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Calculating the volume in closed container?

  1. Jan 28, 2016 #1
    Hi,
    I have a hard time calculating volume in a closed container.
    Let's say, I have a flask with a volume 0.125ml, which is closed, and the pressure inside is 16.87kPa with temperature 294.7k and I did some reaction to create a gas inside the flask. The new pressure in the flask is 87.19kPa and the temperature is now 296.4k. I want to find the volume of the gas this reaction made, but I have no idea how to do it.

    I tried to use combined law (PV/T=PV/T), but since its a closed system, i doesn't work.
    Please help!!!(;_;)
     
  2. jcsd
  3. Jan 28, 2016 #2

    Simon Bridge

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    Welcome to PF;
    Volume of flask minus volume of non-gaseous stuff in it.
    Care to be taken for porosity of reaction products.

    A lot depends on the exact nature of the reaction, and how rigid the flask is.
     
  4. Jan 28, 2016 #3

    BvU

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    Hello tgt, :welcome:

    You want to order what you have available to solve this. (At the moment it's not a 'complete' problem). Perhaps the homework template helps:

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


    2. Relevant equations


    3. The attempt at a solution


    [edit] I leave this to Simon.
     
  5. Jan 28, 2016 #4

    SteamKing

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    A flask with a volume of 0.125 mL is teeny tiny. Did you use tweezers to pick it up?
     
  6. Jan 28, 2016 #5
    Thanks for the replies.
    I was thinking of using modified form of PV/T = PV/T but wasn't working.
    I asked the prof and he said to use
    V = mCaCO3 · %CaCO3 · R/MWCaCO3 x R/(Pf ′/Tf − Pi ′/Ti)
    , which i believe was given but I wansn't aware of.
    Now I can just put in the values I have and get the V!!
    P.S the flask was 0.125L ;)

    Thanks for your help
    best,
     
  7. Jan 28, 2016 #6
    So what was in the flask before generating the gas? The gas produced will have the volume available in the flask. Seems that you have some calcium carbonate. What do you do to it? Heat is up to release CO2? Or you add some reactant (acid maybe).
     
  8. Jan 28, 2016 #7

    Simon Bridge

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    We cannot help you unless you follow advise - you clearly have a specific problem in mind so please follow the advise in post #3.
    Also remember: a lot depends on the exact nature of the reaction, and how rigid the flask is. This means that you have to supply that information.
     
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