An athlete has a large lung capacity, 6.8 L?

An athlete has a large lung capacity, 6.8 L. Assuming air to be an ideal gas, how many molecules of air are in the athlete's lungs when the air temperature in the lungs is 37 ∘C under normal atmospheric pressure?

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  • 1 month ago
    Favorite Answer

    n = PV / RT = (1.00 atm) x (6.8 L) / ((0.082057366 L atm/K mol) x (37 + 273) K) =

    0.26732 moles

    (0.26732 moles) x (6.022 x 10^23 molecules/mole) = 1.6 x 10^23 molecules

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  • Dylan
    Lv 6
    1 month ago

    PV = nRT

    n = PV/(RT)

     = (6.8*101.325)/(8.314*(273.15+37))

     = 0.267 mols

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