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Question related to pressure

  1. Feb 15, 2009 #1
    I came across this thought experiment, and I even think I have a solution for it, but my book insists that I am wrong.

    Consider a completely airtight shere, filled with air. Applying any sort of pressure outside the sphere surely cannot increase the airpressure inside the sphere? (assuming that the volume of the sphere is constant). As I see it, only a change of volume, amount of substance or temperature can change the pressure of the air. I believe that the book incorporates the method of simply adding pressures together. This works fine when asking for a total pressure at a certain depth at sea (total pressure equals the airpressure and the hydrostatic pressure), but surely it cannot work for the "sphere problem"

    Thoughts?
     
  2. jcsd
  3. Feb 15, 2009 #2
    Hello, and welcome to PF.

    I don't really follow your question. I agree with the statement that "only a change of volume, amount of substance or temperature can change the pressure of the air" - PV = nRT!

    Does the problem state that the sphere's volume doesn't change, or that its made of Kryptonite (near infinite elasticity)? Can you provide a sketch of the problem?
     
  4. Feb 15, 2009 #3

    russ_watters

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    Staff: Mentor

    Is this a gage pressure vs atmospheric pressure issue? The last sentence implies that...
     
  5. Feb 16, 2009 #4
    Sorry for being unclear, maybe this sketch helps.

    We have two airtight containers, filled with a gas. Both have the same volume, same temperature, and the pressure inside/outside the containers are equal. They even have exactly the same ammount of substance. (In other words we have to identical containers). We now place a (very) heavy weight on the second container. What is the new pressure inside the container? Is it larger because we have added pressure on the surface of the container or is it exaclty the same as before? (Assume that the second container doesn't undergo any deforamtion as weight is added on it, since this would imply that its volume would change and hence the pressure would change).

    http://img261.imageshack.us/my.php?image=pressurerh5.png
     
  6. Feb 16, 2009 #5
    http://img261.imageshack.us/my.php?image=pressurerh5.png
     
  7. Feb 16, 2009 #6

    stewartcs

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    Science Advisor

    I can't see your sketch, however, based on your description and assumption that no deformation occurs (and that it is a closed adiabatic system), the pressure inside the container would not change.

    CS
     
  8. Feb 16, 2009 #7
    You are correct and so are the above replies.
     
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