Medical Physiology, 3rd Edition


About 30% of total ventilation in a respiratory cycle is wasted ventilating anatomical dead space (i.e., conducting airways)

Total ventilation (image) is the volume of air moved out of the lungs per unit of time:



Here V is the volume of air exiting the lungs during a series of breaths. Note that we are using image differently than in Chapter 27, where image represented flow through an airway at a particular instant in time. A practical definition is that image is the product of tidal volume (TV or VT) and the respiratory frequency (f). Thus, for someone with a tidal volume of 0.5 L, breathing 12 breaths/min,



Because total ventilation usually is reported in L/min, it is sometimes called minute ventilation.

Before an inspiration, the conducting airways are filled with “stale” air having the same composition as alveolar air (Fig. 31-1, step 1); we will see why shortly. During inspiration, ~500 mL of “fresh” atmospheric air (high image/low image) enters the body (step 2). However, only the first 350 mL reaches the alveoli; the final 150 mL remains in the conducting airways (i.e., nose, pharynx, larynx, trachea, and other airways without alveoli)—that is, the anatomical dead space. These figures are typical for a 70-kg person; VT and VD are roughly proportional to body size. During inspiration, ~500 mL of air also enters the alveoli. However, the first 150 mL is stale air previously in the conducting airways; only the final 350 mL is fresh air. By the end of inspiration, the 500 mL of air that entered the alveoli (150 mL of stale air plus 350 mL of fresh air) has mixed by diffusion with the pre-existing alveolar air (see Fig. 31-1, step 3). During expiration (step 4), the first 150 mL of air emerging from the body is the fresh air left in the conducting airways from the previous inspiration. As the expiration continues, 350 mL of stale alveolar air sequentially moves into the conducting airways and then exits the body—for a total of 500 mL of air leaving the body. Simultaneously, 500 mL of air leaves the alveoli. The first 350 mL is the same 350 mL that exited the body. The final 150 mL of stale air to exit the alveoli remains in the conducting airways, as we are ready to begin the next inspiration.


FIGURE 31-1 Ventilation of dead space and alveolar space during a respiratory cycle.

Thus, with each 500-mL inspiration, only the initial 350 mL of fresh air entering the body reaches the alveoli. With each 500-mL expiration, only the final 350 mL of air exiting the body comes from the alveoli. One 150-mL bolus of fresh air shuttles back and forth between the atmosphere and conducting airways. Another 150-mL bolus of stale air shuttles back and forth between the conducting airways and alveoli. Dead-space ventilation (image) is the volume of the stale air so shuttled per minute. Alveolar ventilation (image) is the volume of fresh air per minute that actually reaches the alveoli, or the volume of stale alveolar air that reaches the atmosphere. imageN31-1 Thus, total ventilation—a reflection of the work invested in breathing—is the sum of the wasted dead-space ventilation and the useful alveolar ventilation. In our example,




Inhaled Versus Exhaled Volume

Contributed by Emile Boulpaep, Walter Boron

The rate of O2 consumption (e.g., 250 mL/min) is generally greater than the rate of CO2 production (e.g., 200 mL/min) in individuals consuming a typical Western-pattern diet. That is, the respiratory quotient (RQ) is generally less than unity (see p. 681). As a result, every minute we inhale slightly more air than we exhale (e.g., 250 mL/min − 200 mL/min = 50 mL/min). Because of this difference, in reporting ventilatory volumes, respiratory physiologists have standardized on the volume of exhaled air (VE), which as we have noted is slightly smaller than the volume of inhaled air (VI).

so that the dead-space ventilation is 30% of the total ventilation.

The inset of Figure 31-1 illustrates how inspiration and expiration lead to small fluctuations in the alveolar partial pressures for O2 and CO2, noted above in Figure 30-3B. Throughout the respiratory cycle, blood flowing through the pulmonary capillaries continuously draws O2 out of the alveolar air and adds CO2. Just before an inspiration, alveolar image has fallen to its lowest point, and alveolar image has risen to its highest. During inspiration, a new bolus of inspired fresh air mixes with pre-existing alveolar air, which causes alveolar image to rise and alveolar image to fall. During expiration and until the next inspiration, alveolar image and image gradually drift to the values that we saw at the start of the respiratory cycle. Assuming that functional residual capacity (FRC) is 3 L and that each breath adds 350 mL of fresh air, one can calculate that alveolar image oscillates with an amplitude of ~5 mm Hg, whereas alveolar image oscillates with an amplitude of ~4 mm Hg. imageN31-2


Oscillations in image and image During Breathing

Contributed by Emile Boulpaep, Walter Boron

We will assume that the volume of alveolar air after a quiet expiration is 3000 mL (the FRC), and that this air has a image of 98.4 mm Hg—the nadir of image during our hypothetical respiratory cycle. The subsequent inspiration delivers to the alveoli 350 mL of fresh air (at 500 mL tidal volume less 150 mL of anatomical dead space) that has a image of 149 mm Hg. Thus, after the 350 mL of fresh air mixes with the 3000 mL of pre-existing alveolar air, the alveolar image will be


(NE 31-1)

This value is the zenith of image during our hypothetical respiratory cycle. During the ensuing several seconds, the alveolar image drifts back down to 98.4 mm Hg as O2 diffuses from the alveolar air into the pulmonary-capillary blood. Thus, during a respiratory cycle, alveolar image oscillates from a low of 98.4 mm Hg to a high of 103.6 mm Hg, with a mean image of 101 mm Hg. In other words, alveolar image oscillates around a mean of 101 mm Hg, with the peak and nadir deviating from the mean by ~2.6 mm Hg—an amplitude of ~5 mm Hg.

