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alayham
Forum Senior Topics: 25 Posts: 60 
the compliance of a lung is :0,5 L/Cm H2O. the IPP is 10 Cm H2O what is the new IPP if the person exhaled 1.0 L?? please this question from qablan book i need the answer with explanation ( even mathematical procedure ) thanx alooooooooot  
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Ferritin
Forum Elite Topics: 51 Posts: 315 
compliance =change in volume/change in pressure 0.5=1ltr /change in pressure therefore change in pressure is 2cm h20 now the IPP and i checked from kaplan is normally negative so the new intra pleural pressure is 10+2=8 not once a person is exhaling out 1l in volume he is changing one litre in volume and once the change in pressure is determined the original pressure DECREASES as you have exhaled remember intralvelar pressure becomes positive during expiration as seen by the graph shown in kaplan hope it is clear  
alayham
Forum Senior Topics: 25 Posts: 60 
please now the change in pr is 2 ok? P1P2=2 then 10P2=2 then here is the problem by mathmematics when we shift the number or the letter the signal should be changed ok?? the P2=2+10 then P2 = 12 please i don't know if some thing i missed and i get the different result can u inform me if some thing is wrong??  
Ferritin
Forum Elite Topics: 51 Posts: 315 
lol the new value is p2 so tp explain this to you p2(10)=2 p2+10=2 p2=8 it can work with you formula as well if u consider the change is negative that is it is 2 if you draw a graph it will be on the negative side... i hope thats clear.... remember u r calculating the new calue p2 p1=10 as the question says  
Ferritin
Forum Elite Topics: 51 Posts: 315 
to make this even simpler the new value has to be less negative now as the ipp has to decrease it cant be more negative right if it were i would be inhaling u make ur ipp more negative by inhaling and more postive by exhaling  
alayham
Forum Senior Topics: 25 Posts: 60 
thanx i got it man thanx alot  
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