Question

Question #f8a34

Answer 1

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.