In the case of CO2, the analysis is similar. After a quiet expiration, the 3000 mL of alveolar air has a image of 42.2 mm Hg—the zenith of image during our hypothetical respiratory cycle. The subsequent inspiration delivers to the alveoli 350 mL of fresh air, which has a image of ~0 mm Hg. Thus, after the 350 mL of fresh air mixes with the 3000 mL of pre-existing alveolar air, the alveolar image will be


(NE 31-2)

This value is the nadir of image during our hypothetical respiratory cycle. During the ensuing several seconds, the alveolar image drifts back up to 42.2 mm Hg as CO2 diffuses from the pulmonary-capillary blood to the alveolar air. Thus, during a respiratory cycle, alveolar image oscillates from a high of 42.2 mm Hg to a low of 37.8 mm Hg, with a mean image of 40 mm Hg. In other words, alveolar image oscillates around a mean of 40 mm Hg, with the peak and nadir deviating from the mean by ~2.2 mm Hg—an amplitude of ~4 mm Hg.

The Fowler single-breath N2-washout technique estimates anatomical dead space

In 1948, Ward Fowler introduced an approach for estimating the anatomical dead space, based on the washout of N2 from the lungs. The key concept is that N2 is physiologically inert. After the subject has been breathing room air, the alveolar air is ~75% N2. After a quiet expiration, when lung volume is FRC (Fig. 31-2A, step 1), the subject takes a single normal-sized breath (~500 mL). The inspired air is 100% O2, although, in principle we could use any nontoxic gas mixture lacking N2. The first portion of inspired O2 enters the alveolar spaces (step 2), where it rapidly mixes by diffusion and dilutes the N2 and other gases remaining after the previous breaths of room air (step 3). The last portion of the inspired O2 (~150 mL) remains in the conducting airways, which have a image of zero.


FIGURE 31-2 Fowler technique for measuring anatomical dead space.

The subject now exhales ~500 mL of air (see Fig. 31-2A, step 4). If no mixing occurred between the N2-free air in the most distal conducting airways and the N2-containing air in the most proximal alveolar spaces, then the first ~150 mL of air emerging from the body would have an [N2] of zero (see Fig. 31-2B, red lines). After this would come a sharp transition to a much higher [N2] for the final ~350 mL of expired air. Thus, the expired volume with an [N2] of zero would represent air from the conducting airways (anatomical dead space), whereas the remainder would represent air from the alveoli.

In reality, some mixing occurs between the air in the conducting airways and alveoli, so that the transition is S-shaped (see Fig. 31-2C, red curve). A vertical line drawn through the S-shaped curve imageN31-3 so that area a is the same as area b marks the idealized transition between air from conducing and alveolar airways—as in Figure 31-2B. The expired lung volume at the point of this vertical line is thus the anatomical dead space. In Figure 31-2C, the part of the S-shaped curve with an expired [N2] of zero represents pure dead-space air, the part where [N2] gradually rises represents a mixture of dead-space and alveolar air, and the part where [N2] is high and flat represents pure alveolar air. This plateau is important, because it is during this plateau that one obtains an alveolar gas sample.imageN31-4


Shape of the Single-Breath N2-Washout Curve

Contributed by Emile Boulpaep, Walter Boron

Exhaling vigorously causes turbulence (see p. 617) in the larger airways, further blurring the boundary between dead space and alveolar air. The greater the mixing, the more spread out is the S-shaped transition in Figure 31-2C.

Breath holding not only blurs the boundary (because more time is available for diffusional mixing of N2 in the alveolar spaces with the O2 in the dead space) but also moves the boundary to the left. In fact, if you were to hold your breath infinitely long, all the O2 in your conducting airways would be contaminated with N2. As a result, the N2 profile in Figure 31-2B or C would be a low, stable value from the very first milliliter of exhaled air. That is, there would be no gray area, and it would appear—according to the Fowler technique—as if you did not have any dead space.

Inhaling a large tidal volume of 100% O2 increases the apparent anatomical dead space as measured by the Fowler technique. The reason is that the conducting airways have a finite compliance (see p. 610)—lower than the compliance of alveoli, but still greater than zero. Thus, with a large inhalation (and thus a rather negative intrapleural pressure), the conducting airways dilate somewhat, and this dilation is reflected in a larger-than-normal value for anatomical dead space.


Obtaining a Sample of Alveolar Air

Contributed by Emile Boulpaep, Walter Boron

How do we obtain the alveolar air sample? As noted in our discussion of the Fowler single-breath N2-washout technique on pages 676–677, we ask the subject to make a prolonged expiratory effort. We discard the first several hundred milliliters of expired air—which contains pure dead-space air, a mixture of dead-space and alveolar air, and some pure alveolar air—to be certain that we do not contaminate our sample with air from the conducting airways. We then collect the end-tidal sample of air—pure alveolar air—and assay it for image.

The Bohr expired-[CO2] approach estimates physiological dead space

In principle, we could compute the dead space using any gas whose expiration profile looks like that of N2. Nitrogen is useful because we can easily create an artificial situation in which the subject makes a single inhalation of N2-free air (e.g., a single breath of 100% O2). Another possibility is CO2. Its profile during expiration is similar to that of N2. Moreover, we do not need to use any special tricks to get it to work because room air has practically no CO2. Yet plenty of CO2 is in the alveoli, where it evolves from the incoming mixed-venous blood. After a quiet expiration (Fig. 31-3A, step 1), the image of the alveolar air is virtually the same as the image of the arterial blood (see p. 673), ~40 mm Hg. The subject now inhales a normal tidal volume (~500 mL) of room air, although any CO2-free gas mixture would do. The first portion enters the alveoli (step 2), where it rapidly dilutes the CO2 and other gases remaining after the previous breath (step 3). The rest (~150 mL) remains in the conducting airways, which now have a image of ~0. When the subject now expires (step 4), the first air that exits the body is the CO2-free gas that had filled the conducting airways, followed by the CO2-containing alveolar air. Thus, the idealized profile of expired [CO2] (see Fig. 31-3B, red lines) is similar to the idealized [N2] profile (see Fig. 31-2B). In particular, the volume of expired air at the vertical line in Figure 31-3B is the estimated anatomical dead space.


