The alveolar gas equation

This chapter is most relevant to Section F9(iv) from the 2017 CICM Primary Syllabus, which expects the exam candidates to be able to "understand the common respiratory equations". Though there are no specific CICM primary or Fellowship exam questions which ask about this equation directly, virtually every ABG question requires the candidate to consider the A-a gradient, making the alveolar gas equation essential.

In short, this equation describes the concentration of gases in the alveolus, and thus allows us to make educated guesses as to the effectiveness of gas exchange. One can use this to calculate the tension-based indices of oxygenation, such as A-a gradient or the a/A ratio (which is expressed as a percentage). The ABG machine frequently does this work for you, provided you have entered the FiO₂ and have specified that your sample is "arterial". The result is usually reported as pO₂(a/A).

The physical basis for the alveolar gas equation

The equation is necessary because there is no convenient way to sample alveolar gas. At the same time, it is gas, and therefore can be expected to behave according to some fundamental laws. Specifically, we are talking about Dalton's law, which states that

The total pressure of a mixture of gases is equal to the sum of the partial pressures of all of the constituent gases

In this scenario we can be confident that the total pressure of the mixture of gases in the alveolus is going to be basically the same as atmospheric pressure, as the alternative hypothesis would have us either explode or implode. Let us represent this total pressure as:

 P_atmos  = 760 mmHg.

Next, the pressures of the constituent gases. Of these, some may be predictable. For example, the human body is usually at around 37 degrees, and the gas mixture is  predictably humidified up to 100% humidity by its passage though the airways, such that the absolute water content of the alveolar gas mixture is predictably 44g/L. At this temperature, water vapour exerts a pressure of 47 mmHg (this, by the way, is Gay-Lussac's law). Thus, in a mixture of gases where there is 44g/m³ of water vapour present, the water vapour will exert a partial pressure of 47 mmHg.

That leaves only 713 mmHg of partial pressure for the rest of the gases. This is because the inhaled gas mixture does not have the water vapour added to it like a sum, i.e. the partial pressure of the alveolar gas is not 807 mmHg (760 mmHg + 47 mmHg). Rather, the inspired gas mixture is diluted with the addition of water vapour until the mixture is about 6.2% water vapour and 93.8% original gas mixture. The partial pressure exerted by the remains of the original atmospheric gas mixture is therefore 713 mmHg, and the total alveolar gas mixture can now be represented as:

P_atmos + P_H2O = 760 mmHg

Of this remaining gas mixture, the vast majority would surely have to be nitrogen and oxygen. The human body neither generates nor consumes significant amounts of atmospheric nitrogen, which means the alveolar gas should contain approximately the same amount of nitrogen as the ambient atmosphere. Thus, the nitrogen content of this 713 mmHg should be 78%, and the oxygen content should be 21%, because the Earth titrates its FiO₂ to 21% (or, more precisely, 20.9%).

Why does it do that? Thankfully, this would not be the first ridiculous digression in physiology. For instance,  John F. Nunn has written a chapter (Chapter 1 of Nunn's Respiratory Physiology) about the atmosphere. In it,  he is grateful that greenhouse gases have allowed the existence of surface water for the last 4000 million years, and he laments that the sun "proceeds remorselessly towards becoming a red giant, which will ultimately envelop the inner planets". Unfortunately the rest of the textbook proceeds soberly along a straight and predictable path. A better introduction into the subject would probably be afforded by The Chemical Evolution of the Atmosphere and Oceans, by Heinrich D. Holland.  The author confesses on page 2 that "the range of topics considered in the book is uncomfortably large", and that due to the mass of information "chaos was a continuous threat" during the assembly of the manuscript. In any case, for a monograph written between the years 1968 and 1981, this is a fine work. It is deserving of attention from anybody who has finished with their final CICM exams and still has some enthusiasm for the written word.

