If Patm = 0 mm Hg and Pa|v = - 2 mm Hg, then

Question

If Patm = 0 mm Hg and Pa|v = - 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 (Ptp) is - 2 mm Hg (D) air is flowing into the lungs

Tardigrade Admin · Accepted Answer

If alveolar pressure (Pa|v) is negative with respect to atmospheric pressure (Patm), then driving force for airflow is inward i.e., from the atmosphere into the lungs.

Q. If $P_{a t m} = 0 mm H g$ and $P_{a ∣ v} = - 2 mm H g$ , then

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

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

transpulmonary pressure $(P_{tp})$ is $- 2 mm H g$

air is flowing into the lungs

If alveolar pressure $(P_{a ∣ v})$ is negative with respect to atmospheric pressure $(P_{a t m})$ , then driving force for airflow is inward i.e., from the atmosphere into the lungs.

Q. If Patm​=0mmHg and Pa∣v​=−2mmHg, then