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

Questions 72

ATI LPN

ATI LPN Test Bank

Respiratory System Questions Questions

Question 1 of 5

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

Correct Answer: D

Rationale: 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.

Question 2 of 5

Which of the following will decrease diffusion?

Correct Answer: D

Rationale: Per Fick's law (Rate = A × D × ΔP / d), diffusion decreases if surface area (A) drops (e.g., emphysema destroys alveoli, halving A halves rate), diffusion distance (d) increases (e.g., pulmonary edema doubles d from 0.5 to 1 μm, halving rate), or partial pressure gradient (ΔP) falls (e.g., hypoventilation lowers alveolar PO2). Decreased pressure coefficient' likely means ΔP; reducing it (e.g., from 60 to 30 mmHg) slows diffusion. Increased lung fluid thickens the barrier, adding resistance beyond distance (e.g., protein debris). All factors reduced A, increased d, lowered ΔP independently and collectively cut diffusion, as seen in hypoxemia from edema or fibrosis. Diffusion coefficient (D) is unchanged here. Each aligns with clinical scenarios impairing O2 transfer, making all the above' correct, reflecting multiple pathways to reduced gas exchange efficiency.

Question 3 of 5

Which person would be expected to have the largest PAO2-PaO2 gradient? (A stands for alveolar and a-stands for arterial)

Correct Answer: B

Rationale: The PAO2-PaO2 gradient (alveolar-arterial O2 difference) is normally ~5-10 mmHg due to efficient diffusion. In pulmonary fibrosis, thickened alveolar walls impair O2 transfer, dropping PaO2 (e.g., to 60 mmHg) while PAO2 (~100 mmHg, per alveolar gas equation) stays closer to normal, widening the gradient (e.g., 40 mmHg). During exercise, a normal person's ventilation and perfusion match, keeping the gradient small despite higher O2 use. Anemia lowers O2-carrying capacity, not diffusion, so PaO2 ≈ PAO2, maintaining a normal gradient. At 5000 m, low atmospheric PO2 reduces both PAO2 and PaO2 (e.g., 50 vs. 45 mmHg), keeping the gradient small. Fibrosis's diffusion barrier creates the largest gradient, as O2 struggles to cross, a hallmark of restrictive disease affecting gas exchange, unlike other scenarios.

Question 4 of 5

All of the following lab-values are consistent with Pulmonary fibrosis except?

Correct Answer: D

Rationale: Pulmonary fibrosis, a restrictive disease, stiffens lungs via interstitial scarring. FEV1/FVC is normal or high (≥80%) as FEV1 and FVC drop proportionally true. Vascular resistance rises as fibrosis compresses capillaries true. Peak expiratory flow (PEF), corrected for reduced volume, can be normal or high, as airflow isn't obstructed true. Residual volume (RV) decreases (e.g., from 1.5 L to <1 L) in fibrosis due to stiff lungs limiting all volumes, not increases as in obstructive diseases (e.g., COPD) false. Increased RV contradicts restrictive physiology, where elasticity loss shrinks residual air, making it the inconsistent value, while others align with fibrosis's impact on mechanics and circulation.

Question 5 of 5

A healthy 10-year-old boy Mohammad Emyan breathes quietly under resting conditions. His tidal volume is 400 milliliters and his ventilation frequency is 12/min. Which of the following best describes the ventilation of the upper, middle, and lower lung zones in this boy?

Correct Answer: D

Rationale: In an upright healthy child, gravity gradients affect ventilation. At rest (VT = 400 ml, RR = 12/min), intrapleural pressure is more negative at the apex (~-10 cm H2O) than base (~-2.5 cm H2O), making apices less compliant and bases more so. Diaphragm movement ventilates lower zones most, where compliance and volume change peak. Studies (e.g., West) show lower lobes get ~4 times more ventilation per unit volume than apices. Thus, ventilation is Lower > Middle > Upper. Equal distribution ignores gravity; Upper > Middle > Lower reverses it. VA = (400 - ~120 ml VD) × 12 ≈ 3.36 L/min, mostly basal, making this the best description of regional ventilation in a healthy boy.

Access More Questions!

ATI LPN Basic


$89/ 30 days

ATI LPN Premium


$150/ 90 days

Similar Questions