FIGURE 31-3 Bohr method for measuring physiological dead space.

One could use a CO2 probe to record the expired [CO2] profile during a single-breath CO2 washout, rather than the [N2] profile that Fowler used to measure anatomical dead space. However, because CO2probes did not exist in his day, Christian Bohr imageN31-5 used a single-breath CO2 washout but analyzed the average image in the mixed-expired air (i.e., averaged over the dead space plus expired alveolar air).


Christian Bohr

Contributed by Walter Boron

Christian Bohr (1855–1911), a native of Copenhagen, was an eminent physiologist and also the father of the physicist Niels Bohr. The major contributions of Christian Bohr were the first description of dead space (specifically the physiological dead space) and the description of the Bohr effect. Christian Bohr was a proponent of the hypothesis that the lung actively secretes O2 into the pulmonary-capillary blood. His trainee, future Nobelist August Krogh, imageN31-21 later disproved this hypothesis.


August Krogh

For more information about Schack August Steenberg Krogh (1874–1949) and the work that led to his Nobel Prize, visit (accessed January 2015).

The principle of the Bohr approach is that the amount of CO2 present in the volume of mixed-expired air (VE) is the sum of the CO2 contributed by the volume of air from the dead space (VD) plus the CO2contributed by the volume of air coming from the alveoli (VE − VD). As summarized in Figure 31-3B:

1. The amount of CO2 coming from the dead space is the product of VD and [CO2] in this dead-space air. Because [CO2]D is zero, the area beneath VD is also zero.

2. The amount of CO2 coming from alveolar air is the product of (VE − VD) and alveolar [CO2] and is represented by the rose area in Figure 31-3B.

3. The total amount of CO2 in the mixed-expired air is the product of VE and the average [CO2] in this air and is represented by the hatched area in Figure 31-3B.

Because the rose and hatched areas in Figure 31-3B must be equal, and because the alveolar and expired [CO2] values are proportional to their respective image values, it is possible to show that imageN31-6



This is the Bohr equation. Typically, VD/VE ranges between 0.20 and 0.35. For a VD of 150 mL and a VE of 500 mL, VD/VE would be 0.30. For example, if the alveolar image is 40 mm Hg and the mixed-expired image is 28 mm Hg, then



Equation 31-4 makes good intuitive sense. In an imaginary case in which we reduced VD to zero, the expired air would be entirely from the alveoli, so that



On the other hand, if we reduced the tidal volume to a value at or below the dead-space volume, then all of the expired air would be dead-space air. In this case, image would be zero, and




The Bohr Equation

Contributed by Emile Boulpaep, Walter Boron

The principle of the Bohr approach is that the CO2 concentration of the expired air is the CO2 concentration of the alveolar air—as diluted by the CO2 in the dead space, which contains no CO2. Imagine that we have a volume of expired air (VE). The total amount of CO2 present in this expired air is the sum of the CO2 contributed by the volume of air from the dead space (VD) and the CO2 contributed by the volume of air coming from the alveoli (VE − VD):


(NE 31-3)

Knowing that the amount of the gas is simply the product of the volume and the concentration of the gas in that volume, we can write


(NE 31-4)

Note that the volume of air coming from the alveoli is not the total volume of alveolar air, which might be a couple of liters, but rather that part of the expired volume that came from the alveoli. For example, if the expired volume were 500 mL and the dead space were 150 mL, the air coming from the alveoli (VE − VD) would be 350 mL.

Because [CO2]D is virtually zero, we can drop the “conducting airway” term from Equation NE 31-4. In other words, the hatched area beneath the dashed line in Figure 31-3B is equal to the red area under the solid red line. The simplified version of Equation NE 31-4 is thus


(NE 31-5)

Furthermore, because the CO2 concentration is proportional to the CO2 partial pressure, and because the proportionality constant is the same for the expired air and the alveolar air,


(NE 31-6)

We can rearrange this equation and solve for the ratio VD/VE, which is the fraction of the expired volume that came from the dead space:


(NE 31-7)

This is the Bohr equation. Typically, VD/VE ranges between 0.20 and 0.35. For a VD of 150 mL and a VE of 500 mL, VD/VE would be 0.30. Because the partial pressure of CO2 in the dead-space air is virtually zero, image must be less than image. For example, imagine that the alveolar image were 40 mm Hg and the average image in the expired air were 28 mm Hg:


(NE 31-8)

Equation NE 31-7 makes good intuitive sense. In an extreme hypothetical case in which we reduced VD to zero, the expired air would be entirely from the alveoli, so that


(NE 31-9)

On the other hand, if the tidal volume and the dead space were both 150 mL, then all of the expired air would be dead-space air. Thus, the image would have to be zero, and


(NE 31-10)

If the tidal volume were increased from 500 mL to 600 mL and VD remained at 150 mL, the ratio VD/VE would fall from 0.30 to 0.25. With a greater VE, the alveolar air would make a greater contribution to the mixed expired air, so that the mean expired image would be 30 mm Hg (i.e., closer to image) instead of only 28 mm Hg. Thus, if you wish to compute VD using the Bohr approach, it is best to use an expired volume that is larger than VD, but not too much larger.

Two examples of this principle are of practical importance. During panting, the respiratory frequency is very high, but the tidal volume is only slightly greater than the anatomical dead space. Thus, most of the total ventilation is wasted as dead-space ventilation. If we reduced tidal volume below VD, then in principle there would be no alveolar ventilation at all! During snorkeling, a swimmer breathes through a tube that increases VD. If the snorkeling tube had a volume of 350 mL and the dead space within the body of the swimmer were 150 mL, then a tidal volume of 500 mL would in principle produce no alveolar ventilation. Consequently, the swimmer would suffocate, even though total ventilation was normal! Thus, the fractional dead space (VD/VE) depends critically on tidal volume.

