(a) Calculate the number of molecules in a deep breath of air whose volume is 2.25 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 of 28.98 \(\mathrm{g} / \mathrm{mol}\) ) contained in an adult blue whale's lungs at \(0.0^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm},\) assuming the air behaves ideally.

The Ideal Gas Law formula is given by: \(PV = nRT\) Where: P = pressure (in atm) V = volume (in L) n = number of moles R = Ideal Gas Constant (R = 0.0821 L atm mol^{-1} K^{-1}) T = temperature (in K)

To find the number of molecules in a deep breath, we first need to find the number of moles using the Ideal Gas Law formula. Then, we can multiply the number of moles with the Avogadro's number to find the number of molecules. Given: Volume (V) = 2.25 L Temperature (T) = 37 °C = 37 + 273.15 = 310.15 K Pressure (P) = 735 torr; we need to convert it to atm: \(P = \frac{735 \ torr}{760 \ torr/atm} = 0.967 \ atm\) The Ideal Gas Law formula becomes: \(n = \frac{PV}{RT}\) Plug in the values: \(n = \frac{(0.967 \ atm)(2.25 \ L)}{(0.0821 \ L \ atm \ mol^{-1} K^{-1})(310.15 \ K)}\) \(n \approx 0.098 \, mol\) Now, multiply the number of moles by Avogadro's number to find the number of molecules: Number of molecules = \(n \times N_A\) Number of molecules = \((0.098 \, mol)(6.022 \times 10^{23} \ molecules/mol)\) Number of molecules \( ≈ 5.90 \times 10^{22} \ molecules\)

First, we will find the number of moles of air in the adult blue whale's lungs. Given: Volume (V) = 5.0 x 10^3 L Temperature (T) = 0.0 °C = 0 + 273.15 = 273.15 K Pressure (P) = 1.00 atm Use the Ideal Gas Law formula to find the moles of air: \(n = \frac{(1.00 \ atm)(5.0 \times 10^3 \ L)}{(0.0821 \ L \ atm \ mol^{-1} K^{-1})(273.15 \ K)}\) \(n \approx 224.6 \, mol\) Now, convert the moles of air into mass using the given average molar mass: Mass of air = \(n \times M\) Mass of air = \((224.6 \, mol)(28.98 \, g/mol)\) Mass of air \( ≈ 6.51 \times 10^3 \, g\) #Answer#: (a) There are approximately \(5.90 \times 10^{22} \ molecules\) in a deep breath of air. (b) The mass of air contained in an adult blue whale's lungs is approximately \(6.51 \times 10^3 \, g\).