5.137 a person inhales air richer in o2 and exhales air richer in co2 and water vapor. during each hour of sleep, a person exhales a total of about 300 l of this co2-enriched and h2o-enriched air. (a) if the partial pressures of co2 and h2o in exhaled air are each 30.0 torr at 37.0ãâ°c, calculate the mass (g) of co2 and of h2o exhaled in 1 h of sleep. (b) how many grams of body mass does the person lose in 8 h of sleep if all the co2 and h2o exhaled come from the metabolism of glucose?

228.16 g of body mass is lost in 8 hours of sleep.

From the question, we can obtain the number of moles of each gas as follows;

Partial pressure of each gas= 30 torr or 0.039 atm

Volume of the gases =  300 L

Temperature of the gases = 37.0°C + 273 = 310 K

R = 0.082 atmLmol-1K-1

From PV =nRT

n = PV/RT =  0.039 atm × 300 L/0.082 atmLmol-1K-1  × 310 K

n = 0.46 moles of each gas

Hence;

Mass of CO2 lost in 1 hour =  0.46 moles × 44 g/mol = 20.24 g

Mass of water vapor lost in 1 hour = 0.46 moles × 18 g/mol = 8.28 g

If 20.24 g of CO2 is lost in 1 hour of sleep, in 8 hours of sleep;

8 × 20.24 g = 161.92 g of CO2 is lost.

Similarly, if 8.28 g of water vapor is lost in 1 hour of sleep, in 8 hours of sleep; 8 × 8.28 g = 66.24 g of water vapor is lost

Since both the CO2 and water vapor come from metabolism, total body mass lost in 8 hours of sleep = 161.92 g + 66.24 g  = 228.16 g

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(a) 20.5 g O₂, 8.38 g H₂O; (b) 230 g  

Step-by-step explanation:  

(a) Masses of CO₂ and H₂O  

pV = nRT            Divide each side by RT  

n = (pV)/(RT)  

p = 30.0 torr      Convert to atmospheres  

p = 30.0 × 1/760 = 0.039 47 atm  

V = 300 L  

R = 0.082 06 L·atm·K⁻¹mol⁻¹  

T = (37.0 + 273.15) K = 310.15 K  

n = (0.039 47 × 300)/(0.082 06 × 310.15)  

n = 0.465 mol  

Mass of CO₂ = 0.465 × 44.01 = 20.5 g  

Mass of H₂O = 0.465 × 18.02 = 8.38 g

(b) Mass lost during sleep  

Mass lost in 1 h = 20.5 + 8.38 = 28.9 g  

Mass lost in 8 h = 8 × 28.9     = 230 g