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Proof gas loses 1/273 of its volume - Charle's Law

  1. Nov 29, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that any gas loses 1/273 of its volume at 0 °C, when it is cooled by 1 Celsius degree.

    [itex]V_{1}=?[/itex]

    [itex]V_{2}=?[/itex]

    [itex]T_{1}=0 °C=273 K[/itex]

    [itex]T_{2}=272 K[/itex]

    2. Relevant equations
    I'm assuming I have to use Charle's Law AND starting with any starting arbitrary volume:

    [itex]\dfrac{V_{1}}{T_{1}}=\dfrac{V_{2}}{T_{2}}[/itex]

    3. The attempt at a solution

    So substituting in all the values above and solving for V2 I get:

    [itex]\dfrac{V_{1}}{273}=\dfrac{V_{2}}{272}[/itex]

    [itex]272V_{1}=273V_{2}[/itex]

    [itex]\dfrac{272}{273}V_{1}=V_{2}[/itex]

    [itex]\dfrac{273-1}{273}V_{1}=V_{2}[/itex]

    [itex]\left(\dfrac{273}{273}-\dfrac{1}{273}\right)V_{1}=V_{2}[/itex]

    [itex]\left(1-\dfrac{1}{273}\right)V_{1}=V_{2}[/itex]

    [itex]V_{1}-\dfrac{1}{273}V_{1}=V_{2}[/itex]

    That's as far as I can get and I guess it makes sense? If I start with any [itex]V_{1}[/itex] i subtract that with 1/273th of the initial volume and i get the second volume. Assuming I start with

    [itex]V_{1}=1 L[/itex], the second volume [itex]V_{2}[/itex] would be 0.996 L. The second set of experimental conditions are colder so I expect the volume to decrease following Charle's Law.

    BUT! am I supposed to show explicitly that [itex]\dfrac{1}{273}=V_{2}[/itex] ? How do I show [itex]\dfrac{1}{273}=V_{2}[/itex] explicitly?
     
  2. jcsd
  3. Nov 30, 2011 #2

    Borek

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    Staff: Mentor

    You are almost done. Hint: loses 1/273 volume doesn't mean V2=1/273. Try to express the loss in terms of V1-V2, or V2/V1.
     
  4. Nov 30, 2011 #3
    Volume is proportional to T in Kelvins.
    The volume at 0C (273K) = Vo, the volume at -273C (0K) = 0
    Change in volume ???? Change in temp ????
     
  5. Nov 30, 2011 #4
    The volume at 0K = 0 only holds for ideal gases (hint: nothing is ideal).
     
  6. Dec 1, 2011 #5
    removed
     
    Last edited: Dec 1, 2011
  7. Dec 1, 2011 #6

    Borek

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    Staff: Mentor

    This discussion if off topic, and in no way helpful for the OP. Question doesn't need any information about gas behavior at 0 deg K to be solved.
     
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