dr_sarim
Forum Senior
Topics: 16
Posts: 75
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compliance of lung is 0.5 l/cm H2O and intrapleural pressure is -10 cm H2O.what is the new itrapleural pressure if this person exhaled 1.0 L
can any body explain this.....
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| doc179
Forum Guru
Topics: 67
Posts: 1,217
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ok the ans is -8cm H2O.
compliance is volume change / pressure change.
once we know the pressure difference we can get to know the intrapleural pressure after he exhaled. so C = V1-V2 / P1-P2
V1- V2 = 1 ltr
P1-P2 = 1ltr / 0.5 ltr/ cmH2O = 2 cmH2O
Ithink now you can calculate P2 when you substitute P1 for 10 cmH2O.
But this is what I was confused abt so would tell you wht I thought at that moment. I was confused whether P2 now is -12 or -8 , If he had inhaled one ltr then it would have been -12 but he has exhaled and the intrapleural pressure becomes more positive so it changes from -10 to -8.
hope this helps. GL
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| robin082006
Forum Hero
Topics: 471
Posts: 5,125
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Good explanation, thank doc 179
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