Homework #1

_1. P_I02 = (P_B – P_H20) x F_I02

= 713 x .28 = 199.64

_2. P_AO2 = P_I02 – P_aco2/r

= 199.64 – 48 = 199.64 – (48 x 1.25)

= 199.64 – 60° = 139.64

3. D (A – a) 0₂ = 139.64 – 60 = 79.64 – Is this normal?

^4. a/A = 60/139.64 = 43% ^Normal?

_5. New F_I02 = Desire P_a02 + P_ac02

a/A 8

-------------------------------------

P_B – P_H20

.35 = x + 48

.43 .8

--------------------------

713

 (.35) (713) = x/.43 +60

249.55 = x/.43 +60

-60 -60

(.43) x 189.55 = x/.43 x .43

81.5 or 82 mmHg = x

6. x = 100 + 48 = 232.55 + 60 = 292.55

.43 .8 713 713

-------------------------

(760 –47)

= 41%

KB/rm

Resp/Blaine/Homework#1

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