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how can i solve it? - iwnn
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12-22-2009, 09:31 AM
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
12-22-2009, 10:49 AM
friends please help me.....not that much smart in Math.
12-22-2009, 01:34 PM
no it's not....
ans is -8 cm H2O thanks syed1027 , atleast u tried.. still waiting for any kind help .
12-22-2009, 02:50 PM
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.
12-22-2009, 03:04 PM
-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....
12-22-2009, 03:23 PM
Thank you verryyyyyyyy much yeabfre......you saved my life!!!!!
12-22-2009, 04:41 PM
oh man, so I was right but I didn't finish the damn equation!
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