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Partial Pressure of a Hg-N2 System
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[QUOTE="Chestermiller, post: 4523617, member: 345636"] I'm not quite sure what you did after getting the number of moles of the Hg vapor that were mixed with the 0.718 moles of N2 in the gas at 100 C, but Raoult's law was not the thing to use. N2 is essentially insoluble in the liquid Hg, so the mole fraction of Hg in the liquid phase is 1.0 (i.e., virtually pure Hg). But, getting back to the gas phase, you can calculate the mole fractions of N2 and Hg in the gas phase at 100 C (since you know how many moles of each is present), and you know the total pressure, so you can calculate the partial pressure of each gas (equal to the mole fraction times the total pressure). Because the liquid Hg was pure, its partial pressure for the gas mixture at 100 C is equal to its equilibrium vapor pressure. [/QUOTE]
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Partial Pressure of a Hg-N2 System
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