Assume that an exhaled breath of air consists of 74.8% N2, 15.3% O2, 3.7% CO2, a

Question

Assume that an exhaled breath of air consists of 74.8% N2, 15.3% O2, 3.7% CO2, and 6.2% 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 455mL and its temperature is 37 degrees Celcius , calculate the number of moles of CO2 exhaled.

c) How many grams of glucose (C6H1206) would need to be metabolized to produce this quantity of CO2? (the chemical reaction is the same as that for combustion of C6H12O6).

Explanation / Answer

So, for part (a), multiply each of those percents (as decimals) by the total pressure, and you'll get the partial pressures. For example 0.748 * 0.980 atm = 0.733 atm. That's the partial pressure of Nitrogen. You can do the rest, I hope.

for part (b), assume you're dealing with an ideal gas, and use PV=nRT. The P is the partial pressure for CO2, not the total pressure of 0.980 atm; the V is 455 mL (or 0.455 L) and the T is 37

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