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.

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

it should be all of them because it asks if you included all of those in your answer

explanation: