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Partial Pressure of a Hg-N2 System

  1. Oct 1, 2013 #1
    1. The problem statement, all variables and given/known data

    At 1 atm of pressure a volume of 22 liters of N2 gas is passed in a closed system over a boat containing Hg liquid at 100°C. The flow of N2 is slow to allow the gas to become saturated with mercury. At 20°C and 1 atm, the nitrogen was found to contain 0.0647g of Hg. Calculate the vapor pressure of Hg at 100°C.

    2. Relevant equations

    Raoult's Law (Ideal Solution):
    PHg = XHgHg

    Ideal Gas Law:

    PV = nRT

    3. The attempt at a solution

    My endgame here is to use Raoult's Law to calculate the partial pressure of mercury. To do that, though, I will need the mole fraction XHg and the pressure the mercury would exert if it were alone in the container P°Hg. As far as the latter goes, I suspect I can use the ideal gas law to calculate the pressure of mercury when nitrogen is not involved.

    This is where I run into problems. I'm not sure how to synthesize the information given into the information I need. I could calculate a mole fraction for the system at 20°C, again via the ideal gas law, but I'm not sure how I can relate this to either P°Hg or XHg for the same system at 100°C.

    Any insight anyone can provide would be incredibly helpful and greatly appreciated.
     
  2. jcsd
  3. Oct 1, 2013 #2
    I think that the problem statement meant to say that 0.0647 gm of mercury condensed out of the N2 when the gas was cooled to 20 C. I also think it is safe to say that this was virtually all the Hg that had evaporated into the 22 liters of N2 (assuming that the Hg vapor pressure to 20 C is much lower than the vapor pressure at 100C). How many moles of Hg condensed? How many moles of N2 were in the mixture (it should be on the order of 1 mole)? What was the mole fraction of Hg in the gas phase when it got saturated with Hg at 100C? Treating the gas phase as an ideal gas, what was the vapor pressure of Hg at 100C?
     
  4. Oct 1, 2013 #3
    Okay. I think this was the missing link I needed. Let me make sure though:

    nHg = mHg/MHg, where MHg is the atomic mass of mercury, 200.59 g/mol

    nHg = 3.23 x 10-4 mol

    The number of moles of nitrogen can be calculated using the ideal gas law:

    nN2 = [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 XHg is trivial to calculate. The pressure of Hg in its own container P°Hg is given again by the ideal gas law, using the above nHg, T, and V from previous calculations. Then the partial pressure PHg is just a matter of using Raoult's Law:

    PHg = XHgHg

    I got PHg = 2.015 x 10-7 atm. Is this correct?

    Thank you so much for your assistance with this problem.
     
  5. Oct 2, 2013 #4
    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.
     
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