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SoggyBottoms
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Homework Statement
Air at sealevel is composed of 80% nitrogen and 20% oxygen, but at the top of Mount Everest the composition of air is different. Give an expression for the ratio ([itex]f_O[/itex]) of oxygen in the air as a function of height z above sealevel, assuming the temperature T is constant and the air is an ideal gas. The gravitational acceleration is g and the mass of a molecule of nitrogen is [itex]m_N[/itex] and the mass of a molecule of oxygen is [itex]m_O[/itex].
The Attempt at a Solution
The potential energy of a molecule of oxygen is [itex]m_O g z[/itex], so the probability of finding it at height z is: [itex]P_{mO}(z) \propto e^{-\frac{m_O g z}{k_B T}}[/itex]. Similarly for nitrogen: [itex]P_{mN}(z) \propto e^{-\frac{m_N g z}{k_B T}}[/itex]. So the ratio is simply [itex]f_O = \frac{e^{-\frac{m_O g z}{k_B T}}}{e^{-\frac{m_N g z}{k_B T}}} = e^{-\frac{(m_O - m_N)gz}{k_B T}}?[/itex]