# Question f8a34

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.

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

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,

${P}_{\text{oxygen") + P_("H₂O") = P_("atm}}$

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

${P}_{\text{oxygen") = P_("atm") - P_("H₂O}}$ = (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; ${V}_{2}$ = ?

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

V_2 = V_1 × P_1/P_2 # = 1.874 L × $\left(742 \text{ Torr")/(760" Torr}\right)$ = 1.83 L

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

Rate of consumption is $\left(1.83 \text{ L")/(6" min}\right)$ = 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.