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Homework Help: Thermo Question

  1. Jan 26, 2013 #1
    Thermo Question!!

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

    CCl3F and CCl2F2 have been linked to ozone depletion in Antarctica. As of 1994, these gases were found in quantities of 261 and 509 parts per trillion (1012) by volume (World Resources Institute, World resources 1996–97). Compute the molar concentration of these gases under conditions typical of (a) the mid-latitude troposphere (10°C and
    1.0 atm) and (b) the Antarctic stratosphere (200 K and 0.050 atm).

    2. Relevant equations

    3. The attempt at a solution
    Okay ppt is hurting my head. Basically ppt = 1ng/L

    CCl3F MW = 137.37 g/mol

    So I did (137.37 g/mol)/(251g/1x10^9L )to get mol/L

    But I end up getting 0.526 nL/mol..

    So when I go to plug in n = pV/RT to solve for my molecular weight concentration I don't know how the units are going to cross out... because its 0.526 nL/mol..
    Can I just multiply 0.526(1x10^-9nL= to get L? Or is that against multiplication rules. I've never really dealt with ppt.
  2. jcsd
  3. Jan 26, 2013 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Thermo Question!!

    Hello Workout,

    Just stick to the definition of ppt given in your problem statement. 1 part per trillion is exactly what is says. It's just a RATIO. So, for every trillion molecules of air*, there is 1 molecule of the chloro-fluoro-carbon, or whatever that is (I'm not a chemist).

    So what you have to do is use the ideal gas equation to compute how many moles of air you have in a volume of 1 litre under each set of atmospheric conditions. Based on that number of moles of air, you can figure out the number of moles of the CFC in question (becuase you know the ratio in ppt). This will give you, directly, the number of mol/L of the CFC.

    *note: it was 1 part per trillion by volume, not by mass. Since, in a gas at a given temperature and pressure, the volume occupied is proportional to the number of particles, this means that it's one litre for every trillion litres, or equivalently, one particle for every trillion particles. In contrast, if it had been one ppt BY MASS, then it would not be one particle for every trillion particles, it would have been one gram for every trillion grams. If the substance is heavier than air, this would mean *fewer* than one particle for every trillion particles.
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