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Air has an average molar mass of 29.0 g/mol. The density of air at 1.00 atm and 30 °C is:?

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Air has an average molar mass of 29.0 g/mol. The density of air at 1.00 atm and 30 °C is:?

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  1. I think that the volume of 1 mole of a gas at 1 atm and 0C is 22.4 L. So at 30C it would be 303/273 X 22.4 = 24.9 L. Density would be 29.0/24.9 = 1.16 g/L


  2. We can use the Ideal Gas Law (PV = nRT) to solve this problem, but we'll have to first determine the volume of air, V.

    With the information given:

    moles = mass / (molar mass)

    => mass = moles * molar mass, which I'll rewrite as:

    m = n * M

    density = mass / volume

    => volume = mass / density, which I'll rewrite as:

    V = m / d

    From above, we know that m = nM, so:

    V = nM / d

    Since PV = nRT:

    P * (nM / d) = nRT

    => nPM / d = nRT

    => PM / d = RT

    => d = PM / RT

    d = PM / RT

    where,

    d = Density in g/L

    P = Pressure = 1.00 atm

    M = Molar Mass = 29.0 g/mol

    R = Ideal Gas Constant = 0.0821 L*atm/K*mol

    T = Temperature = 30°C = 303.15 K

    => d = (1.00 * 29.0) / (0.0821 * 303.15)

    => d = 1.165 g/L

    Therefore, the density of the air is: 1.165 g/L

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