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24 July, 23:28

The adult blue whale has a lung capacity of 5.0*103 L. Calculate the mass of air (assume an average molar mass 28.98 g/mol) contained in an adult blue whale's lungs at 0.2 ∘C and 1.04 atm, assuming the air behaves ideally.

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  1. 25 July, 00:35
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    6.7 kg

    Explanation:

    V = 5000 L = 5000 x 10^-3 m^3 = 5 m^3

    T = 0.2 degree C = 273.2 K

    P = 1.04 atm = 1.04 x 1.01 x 10^5 Pa = 1.0504 x 10^5 Pa

    R = 8.314 in SI system of units

    Use the ideal gas equation. Let n be the moles of air occupied in the lungs of whale.

    P V = n R T

    n = P V / R T

    n = (1.0504 x 10^5 x 5) / (8.314 x 273.2) = 231.22

    the mass of one mole of air is 28.98 g

    So, the mass of 231.22 moles of air = 231.22 x 28.98 = 6700.88 g = 6.7 kg
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