ATI LPN
Respiratory System Questions Questions
Question 1 of 5
At the end of normal quiet expiration, just before the start of inspiration, the lungs are said to be in:
Correct Answer: C
Rationale: Functional residual capacity (FRC) is the volume of air in the lungs at the end of a normal, quiet expiration, typically around 2.5-3 liters in adults. It's the resting state where the inward elastic recoil of the lungs balances the outward recoil of the chest wall, with no active muscle effort. Residual volume (RV) is the air left after maximal expiration (~1-1.5 L), not quiet expiration. Expiratory reserve volume (ERV) is the additional air that can be forcibly exhaled after a normal expiration (~1-1.5 L), not the resting volume itself. Inspiratory reserve volume (IRV) is the extra air inhaled beyond a normal breath (~2-3 L), relevant during inspiration, not expiration. Total lung capacity (TLC) is all lung volumes combined (~6 L), far exceeding the resting state. FRC represents the equilibrium point before inspiration begins, maintaining alveolar patency and efficient gas exchange, distinguishing it from other volumes tied to maximal efforts or different respiratory phases.
Question 2 of 5
A person breathes into and from a spirometer (volume 12 liters) containing 10% helium gas mixture. After equilibration, helium concentration of expired gas was found to be 6.67%. His ERV is 4.2 liters. What is his residual volume? (Hint: V1C1=V2C2)
Correct Answer: C
Rationale: Residual volume (RV) is the air left after maximal expiration, not measurable by spirometry but calculable via helium dilution. Here, a 12 L spirometer with 10% helium (C1 = 0.1) equilibrates with lung volume (initially FRC). Post-equilibration, expired gas is 6.67% helium (C2 = 0.0667). Using V1C1 = V2C2 (helium conservation), V1 = 12 L, C1 = 0.1, C2 = 0.0667: 12 × 0.1 = V2 × 0.0667, so 1.2 = V2 × 0.0667, V2 = 1.2 / 0.0667 ≈ 18 L. V2 is total gas volume (spirometer + FRC). FRC = V2 - V1 = 18 - 12 = 6 L. FRC = ERV + RV, and ERV = 4.2 L, so RV = FRC - ERV = 6 - 4.2 = 1.8 L = 1800 ml. This assumes equilibration at FRC (post-normal expiration), common in such problems. The 1800 ml matches helium dilution principles, where dilution reflects unexpired lung volume, confirming RV amidst the options.
Question 3 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 4 of 5
Compared to a normal individual, a patient with idiopathic pulmonary fibrosis, one of the following is expected to be more than normal?
Correct Answer: B
Rationale: Idiopathic pulmonary fibrosis (IPF) scars the lung interstitium, reducing elasticity and volumes. Total lung capacity (TLC) decreases (e.g., from 6 L to 4 L) as stiff lungs resist expansion. FEV1 and FVC both drop due to restricted capacity, though their ratio (FEV1/FVC) stays normal or high (≥80%). Arterial PO2 (PaO2) falls (e.g., from 75-100 mmHg to 60 mmHg) due to impaired diffusion across thickened alveoli, causing hypoxemia. However, total pulmonary vascular resistance (PVR) increases as fibrosis compresses and obliterates capillaries, narrowing the vascular bed and raising resistance to blood flow. This can strain the right heart, potentially leading to cor pulmonale, a known IPF complication. Among these, only PVR exceeds normal levels, reflecting the disease's vascular impact, while volumes and oxygenation decline, aligning with IPF's restrictive pattern and distinguishing it from healthy physiology.
Question 5 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.