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given that the compliance of an individual's lung is 0.5 l/cm H2O and mean intrapleural pressure is -10 cm H2O , What is the new intrapleural pressure if this person exhaled 1.0 L?
please help me................thanks
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friends please help me.....not that much smart in Math.
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samomcos
please help me to do it.
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shot in the dark but is it 2?
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no it's not....
ans is -8 cm H2O
thanks syed1027 , atleast u tried..
still waiting for any kind help .
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Compliance is calculated as change in vol/ change in pressure [ v/p].... so the compliance in the above case is 0.5 ml/cm H2O .....i.e v/p=0.5=1/2 hence from this we can derive that for an x ml vol change the pressure changes by a factor of 2x. the patient exhaled 1L....the p should change by 2cm H2O. since he is exhaling the intapleural p is getting more positive.......-10 +2= -8 cmH2O.
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-8 cmh2o is the new intra pleural pressure after exhale.....
c= v/p1-p2 , c = compliance and v = volume of air , p1,p2= intra pleural pressure....
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Thank you verryyyyyyyy much yeabfre......you saved my life!!!!!
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iwnn, U R very well come !
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oh man, so I was right but I didn't finish the damn equation!