its
radius. The average radius for the Earth may be taken as 6371 km.
Another way of looking at the problem is to say that the mass of the
Earth’s atmosphere is equal to that of a pool of mercury approximately 76
cm deep and covering the entire surface of the globe. The Christmas tree
ornament to end all Christmas tree ornaments!
Problem 1. Calculate the mass of the Earth’s atmosphere.
Problem 2. Make a guess as to the average depth and extent of the
oceans and calculate their approximate mass.
Problem 3. The volume of a sphere is given by 4/3 π r3. Assume a
reasonable value for the average density of the solid Earth and hence
find its approximate mass.
Problem 4. Can you imagine how the answers to Problems 2 and 3
are more reliably estimated?
Problem 5. To what would the gravitational mass calculated by
Newton’s gravitational law correspond?
Problem 6. The planet Venus has a radius of 6161 km. Space probes
have determined that the atmospheric pressure at its surface is about 96
atmospheres and the temperature about 477 °C. What is the mass of the
Venusian atmosphere?
Problem 7. Why is it that if the barometer falls in one part of the
world, it must rise somewhere else?
(Answers to problems 1, 2, 3, and 6 can be found on The Puzzle Page, page 16)
REFERENCES
Lawrence, Richard M. et al. Space Resources for Teachers—Chemistry. NASA, Washington, D.C.;
1971.
Campbell, J.A. Journal of Chemical Education. Vol. 49, No. 9, p. 624.