Respuesta :
Answer:
(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
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(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
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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