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If P(atm) = 0 mm Hg and P(alv) = -2 mm H...

If `P_(atm) = 0 mm Hg` and `P_(alv) = -2` mm Hg, then

A

It is the end of the normal inspiration and there is no airflow

B

it is the end of the normal expiration and there is no airflow

C

transpulmonary pressure `(P_(tp))` is `-2` mm Hg

D

air is flowing into the lungs.

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the given pressures and their implications on airflow into the lungs. ### Step-by-Step Solution: 1. **Identify the Given Pressures**: - Atmospheric Pressure (P_atm) = 0 mm Hg - Alveolar Pressure (P_alv) = -2 mm Hg 2. **Understand the Relationship Between Pressures**: - The pressure difference between the atmospheric pressure and the alveolar pressure determines the direction of airflow. - If the alveolar pressure is lower than the atmospheric pressure, air will flow into the lungs. 3. **Calculate the Pressure Difference**: - Pressure difference (ΔP) = P_atm - P_alv - ΔP = 0 mm Hg - (-2 mm Hg) = 0 mm Hg + 2 mm Hg = 2 mm Hg 4. **Determine the Direction of Airflow**: - Since the pressure difference (ΔP) is positive (2 mm Hg), it indicates that the atmospheric pressure is greater than the alveolar pressure. - This means that air will flow from the atmosphere (higher pressure) into the alveoli (lower pressure). 5. **Conclusion**: - Therefore, with P_atm = 0 mm Hg and P_alv = -2 mm Hg, air is flowing into the lungs. ### Final Answer: - Air is flowing into the lungs. ---
To solve the problem, we need to analyze the given pressures and their implications on airflow into the lungs. ### Step-by-Step Solution: 1. **Identify the Given Pressures**: - Atmospheric Pressure (P_atm) = 0 mm Hg - Alveolar Pressure (P_alv) = -2 mm Hg ...
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