It is important to recognize that although the Fowler and Bohr methods yield about the same estimate for VD in healthy individuals, the two techniques actually measure somewhat different things. The Fowler approach measures anatomical dead space—the volume of the conducting airways from the mouth/nose up to the point where N2 in the alveolar gas rapidly dilutes inspired 100% O2. The Bohr approach, on the other hand, measures the physiological dead space—the volume of airways not receiving CO2 from the pulmonary circulation, and, therefore, not engaging in gas exchange. In a healthy person, the anatomical and physiological dead spaces are identical—the volume of the conducting airways. However, if some alveoli are ventilated but not perfused by pulmonary-capillary blood, these unperfused alveoli, like conducting airways, do not contain CO2. The air in such unperfused alveoli, known as alveolar dead space, contributes to the physiological dead space:



The Fowler and Bohr methods could yield very different results in a patient with a pulmonary embolism, a condition in which a mass such as a blood clot wedges into and obstructs part or all of the pulmonary circulation. Alveoli downstream from the embolus are ventilated but not perfused; that is, they are alveolar dead space (see Fig. 31-3C). Thus, the Bohr method—but not the Fowler method—could detect an increase in the physiological dead space caused by a pulmonary embolism.

Alveolar ventilation is the ratio of CO2 production rate to CO2 mole fraction in alveolar air

One way of computing alveolar ventilation is to subtract the dead space from the tidal volume and multiply the difference by the respiratory frequency (see Equation 31-3). We can also calculate image from alveolar image. The body produces CO2 via oxidative metabolism at a rate of ~200 mL/min. In the steady state, this rate of CO2 production (image) must equal the rate at which the CO2 enters the alveoli, and the rate at which we exhale the CO2. Of course, this 200 mL/min of exhaled CO2 is part of the ~4200 mL of total alveolar air that we exhale each minute. Therefore, the exhaled 200 mL of CO2 is ~5% of the exhaled 4200 mL of alveolar air:



Rearranging the foregoing equation and solving for image yields



The equation above the brace is true only when we measure all parameters under the same conditions. This obvious point would hardly be worth noting if respiratory physiologists had not managed, by historical accident, to measure the two volume terms under different conditions:

1. Body temperature and pressure, saturated with water vapor (BTPS; see p. 594) for image.

2. Standard temperature and pressure/dry (STPD; see p. 594) for image.

Thus, in Equation 31-10 we introduce a constant k that not only indicates that alveolar image (image; measured at 37°C) is proportional to the mole fraction of CO2 in alveolar air, but also accounts for the different conditions for measuring the parameters. imageN31-7




The Conversion Factor 0.863

Contributed by Emile Boulpaep, Walter Boron

Unfortunately for the student of respiratory physiology, the way customs have evolved for measuring respiratory volumes and partial pressures is a case study in mixed conventions. In a rational world, all parameters would be measured under a consistent set of conditions. In the real world of respiratory science, however, the parameters you (and clinicians) will need to compute alveolar ventilation, alveolar image, and alveolar image are measured under at least three very different sets of conditions. As a result, the student is faced with “correction factors” such as 0.863. Our advice to the student, when working numerical problems, is to insert the laboratory data directly into the correct equation—which you ought to intuitively understand—and use the proper correction factor. You will get the right answer. Before explaining the origin of 0.863 in Equation 31-11, we will examine the system of units in which clinical laboratories report each of the three terms in that equation.

First, image, the alveolar ventilation, is reported in the same units as image (the rate at which air is expired from the lungs). If you collect all the warm, moist air that a subject exhales over a period of 1 minute, the laboratory will report that volume BTPS (body temperature and pressure, saturated with water vapor), precisely the same volume that the sample of expired air would have occupied in the warm, moist confines of the lung (see Box 26-3).

Second, image, the rate at which the body produces CO2, is measured at STPD (standard temperature and pressure/dry). Thus, if you collect a sample of warm, moist air expired over a period of 1 minute and give it to a clinical laboratory for analysis, the laboratory will report the image value as the volume that the dry CO2 in the sample would occupy at 0°C. The rationale is that this is the way chemists treat gases.

Third, image, the partial pressure of CO2 in the alveoli, is reported in BTPD (body temperature and pressure/dry). If you obtain a sample of alveolar air and send it to a clinical laboratory, it will take the warm, moist air you send it and keep it at 37°C but remove the H2O. Of course, if the sample is in a rigid container and if PB is 760 mm Hg, this removal of water vapor will lower the total pressure of the sample by 47 mm Hg (i.e., the vapor pressure of water at 37°C) to 713 mm Hg. To keep the pressure constant, the laboratory will allow the sample volume to decrease to a volume that is 713/760 of the original volume. This, however, means that the mole fraction of CO2 increases by the fraction 760/713. Thus, the image in the warm, moist alveolar air is actually lower than the reported BTPD; the BTPS value of image is the reported value multiplied by (713/760) or ~0.983. The real tragedy of the convention for reporting image at BTPD is that the same laboratory will report arterial image as BTPS.

In deriving the factor 0.863, we begin with a rearrangement of Equation 31-9, using a consistent set of units (i.e., BTPS):


(NE 31-11)

Because image is actually reported in STPD, our first task is to convert image (STPD) to image (BTPS). Because standard temperature is 0°C and body temperature is 37°C, we must multiply image (STPD) by the ratio of temperatures in degrees Kelvin, the factor [(273 + 37)/273]. Furthermore, as noted in the previous paragraph, when we add water vapor to a dry sample of a gas and then expand the volume to keep the total pressure fixed at PB, the partial pressure of that gas will decrease to (713/760) of its dry partial pressure. Thus,


(NE 31-12)

Our final task is to replace (%CO2 BTPS)A with image (BTPD). Because image (BTPS) = (%CO2 BTPS)A × PB,


(NE 31-13)

Because image is reported in BTPD, we must multiply it by the factor (713/760). Thus,


(NE 31-14)

Substituting the expression for (%CO2 BTPS)A in Equation NE 31-14 into Equation NE 31-12, we have


(NE 31-15)

In the above equation, both image and image are reported in mL/min. Because it is customary to report image in L/min, the factor 863 becomes 0.863:


(NE 31-16)

The above is the alveolar ventilation equation.

