Respiratory physiology question

[Soleus]

---

Members do not see this ad.

So I was going through a Respiratory Phys book and, without giving the exact question (I will paraphrase), I was unable to figure out how the authors came to the correct answer. I have the correct answer but don't understand the reasoning.

Question:A man has an arterial PCO2 of 40 mm Hg and takes an overdose of an opiate which decreases his alveolar ventilation by half but his CO2 output remains the same. His respiratory exchange ratio is 0.8 and, thus, his arterial PO2 is 50 mm Hg. How much does the FIO2 concentration have to be raised to return the arterial PO2 to its original level?

Answer: 7

Is this answer arrived at simply by going by the "rule of 7's" to estimate how much the FIO2 will have to increased to compensate or is there another formula to plug this into that I'm missing? I know the alveolar gas equation can be used for estimation of alveolar ventilation, but are they using this equation and presuming a similar arterial O2 will apply to the formula also? Thanks in advance.

 

[pie944]

Solve for the equation
FAO2 = FiO2(PB-Ph20) - PaCO2/RQ

Initially
FAO2 = 0.21(760-47) - 40/0.8
FAO2 ~ 100

Pops a few pills(maybe shoots it up)
FAO2 = .21(760-47) - 80/0.8
FAO2 = 50

Now you want to know how much oxygen supplementation will it take to get back to a FAO2 ~ 100.

Old School math.

100 = x(760-47) - 80/0.8
100 = 713x - 100
200 = 713x
x = 200/713
x ~ 0.28

So you need an FiO2 of ~ 28%, or an increase of 7% over room air.

I know they are asking for arterial oxygen, I assume that either they meant alveolar or they are ignoring any gradient that might exist.

Hope this helps.

 

[RxBoy]

Very nice math and explanation.

Anyways its a pretty tough question but it does tie in a lot of important physiologic principles. Of course respiratory formulas assume lungs are in a steady state but in reality they are in a dynamic state (alvelor vent, perfusion, shunting, dead spacing constantly changing). But for simplicity I just wanted to add:

Pops a few pills(maybe shoots it up)
FAO2 = .21(760-47) - 80/0.8
FAO2 = 50

I quoted the above because it doesn't explain where the PaCO2 of 80 came from.

PaCO2 is proportional to VCO2/VA. Where VCO2 is the CO2 produced, VA is the alveolar ventilation (ventilation involved in gas exchange). In a steady state if VCO2 was held constant (the part about CO2 output remaining the same) but VA was halved as this question states then PaCO2 would eventually double hence 40 divided by 0.5 = 80. Solving like above the PA02 would equal 50 and since the question states the PaO2=50 it means there is no A-a gradient. This drug addict must have the healthiest lungs in the word.

In reality PaCO2 would actually take quite some time for it to double, it wouldn't instantly double the second your VA decreased.