black_horse
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See the original equation is: (alvPO2=inspiredPO2 – dec. in PO2)
We are able to replace the dec. in PO2 by the inc. in PCO2(40-0=40) only when R=1 so that the change in oxygen conc. in the alveoli equals the change in CO2 conc.(in other words the change in alv. PO2 equals the change in alv. PCO2), but when R≠1 then the equation is no longer right and we must include R in it.
I hope this makes sense.
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| emhava
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It makes sense. I thought at it but I dismissed it because I thought that there is no inc. in PCO2 since it always stays around 40 and never goes down to 0. So actually there is no 40-0. That's in fact my problem  ...
But I admit that your explanation must be it 
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| black_horse
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You are right, the air in the respiratory zone remains stable, but what I meant is the change in PO2 and PCO2 of the inspired air as it moves from the conducting zone to the respiratory zone and undergoes gas exchange,
PCO2 inc. from 0 in the airways to 40 in the alveoli (inc. 40 mmHg) and PO2 dec. from 150 in the airways to 100 in the alveoli (dec.50 mmHg)
Now see the ratio between (inc. in PCO2/dec. in PO2) = 40/50=0.8, that's normal R (40 mmHg PCO2 in exchange of 50 mmHg PO2)
remember the original equation (alv.PO2 = inspired PO2 – dec. in PO2)
we can simply replace the dec. in PO2 by the inc. in PCO2(=PCO2)/R
GL
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| emhava
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Thank you!
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| SmokyWaters
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I am sorry but I differ with this theory...
R isnt a correction factor first of all...
seconldy....
R is an experimentally calculated coefficent which means
1 mol of Oxygen produces 0.8 mol of CO2
so when we are equating the whole thing... we have to divide the PCO2 by this factor... so to get an equimolar value ...
tnx
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