forumdoctor
Forum Senior
Topics: 14
Posts: 123
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Given that the complainace of an individual lung is 0.5 L/cm H2O and mean intrapleural pressure is -7 cm H2O, what is the volume of exhaled gas if intrapleural pressure rose to -5 cm H20?
Ans is 1 liter itsims.
can any one explain,
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| forumdoctor
Forum Senior
Topics: 14
Posts: 123
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Still no one................................................
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Forum Senior
Topics: 5
Posts: 135
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ok so we know that compliance =dv/dp i.e change in volume/change in pressure by its definition...so just substituting
.5=dv/-5-(-7)
thus dv=1 L per cm of H2O
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