The Wikipedia uses the value of 5.14×1018 kg (5 140 gigatonnes) for the mass of the Earth's atmosphere. Every 22. 4 liters will weigh about 30g (1 g-mol air = about 30 g = 22.4 liters)
There would be 22.4 x (5.14 ×1021/ 30) liters of air (about 4 x 1021 liters) at STP (Standard Temperature and Pressure).
However, the Earth's atmosphere varies in temperature and pressure as you travel from 0 miles to 60 miles above sea level. Pressure is noticeably different in the mountains (a few miles above sea level) so that cooking instructions must be changed. This is even more pronounced by the time one reaches the stratosphere at 5-10 miles altitude.
Moreover, the air we breathe does not circulate much above the tropopause. So a closer, though still incorrect, approximation to your question (using the same Wikipedia numbers) would be to assume the air we breathe goes up 11 km in altitude (tropopause varies from 7-17 km above sea level; using a lower than middle number to account for some of the drop in pressure and for mountains) over 197 million square miles of Earth's surface area. This makes a total of
1.3x10^9 mi^3 (cubic miles) [WolframAlpha.com]
of space, or
5.6 × 1021 litersup to the tropopause. Air has stopped acting as we understand air near the ground to act by this height; but let's compute the height at the exobase (the height at which air ceases to act like a fluid because the particles are so far apart). At 60 miles times 197 million square miles that would total4.9x10^22 liters.
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