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Ideal gas law and moles of air

  1. Nov 19, 2007 #1
    [SOLVED] Ideal gas law

    Please help. I don't know what to do.
    The air temperature and pressure in a laboratory are 23°C and 1.2 atm. A 1.67 L container is open to the air. The container is then sealed and placed in a bath of boiling water. After reaching thermal equilibrium, the container is open. How many moles of air escape?
    Last edited: Nov 19, 2007
  2. jcsd
  3. Nov 19, 2007 #2


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    You could start by quoting the ideal gas law.
    How much molecules of air are in the container at first?
    What does it mean to have thermal equilibrium in the vessel when it's in the boiling water?
    What will happen after you take it out and open it (what does the ideal gas law say about this)?
  4. Nov 19, 2007 #3
    I did the first part using PV=nrt, and found the # of moles to be 0.0825. But, I don't know what is different after the container is open.
  5. Nov 19, 2007 #4
    Now warm up your air till 100°C (or even more at 1,2 bar) and calculate the pressure at constant volume and constant n.

    What happens when you open the container?

    I have a feeling that P will drop to 1,2 bar, but T will drop too. I would say, look at your T-s diagram using isenthalpic or isentropic expansion.

    Then you have P and T and you can calculate n.
    Last edited: Nov 19, 2007
  6. Nov 19, 2007 #5
    I don't understand. I am looking for moles of air that escape. how can it be constant?
  7. Nov 19, 2007 #6
    Because at first the container is closed. They just can't escape!
  8. Nov 19, 2007 #7
    I tried that answer. It is wrong.
  9. Nov 19, 2007 #8
    What did you do exactly?

    You have to calculate n2 and substract n1-n2=....

    So, what's your initial pressure?
    Last edited: Nov 19, 2007
  10. Nov 19, 2007 #9
    I just calculated n1. how do i find n2? I am confused. You said moles were constant.
  11. Nov 19, 2007 #10
    Okay here goes:

    You put the air inside the thing. Then you warm it up. Everything is closed you know!
    Now you warm it up. What will happen first?

    You didn't open the container yet.

    You will find that V is constant because you didn't open the container and the container isn't shrinking or anything.

    n is also constant because the air can't escape.

    T will go up and therefore P will go up. Now I ask you to calculate this P. You'll have to calculate T first of course!! And it isn't 100°C... it's higher, search it up. What's the boiling point of water at 1,2 bar?
    Last edited: Nov 19, 2007
  12. Nov 19, 2007 #11
    I got 153182.03 Pa.
  13. Nov 19, 2007 #12
    Looking normal.

    Now second step:

    You open the container.

    n2 will become lower = the amount of molecules inside the container.
    P will become 1,2 bar again.
    T will become lower. Calculate the T from your T-s diagram.
    V = cte

    n1-n2 = the amount of molecules that went away.

    Now how to read this T-s diagram? You start from ~100°C and 1,5 bar and you go to 1,2 bar. I would try isentropic expansion. So go vertically down till 1,2 bar and look at the T.
    Last edited: Nov 19, 2007
  14. Nov 19, 2007 #13
    If it is thermal equilibrium, wouldn't T be the same. Or is it room temperature?
  15. Nov 19, 2007 #14
    Oh you mean it stays at 100°C? Yeah maybe they meant that.

    So then you would have 1,2 bar, 100°C. => n2 = ....
    n1 = 0,08

    (then I wonder, why did they seal it then?)
    Last edited: Nov 19, 2007
  16. Nov 19, 2007 #15
    Thanks for your help.
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