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3 April, 16:22

The adult blue whale has a lung capacity of 5.0*103 L5.0*103 L. Calculate the mass of air (assume an average molar mass 28.98 g/molg/mol) contained in an adult blue whale's lungs at 0.2 ∘C∘C and 1.07 atmatm, assuming the air behaves ideally.

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  1. 3 April, 16:39
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    The mass of the air is 6920.71g

    Explanation:

    Step 1:

    Data obtained from the question. This includes the following:

    Volume (V) = 5.0x10^3 L

    Molar Mass of air (M) = 28.98 g/mol

    Temperature (T) = 0.2°C

    Pressure (P) = 1.07 atm

    mass air (m) = ?

    Number of mole (n) = ?

    Recall:

    Gas constant (R) = 0.082atm. L/Kmol

    Step 2:

    Conversion of celsius temperature to Kelvin temperature.

    K = °C + 273

    °C = 0.2°C

    K = °C + 273

    K = 0.2°C + 273

    K = 273.2 K

    Therefore, the temperature (T) = 273.2 K

    Step 3:

    Determination of the number of mole of air.

    Applying the ideal gas equation PV = nRT, the number of mole n, can be obtained as follow:

    PV = nRT

    1.07 x 5.0x10^3 = n x 0.082 x 273.2

    Divide both side by 0.082 x 273.2

    n = (1.07 x 5.0x10^3) / (0.082 x 273.2)

    n = 238.81 moles

    Step 4:

    Determination of the mass of air. This is illustrated below:

    Number of mole of air = 238.81 moles

    Molar Mass of air = 28.98 g/mol

    Mass of air =.?

    Mass = number of mole x molar Mass

    Mass of air = 238.81 x 28.98

    Mass of air = 6920.71g
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