Sem 1, Pulmonary Mechanics Practice Questions

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  • 1/14 Questions

    You are called to examine a premature early born infant. He is suffering from breathlessness and is cyanotic.  Your diagnosis is “infant respiratory distress syndrome, IRDS). After treatment with artificial surfactant, the situation is under control. What did surfactant do?

    • It decreased the lung compliance
    • It decreased surface tension in the lung
    • It increased the force of respiratory muscles
    • It increased the activity of inspiratory motoneurons
    • It increased the concentration of erythrocytes in the blood
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About This Quiz

In the human body, the word “pulmonary” is related to the lungs and the respiratory system, and is mainly linked to the process of having air move in and out of the lungs. What do you know about these mechanics?

Sem 1, Pulmonary Mechanics Practice Questions - Quiz

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  • 2. 

    A person connected to a spirometer has been asked to take a maximum breath in, hold it and then exhale as hard and fast as he can (forced vital capacity manoeuvre). The expiratory volume in 1 second (FEV1) and forced vital capacity (FVC) have been measured. The value of FEV1/FVC is significantly below 80 % by

    • A healthy person

    • Lung obstructive disease

    • Lung restrictive disease

    • Lung fibrosis

    Correct Answer
    A. Lung obstructive disease
    Explanation
    The value of FEV1/FVC being significantly below 80% indicates a lung obstructive disease. In a healthy person, the FEV1/FVC ratio is typically above 80%. Lung obstructive diseases, such as asthma or chronic obstructive pulmonary disease (COPD), cause airflow limitation and result in a decreased FEV1/FVC ratio. This is because the obstruction in the airways makes it difficult for the person to exhale forcefully, leading to a lower FEV1 value compared to the FVC value. Therefore, the given answer of lung obstructive disease is correct.

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  • 3. 

    What is the volume of one mol ideal gas under STPD condition:

    • 10.5 L

    • 15.2 L

    • 22.4 L

    • 31.0 L

    • 36.4 L

    Correct Answer
    A. 22.4 L
    Explanation
    The volume of one mole of an ideal gas under STPD (Standard Temperature and Pressure, 0 degrees Celsius and 1 atmosphere) conditions is 22.4 L. This is known as the molar volume of a gas at STPD and is a result of the ideal gas law, which states that at STPD conditions, one mole of any ideal gas occupies the same volume. Therefore, the correct answer is 22.4 L.

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  • 4. 

    A patient has a skeletal abnormality (Kyphoscoliosis). You are interested to verify the compliance of his chest wall. You are measuring changes in lung volume by a spirometer. Which pressure change you have to measure in addition to ΔV to be able to determine the compliance of his chest wall?

    • Changes in alveolar pressure (ΔPA)

    • Changes in intra pleural pressure (ΔPPl)

    • Changes in airway pressure (ΔPaw)

    • Changes in alveolar O2-partial pressure (ΔPAO2)

    • Changes in inspired O2-partial pressure (ΔPIO2)

    Correct Answer
    A. Changes in intra pleural pressure (ΔPPl)
    Explanation
    To determine the compliance of the patient's chest wall, you would need to measure changes in intra pleural pressure (ΔPPl). Compliance is a measure of how easily the chest wall can expand and contract. Intra pleural pressure is the pressure within the pleural cavity, which is the space between the lungs and the chest wall. By measuring the changes in intra pleural pressure, you can assess how the chest wall responds to changes in lung volume, and thus determine its compliance.

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  • 5. 

    A gas mixture of several dry gases has a total pressure of 200 mm Hg and a total gas amount of 10 Mol. What is the partial pressure of one gas that has an amount 3 Mol?

    • 20 mm Hg

    • 30 mm Hg

    • 40 mm Hg

    • 50 mm Hg

    • 60 mm Hg

    Correct Answer
    A. 60 mm Hg
    Explanation
    The partial pressure of a gas is directly proportional to its amount. In this case, the gas with 3 Mol has a partial pressure of 60 mm Hg because it represents 3/10 or 30% of the total gas amount (10 Mol). Therefore, the partial pressure of this gas is 30% of the total pressure (200 mm Hg), which equals 60 mm Hg.

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  • 6. 

    By a routine examination of respiratory functions a spirometer is used to measure lung volumes. Which lung volume or lung capacity can not be determined by this method?

    • Inspiratory reserve volume (IRV)

    • Inspiratory capacity (IC)

    • Expiratory reserve volume (ERV)

    • Vital capacity (VC)

    • Total lung capacity (TLC)

    Correct Answer
    A. Total lung capacity (TLC)
    Explanation
    A spirometer is a device used to measure lung volumes, which includes inspiratory reserve volume (IRV), inspiratory capacity (IC), expiratory reserve volume (ERV), and vital capacity (VC). However, it cannot directly measure total lung capacity (TLC). TLC is the maximum amount of air that the lungs can hold, including the residual volume (the air that remains in the lungs after a maximum exhalation). Since a spirometer cannot measure the residual volume, it cannot determine the total lung capacity.

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  • 7. 

    A patient inhales 2 L of air and holds his breath. At the beginning of inspiration his lung transmural pressure is – 5 cm H2O. During breath holding his lung transmural pressure is – 10 cm H2O. What is his lung compliance in L / cm H2O?

    • 0.1

    • 0.2

    • 0.3

    • 0.4

    • 0.5

    Correct Answer
    A. 0.4
    Explanation
    in this question values given for transmural prssure of -5 and -10
    should be replaced by +5 and +10 repectively.

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  • 8. 

