Question

A student is pedaling a stationary bicycle at the U of M Rec Center. Her alveolar PO2 = 115 mm Hg and her oxygen consumption is 1.0 liter/min. If the inspired PO2 = 150 mm Hg and the barometric pressure is 747 mm Hg, what is her alveolar ventilation?

Answer 1

This is an alveolar gas equation question and it is used to approximate the partial pressure of oxygen in the alveolus (PAO2):

The equation states;

PₐO₂ = P₁O₂ - (PₐCO₂/R) = [(PB - PH₂O) × F₁O₂ - (PₐCO₂/R)] ...................Eqn 1

where

PₐO₂ = Alveolar partial pressure of O2 = 115mmHg

P₁O₂ = Inspired partial pressure of O2 = 150mmHg

PB = barometric pressure,

PH₂O = Water vapor pressure (usually 747 mmHg),

F₁O₂ = fractional concentration of inspired oxygen,

and R = gas exchange ratio. (Usually around 0.8)

FₐO₂ = Fraction of alveolar O₂

(O₂) = 1L/min = 1dm³

From eqn 1. we have

PₐCO₂ = (P₁O₂ - PₐO₂)/R

= (150 - 115)x0.8

PₐCO₂ = 28mmHg

Similarly from Eqn 1, we have

F₁O₂ = (PₐO₂ + PₐCO₂/R)/(PB - PH₂O)

F₁O₂ = (115 + (28/.8))/(747 - 47)

F₁O₂ = 0.21

Now to find the Alveolar Ventilation A, we will use this equation;

O₂ = A(F₁0₂ - FₐO₂) .................Eqn 2

But FₐO₂ = PₐO₂/(PₐO₂ + PₐCO₂)

FₐO₂ = 115/ (115+28) = 0.8

A = O₂/(F₁0₂ - FₐO₂)

A = 0.001/(0.21 - 0.8)

A = 0.00169m³/min

Hence, the aveolar ventilation is 0.00169m³/min