For example, if image is 200 mL/min and image is 40 mm Hg, then


(NE 31-17)

This is the alveolar ventilation equation, which we can use to compute image. We determine image by collecting a known volume of expired air over a fixed time period and analyzing its CO2 content. For a 70-kg human, image is ~200 mL/min. We can determine image by sampling the expired air at the end of an expiration—an alveolar gas sample (see pp. 676–677). imageN31-4 This end-tidal image (see Fig. 31-2C) is ~40 mm Hg. In practice, clinicians generally measure arterial image (image), and assume that alveolar and arterial image are identical (see p. 673). Inserting these values into the alveolar ventilation equation, we have



For the sake of simplicity—and consistency with our example in Equation 31-3—we round this figure to 4.2 L/min.

Although it is usually safe to regard image as being constant for a person at rest, a clinical example in which image can increase markedly is malignant hyperthermia (see Box 9-2), which is associated with increased oxidative metabolism. One hallmark of this clinical catastrophe is that the increase in image leads to an increase in image even though image is normal.

Alveolar and arterial image are inversely proportional to alveolar ventilation

Viewing Equation 31-11 from a different perspective illustrates one of the most important concepts in respiratory physiology: Other things being equal, alveolar image is inversely proportional to alveolar ventilation. This conclusion makes intuitive sense because the greater the image, the more the fresh, inspired air dilutes the alveolar CO2. Rearranging Equation 31-11 yields



In other words, if CO2 production is fixed, then doubling image causes image to fall to half of its initial value. Conversely, halving image causes image to double. Because, arterial image is virtually the same as alveolar image (see p. 673), changes in image affect both image and image.

The blue curve in Figure 31-4 helps illustrate the principle. Imagine that your tissues are producing 200 mL/min of CO2. In a steady state, your lungs must blow off 200 mL of CO2 each minute. Also, imagine that your lungs are exhaling 4200 mL/min of alveolar air. Because the 200 mL of expired CO2 must dissolve in the 4200 mL of exhaled alveolar air (center red point in Fig. 31-4), Equation 31-13 tells us that your image (and thus image) must be ~40 mm Hg.


FIGURE 31-4 Dependence of alveolar CO2 and O2 on alveolar ventilation. As alveolar ventilation increases, alveolar image and image approach their values in inspired air. At extremely low or high image, where O2 consumption and CO2 production do not remain constant, these idealized curves are no longer valid.

What would happen if the excitement of reading about respiratory physiology caused your alveolar ventilation to double, to 8400 mL/min? This is an example of hyperventilation. You could double image either by doubling respiratory frequency or by doubling the difference between tidal volume and dead space, or a combination of the two (see Equation 31-3). Immediately on doubling image, you would be blowing off at twice the previous rate not only alveolar air (i.e., 8400 mL/min) but also CO2 (i.e., 400 mL/min). Because your body would continue to produce only 200 mL/min of CO2—assuming that doubling image does not increase image—you would initially blow off CO2 faster than you made it, causing CO2 levels throughout your body to fall. However, the falling image of mixed-venous blood would cause alveolar image to fall as well, and thus the rate at which you blow off CO2 would gradually fall. Eventually, you would reach a new steady state in which the rate at which you blow off CO2 would exactly match the rate at which you produce CO2 (i.e., 200 mL/min).

When you reach a new steady state, the image values in your mixed-venous blood, arterial blood, and alveolar air would be stable. But what would be the image? Because each minute you now are blowing off 8400 mL of alveolar air (i.e., twice normal) but still only 200 mL of CO2 (right red point in Fig. 31-4), your image must be half normal, or ~20 mm Hg. Not only does the hyperventilation cause alveolar image to fall by half, it also causes arterial image to fall by half. Thus, hyperventilation leads to respiratory alkalosis (see p. 634). This respiratory alkalosis causes the arterioles in the brain to constrict (see p. 559), reducing blood flow to the brain, which causes dizziness.

What would happen if, instead of doubling alveolar ventilation, you halved it from 4200 to 2100 mL/min? This is an example of hypoventilation. At the instant you began hypoventilating, the volume of CO2expired per unit time would fall by half, to 100 mL/min, even though CO2 production by the tissues would remain at 200 mL/min. Thus, CO2 would build up throughout the body, causing image to rise. To what value would image have to increase before you would reach a new steady state? Because each minute you must exhale 200 mL of CO2 but this can be diluted in only 2100 mL, or half the usual amount of alveolar air (left red point in Fig. 31-4), the alveolar [CO2] must double from ~40 to 80 mm Hg. Of course, this doubling of alveolar image is paralleled by a doubling of arterial image, leading to a respiratory acidosis (see p. 633).

Therefore, the steady-state alveolar image is inversely proportional to alveolar ventilation. The higher the image, the lower the image. If image were infinitely high, then image would theoretically fall to zero, the image of inspired air.

Alveolar and arterial image rise with increased alveolar ventilation

As illustrated by the red curve in Fig. 31-4, increases in alveolar ventilation cause alveolar image to rise and—at an infinite image—approach the inspired image of ~149 mm Hg.

Although alveolar image obviously depends critically on image, it is also influenced to a lesser extent by alveolar gases other than O2—namely, H2O, N2, and CO2. The partial pressure of H2O at 37°C is 47 mm Hg, and image will not change unless body temperature changes. The partial pressures of all of the other gases “fit” into what remains of barometric pressure (PB) after image has claimed its mandatory 47 mm Hg. We can think of N2 as a “spectator molecule,” because it is not metabolized; image is whatever it has to be to keep the total pressure of the dry alveolar air at 760 − 47 = 713 mm Hg. That leaves us to deal with CO2.

