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  • Apr 26, 2009, 04:50 AM
    roberts61
    mass of air
    The mass of 1 m" of dry air at standard temperature and pressure is 1.29 kg;
    therefore the mass of dry air in a room 8 x 6 x 4 m, at an atmospheric pressure of
    96 kPa and a temperature of 20°C, is:
    400410
    A 195 kg
    B 201 kg
    C 206 kg
    D 219 kg

    if anyone can answer this rubic cube of a question, please include the calculations
    cheers
  • Apr 26, 2009, 05:01 AM
    Perito

    The mass of 1 cubic meter (?) of air is 129 kg. The mass of air is proportional to its volume.

    The "combined gas law" (combining Boyle's, Charles, and Gay-Lussac's laws) is:



    at STP,
    P1 = 1 atm = 101.32500 KPa
    P2 = 96 KPa
    V1 is unknown for the volume specified.
    V2 is given, 8 x 6 x 4 m = 192 m^3
    T1 at STP is 273.15 Kelvins (0 °C).
    T2 is 293.15 Kelvins (20 °C)

    Solve the equation for V1 (the volume at STP). Set up a ratio equation wherein 1 cubic meter is to 129 Kg as V2 cubic meters is to the unknown weight. Solve for the unknown weight.

    I've set it up for you. You can do the math.

    This should have been put under Chemistry or possibly Physics.
  • Apr 27, 2009, 02:26 AM
    roberts61
    Quote:

    Originally Posted by Perito View Post
    The mass of 1 cubic meter (?) of air is 129 kg. The mass of air is proportional to its volume.

    The "combined gas law" (combining Boyle's, Charles, and Gay-Lussac's laws) is:



    at STP,
    P1 = 1 atm = 101.32500 KPa
    P2 = 96 KPa
    V1 is unknown for the volume specified.
    V2 is given, 8 x 6 x 4 m = 192 m^3
    T1 at STP is 273.15 Kelvins (0 °C).
    T2 is 293.15 Kelvins (20 °C)

    Solve the equation for V1 (the volume at STP). Set up a ratio equation wherein 1 cubic meter is to 129 Kg as V2 cubic meters is to the unknown weight. Solve for the unknown weight.

    I've set it up for you. You can do the math.

    This should have been put under Chemistry or possibly Physics.

    Perito
    thanks for your help, I've still not figured out the answer yet, but I'm sure I will.
    thanks again
    peter
  • Mar 23, 2016, 11:02 AM
    SolveWatt
    D.) 219 = 1.29 x 273/293 x 96/101 x 192

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