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Mass of air

  1. Nov 27, 2008 #1
    1. The problem statement, all variables and given/known data

    A house has a volume of 840 m^3.
    What is the total mass of air inside the house at 20 degrees celcius? Assume that the pressure is 1.00 atm.

    2. Relevant equations
    Charles law Volume is directly proportional to Temperature
    1 mol of gas at 0 degrees celcius has V = 22.4L


    3. The attempt at a solution
    using charles law i determined that the new volume at 20 degrees celcius would be = (20*22.4 L ) or (22.4 *10^-3 m^3)*(2) = .448 m^3

    number of moles = 840 m^3 / .448 m^3 = 1875 mol
    1 mol of air has a mass of about 29 g = .029 kg

    so m= (1875 mol) (.029kg/mol) = 54.37 kg

    my method apparenlt is faulty because the answer is rejected
    Please help me solve this
     
  2. jcsd
  3. Nov 27, 2008 #2
    I don't think Charles law applies since the house has fixed volume. You just need to use the ideal gas law PV = nRT
     
  4. Nov 27, 2008 #3
    so using PV=nRT

    n = PV / RT = (1.013*10^5 Pa)*(840 m^3) / (8.314)(293.15K) = 3.49131*10^4

    then 3.49131*10^4 mol (0-029 kg/mol)= 1012 kg

    does that seem correct?
     
  5. Nov 27, 2008 #4
    Looks about right, the density of air at sea level (20 C) is 1.2Kg/m^3 IIRC.
     
  6. Nov 27, 2008 #5
    I used that the assumption which was also used in one fo the examples in my book that air is 20 % O and 80% N ..so adding together the percent compositions yielded 29 g ..

    Why did you offer me the density figure of air ?
     
  7. Nov 27, 2008 #6
    Just saying that your density would be about 1.2Kg/m^3, just as a sanity check.
     
  8. Nov 27, 2008 #7
    oh ok. I was making this problem much more complicated than it turned out. Thanks for leading me through !
     
  9. Nov 27, 2008 #8
    uh oh. There's a follow up question : If the temperature drops to -19 C, what mass of air enters or leaves the house?
    it wants delta m
    I used the same equation and substituted T with delta T which would be -39 K ..
    but thats not working ...
     
  10. Nov 28, 2008 #9
    P1V1/T1 = P2V2/T2

    Volume is constant so, you know P1 and T1 and T2, so work out P2. Then use P2 to find out how much air is still in the house at P2, T2, V using PV=nRT remember that n=M/m
     
  11. Nov 29, 2008 #10
    so here it goes :

    P1 = 1.013*10^5 Pa
    T1 = 293.15 K
    T2 = 254.15 K

    P1/T1 = P2/T2
    and i got P2= 8.78233*10^4
    then: PV=nRT
    (8.78233*10^4 Pa)(840m^3)=m(8.314)(254.15K)
    n= 3.49131*10^4

    im confused here after... please explain how to incorporate n=M/m

    can i not just multiply n by 0.029 kg which is mass of 1 mol of air
    and then subtract that amount from the inital mass to get delta m ?
     
  12. Nov 29, 2008 #11
    Doesn't that formula assume that n is constant? i.e. closed system.

    I think you just assume that the pressure is constant, but haven't do any thermodynamics for a while so you have constant pressure and constant volume for there it's rather easy to find the change in n.
     
  13. Nov 29, 2008 #12
    hmmm ... so there was no need to find P2?

    And yes I looked up in my notes the ideal gas law is only valid if the number of particles (n) stays constant
     
    Last edited: Nov 29, 2008
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