Pulmonary 4: Ventilation

Question 1

Describe Boyle’s Law.

Answer 1

Pressure volume law
volume of a given amount of gas varies inversely with the applied pressure when temperature and mass are constant.

V is proportional to 1/P …V1P1=V2P2

Question 2

What is Charles’ Law?

Answer 2

temperature-volume law
Volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature.

V proportional to T

Question 3

Describe Gay- Lussac’s Law.

Answer 3

(pressure-temperature law)
The pressure of a given amount of gas at constant volume is directly proportional to the Kelvin temperature

P is proportional to T

Question 4

Describe Avogadro’s Law.

Answer 4

(volume-amount law)

If the amount of gas in a container is increased, the volume increases

Question 5

Describe the combined gas law.

Answer 5

The ratio between pressure-volume product and the temperature remains constant.

V is proportional to T/P

(P1V1)/T1 = (P2V2)/T2

Question 6

Describe the ideal gas law.

Answer 6

The state of an amount of gas is determined by its pressure, volume and temperature.

PV=nRT

Question 7

Describe Dalton’s Law of Partial pressures.

Answer 7

the total pressure of a mixture of non-reacting gases is the sum of their individual partial pressures

Ptot=P1+P2+P3

Question 8

Describe Amagat’s Law of Partial volumes

Answer 8

The volume of a gas mixture is equal to the sum of the component volumes of each individual component.

Vtot=V1+V2+V3

Question 9

Describe Henry’s Law of gas solubility.

Answer 9

The concentration of a solute gas in a solution is directly proportional to the partial pressure of that gas above the solution.

C=KhP

Question 10

Describe the composition of air in the airways:

ambient air:
N2, O2, water vapor, argon, CO2

atmospheric/barometric pressure at sea level

Answer 10

79% N2 
21% O2
1% water vapor
.1% argon
.04% CO2

N2>O2»>CO2, argon…

atmospheric pressure at sea level: 760mmHg

Question 11

What are gas fractions? Partial pressures? Provide partial pressure values for N2 and O2 in atmospheric air.

Answer 11

gas fractions: sum of individual gas fractions =1
(1=FN2 +FO2 +…)

Partial pressure:
sum of partial pressures is equal to total pressure
(Patm= 760mmHg= PN2 +PO2 +…)

Pgas=Fgas xPatm

79% N2, partial pressure of N2= 600mmHg
P(N2)= 760mmHg x .79= 600mmHg.

Partial pressure of O2 in atmospheric air:
P(O2)=760mmHg x .21=160mmHg

Question 12

Describe what happens when air enters the respiratory system by ventilation.

What is the value of water vapor pressure?

Answer 12

When air enters respiratory system by ventilation, besides being filtered of airborne particles and dust, it is warmed to body temperature (37 degrees C) and completely humidified (100%).

Water vapor pressure is 47mmHg at 37 degrees C and “dilutes” the other gases.

Question 13

What is ventilation? (equation)

What is V in adults and children?

Answer 13

V= f x Vt
=15/min x 500mL =7.5L/min
children (3-5 mL/kg)

Question 14

Ambient air has 79% N2, P(N2)= 600mmHg
21% O2 =P(O2)=160mmHg.

When inhaled, describe Ptrachea(O2) and Ptrachea(N2). Do the calculations.

Answer 14

Ptrachea(O2)= (Patm-Pwater)xF(O2)
=150mmHg

Ptrachea(N2)=(Patm-Pwater)xF(N2)
Ptrachea(N2)=563mmHg

Question 15

What is the alveolar gas equation?
Solve

What is R?
Normally? fatty acid? carbohydrate?

What are the average PA(O2), PA(CO2) and PA(N2) values?

Answer 15

PA(O2)=PI(O2)-PA(CO2)/R

= (Patm-Pwater) x FiO2 -PA(CO2)/R
=102mmHg

R= respiratory quotient= excreted CO2/O2 taken up
Normal =.8
(fatty acid: 0.7/carbohydrate:1)

on average PA(O2)=102mmHg
PA(CO2)=40-45mmHg. (these fluctuate with respiratory cycle a bit but P(N2) is always 563mmHg bc human body neither produces or consumes nitrogen gas.

Question 16

What is the alveolar PO2 (PAO2) if you breath 100% oxygen and PA(CO2) is 45mmHg (R=0.8)?

