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The Isothermal Atmosphere

The gravitational potential energy of a gas molecule of mass  m  at an altitude  h  above sea level is given approximately by   $\varepsilon = m g h$,  where  g = 9.81 m/s2.  Here we neglect the decrease of  g  with altitude, which is a good approximation over a few dozen miles. Next we pretend that the temperature of the atmosphere does not vary with altitude, which is untrue, but perhaps relative to 0 K it is not all that silly, since the difference between the freezing (273.15 K) and boiling (373.15 K) points of water is less than 1/3 of their average. For convenience we will assume that the whole atmosphere has a temperature  T = 300 K  (a slightly warm ``room temperature''). In this approximation, the probability   ${\cal P}(h)$  of finding a given molecule of mass  m  at height  h  will drop off exponentially with  h:

\begin{displaymath}{\cal P}(h) \; = \; {\cal P}(0) \; \exp \left( - {mgh \over \tau} \right)
\end{displaymath}

Thus the density of such molecules per unit volume and the partial pressure  pm  of that species of molecule will drop off exponentially with altitude  h:

\begin{displaymath}p_m(h) \; = \; p_m(0) \; \exp \left( - {h \over h_0} \right)
\end{displaymath}

where  h0  is the altitude at which the partial pressure has dropped to  1 / e  of its value  pm(0)  at sea level. We may call  h0  the ``mean height of the atmosphere'' for that species of molecule. A quick comparison and a bit of algebra shows that

\begin{displaymath}h_0 = {\tau \over m g}
\end{displaymath}

For oxygen molecules (the ones we usually care about most)   $h_0 \approx 8$ km. For helium atoms   $h_0 \approx 64$ km and in fact He atoms rise to the ``top'' of the atmosphere and disappear into interplanetary space. This is one reason why we try not to lose any helium from superconducting magnets etc. - helium is a non-renewable resource!


next up previous
Next: How Big are Atoms? Up: The Boltzmann Distribution Previous: The Boltzmann Distribution
Jess H. Brewer
1998-11-22