Question #f8a34

1 Answer
Ernest Z.
Apr 30, 2014

The oxygen consuming speed is 305 mL/min.

This is an example of a disguised Boyle's Law problem involving partial pressures.

#P_1V_1=P_2V_2#

The "trick" is to calculate the partial pressure of the dry oxygen. Both water vapour and the oxygen itself contribute to the pressure of the "wet" gas.

I assume that the atmospheric pressure is 1 atm = 760 Torr.

According to Dalton's Law of Partial Pressures,

#P_("oxygen") + P_("H₂O") = P_("atm")#

At 20 °C, the vapour pressure of water is 17.5 torr.

#P_("oxygen") = P_("atm") - P_("H₂O") # = (760 – 17.5) torr = 742 torr

#T_1# = (20 + 273.15) K = 293 K

Summary:
#P_1# = 742 Torr; #V_1# = 1.874 L
#P_2# = 760 Torr; #V_2# = ?

#P_1V_1=P_2V_2#

#V_2 = V_1 × P_1/P_2 # = 1.874 L × #(742" Torr")/(760" Torr")# = 1.83 L

The patient is consuming 1.83 L of oxygen in 6 min.

Rate of consumption is #(1.83" L")/(6" min")# = 0.305 L O₂/min = 305 mL O₂/min

For comparison, a "normal" adult breathes about 6 L of air or 1 L of O₂/min.

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