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Treating air as an ideal gas of diatomic molecules, calculate how much heat is required to raise the temperature of the air in an 7.47 m by 14.9 m by 2.87 m room from 19.9°C to 21.7°C at 101 kPa. Neglect the change in the number of moles of air in the room

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  1. The volume of the room is
    7.47x14.9x2.87 = 319.4 m^3
    = 319.4*10^6 liters.

    If you assume the number of moles in the room remains the same, you are talking about a constant-volume heating. The specific heat for that, for a diatomic molecule, is (5/2)*R = 4.967 cal/mole C

    The number of liters per mole at 19.9 C and 101 kPa (1 atm) is

    22.4*(293.1/273.3) = 24.0 l/mol

    Number of moles = 319.4*10^6 l/24.0 l/mol = 13.3*10^6 mol

    heat required to raise temperature 1.8 C
    = 1.8 C*4.967 cal/mole C*13.3*10^6 mole = ___

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    posted by drwls
  2. The volume of the room is
    7.47x14.9x2.87 = 319.4 m^3
    = 319.4*10^3 liters.
    not 319.4*10^6 liters.

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    posted by swcb

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