During a person's typical breathing cycle, the \(\mathrm{CO}_{2}\) concentration in the expired air rises to a peak of 4.6\(\%\) by volume. (a) Calculate the partial pressure of the CO \(_{2}\) in the expired air at its peak, assuming 1 atm pressure and a body temperature of \(37^{\circ} \mathrm{C}\) (b) What is the molarity of the \(\mathrm{CO}_{2}\) in the expired air at its peak, assuming a body temperature of \(37^{\circ} \mathrm{C} ?\)

We know the peak concentration of \(\mathrm{CO}_2\) is 4.6% by volume. Since the total pressure of the air is given as 1 atm, we can calculate the partial pressure of \(\mathrm{CO}_2\) using Dalton's law of partial pressures: Partial pressure of \(\mathrm{CO}_2 = \text{Fraction of }$$\mathrm{CO}_2 \times \text{Total pressure}\) Partial pressure of \(\mathrm{CO}_2 = 0.046 \times 1 \: \mathrm{atm} = 0.046 \: \mathrm{atm}\)

The body temperature is given as \(37^{\circ} \mathrm{C}\). We need to convert this value to Kelvin since gas laws use Kelvin for temperature. The formula to convert Celsius to Kelvin is: \(K = ^{\circ}\mathrm{C} + 273.15\) \(T = 37^{\circ} \mathrm{C} + 273.15 = 310.15 \: \mathrm{K}\)

We will use the ideal gas law to calculate the molarity of \(\mathrm{CO}_2\). The ideal gas law is: \(PV = nRT\) We need to solve for n, the number of moles of \(\mathrm{CO}_2\). We have the partial pressure of \(\mathrm{CO}_2\) from step 1 (P = 0.046 atm) and the temperature from step 2 (T = 310.15 K). The ideal gas constant, R, is equal to 0.0821 L atm/mol K. The volume of the expelled air is not given and cannot be canceled out, so we will solve for the molar concentration (n/V) instead: \((n/V) = \frac{P}{RT}\) Molarity of \(\mathrm{CO}_2 = \frac{0.046 \: \text{atm}}{(0.0821 \: \mathrm{L \: atm/mol \: K})(310.15 \: \mathrm{K})} = 0.0018 \: \text{mol/L}\) The partial pressure of \(\mathrm{CO}_2\) in the expired air at its peak is 0.046 atm, and the molarity of the \(\mathrm{CO}_2\) in the expired air at its peak is 0.0018 mol/L.