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Pv=nRT Thermodynamics Question

  1. Feb 7, 2015 #1
    I managed to get the correct answer, however I don't know if my logic was sound.

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

    At ordinary temperature nitrogen tetroxide is partially dissociated (broken up) into nitrogen oxide.

    Into an evacuated flask of 250 cm^3 volume, 0.86g of liquid N2O4 at 0 C is introduced. When the temperature in the bulb has risen to 27 C the liquid has all vaporized and the pressure is 1120mm of mercury. What percent of the nitrogen tetroxide has dissociated in this process?

    2. Relevant equations

    PV=nRT

    3. The attempt at a solution

    First thing, I decided to convert .86 grams of N2O4 to moles. Found it to be 0.00935 moles

    Then, using PV=nRT, determined the total number of "gas" moles in the flask at 27 C (300K). Found this to be 0.0149 moles.

    0.0149 - 0.00935 = 0.00555 moles added ?
    (0.00555/0.00935) x 100% = 59.3%

    Initially this all made sense to me, but now I go back and read the question, it says all my liquid was vaporized (consequently expanded into the flask). So shouldn't my %dissociation be 100%?

    In addition, if my method was correct, what does the 0.00555 actually mean? Is that the amount of N2O4 that didn't/did turn into gas?

    Any help appreciated.
     
  2. jcsd
  3. Feb 7, 2015 #2

    gneill

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    Staff: Mentor

    Presumably there will be an equilibrium of N2O4 as a gas and NO as a gas.

    Edit: @SPhy : On second thought, there must be more species in the flask than just those two alone. What are the possible products when N2O4 dissociates? (equations must balance).
     
    Last edited: Feb 7, 2015
  4. Feb 7, 2015 #3
    For the equation to balance we must require 2NO2 .

    So the number of moles of gas that occupy the flask contain both gases? Thus the .00555 value is actually the disassociated amount?
     
  5. Feb 7, 2015 #4

    gneill

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    Staff: Mentor

    The problem statement indicates that one of the products is NO. So, how many NO's can you get from one N2O4. What's left over?
     
  6. Feb 7, 2015 #5
    If it dissociates to 2NO2 molecules, then when a molecule of N2O4 dissociates, one molecule disappears and two new molecules are formed, so the net is one extra molecule for every molecule of N2O4 that dissociates. So the 0.00555 moles represents the number of N2O4 moles that dissociated (out of the original 0.00935 moles). In the end, the container holds 0.0038 moles of N2O4 and 0.0111 moles of NO2.

    Chet
     
  7. Feb 7, 2015 #6
    Ahhh! Got it, thanks!
     
  8. Feb 7, 2015 #7

    gneill

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    Staff: Mentor

    Hi Chet. The difficulty I see with the problem statement is that they specifically name Nitrogen Oxide (NO) as a product. That leaves one to wonder what happens to the extra oxygens. At room temp they might remain as individual O's or combine to O2.
     
  9. Feb 7, 2015 #8
    Hi Greg. I'm pretty sure that the products of N2O4 dissociation are 2 NO2, and I had the sense that the OP implied that he was told assume this. But, you're right about what you are saying. If there's any O2 formed, this changes the total number of moles formed.

    Chet
     
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