After the end of a normal inspiration, the volume of air in the lungs is about 2.8 L. Normally quiet inspiration is driven by a pressure difference of about 2 mm Hg. The air in the lungs is at 37C and after normal expiration it is at atmospheric pressure. Quiet inspiration is driven by the expansion of the chest cavity by contraction of the diaphragm, which expands the air in the lungs. How much is the air expanded to produce an decrease of 2 mmHg in pressure

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-05-13T04:19:34+02:00

Answer:

The Volume of the lungs that would produce 2 mmHg pressure decrease is

[tex]V_2 = 2.81 \ L[/tex]

Explanation:

From the question we are told that

The volume of air in the lungs is [tex]V = 2.8 \ L[/tex]

The pressure difference for quit normal inspiration is [tex]P = 2 \ mmHg[/tex]

The temperature of air in the lungs [tex]T = 37^oC[/tex]

The pressure after normal expiration is at [tex]T = 760 \ mmHg[/tex]

From ideal gas law we have that

[tex]PV= nRT[/tex]

Now since nRT is constant we have that

[tex]P_1 V_1 = P_2 V_2[/tex]

As the pressure decreased by 2 mmHg the volume becomes

[tex]V_2 = \frac{P_1 V_1}{P_2}[/tex]

[tex]V_2 = \frac{2.8 * 760}{758}[/tex]

[tex]V_2 = 2.81 \ L[/tex]