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Vapor Pressure Equilibrium

  1. Nov 17, 2013 #1

    Qube

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    Gold Member

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

    If 5.00 mL of liquid carbon tetrachloride (CCl4, density = 1.587 g/mL) was injected into a sealed 5.00 L flask at 30.0°C, what volume (if any) of the CCl4 would remain as liquid after equilibrium is reached? (the vapor pressure of CCl4 at 30.0°C is 143.0 mmHg)

    2. Relevant equations

    Divide mm Hg by 760 to yield pressure in atm.

    PV=nRT

    Vapor pressure = pressure of liquid solution at equilibrium.

    3. The attempt at a solution

    VP = 0.188 atm = 143 / 760. I know this.

    P(solution) = nRT/V.

    Do I set this equal to the VP? This seems rather unlikely as I'd have to solve for two variables at once - the volume of the liquid solution - and the moles of the liquid left.
     
  2. jcsd
  3. Nov 18, 2013 #2

    Borek

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    Staff: Mentor

    Calculate how much gaseous CCl4 would be present assuming saturated vapor. If it is lower than the amount present - there will be some liquid left. If it is higher than the amount present - all CCl4 will evaporate and there will be no saturated vapor.
     
  4. Nov 18, 2013 #3

    Qube

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    Gold Member

    Using P=nRT/V I found the moles of gaseous CCl4 present assuming saturated vapor. I found there was 0.0377 moles of CCl4(g).

    I then found the grams of CCl4 added to the container by multiplying milliliters by density. I then found the moles of CCl4 present by dividing grams by molar mass. Subtracting the gaseous number of moles from the total number of moles injected we can find the number of moles of CCl4(l). That number is 0.014 moles.

    We can then convert moles back into grams and back into volume given density.

    Question:

    1) Vapor pressure of gaseous CCl4 = nRT/V. What should my value for V be? Container volume, or free volume - 5 milliliters subtracted from 5 liters?

    I'm guessing the amount of CCl4 injected is purposely small so it doesn't matter on a multiple-choice test, and that it's sorta hard to guess the free volume given that we're supposed to be finding the amount of volume of liquid CCl4 remaining.
     
  5. Nov 18, 2013 #4

    Borek

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    Staff: Mentor

    To be exact - neither. It should be 5L minus the volume of the liquid left (which, as you already know, is lower than 5 mL). But error you are making using 5L is negligible.
     
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