« on: 16/04/2020 03:35:49 »
Even in a sealed container with no temperature gradient, you won't get a useful degree of separation.What do you think will happen if I put a mixture of Helium, Nitrogen, and SO2 with equal volume and pressure inside a 10 meters vertical pipe. Will we get the same composition between top and bottom part?
That will depend on the temperature, pressure, and gravity.
But assuming that you mean Earth gravity (9.8 ms2), 1 atm, and 300 K (27 °C), then the difference in gravitational potential energy between having 1 mole of He at the very top of the 10 m tube and 1 mole of SO2 at the very bottom, and having 1 mole of He at the very bottom of the 10 m tube and 1 mole of SO2 at the very top is 9.810(0.0640.04) = 5.88 J/mol. This could be viewed as the maximal enthalpy of gravitational un-mixing in a 10 m column (being favorable by 5.88 J/mol, so ΔH > 5.88 J/mol).
But there is also an entropic cost. http://hyperphysics.phy-astr.gsu.edu/hbase/Therm/entropgas.html
The change of entropy of an ideal monoatomic gas with change in volume is (ideal gas is a good model for He, not so great for SO2, but doesn't really matter here... the errors will cancel):
ΔS = Nk*ln(Vf/Vi)
so if we had 1 mole of each gas being restricted to 1/2 the volume of the tube (from having had the whole tube), then for each mole of gas we would have:
ΔS = Nk*ln(1/2) = 5.76 JK1mol1
and therefore ΔS = 11.52 JK1mol1 (because we are restricting two moles of gas, the He and the SO2)
thus at 300 K:
ΔGun-mixing = 5.88 Jmol1 (300 K)(11.52 JK1mol1 ) = +3450 J/mol
(recall that ΔG must be negative for a process to be spontaneous)
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