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Homework Help: Thermal Physics

  1. Nov 13, 2008 #1
    1. The problem statement, all variables and given/known data
    A steel container contains 300g of ideal gas at a pressure of 1.35 x 106 Pa and temperature of 77oC. When the tank is checked later, the temperature is dropped to 22oC and the pressure has fallen to 8.7 x 105 Pa. How many grams of gas leaked out of the tank?


    2. Relevant equations
    PV = NkT


    3. The attempt at a solution
    first, converting: 77oC = 350.15K and 22oC = 295.15K

    The problem is, i do not know how to link mass to the equation of PV = NkT
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 13, 2008 #2

    Hootenanny

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    Hint: Can you relate N - the number of particles to the mass?
     
  4. Nov 14, 2008 #3
    Number of particles = Total mass / mass per molecule?

    But i do not know the mass per molecule...
     
  5. Nov 14, 2008 #4
    I really need an answer quick, my holidays are ending soon and this is my holiday assignment question that i only stuck with..
     
  6. Nov 14, 2008 #5
    HINT 2 : Avogardro number.......
     
  7. Nov 15, 2008 #6
    I tried it again, closest i could get was,


    PV = nRT

    n = Mtotal / Mpermole = Mtotal/(mNA)

    Where m = mass per molecule..
     
  8. Nov 15, 2008 #7
    Hey

    Assume that each particle have the mass [tex]m_0[/tex]. Can you now relate the mass to the number of particles N?
    Use this relations to express N in terms of m. Now you can relate the mass after [tex]m_2[/tex] to the mass before [tex]m_1[/tex] and this relation is independent of [tex]m_0[/tex] and only depends on known parameters.
    Hopefully this can help you to solve the problem.
     
  9. Nov 15, 2008 #8
    Ok i solved it!!! :D:D The crux of this problem is that, the volume before and after do not change, and i managed to equate mass/molecule.

    PV = NkT, PV = (Mtotal/mo)kT

    After the reaction,

    P'V' = N'kT', P'V' = (M'total/mo)kT

    equating mo = mo,

    I get

    MtotalT/V = M'totalT'/V'

    So i solved for M', giving 229.36 g. So, the mass that leaked = (300 - 229.3) g = 70.6g

    I forgot that V was constant, that was perplexing me throughout!!
     
  10. Nov 15, 2008 #9

    Hootenanny

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    Good job :approve:
     
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