Because inspired air contains virtually no CO2, we can think of the alveolar CO2 as coming exclusively from metabolism. However, image depends not only on how fast the tissues burn O2 but also on the kind of fuel they burn. If the fuel is carbohydrate, then the tissues produce one molecule of CO2 for each O2 consumed (see p. 1188). This ratio is termed the respiratory quotient (RQ):



In this example the RQ is 1, which is a good place to start when considering how alveolar image affects alveolar image. If we consider only the dry part of the inspired air that enters the alveoli (see Table 26-1), then image is 713 × 0.78 = 557, image is 713 × 0.2095 = 149 mm Hg, and image is ~0. As pulmonary-capillary blood takes up incoming O2, it replaces the O2 with an equal number of CO2 molecules in the steady state (RQ = 1). Because the exchange of O2 for CO2 is precisely 1 for 1, alveolar image is what is left of the inspired image after metabolism replaces some alveolar O2 with CO2 (image = 40 mm Hg):



A typical fat-containing Western-pattern diet produces an RQ of ~0.8 (see p. 1188), so that 8 molecules of CO2 replace 10 molecules of O2 in the alveolar air. This 8-for-10 replacement has two consequences. First, the volume of alveolar air falls slightly during gas exchange. Because the non-H2O pressure remains at 713 mm Hg, this volume contraction concentrates the N2 and dilutes the O2. Second, the volume of expired alveolar air is slightly less than the volume of inspired air.

The alveolar gas equationimageN31-8 describes how alveolar image depends on RQ:






The Alveolar Gas Equation

Contributed by Emile Boulpaep, Walter Boron

The alveolar gas equation allows one to compute the ideal alveolar image (image) from one's knowledge of the alveolar image (image) as well as two other parameters, the mole fraction of O2 in the dry inspired air (image) and the RQ. We begin by deriving an expression for RQ, which, as described on page 1188, is defined as the ratio of the rate of metabolic CO2 production (image) to the rate of O2 consumption (image):


(NE 31-18)

For RQ values less than unity, the body consumes more O2 than it produces CO2, so that the inspired alveolar ventilation (image) must be greater than the expired alveolar ventilation (image). Because the metabolically produced CO2 must all appear in the expired alveolar gas, image is the fraction of the expired alveolar ventilation that is CO2 gas (image):


(NE 31-19)

Similarly, the consumed O2 must all enter the body via the lungs; image is the difference between (1) the amount of O2 entering the alveoli during inspirations over a certain period of time, and (2) the amount of O2 exiting the alveoli during expirations over that same time interval:


(NE 31-20)

Here, image is the mole fraction of O2 in the inspired air, and image is the mole fraction of O2 in the expired alveolar air. Substituting the expressions for image (see Equation NE 31-19) and image (see Equation NE 31-20) into the definition of RQ (see Equation NE 31-18), we have


(NE 31-21)

We now need an expression for the ratio image. In the derivation of this expression, the fundamental assumption is that the N2 gas in the lungs is not metabolized. Thus, in the steady state, the amount of N2entering the alveoli with inspirations over a certain period of time is the same as the amount of N2 leaving the alveoli with the expirations over the same period. In other words,


(NE 31-22)

Here, image is the mole fraction of N2 in this inspired gas, and image is the mole fraction of N2 in the expired alveolar air. For the rest of this derivation, we will consider only dry gases; the laboratory reports all partial pressure (P) and mole fraction (F) values in terms of the “dry” gas (see Box 26-1). Thus, considering only the dry inspired air, the mole fractions of N2, O2, and CO2 must sum to 1:


(NE 31-23)

Because the inspired air contains virtually no CO2 (i.e., image ≅ 0),


(NE 31-24)

Similarly, considering only the dry expired air, the mole fractions of N2, O2, and CO2 must likewise sum to 1:


(NE 31-25)

If we now substitute the expressions for image (see Equation NE 31-24) and image (see Equation NE 31-25) into Equation NE 31-22, we have


(NE 31-26)

Solving for the ratio of inspired to expired alveolar ventilation yields


(NE 31-27)

We now have the ratio (image) that we needed for Equation NE 31-21. Substituting Equation NE 31-27 into Equation NE 31-21, we get


(NE 31-28)

If we now solve Equation NE 31-28 for image, we have


(NE 31-29)

Realizing that the “dry” partial pressure (P) is the product of mole fraction (the above F terms) and (PB − 47), we can multiply Equation NE 31-12 through by (PB − 47) to arrive at our final equation, which expresses quantities in terms of partial pressures:


(NE 31-30)

Remember, all the partial-pressure and mole-fraction values in the above equation refer to dry gases. Thus, image is image × (PB − 47).

On page 681, we provide a simplified version of Equation NE 31-30:


(NE 31-31)

Why is this equation a reasonable approximation? Examination of Equations NE 31-30 and NE 31-31 shows that they are identical except that the term in brackets in Equation NE 31-30 is replaced by 1/RQ in Equation NE 31-31. For Equation NE 31-30, we can rearrange the term in brackets as follows:


(NE 31-32)

Under physiological conditions—when image is 0.21 and RQ is 0.8—the above expression evaluates to 0.958/RQ, which is quite close to 1/RQ … which is to say that Equation NE 31-31 is a reasonable approximation of Equation 31-30 under these conditions.

When RQ is unity, the term in brackets in Equation 31-30 evaluates to unity, and Equation NE 31-30 simplifies to Equation 31-16 in the text:


(NE 31-33)

image is the fraction of inspired dry air that is O2, which is 0.21 for room air (see Table 26-1). Note that when RQ is 1, the term in parentheses becomes unity, and Equation 31-17 reduces to Equation 31-16. The term in parentheses also becomes unity, regardless of RQ, if image is 100% (i.e., the subject breathes pure O2)—in this case, no N2 is present to dilute the O2.

