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26 December, 16:12

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?

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  1. 26 December, 16:32
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    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
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