    Physiological dead space calculated by Bohr equation is:

    • Always greater than anatomical dead space

    • Always smaller than anatomical dead space

    • Always equal to anatomical dead space

    • Either greater or equal to anatomical dead space

    • Either smaller or equal to anatomical dead space

    Correct Answer
    A. Either greater or equal to anatomical dead space
    Explanation
    The Bohr equation is used to calculate physiological dead space, which refers to the portion of the tidal volume that does not participate in gas exchange. Physiological dead space includes both anatomical dead space (the space in the airways where no gas exchange occurs) and any additional non-functional alveoli. Therefore, physiological dead space can be equal to or greater than anatomical dead space, depending on the presence of non-functional alveoli.

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  • 9. 

    The compliance of the respiratory system is changing with lung volume. Close to which lung volume has the respiratory system it´s greatest compliance?

    • FRC

    • RV

    • TLC

    • Minimal lung volume (below RV)

    Correct Answer
    A. FRC
    Explanation
    The respiratory system has its greatest compliance close to the Functional Residual Capacity (FRC). Compliance refers to the ability of the lungs to expand and contract. FRC is the volume of air present in the lungs at the end of a normal expiration, and it represents the equilibrium point where the elastic recoil of the lungs is balanced by the outward recoil of the chest wall. At this volume, the respiratory system is most compliant, meaning it can easily expand and contract to accommodate changes in lung volume.

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  • 10. 

    A Patient consults the doctor and complains over breathlessness. The doctor thinks he is suffering from lung emphysema. Therefore he decides to measure the residual volume of the lung. After a maximal expiration, the Patient is connected with a bag containing a mixture of 2 liters air ( Vbag = 2 l) with 2% He (FHe = 0.02). Helium concentration decreases to 1% (FHe = 0.01) after mixing of lung volume with the volume of the bag. What is the residual volume of the patient?

    • 0.5 L

    • 1.0 L

    • 1.5 L

    • 2.0 L

    • 3.0 L

    Correct Answer
    A. 2.0 L
    Explanation
    The residual volume of the patient is 2.0 L. This can be determined by calculating the difference in helium concentration before and after mixing the lung volume with the volume of the bag. The bag initially contains 2 liters of air with 2% helium (0.02), but after mixing with the lung volume, the helium concentration decreases to 1% (0.01). This decrease in helium concentration indicates that the lung volume is equal to the volume of the bag, which is 2.0 L. Therefore, the residual volume of the patient is 2.0 L.

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  • 11. 

    Some of diseases cause injury of alveolar surface (gas exchange region). Which type of the lung cells repairs this membrane?

    • Alveolar epithel cell type I

    • Alveolar epithel cell type II

    • Alveolar macrophage

    • Brush cell

    • Clara cell

    Correct Answer
    A. Alveolar epithel cell type II
    Explanation
    Alveolar epithelial cells type II are responsible for repairing the injured alveolar surface. These cells have the ability to differentiate into type I cells, which are the main cells involved in gas exchange. Type II cells also produce surfactant, a substance that helps to reduce surface tension in the alveoli and prevent their collapse. When the alveolar surface is injured, type II cells play a crucial role in the repair process by proliferating and differentiating into type I cells, thus restoring the gas exchange membrane.

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  • 12. 

    A patient is suffering from obstructive lung disease.  Values of compliance for his chest wall and respiratory system are known as: Chest wall compliance (CCW) = 200 ml / cm H2O Respiratory system compliance (CRW) = 100 ml / cm H2O What is his Lung compliance (CL) in ml / cm H2O?  

    • 50

    • 100

    • 200

    • 300

    • 400

    Correct Answer
    A. 200
    Explanation
    The lung compliance (CL) is the sum of the chest wall compliance (CCW) and the respiratory system compliance (CRW). In this case, the CCW is 200 ml / cm H2O and the CRW is 100 ml / cm H2O. Therefore, the lung compliance (CL) would be 200 ml / cm H2O + 100 ml / cm H2O = 300 ml / cm H2O.

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  • 13. 

    By which combination of the following respiratory pattern is the alveolar ventilation (VA) the greatest?                                     Tidal volume (VT)                Respiratory frequency (f)                                                 (L)                                    (breath / min)

    • 0.25 24

    • 0.5 12

    • 1.0 6

    • 2.0 3

    Correct Answer
    A. 2.0 3
    Explanation
    The alveolar ventilation (VA) is the volume of fresh air that reaches the alveoli per minute. It is calculated by multiplying the tidal volume (VT) by the respiratory frequency (f). In this question, the options given are different combinations of VT and f. To maximize VA, we need to maximize both VT and f. Among the given options, the combination of 2.0 L VT and 3 breaths/min f will result in the greatest VA. This is because it has the highest product of VT and f, leading to a larger volume of fresh air reaching the alveoli per minute.

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  • 14. 

    Force of respiratory muscles is dependent on lung volume. At which lung volume have the expiratory muscles their greatest force?

    • Minimal lung volume (resting volume of the lung)

    • RC

    • FRC

    • 60% of VC

    • TLC

    Correct Answer
    A. TLC
    Explanation
    The expiratory muscles have their greatest force at total lung capacity (TLC). This is because TLC represents the maximum amount of air that the lungs can hold, and therefore the maximum amount of air that can be forcefully exhaled. At TLC, the lungs are fully expanded, allowing for maximum contraction and force generation by the expiratory muscles.

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  • Current Version
  • Mar 20, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Nov 05, 2011
    Quiz Created by
    Chachelly

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