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Chapter 13: Problem 55

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

Short Answer

Expert verified
The partial pressure of CO鈧� at the peak concentration during a typical breathing cycle is 0.046 atm. Given a body temperature of 37掳C, the molarity of CO鈧� in air at this point is 0.00181 mol/L.

Step by step solution

01

Calculate the partial pressure of CO2

We are given that the concentration of CO2 is 4.6% by volume. Since the total pressure (air pressure) is 1 atm, we can calculate the partial pressure of CO2 by multiplying its fraction by the total pressure: Partial Pressure of CO2 = Mole fraction of CO2 脳 Total air pressure = 4.6% 脳 1 atm = 0.046 脳 1 atm = 0.046 atm
02

Convert the given temperature to Kelvin

The body temperature is given in Celsius, but we need to convert it to Kelvin for the ideal gas law calculation: Body Temperature = 37掳C + 273 = 310 K
03

Calculate the molarity of CO2 using the Ideal Gas Law

The ideal gas law is given by: PV = nRT Where P is pressure, V is volume, n is the number of moles, R is the gas constant (0.0821 L鈰卆tm/mol鈰匥), and T is the temperature in Kelvin. We can rearrange this formula to find the molarity, which is the number of moles (n) per volume (V): Molarity (M) = n/V = P/RT Now, we plug in the values to calculate the molarity of CO2: M = (0.046 atm) / (0.0821 L鈰卆tm/mol鈰匥)(310 K) = 0.001805 mol/L
04

Round off the molarity of CO2

Finally, we round off the molarity of CO2 to an appropriate number of significant figures: Molarity of CO2 = 0.00181 mol/L (rounded to 4 significant figures) So, the molarity of CO2 in air at this point is 0.00181 mol/L

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