A simplified version of Equation 31-17 is nearly as accurate:



The concepts developed in the last two sections allow us to compute both alveolar image and alveolar image. The approach is first to use Equation 31-13 to calculate image from image and image, and then use Equation 31-17 to calculate image from image and RQ. Imagine that we first found that image is 40 mm Hg and that we know that RQ is 0.8. What is image?



By default, the partial pressure of N2 and other gases (e.g., argon) is PB − image − image or 713 − 40 − 101 = 572 mm Hg. For simplicity, we round down this image to 100 mm Hg in our examples.

Because of the action of gravity on the lung, regional ventilation in an upright subject is normally greater at the base than the apex

Until now, we have assumed that all alveoli are ventilated to the same extent. We could test this hypothesis by using an imaging technique for assessing the uniformity of ventilation. Imagine that a subject who is standing up breathes air containing 133Xe. Because Xe has very low water solubility, it (like He and N2) has a very low diffusing capacity (see pp. 661–663) and—over a short period—remains almost entirely within the alveoli.

Imaging the 133Xe radioactivity immediately after a single breath of 133Xe provides an index of absolute regional ventilation (Fig. 31-5A). However, [133Xe] might be low in a particular region either because the alveoli truly receive little ventilation or because the region has relatively little tissue. Therefore, we normalize the absolute data to the maximal regional alveolar volume. The subject continues to breathe the 133Xe until [133Xe] values stabilize throughout the lungs. When the subject now makes a maximal inspiratory effort (VL = total lung capacity [TLC]), the level of radioactivity detected over any region reflects that region's maximal volume. Dividing the single-breath image by the steady-state image at TLC yields a ratio that describes regional ventilation per unit volume.


FIGURE 31-5 Distribution of ventilation. (Data from West JB: Respiratory Physiology—The Essentials, 4th ed. Baltimore, Williams & Wilkins, 1990.)

This sort of analysis shows that alveolar ventilation in a standing person gradually falls from the base to the apex of the lung (see Fig. 31-5B). Why? The answers are posture and gravity. In Chapter 27 we saw that, because of the lung's weight, intrapleural pressure (PIP) is more negative at the apex than at the base when the subject is upright (see Fig. 31-5C). The practical consequences of this PIP gradient become clear when we examine a static pressure-volume diagram not for the lungs as a whole (see Fig. 27-5), but rather for a small piece of lung (see Fig. 31-5D). We assume that the intrinsic mechanical properties of the airways are the same, regardless of whether the tissue is at the base or at the apex. At the base, where PIP might be only −2.5 cm H2O at FRC, the alveoli are relatively underinflated compared to tissues at the apex, where PIP might be −10 cm H2O and the alveoli are relatively overinflated. However, because the base of the lung is underinflated at FRC, it is on a steeper part of the pressure-volume curve (i.e., it has a greater static compliance) than the overinflated apex. Thus, during an inspiration, the same ΔPIP (e.g., 2.5 cm H2O) produces a larger ΔVL near the base than near the apex. Keep in mind that it is the change in volume per unit time, not the initial volume, that defines ventilation.

The relationship between ventilation and basal versus apical location in the lung would be reversed if the subject hung by the knees from a trapeze. A person reclining on the right side would ventilate the dependent lung tissue on the right side better than the elevated lung tissue on the left. Of course, the right-to-left PIP gradient in the reclining subject would be smaller than the apex-to-base PIP gradient in the standing subject, reflecting the smaller distance (i.e., smaller hydrostatic pressure difference). Subjects under microgravity conditions (see p. 1233), such as astronauts aboard the International Space Station, experience no PIP gradients, and thus no gravity-dependent regional differences in ventilation.

Restrictive and obstructive pulmonary diseases can exacerbate the nonuniformity of ventilation

Even in microgravity, where we would expect no regional differences in ventilation, ventilation would still be nonuniform at the microscopic or local level because of seemingly random differences in local static compliance (C) and airway resistance (R). In fact, such local differences in the ventilation of alveolar units are probably more impressive than gravity-dependent regional differences. Moreover, pathological changes in compliance and resistance can substantially increase the local differences and thus the nonuniformity of ventilation. imageN31-9


Detecting Nonuniformity of Ventilation

Contributed by Emile Boulpaep, Walter Boron

We have already seen how 133Xe scanning can be useful in detecting the physiological nonuniformity of ventilation. Obviously, because of limitations in scanning technology, 133Xe scans are helpful only if sufficiently large regions of the lung have sufficiently large differences in ventilation. However, a 133Xe scan might not detect a nonuniformity of ventilation in which many small, well-ventilated airways are intermingled with many, small poorly ventilated airways. Two other approaches that we have mentioned in other contexts would detect such a nonuniformity of ventilation.

The first is the single-breath N2-washout technique, which we introduced in our discussion of the Fowler method for measuring anatomical dead space (see p. 675). In this approach, the subject inhales one breath of 100% O2, and we assume that the inspired 100% O2 distributes evenly throughout all the alveoli of the lungs. If ventilation is indeed uniform, the inspired O2 dilutes alveolar N2 uniformly in all regions of the lung. Thus, when the subject exhales, the air emerging from the alveolar air spaces should have a uniform [N2], and the plateau of the single-breath N2 washout should be flat, as shown by the red curve in Figure 31-2C labeled “Pure alveolar air.”

However, if the alveoli are unevenly ventilated, the inhaled 100% O2 will not be distributed uniformly throughout the lungs and therefore will not uniformly dilute the pre-existing alveolar N2. Regions of the lung that are relatively hypoventilated will receive relatively less 100% O2 during the single inspiration, so that they will be relatively poor in O2 but rich in N2. Conversely, hyperventilated regions will receive relatively more of the inhaled 100% O2 and hence will be O2 rich and N2 poor. During the expiration, we no longer observe a plateau for [N2]. Why? After exhalation of the anatomical dead space, the first alveolar air out of the lungs is dominated by the O2-rich/N2-poor gas coming from the relatively hyperventilated airways—which inflate and deflate relatively quickly. As the expiration continues, the alveolar air gradually becomes dominated more and more by the O2-poor/N2-rich gas from the hypoventilated airways, which inflate and deflate relatively slowly. Because of this shift from hyperventilated to hypoventilated regions, the [N2] gradually creeps upward—that is, there is no clear plateau—in subjects with a sufficiently high nonuniformity of ventilation.