Anyway: to digress no further, let us again return to the idea that the alveolar gas mixture should contain similar proportions of oxygen and nitrogen to the atmosphere.  In other words,

PN₂ + PO₂ = 713 mmHg

Or rather, 

PN₂ + PO₂ + P_H2O = 760 mmHg

The partial pressure of nitrogen is never directly measured because nitrogen is boring and is of no interest to anyone, except in the context of trying to measure dead space by the Fowler method. There is about as much nitrogen in the body fluids as there is in the alveolar gas mixture, and it is neither created nor consumed. The nitrogen concentration is therefore left out of the equation. Oxygen, on the other hand, is both interesting clinically, and variable in its concentration. Assuming that there is nothing in the inspired gas mixture other than oxygen, nitrogen and water vapour, the equation that describes the concentration of oxygen in while ignoring the unknown amount of nitrogen looks like this:

PO₂ = FiO₂  ×  (P_atmos - P_H2O) 

This gives us 149.73 mmHg as the partial pressure of oxygen, and that would be the oxygen concentration of the inspired gas mixture. However, the alveolar gas also contains CO₂. This also needs to be incorporated into the gas equation. Like water vapour, it is incorporated into the partial pressure fractions, which means it eats into the pressures of oxygen and nitrogen equally. If one were breathing 100% FiO₂, the PCO₂ would be subtracted directly from the PO₂, i.e.  a 10mmHg increase in PCO₂ would result in a 10mmg Hg decrease in PO₂. But while breathing room air, all fractions are reduced proportionally, i.e. that 10mmHg increase in PCO₂ causes a 7.9 mmHg decrease in PN₂ and a 2.1 mmHg decrease in PO₂, preserving this relationship:

PO₂ + P_H2O  + PCO₂  + PN₂= 760 mmHg

Or, incorporating FiO₂, considering we do not know what the PO₂ is:

FiO₂  ×  (P_atmos - P_H2O) + PCO₂  + PN₂ = 760 mmHg

Or

FiO₂  ×  (P_atmos - P_H2O) + PCO₂ = (760 mmHg - PN₂)

 

But we do not know the PN₂, nor do we care; as we are mainly interested in the alveolar oxygen concentration. We can therefore eliminate nitrogen from our thinking completely, and rearrange the equation so it produces the PAO₂ as the result:

FiO₂  ×  (P_atmos - P_H2O) - PCO₂ = PAO₂

This still leaves the alveolar CO₂ concentration as an unknown. It would be inconvenient to sample it from the alveolar capillary blood (with which alveolar CO₂ should be in reasonably good equilibrium), but we can sample it from the arterial blood, and introduce it into the equation with an respiratory quotient (RQ) modifier which represents the extra CO₂ that is added during the passage of arterial blood through the tissues. For most normothermic people with a conventional diet this RQ modifier is around 0.8. 

This gives us the modern form of the equation:

On room air and at sea level, we can assume certain constants:

PAO₂ = (0.21 × (760 - 47)) - (PaCO₂ × 1.25)

Thus:

PAO₂ = (149 - (PaCO₂ × 1.25)

Thus, the patient with a relatively normal PaCO2 (say, 40) :

PAO₂ = (149 - 50)

So, a normal person should have a PAO₂ of around 99 mmHg.

Or, for a patient with normal PaCO₂ and an increased FiO₂:

PAO₂ = (FiO2 × 713) - 50

Of course, it is possible to have a strange respiratory quotient, but for this we would need to measure the total body VO₂ and VCO₂, which can only be accomplished by means of indirect calorimetry.

So, what should your PAO₂ be at any given FiO₂? In mmHg, the values are as follows:

FiO2 21%	100
FiO2 30%	164
FiO2 40%	235
FiO2 50%	307
FiO2 60%	378
FiO2 70%	449
FiO2 80%	520
FiO2 90%	592
FiO2 100%	662

In a nutshell, one can say that for every 10% increase in FiO₂, the PAO₂ will rise by about 71-72 mmHg.

Alveolar hypoventilation as the cause of hypoxia

Considering the limitations of the tension-based indices of oxygenation, the only role for the alveolar gas equation is said to be the detection of alveolar hypoventilation as the cause of hypoxia. Medical students and junior critical care trainees are captured during a highly susceptible impressionable age and indoctrinated with the idea that, while breathing room air, you can hypoventilate and accumulate so much CO₂ in the alveolar gas mixture that the Daltonian balance of partial pressures cannot contain enough O₂ to support normoxia. In other words, CO₂ displaces O₂ from the gas mixture, producing hypoxia - or so the fable reads. However, this is an inversion of the real causal relationship. The PAO₂ drops not because of Dalton's law, but because oxygen is being consumed by being incorporated into CO₂, and then the CO₂ is not being exhaled effectively (because the minute ventilation has dropped). This is beautifully expressed by a letter from David A. Story from the Alfred in Melbourne to the editor of Anaesthesiology (1996), which was followed by a response by John F. Nunn. Considering the value of this exchange, it is curious that the concepts these authors had outlined are not better popularised. To hopefully improve its representation, what follows is an attempt to paraphrase and expand these excellent comments. 

