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Change in pressure using PV=nRT

  1. Nov 17, 2008 #1
    1b]1. Gas is confined in a tank at a pressure of
    6.5 atm and a temperature of 9.4◦C.
    If half of the gas is withdrawn and the
    temperature is raised to 62.4◦C, what is the
    new pressure in the tank? Answer in units of

    2. PV=nRT
    Ti= 9.4 C + 273= 282.4K
    Tf= 62.4 C +273= 335.4K

    3. PV= nRT, n and R are constants, therefore I used PV= T
    So, initially, 6.5atmVi= 282.4K giving, Vi = 43.446

    I then used (1/2) 43.446 as Vf and solved for Pf
    Pf= Tf/ Vf=> 15.439 atm

    But this isn't right, I know. What am I missing?? .
  2. jcsd
  3. Nov 17, 2008 #2


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    My thought would be that "half of the gas is withdrawn" means that the number of particles n is halved, and the volume is kept constant, not the other way around.
  4. Nov 17, 2008 #3
    It doesn't mention the volume has been reduced by a half, the amount of gas has been reduced by a half so you need to use 0.5n.
  5. Nov 17, 2008 #4
    Thanks to the both of you!
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