Question

3. An athlete at high performances inhales 4.0 L of air at 1.0 atm and 298 K at a respiration rate of 40
breaths per minute.
a. If the exhaled and inhaled air contain 15.3% and 20.9% oxygen by volume, respectively, how many
moles of oxygen per minute are absorbed by the athlete’s body?
b. Assume that inhaled and exhaled air contain 0.5% and 6.2% water vapor by volume, respectively. If the
athlete performs for one hour, how much water (mL) will he have to drink to compensate for what he lost
through breathing?

FOR PART B I GOT 547.2 LITER BUT I DON'T THINK THAT IS RESONABLE. PLEASE HELP PART B

Answer 1

Volume of air inhaled = 4 L

Pressure P = 1 atm

Temperature T = 298 K

Respiration rate = 40 breaths/minute

Part a

Volume of oxygen = number of breaths x volume of air x (fraction inhaled - fraction exhaled)

= 40 breaths/min x 4L x (0.209 - 0.153)

= 8.96 L/min

Moles of O2 absorbed = pressure x volume of O2 / (R x T)

= 1 atm x 8.96 L/min / (0.0821 L-atm/mol-K x 298K)

= 0.3662 mol/min

Part b

Volume of H2O lost in an hour

= number of breaths x volume of air x (fraction exhaled - fraction inhaled)

= 40 breaths/min x 4L x (0.062 - 0.0050)

= 9.12 L/min x 60min/hr

= 547.2 L/hr

547.2 L/hr water to drink to compensate for what he lost through breathing.