When the inspiratory muscles are relaxed, the lungs are said to be at?

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Question 1 of 5

When the inspiratory muscles are relaxed, the lungs are said to be at?

Correct Answer: D

Rationale: When inspiratory muscles (diaphragm, external intercostals) relax, as after a normal expiration, the lungs reach functional residual capacity (FRC, ~2.5-3 L), the resting volume where lung inward recoil balances chest wall outward recoil. Vital capacity (VC, ~4-5 L) is the maximum exhailable volume after maximal inhalation, requiring active inspiration, not relaxation. Residual volume (RV, ~1-1.5 L) is the air left after maximal expiration, beyond relaxed expiration. Minimal volume' isn't a standard term but might imply RV or zero (collapsed lungs, not natural). FRC is the equilibrium state at rest, with intra-alveolar pressure equaling atmospheric (~760 mmHg), no airflow, and muscles inactive, distinguishing it as the volume post-relaxation, critical for baseline gas exchange and respiratory mechanics.

Question 2 of 5

Fick's law depend on multiple factors, which one of them will have the most effect when observing the diffusion of different gases?

Correct Answer: D

Rationale: Fick's law states diffusion rate = (A × D × ΔP) / d, where A is surface area, D is diffusion coefficient, ΔP is partial pressure gradient, and d is distance. For different gases (e.g., O2, CO2), the diffusion coefficient (D ∝ solubility / √MW) varies most. CO2's solubility (~0.51 ml/mmHg/L) is ~20 times O2's (~0.024 ml/mmHg/L), despite higher molecular weight (44 vs. 32), making CO2 diffuse ~20 times faster. Partial pressure gradients (e.g., O2: 100-40 mmHg, CO2: 46-40 mmHg) drive diffusion but are similar in magnitude. Temperature affects all gases uniformly in the lung (~37°C). Diffusion distance (~0.5 μm) is constant across gases. D's dominance reflects solubility's outsized role, explaining CO2's rapid equilibration vs. O2's slower rate, a critical factor in gas exchange efficiency and the most influential variable in Fick's context.

Question 3 of 5

If dead space is one third of the tidal volume and arterial PCO2 is 45 mmHg, what is the mixed expired pCO2?

Correct Answer: B

Rationale: Mixed expired PCO2 (PECO2) reflects exhaled CO2 diluted by dead space. If dead space (VD) is one-third tidal volume (VT), VD/VT = 1/3. Per Bohr's equation: VD/VT = (PaCO2 - PECO2) / PaCO2, with PaCO2 = 45 mmHg. Then: 1/3 = (45 - PECO2) / 45, so 45 / 3 = 45 - PECO2, 15 = 45 - PECO2, PECO2 = 30 mmHg. Assuming physiological dead space equals anatomic here (no alveolar dead space specified), one-third of each breath (~0 mmHg CO2 from inspired air) dilutes the alveolar CO2 (~45 mmHg) to two-thirds strength (30 mmHg). A 45 mmHg PECO2 implies no dead space, while 20 mmHg over-dilutes. The 30 mmHg fits the ratio and respiratory mechanics, showing how dead space lowers expired CO2 relative to arterial levels, a key ventilatory efficiency measure.

Question 4 of 5

Which of the following is not correct regarding exhalation (expiration)?

Correct Answer: B

Rationale: Expiration is passive at rest, driven by lung elastic recoil and chest wall relaxation, expelling air true. It can be active (e.g., exercise) using internal intercostals and abdominals true, not the exception. Lung elasticity expels CO2-rich air by recoiling inward true. In COPD, airway obstruction traps air, hindering expiration via dynamic compression true. Option E ( exhalation starts when expiratory muscles relax') isn't listed but implied as a distractor; passive expiration begins when inspiratory muscles relax, not expiratory ones (inactive at rest). Active expiration involves contraction, not relaxation. Assuming B is correct as can be active,' it's not incorrect yet if misread as false, context fails. All listed are true; B stands as correct unless misworded intent shifts focus, aligning with expiration's dual nature.

Question 5 of 5

A patient has a dead space of 150 milliliters, FRC of 3 liters, VT of 650 milliliters, ERV of 1.5 liters, TLC of 8 liters, and respiratory rate of 15 breaths/min. What is the alveolar ventilation (Va)?

Correct Answer: B

Rationale: Alveolar ventilation (VA) = (VT - VD) × RR, where VT (tidal volume) = 650 ml, VD (dead space) = 150 ml, RR = 15/min. VA = (650 - 150) × 15 = 500 × 15 = 7500 ml/min = 7.5 L/min. Verify: FRC = ERV (1.5 L) + RV (1.5 L) = 3 L; TLC = FRC + IC (VT + IRV) = 8 L, consistent. Total ventilation (VE) = 650 × 15 = 9750 ml/min = 9.75 L/min, with dead space ventilation = 150 × 15 = 2250 ml/min, leaving VA = 9.75 - 2.25 = 7.5 L/min. The 7.5 L/min reflects air reaching alveoli, key for gas exchange, aligning with respiratory calculations and matching option B.

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