After the end of a normal inspiration, the volume of air in the lungs is about 2

Question

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 37 degree C 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 that expands the air in the lungs. A. How much is the air expanded to produce a decrease of 2 nu n Hg in pressure? Use the ideal gas equation. PV = nRT, where P is the pressure. V is the volume, n is the number of moles. R =0.082 L atm mol^-1 degree K^-1 is the gas constant and t is the absolute temperature.

Explanation / Answer

Nomal atmospheric pressure = 760 mmHg

Decrease in pressure = 758 mmHg

Initial volume = 2.8 L

So the expanded new volume for the given decrease in pressure would be V2,

with R, T and n be constant, we get,

P1V1 = P2V2

V2 = 2.8 x 760/758 = 2.81 L

So the volume expansion seen for reduction of pressure by 2 mmHg would be 2.81 L

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