A second test for unevenness of ventilation is the 7-minute N2 washout. We saw in Chapter 26 how to compute lung volume from the volume of distribution of N2 (see p. 602). The general approach is for a subject to inhale 100% O2, allow that O2 to dilute the pre-existing alveolar N2, and then exhale into a collection container. If this breathing pattern is continued for a standard period of 7 minutes, and if ventilation is evenly distributed, virtually all of the pre-existing N2 washes out of the lungs (see Fig. 26-9B). We already learned that, from the amount of N2 washed into the collection container, we can compute VL. However, we can also use this experiment to assess the evenness of ventilation. In a normal individual, the mean alveolar [N2] in the expired air is <2.5% after 7 minutes of O2 breathing. However, if some airways are poorly ventilated, their N2 will not be washed out as well after 7 minutes of O2 breathing, so that the [N2] in these hypoventilated airways may be substantially greater than 2.5%. Because these hypoventilated airways contribute to the total expired alveolar air, the mean expired alveolar [N2] after 7 minutes will be elevated. Obviously, the degree of elevation depends on the volume of hypoventilated airways and the extent of their hypoventilation.

Restrictive Pulmonary Disease

As discussed in Box 27-1, restrictive pulmonary diseases include disorders that decrease the static compliance of alveoli (e.g., fibrosis) as well as disorders that limit the expansion of the lung (e.g., pulmonary effusion). Figure 31-6A shows a hypothetical example in which R is normal and disease has halved the static compliance of one lung but left the other unaffected. imageN31-10 Thus, for the usual change in PIP, the final volume change (ΔV) of the diseased lung is only half normal, so that its ventilation is also halved. Because the ventilation of the other unit is normal, decreased local compliance has increased the nonuniformity of ventilation.


FIGURE 31-6 Pathological nonuniformity of ventilation.


Effect of Changes in Compliance on the Time Constant Governing Changes in Lung Volume

Contributed by Emile Boulpaep, Walter Boron

On page 622, we saw that, during inspiration and expiration, the time course of lung volume (VL) is approximately exponential. The time constant (τ) for the change in VL (ΔVL) is the time required for the change in VL to be ~63% complete. Moreover, imageN27-10 explains why τ is the product of airway resistance (R) and alveolar compliance (C):

image (NE 31-34)

Decreased Compliance

In the example discussed under restrictive pulmonary disease on page 683, we decreased the compliance of one lung by half. Of course, other things being equal, the ΔV of the affected lung will be half normal, as illustrated in Figure 31-6A. What is not so obvious is that τ of the affected lung will also be half normal. In other words, the lung with half-normal compliance will achieve its final volume twice as fast as normal. The reason is that, with a normal airway resistance, the inhaled air—at any instant in time—will flow at a more-or-less normal rate, so that the affected lung achieves its half-normal ΔV earlier than the normal lung.

Although it might seem that the decreased τ in this example is a good thing, the problem is that the reduced ΔV translates to less ventilation, and that is not a good thing. As described in the text, this reduced ventilation increases the nonuniformity of ventilation, which in turn—as discussed below in this chapter—tends to lead to hypoxia and respiratory acidosis.

Increased Compliance

What would be the effect of increasing the static compliance of one lung? A disease that increases the static compliance is emphysema (see Fig. 27-5). Other things being equal, an increase in compliance will cause the ΔV of the affected lung to be greater than normal. At the same time, the τ of the affected lung will also be increased. The reason is that, with a normal airway resistance, the inhaled air—at any instant in time—will flow at a more-or-less normal rate, so that the affected lung achieves its greater-than-normal ΔV later than a normal lung. If τ is too large, then the time that the person allows for inspiration may not be long enough for the affected lung to increase its volume to the level that it could achieve if the inspiration were infinitely long. As a result, the inspiration will be truncated, and the true ΔV for the affected lung will be less than normal, thereby exacerbating the unevenness of ventilation. Moreover, the greater the respiratory frequency, the greater the truncation of the inspiration, and thus the greater the exacerbation of the unevenness of ventilation.

Figure 27-15A illustrates the effect of increased airway resistance on the time course of VL, and Figure 27-15B illustrates the effect on the change in lung volume, which is proportional to dynamic compliance. In the example in Figure 27-15, the static compliance was normal and the final ΔV was the same for the normal and affected lung. The example we are discussing in this webnote is just the opposite (i.e., a normal airway resistance but altered compliance). Nevertheless, the principle of how increases in τ affect ventilation is the same: the greater the τ, the greater the chance that increased respiratory frequency will lead to a truncated inspiration and thus a greater unevenness of ventilation.

Although we have focused on inspiration in these examples, the same principles apply to expiration, namely, the greater the τ, the greater the chance of a truncated expiration.

Obstructive Pulmonary Disease

As discussed in Box 27-2, obstructive pulmonary diseases include disorders (e.g., asthma, chronic obstructive pulmonary disease [COPD]) that increase the resistance of conducting airways. Scar tissue or a local mass, such as a neoplasm, can also occlude a conducting airway or compress it from the outside. Even if the effect is not sufficiently severe to increase overall airway resistance, a local increase in R causes an increase in the time constant τ imageN31-10 for filling or emptying the affected alveoli (see Fig. 31-6B, lower curve). An isolated increase in R will not affect ΔV if sufficient time is available for the inspiration. However, if sufficient time is not available, then alveoli with an elevated τ will not completely fill or empty, and their ventilation will decrease. Of course, the mismatching of ventilation between the two units worsens as respiratory frequency increases, as we saw in our discussion of dynamic compliance (see p. 624).