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Nursing QUESTION #9225
Question 1
A Charge Nurse notes that a mechanically ventilated patient has a tidal volume (V\(_T\)) set at \(700 \text{ mL}\) and a respiratory rate (RR) of \(14 \text{ breaths/min}\). What is the minute ventilation \(\dot{V}_E\)?
  • \(9.4 \text{ L/min}\)
  • \(9.8 \text{ L/min}\)✔️
  • \(10.2 \text{ L/min}\)
  • \(8.6 \text{ L/min}\)
Correct Answer Explanation
Correct Answer: B — \(9.8 \text{ L/min}\)

Formula for Minute Ventilation:
\[\dot{V}_E = V_T \times RR\]
\[\dot{V}_E = 700 \text{ mL} \times 14 = 9800 \text{ mL/min} = 9.8 \text{ L/min}\]

Clinical significance:
  • Normal \(\dot{V}_E\) = \(5 - 8 \text{ L/min}\)
  • This patient's \(\dot{V}_E\) of \(9.8 \text{ L/min}\) is elevated, risking hyperventilation and respiratory alkalosis
  • Lung-protective ventilation targets \(V_T = 6 \text{ mL/kg}\) ideal body weight (IBW) to prevent ventilator-induced lung injury (VILI)
  • If IBW = 70 kg, then target \(V_T = 6 \times 70 = 420 \text{ mL}\)
The Charge Nurse should alert the intensivist to review ventilator settings.