Question 1
Place the following steps for normal inhalation in order: (1) decrease in intrapleural pressure to 754 mmHg (from -4 mmHg to -6 mmHg). (2) flow of air from higher to lower pressure (inhalation). (3) lung size increases. (4) decrease in intra-alveolar pressure to 759 mmHg (-1 mmHg). (5) contraction of the diaphragm + external intercostals muscles
- A. 5,2,3,4,1
- B. 1,3,4,5,2
- C. 5,4,3,2,1
- D. 5,1,3,4,2
Show answer and rationale
Correct answer: D — 5,1,3,4,2
Normal inhalation follows a mechanical sequence. (5) Contraction of the diaphragm and external intercostals starts it, expanding the thoracic cavity. (1) This lowers intrapleural pressure (IPP) from -4 mmHg (756 mmHg) to -6 mmHg (754 mmHg), increasing transpulmonary pressure. (3) Lung size increases as the lungs expand with the chest wall. (4) Intra-alveolar pressure drops to 759 mmHg (-1 mmHg) as volume rises (Boyle's law), creating a gradient from atmospheric pressure (760 mmHg). (2) Air flows in from higher to lower pressure. The order 5,1,3,4,2 reflects causality: muscle action lowers IPP, expands lungs, drops alveolar pressure, and drives airflow. Alternatives disrupt this: 5,2,3,4,1 puts flow before pressure changes; 1,3,4,5,2 starts with IPP drop without muscle action; 5,4,3,2,1 misplaces alveolar pressure before lung expansion. The correct sequence mirrors respiratory physiology's step-by-step process.