Consider: the rate of oxygen consumption, VO₂, is described by this Fick equation: 

VO₂  =  V_A × (FiO₂ - FAO₂)

where:

V_A is alveolar ventilation

FiO₂ is the fraction of inspired O₂ 

FAO₂ is the fraction of alvolar (i.e. exhaled) O₂ 

It's not rocket science: the minute volume multiplied by the difference between the amount of inhaled oxygen and exhaled oxygen is the total oxygen consumption (and if oxygen was not being consumed, that difference would be zero). From this equation, it follows that if the VO₂ remains stable (as it does, because organs have got to keep metabolising stuff), and the FiO₂ remains stable, then then the fraction of alveolar oxygen must decrease with decreasing V_A:

It would be possible to end the explanation of hypoventilation-related hypoxia here, as by itself this is sufficient and self-evident, but we have yet to see how CO2 also increases. To elaborate, CO₂ consumption (VCO₂) is also described by a similarly straightforward equation:

VCO₂ = V_A  × PACO₂ / (P_atmos - P_H2O)

where:

P_atmos is the total atmospheric pressure

P_H2O is the partial pressure of water vapour

PACO₂ is the partial pressure of alveolar CO₂

VCO₂ is the total CO₂ production

V_A is alveolar ventilation

Yes, the amount of CO₂ produced and exhaled per unit time is logically just the minute volume times the fractional concentration of CO² in the alveolar gas mixture. And you can also rearrange this to yield PACO₂:

PACO₂  =  (P_atmos - P_H2O )  × (FiCO₂ + VCO₂  / V_A)

The FiCO₂ usually being zero, 

PACO₂  =  713   × (VCO₂  / V_A)

The bottom line is that the alveolar ventilation is an essential component of calculating both VO₂ and VCO₂. Why is this interesting? Well: VO₂ and VCO₂ are interrelated: for every unit of oxygen consumed, a certain amount of CO2 is produced.  This relationship is represented as the respiratory quotient (RQ):

 RQ = VCO₂ / VO₂

where:

VCO₂ is CO₂ production

VO₂ is O₂ consumption

This means that the rates of oxygen consumption CO₂ production and alveolar ventilation are all inextricably linked. For example, if you make the RQ equation all about VCO₂, 

VCO₂  =  RQ × VO₂

you can then represent the VCO₂ using the Fick relationship mentioned above, and produce an equation that contains all three variables:

 V_A  × PACO₂ / (P_atmos - P_H2O) =  RQ × VO₂ 

This means you cannot alter the V_A without changing the VCO₂ and the VO₂ in opposite directions, to a magnitude proportional to the RQ.  Observe what happens when we plug some standard variables into this equation:

2852 × 50 / (760-47) = 0.8 × 250,

where

V_A = 2852 ml/min

VO₂ = 250 ml/min 

RQ = 0.8

P_H2O = 47 mmHg

If VCO₂ obeys the rules, it should be 200 ml/min (because VCO₂  =  RQ × VO₂)

Thus, PACO₂  =  713   × (VCO₂  / V_A) = 50 mmHg

These numbers would then give us a PAO₂ of 87.23 mmHg if they were plugged into Nunn's universal air equation, which describes the relationship of PAO₂ to alveolar ventilation oxygen consumption and barometric pressure: 

 PAO₂  =  (P_atmos - P_H2O )  × (FiO₂ - VO₂  / V_A)

= 713 × (0.21 - 250/2852)

= 87.23 mmHg

And, plugging these equations into a spreadsheet, we can produce a totally unrealistic graph that tracks the relationship of PAO₂ and PACO₂ for a respiratory quotient of 0.8, all the way down to a completely ludicrous PAO₂ of 0 mmHg:

Yes, these gas mixtures do happen to obey Dalton's law, but that is probably because they all do. The point is that Dalton's law is not the reason the PAO₂ is dropping here.

To summarise: