Answer 1:
Interesting question! What you are talking
about is also referred to as the "mean free
path," which, as you've said, is the average
distance a gas molecule can freely travel before
hitting another one. I have detailed the equations
below required for finding the mean free path as a
function of altitude, however, I suggest you look
at all of the provided links so that you know what
every variable means and what assumptions these
equations make. mean_free_path = k_B * T /
( * d2 * sqrt(2) *
pressure)where k_B is Boltzmann's
constant, T is temperature, d2 is the
diameter of the gas molecule squared, sqrt(2) is
the square root of 2, and pressure as a function
of altitude is equal to: pressure as a
function of altitude = p_at_sealevel * (1 - (L *
height_above_sealevel) / T_at_sea_level) ((g
* MM)/(R * L))
Please note
that ((g * MM)/(R * L)) is an
exponent
where p_at_sealevel and
T_at_sealevel are the pressure and temperature of
air at sea level, g is the acceleration due to
gravity, MM is the molar mass of the gas, R is the
gas constant, and L is the temperature lapse
rate. Depending on what you're studying, you
may want to plug in average values of the molar
masses and molecular diameters where
required. For the molecular diameters, you
can either look them up or estimate them in the
following way: Take the density of a gas and
divide it by the molar mass to get the molar
density or moles per volume of the gas. Take the
inverse of the molar mass times Avogadro's number
to get the volume of a single gas molecule. Then
by reasonably assuming that the gas molecule is a
sphere, you can solve for the gas molecule's
radius with the following
relationship: radius = ((3/4pi) *
MM*(Avogadro's # / density_of_gas))
(1/3)There's a lot of tricky units in
all of the above equations and dealing with
Boltzmann's constant is always a pain, so please
work carefully and patiently and you'll get it!
Compare the answers you get with Wikipedia, which
says that gas at room temperature and pressure
will have a mean free path of about 70 nanometers.
It took me about 5 or 6 tries to verify
Wikipedia's answer, so please remember that
science requires patience! |