(a) Calculate the number of molecules in a deep breath of air whose volume is \(2.25 \mathrm{~L}\) at body temperature, \(37{ }^{\circ} \mathrm{C},\) and a pressure of 735 torr. (b) The adult blue whale has a lung capacity of \(5.0 \times 10^{3} \mathrm{~L} .\) Calculate the mass of air (assume an average molar mass \(28.98 \mathrm{~g} / \mathrm{mol}\) ) contained in an adult blue whale's lungs at \(0.0{ }^{\circ} \mathrm{C}\) and 1.00 atm, assuming the air behaves ideally.

First, we need to find the amount of substance (in moles) of the air in the deep breath. We will use the Ideal Gas Law, which states: \(PV=nRT \) Where: P - pressure V - volume n - the amount of substance (in moles) R - the ideal gas constant T - temperature The exercise gives us the pressure in torr, so we need to convert it to atmospheres since the value of R we will use is in L*atm/mol*K: \(1 \mathrm{~atm} = 760 \mathrm{~torr} \) Therefore, \(P = \frac{735 \mathrm{~torr}}{760 \frac{\mathrm{torr}}{\mathrm{atm}}} = 0.9671 \mathrm{~atm} \) We have the volume in liters, but we need the temperature in Kelvin: \(T = 37^{\circ} C + 273.15 = 310.15 \mathrm{~K} \) We will use the value of R in L*atm/mol*K: \(R = 0.08206 \mathrm{~L*atm/mol*K} \) Now we can solve the Ideal Gas Law for n:

\(n = \frac{PV}{RT} \) \(n = \frac{(0.9671 \mathrm{~atm})(2.25 \mathrm{~L})}{(0.08206 \mathrm{~L*atm/mol*K})(310.15 \mathrm{~K})} \) \(n \approx 0.09118 \mathrm{~mol} \)

Finally, we can use Avogadro's number to find the number of molecules in this volume of air: Number of molecules = n x Avogadro's number Number of molecules = \(0.09118 \mathrm{~mol} * 6.022 \times 10^{23} \frac{\mathrm{molecules}}{\mathrm{mol}} \) Number of molecules \( \approx 5.48 \times 10^{22} \) molecules So, there are approximately \(5.48 \times 10^{22}\) molecules in a deep breath of air.

In part (b), we need to find the mass of air in the blue whale's lungs. First, we will find the amount of substance (in moles) of the air in the whale's lungs. We have the volume, temperature, and pressure: \(V = 5.0 \times 10^3 \mathrm{~L} \) \(T = 0.0^{\circ} C + 273.15 = 273.15 \mathrm{~K} \) \(P = 1.00 \mathrm{~atm} \) Now we can solve the Ideal Gas Law for n:

\(n = \frac{PV}{RT} \) \(n = \frac{(1.00 \mathrm{~atm})(5.0 \times 10^3 \mathrm{~L})}{(0.08206 \mathrm{~L*atm/mol*K})(273.15 \mathrm{~K})} \) \(n \approx 223.2 \mathrm{~mol} \)

Now we can use the given average molar mass to find the mass of air in the blue whale's lungs: mass = n x average molar mass mass = \( (223.2 \mathrm{~mol})\times(28.98 \mathrm{~g/mol}) \) mass \( \approx 6.47 \times 10^3 \) g So, the mass of air in an adult blue whale's lungs is approximately \(6.47 \times 10^3\) grams (or 6.47 kg).