# Alveolar Ventilation

## Objectives

• 1.

Understand the composition of gases from ambient air to the alveoli.

• 2.

Describe the alveolar air equation and its use.

• 3.

Define the alveolar carbon dioxide equation and the relationship between alveolar ventilation and arterial P co 2 .

• 4.

Describe the distribution of ventilation at the apex and at the base in upright individuals.

• 5.

Describe how the nitrogen washout test can be used to examine the distribution of ventilation.

• 6.

Define anatomic and physiologic dead space and their measurement.

• 7.

Outline the effects of aging on lung growth, lung volumes, elastic recoil, expiratory muscle strength, airway closure, and the diffusion capacity for carbon monoxide.

Ventilation is the process by which fresh gas moves in and out of the lung. Alveolar ventilation is the process by which gas moves between the alveoli and the external environment. It includes both the volume of fresh air entering the alveoli and the (similar) volume of alveolar air leaving the body. Minute (or total) ventilation (MV), also referred to as <SPAN role=presentation tabIndex=0 id=MathJax-Element-1-Frame class=MathJax style="POSITION: relative" data-mathml='V˙’>V˙

V ˙
E, is the volume of air that enters or leaves the lung per minute and is described by:

<SPAN role=presentation tabIndex=0 id=MathJax-Element-2-Frame class=MathJax style="POSITION: relative" data-mathml='MV=f×VT’>MV=f×VT
MV=f×VT

MV = f × V T
where f is the frequency or number of breaths per minute, and V t (also expressed as TV) is the tidal volume or volume of air inspired (or exhaled) per breath. V t varies with age, sex, body position, and metabolic activity. In an average-sized adult, V t is 500 mL; in children, V t is 3 to 5 mL/kg of body weight.

## Composition of a Gas Mixture

The process of respiration brings ambient (or atmospheric) air to the alveoli where oxygen in the ambient air is taken up and carbon dioxide, produced by tissue metabolism, is excreted. Ambient air is a gas mixture composed of nitrogen and oxygen with minute quantities of carbon dioxide, argon, and other inert gases. The composition of a gas mixture can be described either in terms of gas fractions or as the corresponding partial pressures. Because ambient air is a gas, it obeys the gas laws (see Appendix C to review Boyle’s law and Dalton’s law).

Using the gas laws and applying them to ambient air, two important principles arise ( Box 5.1 ). The first is that when viewed in terms of gas fractions (F), the sum of the individual gas fractions of nitrogen (F n), oxygen (F o 2 ), and argon and other gases (Fargon and other gases) must equal 1; that is, for ambient air,

<SPAN role=presentation tabIndex=0 id=MathJax-Element-3-Frame class=MathJax style="POSITION: relative" data-mathml='1.0=FN+FO2+Fargonandothergases’>1.0=FN+FO2+Fargonandothergases
1.0=FN+FO2+Fargonandothergases

1.0 = F N + F O 2 + F argon and other gases

BOX 5.1

Gas Pressure Units and Their Conversion

Respiratory pressures can be expressed in units of torr, mm Hg, or cm H 2 O. Torr is a unit of pressure equal to 1 mm Hg. Pressures used in conjunction with gases in gaseous or liquid phases are usually expressed in units of torr or mm Hg with a vacuum as the reference point. In contrast, pressure gradients, differences in the pressure surrounding a structure with ambient barometric pressure as a reference, are usually expressed in cm H 2 O. To convert, 1 mm Hg = 1.35 cm H 2 O.

When viewed using Boyle’s gas law, the sum of the partial pressures (in mm Hg) or tensions (in torr) of a gas must be equal to the total pressure. (mm Hg also can be expressed as torr, named for Evangelista Torricelli, the inventor of the barometer. The two terms are equal and interchangeable.) Thus at sea level, where the total pressure (also known as the barometric pressure or P b ) is 760 mm Hg, the partial pressure of the individual gases in ambient air is:

<SPAN role=presentation tabIndex=0 id=MathJax-Element-4-Frame class=MathJax style="POSITION: relative" data-mathml='PB=PO2+PN2+Pargonandothergases’>PB=PO2+PN2+Pargonandothergases
PB=PO2+PN2+Pargonandothergases

P B = P O 2 + P N 2 + P argon and other gases

<SPAN role=presentation tabIndex=0 id=MathJax-Element-5-Frame class=MathJax style="POSITION: relative" data-mathml='760mmHg=PO2+PN2+Pargonandothergases’>760mmHg=PO2+PN2+Pargonandothergases
760mmHg=PO2+PN2+Pargonandothergases

760 mm Hg = P O 2 + P N 2 + P argon and other gases

The second important principle is that the partial pressure of a gas (P gas ) is equal to the fraction of gas in the gas mixture (F gas ) times the total or ambient (barometric) pressure (an application of Dalton’s law).

<SPAN role=presentation tabIndex=0 id=MathJax-Element-6-Frame class=MathJax style="POSITION: relative" data-mathml='Pgas=Fgas×PB’>Pgas=Fgas×PB
Pgas=Fgas×PB

P gas = F gas × P B

Ambient air is composed of approximately 21% oxygen and 79% nitrogen. Therefore the partial pressure of oxygen in ambient air ( Po 2 ) is:

<SPAN role=presentation tabIndex=0 id=MathJax-Element-7-Frame class=MathJax style="POSITION: relative" data-mathml='PO2=FO2×PB=0.21×760mmHg=159mmHg(or159torr)’>PO2=FO2×PB=0.21×760mmHg=159mmHg(or159torr)
PO2=FO2×PB=0.21×760mmHg=159mmHg(or159torr)

P O 2 = F O 2 × P B = 0.21 × 760 mm Hg = 159 mm Hg ( or 159 torr )

This is the oxygen tension (i.e., the partial pressure of oxygen) at the mouth at the start of inspiration. The oxygen tension at the mouth can be altered in one of two ways: by changing the fraction of oxygen or by changing the barometric (atmospheric) pressure. For example, if the fraction of oxygen is increased through the administration of supplemental oxygen, the partial pressure of oxygen will be increased. On the other hand, if the barometric (atmospheric) pressure is decreased—for example, by high altitude—the partial pressure of ambient oxygen will decrease.

