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Partial Pressure of a Hg-N2 System
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[QUOTE="wmrunner24, post: 4523265, member: 215311"] Okay. I think this was the missing link I needed. Let me make sure though: n[SUB]Hg[/SUB] = m[SUB]Hg[/SUB]/M[SUB]Hg[/SUB], where M[SUB]Hg[/SUB] is the atomic mass of mercury, 200.59 g/mol n[SUB]Hg[/SUB] = 3.23 x 10[SUP]-4[/SUP] mol The number of moles of nitrogen can be calculated using the ideal gas law: n[SUB]N[SUB]2[/SUB][/SUB] = [itex]\frac{PV}{RT}[/itex] = 0.718 mol, with P = 1 atm, V = 22 L, T = 373 K. I think you were right about the number of moles being ≈1, but because these are not exactly STP conditions, we get a number that is slightly lower. From here, the mole fraction X[SUB]Hg[/SUB] is trivial to calculate. The pressure of Hg in its own container P°[SUB]Hg[/SUB] is given again by the ideal gas law, using the above n[SUB]Hg[/SUB], T, and V from previous calculations. Then the partial pressure P[SUB]Hg[/SUB] is just a matter of using Raoult's Law: P[SUB]Hg[/SUB] = X[SUB]Hg[/SUB]P°[SUB]Hg[/SUB] I got P[SUB]Hg[/SUB] = 2.015 x 10[SUP]-7[/SUP] atm. Is this correct? Thank you so much for your assistance with this problem. [/QUOTE]
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Partial Pressure of a Hg-N2 System
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