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Homework Help: Once more with feeling! another ideal gas question.

  1. Feb 7, 2006 #1
    [SOLVED] once more with feeling! another ideal gas question.

    the pressure inside a sealed can which is kept in a deep freeze at -23C is 0.8x10^5 N/m^2. If the can is placed in water of temperature 27C:

    calculate the increase in pressure inside the can.

    initially, pressure/temperature = 0.8x10^5 / 273-23
    afterwards, pressure 2 divided by (273+27) must equal 320 also. this gives pressure 2 as 96000Pa
    increase in pressure = 0.8x10^5 - 96000
    = -16000
    but if the temperature increases, surely the pressure has to increase? since they are directly proportional??
    or have i taken them away the wrong way round?
  2. jcsd
  3. Feb 7, 2006 #2


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    The pressure HAS increased. 96,000 is larger than 80,000 after all.
  4. Feb 7, 2006 #3
    I think you need much more information to solve this. We have to find the finla temperature of the air in the can to solve the following.
  5. Feb 7, 2006 #4
    so i guess i was taking them away the wrong way around. :yuck: man i'm stupid.
  6. Feb 7, 2006 #5
    Or I think it has to be assumed that the temperature of the system becomes 300K when put in water nadwas 250K initially. Now use the same relation as you said - P1/T1 = P2/T2.
    You have mistook the sign I would say. You made a silly mistake. Your problem solving is okay.
  7. Feb 7, 2006 #6
    yeah that was the equation i was used, just in some weird convoluted form
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