At the start of inspiration, ambient gases are brought into the airways, where they become warmed to body temperature and humidified. By the time the inspired gas reaches the larynx, it has become saturated with water vapor; water vapor is a gas that exerts a partial pressure equal to 47 mm Hg at body temperature (see Appendix C for the reason). Because the total pressure remains constant at the barometric pressure, water vapor dilutes the pressure exerted by the other gases. This can be best understood by considering that humidification of a liter of dry gas in a container at 760 torr would increase its total pressure to 807 torr (i.e., 760 torr + 47 torr). In the body, however, barometric pressure remains constant, and therefore the gas simply expands, according to Boyle’s law. As a result, the partial pressures of the other gases in the 1 L of gas at 760 torr are diluted by the added water vapor pressure. To calculate the partial pressure of a gas in a humidified mixture, the water vapor partial pressure must be subtracted from the total barometric pressure. Thus in the conducting airways, the partial pressure of oxygen is:

<SPAN role=presentation tabIndex=0 id=MathJax-Element-8-Frame class=MathJax style="POSITION: relative" data-mathml='PtracheaO2=(PB−PH2O)×FO2=760−47mmHg×0.21=150mmHg’>PtracheaO2=(PBPH2O)×FO2=76047mmHg×0.21=150mmHg
PtracheaO2=(PB−PH2O)×FO2=760−47mmHg×0.21=150mmHg

P trachea O 2 = ( P B − P H 2 O ) × F O 2 = 760 − 47 mm Hg × 0.21 = 150 mm Hg
and the partial pressure of nitrogen is:
<SPAN role=presentation tabIndex=0 id=MathJax-Element-9-Frame class=MathJax style="POSITION: relative" data-mathml='PtracheaN2=760−47mmHg×0.79=563mmHg’>PtracheaN2=76047mmHg×0.79=563mmHg
PtracheaN2=760−47mmHg×0.79=563mmHg

P trachea N 2 = 760 − 47 mm Hg × 0.79 = 563 mm Hg

Note that the total pressure has remained at 760 mm Hg (150 + 563 + 47 mm Hg). Water vapor pressure, however, has reduced (diluted) the partial pressures of oxygen and nitrogen. The conducting airways do not participate in gas exchange, and therefore the partial pressures of oxygen, nitrogen, and water vapor remain unchanged in the airways until the gas reaches the alveolus.

## Alveolar Gas Composition

At functional residual capacity, the alveoli contain 2.5 to 3 L of gas; an additional 350 mL of gas will enter and leave the alveoli with each breath ( Fig. 5.1 ). Diffusion of O 2 from the alveoli into the pulmonary capillary blood and of CO 2 from the pulmonary capillary blood into the alveoli is a continuous process. Each minute at rest, on average, 300 mL of O 2 are taken up and 250 mL of CO 2 are removed by alveolar ventilation. Thus the partial pressures of O 2 and CO 2 in the alveolar air are determined by alveolar ventilation, by O 2 consumption, and by CO 2 production. The process by which oxygen is taken up and carbon dioxide is removed by the pulmonary capillary blood is described in Chapter 8 .

Here we describe only alveolar ventilation. At the end of inspiration with the glottis open, the total pressure in the alveolus is atmospheric; the same gas laws apply, namely the partial pressures of the gases in the alveolus must equal the total pressure, which in this case is atmospheric. The composition of the gas mixture however is changed and can be described as:

<SPAN role=presentation tabIndex=0 id=MathJax-Element-10-Frame class=MathJax style="POSITION: relative" data-mathml='1.0=FO2+FN2+FH2O+FCO2+Fargonandothergases’>1.0=FO2+FN2+FH2O+FCO2+Fargonandothergases
1.0=FO2+FN2+FH2O+FCO2+Fargonandothergases

1.0 = F O 2 + F N 2 + F H 2 O + F CO 2 + F argon and other gases

Nitrogen and argon are inert gases, and therefore the fraction of these gases in the alveolus does not change. The fraction of water vapor also does not change because the gas is already fully saturated and at body temperature by the time the gas reaches the trachea. As a consequence of gas exchange, the fraction of oxygen in the alveolus decreases and the fraction of carbon dioxide in the alveolus increases. Because of the changes in the fractions of oxygen and carbon dioxide, the partial pressures exerted by these gases also change. The partial pressure of oxygen in the alveolus ( Pao 2 ) is given by the alveolar gas equation , sometimes also called the ideal alveolar oxygen equation:

<SPAN role=presentation tabIndex=0 id=MathJax-Element-11-Frame class=MathJax style="POSITION: relative" data-mathml='PAO2=PIO2−PACO2R=(PB−PH2O)×FIO2−PACO2R’>PAO2=PIO2PACO2R=(PBPH2O)×FIO2PACO2R
PAO2=PIO2−PACO2R=(PB−PH2O)×FIO2−PACO2R

P AO 2 = P IO 2 − P ACO 2 R = ( P B − P H 2 O ) × F IO 2 − P ACO 2 R