Boltzmann distribution of two different gases

AI Thread Summary
In a uniform gravitational field, the density distribution of two different ideal gases with varying molar masses will change with height according to the Boltzmann distribution, expressed as exp(-m*g*h/T). This indicates that the density is proportional to the molecular mass, with heavier gases exhibiting a steeper density gradient. The discussion highlights that the assumption of superposition implies no interaction between the gas molecules, raising questions about pressure effects. Additionally, the equilibrium state is influenced by the total amount of gas present, necessitating a normalization factor. Understanding these dynamics is crucial for analyzing gas behavior in gravitational fields.
oguz
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hi everyone,

consider two different masses of ideal gases with different molar masses, we're putting them in a uniform gravitational field and wait until they come to their equilibrium states. how would the density distribution change with height in this case?

( i came out with this question while working on a problem in "problems on general physics" by irodov. the answer to the problem seemed to be merely superposing two different distributions but doesn't this imply that molecules of both gases do not interact? if they don't, I'm asking should not they be stressing pressure on each other?)
 
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oguz said:
hi everyone,

consider two different masses of ideal gases with different molar masses, we're putting them in a uniform gravitational field and wait until they come to their equilibrium states. how would the density distribution change with height in this case?

( i came out with this question while working on a problem in "problems on general physics" by irodov. the answer to the problem seemed to be merely superposing two different distributions but doesn't this imply that molecules of both gases do not interact? if they don't, I'm asking should not they be stressing pressure on each other?)

ISTR this problem is discussed in the Feynman lectures, where it is stated that the density in equilibrium is proportional to the Boltzmann probability, i.e. exp(-m*g*h/T), where m is the molecular mass (T is in energy units; use kB for unit conversion). (This is assuming an isothermal atmosphere, which is not in mechanical equilibrium.) There will be a normalization factor that depends on the total amount of has. The higher molecular weight gas has a higher rate of density change with height.

BBB
 
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