Question

Calculate the number of molecules in a deep breath of air whose volume is 2.35 L at body temperature, 37 ∘C, and a pressure of 735 torr. Express the answer in molecules to three significant figures. NN = nothing molecules Request Answer Part B 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 0.5 ∘C and 1.07 atm, assuming the air behaves ideally. Express the answer in kilograms to two significant figures. mm = nothing kg

Answer 1

Answer:

Part A. 5.36x10²³ molecules of air

Part B. 6.9kg  

Explanation:

Part A.

To calculate the number of molecules of the air we need first find the number of moles of air using the equation of ideal gas law:

$PV = nRT$    (1)

where P: is the pressure, V: is the volume, n: is the number of moles of the gas, R: is the gas constant and T: is the temperature

$n = \frac{PV}{RT} = \frac{735torr \cdot 1atm/760torr \cdot 2.35L}{0.082 Latm/Kmol \cdot (37 + 273)K} = 0.089 moles$

Now by using the Avogadro's number we can find the number of molecules of air:

$number of molecules = \frac{6.022 \dot 10^{23}}{1mol} \cdot 0.089moles = 5.36 \cdot 10^{22} molecules$

Part B.

Similarly, to calculate the mass of air first we need to detemine the number of moles using equation (1):

$n = \frac{PV}{RT} = \frac{1.07atm \cdot 5.0\cdot 10^{3}L}{0.082 Latm/Kmol \cdot (0.5 + 273)K} = 238.55 moles$

So, the mass of air is:

$m = moles \cdot M$

where M: is the average molar mass of air

$m = 238.55moles \cdot 28.98g/mol = 6.9 kg$

I hope it helps you!