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Author6 Posts
  #1

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






  #2

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


  #3

please now the change in pr is 2 ok?

P1-P2=2 then
-10-P2=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??





  #4

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


  #5

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


  #6

thanx

i got it man

thanx alot







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