A patient is receiving mechanical ventilation with a fractional inspired oxygen (FɪO₂)of 0.85 and a positive end-expiratory pressure (PEEP)of 5 cm H₂O.His arterial partial pressure of oxygen (PₐO₂)is 68 mm Hg,arterial oxygen saturation (SₐO₂)is 88%,and partial pressure of end-tidal carbon dioxide (PₑₜCO₂)is 32 mm Hg.Over the next few minutes his PEEP is titrated resulting in the following data:
$\begin{array} { | l | l | l | l | l | }
\hline \text { Time } & \mathrm { F } _ { 1 } \mathrm { O } _ { \mathbf { 2 } } & \begin{array} { l }
\mathrm { PEEP } ( \mathrm { cm } \
\left. \mathrm { H } _ { \mathbf { 2 } } \mathrm { O } \right)
\end{array} & \mathrm { S } _ { \mathrm { P } } \mathrm { O } _ { \mathbf { 2 } } ( \% ) & \mathrm { P } _ { \mathbf { z t } } \mathrm { CO } _ { \mathbf { 2 } } ( \mathrm { mm } \mathrm { Hg } ) \
\hline 0600 & 0.85 & 5 & 88 & 30 \
\hline 0630 & 0.85 & 8 & 88 & 30 \
\hline 0650 & 0.85 & 10 & 90 & 32 \
\hline 0720 & 0.80 & 12 & 93 & 34 \
\hline 0740 & 0.80 & 15 & 90 & 25 \
\hline
\end{array}$
At 0740 the single-breath carbon dioxide (SBCO₂)curve shifted to the right.What action should the respiratory therapist take at this time?
A) Increase the FɪO₂ to 0.90.
B) Reduce the set tidal volume.
C) Continue to increase the PEEP.
D) Reduce the PEEP to 12 cm H₂O.

Question

A patient is receiving mechanical ventilation with a fractional inspired oxygen (FɪO₂)of 0.85 and a positive end-expiratory pressure (PEEP)of 5 cm H₂O.His arterial partial pressure of oxygen (PₐO₂)is 68 mm Hg,arterial oxygen saturation (SₐO₂)is 88%,and partial pressure of end-tidal carbon dioxide (PₑₜCO₂)is 32 mm Hg.Over the next few minutes his PEEP is titrated resulting in the following data:
$\begin{array} { | l | l | l | l | l | } 
\hline \text { Time } & \mathrm { F } _ { 1 } \mathrm { O } _ { \mathbf { 2 } } & \begin{array} { l } 
\mathrm { PEEP } ( \mathrm { cm } \
\left. \mathrm { H } _ { \mathbf { 2 } } \mathrm { O } \right)
\end{array} & \mathrm { S } _ { \mathrm { P } } \mathrm { O } _ { \mathbf { 2 } } ( \% ) & \mathrm { P } _ { \mathbf { z t } } \mathrm { CO } _ { \mathbf { 2 } } ( \mathrm { mm } \mathrm { Hg } ) \
\hline 0600 & 0.85 & 5 & 88 & 30 \
\hline 0630 & 0.85 & 8 & 88 & 30 \
\hline 0650 & 0.85 & 10 & 90 & 32 \
\hline 0720 & 0.80 & 12 & 93 & 34 \
\hline 0740 & 0.80 & 15 & 90 & 25 \
\hline
\end{array}$
At 0740 the single-breath carbon dioxide (SBCO₂)curve shifted to the right.What action should the respiratory therapist take at this time?
A) Increase the FɪO₂ to 0.90.
B) Reduce the set tidal volume.
C) Continue to increase the PEEP.
D) Reduce the PEEP to 12 cm H₂O.

Quizplus · Accepted Answer

The increase in positive end-expiratory pressure (PEEP)to 15 cm H₂O seems to have decreased pulmonary perfusion because of overinflation of the alveoli.This is evident by the decrease in the partial pressure of end-tidal carbon dioxide (PₑₜCO₂)to 25 mm Hg and the right shift in the single-breath carbon dioxide (SBCO₂)curve.Increasing the fractional inspired oxygen (FɪO₂)will not address this problem.Reducing the set tidal volume will increase the PₑₜCO₂ but will not improve the pulmonary circulation.Continuing to increase the PEEP will further reduce pulmonary perfusion and cause more dead space.Reducing the PEEP back to 12 cm H₂O will optimize the PEEP and reduce overinflation.

A Patient Is Receiving Mechanical Ventilation with a Fractional Inspired