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Air pressure in a container then released

  1. Jun 30, 2009 #1
    1. The problem statement, all variables and given/known data

    The air temperature and pressure in a laboratory are 22°C and 1.3 atm. A 2.05 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?


    2. Relevant equations

    ni=PiVi/RTi

    nf=PfVf/RTf



    3. The attempt at a solution

    I easily found ni by plugging in for the top equation (131690 Pa*.00205 m^3) / (8.31 * 295K) and got .110 mols

    then I found the nf by doing the same thing except assuming that Pf is 1 atm (101300 Pa) and that the Tf is 100 deg C(273 K), with constant volume. I got .066 Then I subtracted the ni-nf to get how many mols are left. Why wouldn't this work?
     
  2. jcsd
  3. Jun 30, 2009 #2
    I even tried the final pressure (Pf) the same as the intital pressure (Pi) of 131690 and it still didn't work.
     
  4. Jun 30, 2009 #3
    any ideas?
     
  5. Jul 1, 2009 #4

    LowlyPion

    User Avatar
    Homework Helper

    The final pressure in the vessel will be equal after letting the air escape.

    Consider then how many moles will be in the container when the pressure equalizes.

    With P and V the same then don't you have ...

    nf * Tf = ni * Ti

    nf = .110 moles * (295/373)

    The difference is how much escapes.
     
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