santaclara
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this is from kaplan physiology.
given that the compliance of and individual's lung is 0.5L/cm H2O and intrapleural pressur is -10cm H2O.what is new intrapleural pressure if this person exhaled 1.0L?
pls reply me.
thanx.
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| mani
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the given compliance means that if person exhale 0.5L of air, ,his pressure will change by 1 cm H2O. as he exhales 1L, so pressure will change and 2 com H2O. so new pressure will be -8cm of H2O
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| santaclara
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I DON'T UNDERSTAND CAN U DO BY FORMULA PLS.
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| mani
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i thought i made it simplem if u read my post one more time that may help.
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| santaclara
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i understan dear what do u mean. if he exhales 1L, so pressure will change and 2 cm H2O and we minus the (10-2=8).
i think i m right.
but when w ekeep according to formula that i don't know how to put it in formula that what i m asking.
and also same type of Q is also before that on page 244 of physio kaplan.and according to formula that we can put it.
so pls if u can explainaccording to formula pls do it for me.
thanx.
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| hiwa
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compliance(0.5) =change in volume(1L)/change in pressure(?)
change in pressure(?)=change in volume(1)/comliance(0.5)=2
but we dont want the change in pressure, we want the pressure now
because it is exhaling so the intrapleural pressure become less negative so we have to abstract the 2 from the 10 to get less negative -8
10-2=8
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| santaclara
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thanx for reply.
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| mani
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i hope santaclara that hiwa's explanation is quite clear
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