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Problem: Assume that an exhaled breath of air consists of 74.7 % N2, 15.4 % O2, 3.8 % CO2, and 6.1 % water vapor.(a) If the total pressure of the gases is 0.985 atm, calculate the partial pressure of each component of the mixture.(b) If the volume of the exhaled gas is 455 mL and its temperature is 37 °C, calculate the number of moles of CO2 exhaled.(c) How many grams of glucose (C6H12O6) would need to be metabolized to produce this quantity of CO2? (The chemical reaction is the same as that for combustion of C6H12O6. See Section 3.2. in the textbook.)

FREE Expert Solution

(a) 

PO2 = XO2PTotalPO2 =(3.8100)(0.985 atm)

PO2 = 0.03743 atm


(b)

PV = nRTn = PVRTn = (0.03743 atm)(455 mL×10-3 L1 mL)(0.08206 atm·Lmol·K)(37°C+273.15)K

n = 6.6916x10-4 mol


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Problem Details

Assume that an exhaled breath of air consists of 74.7 % N2, 15.4 % O2, 3.8 % CO2, and 6.1 % water vapor.

(a) If the total pressure of the gases is 0.985 atm, calculate the partial pressure of each component of the mixture.

(b) If the volume of the exhaled gas is 455 mL and its temperature is 37 °C, calculate the number of moles of CO2 exhaled.

(c) How many grams of glucose (C6H12O6) would need to be metabolized to produce this quantity of CO2? (The chemical reaction is the same as that for combustion of C6H12O6. See Section 3.2. in the textbook.)

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