Answer 16

PIO2= (760-47mmHg) x 1 (100% oxygen)= 713 mmHg

PAO2= 713mmHg-45mmHg/.8=657mmHg

Question 17

What is the PAO2 if you breath 40% oxygen

and the PACO2 is 60 mmHg (R = 0.8)?

Answer 17

PIO2= (760 mmHg – 47 mmHg) x 0.4
= 285 mmHg

PAO2 = 285 mmHg – 60 mmHg/0.8
= 210 mmHg

Question 18

How is the fraction of alveolar CO2 determined?

Answer 18

by metabolism and rate of elimination (alveolar ventilation)

CO2 production (VCO2)= alveolar ventilation (VA) x FA(CO2)

alveolar PCO2 is inversely proportional to alveolar ventilation and direclty proportional to CO2 production.

PA(CO2) proportional to 1/VA
PA(CO2) proportional to V(CO2)

(slide 12 for additional eq.)

Question 19

In the resting state what is the body’s production rate of CO2?

How is alveolar ventilation adjusted to keep alveolar PACO2 near 40mmHg?

If someone hyperventilates to 10L/min what will occur to PACO2?

Answer 19

resting state body has CO2 production of about 250mL/min
Alveolar ventilation is adjusted to 5L/min to keep alveolar PACO2 near 40mmHg.

If hyperventilate to 10L/min, the increased alveolar ventilation will blow off excess CO2 and reduce PACO2 to 20mmHg

Question 20

How will a 50 percent decrease in ventilation at rest (5L/min to 2.5) affect alveolar PCO2?

Show this on a graph of alveolar ventilation to alveolar PCO2. Label hypoventilation and hyperventilation. show resting/mild exercise.

(What two main things determine alveolar CO2?)

Answer 20

50 percent decrease in ventilation at rest will result in doubling of PCO2.

During exercise, CO2 production is increased and to maintain a normal PCO2, ventilation must increase.

Slide 13.

alveolar CO2 is determined by metabolism and ventilation (rate of elimination)

Question 21

Describe the distribution of inspired air into the vertical lung (based on radioactive xenon studies of inhaled Xe133).

How would this look in a graph (distance on horizontal axis lower zone-upper zone) and ventilation/unit volume on vertical axis).

Answer 21

from studies, inhaled Xe133 is distributed more to the base of the lung (near the diaphragm) than the apex of the lung (near the clavicle)

Question 22

What are the two reasons for inspired air being preferentially shunted to lung base?

Answer 22

First, the 5 lobes of the lung form a triangular structure with the widest part at the base and the narrowest part at the apex. Thus anatomically the base of lung has more alveoli to receive more air.

Second, at FRC the base of the lung is more compliant than the apex of the lung. Thus, physiologically, equal changes in intrapleural pressure at both the base and apex produces a greater volume change in the base (more compliant) than apex (stiffer).

Question 23

How does suspension of the lung in upright position affect pleural pressure and translung pressure?

Where is this effect greatest and least between TLC, RV, and FRC?

(How is inspired air distributed differentially to lung units?

Answer 23

(Slide 16)
Bc of suspension of the lung in the upright position, the pleural pressure (Ppl) and translung pressure (PL) of units at the apex will be greater (more negative than those at the base…

These lung units will be larger at any lung volume than units at the base. The effect is greatest at residual volume, less at functional residual capacity (FRC), and disappears totally at TLC.

(Also bc of their location on the pressure volume curve, inspired air will be differentially distributed to these lung units; the lung units at the apex are less compliant (=smaller increase in volume at any given pressure change) and thus will receive a smaller proportion of the inspired air than the lung units at the base, which are more compliant (reside at a steeper part of the pressure-volume curve, =larger increase in volume at any given pressure change)

Question 24

Describe the regional differences in ventilation due to gravitational effects at tidal volume and residual volume by two graphs. (draw graph comparing intrapleural pressure to volume and show where on curve is Vt and RV)

Is intrapleural pressure more or less negative at the base of lung? What are the implications therefore for its resting state and inspiration as compared to apex.

What is the situation at very low lung volumes?

Answer 24

Slide 15.
(Explanation of the regional differences of ventilation down the lung.)
Because of the weight of the lung, the intrapleural pressure is less negative at the base
than at the apex. As a consequence, the basal lung is relatively compressed in its resting state but expands more on inspiration than the apex.

Situation at very low lung volumes. Now intrapleural pressures are less
negative, and the pressure at the base actually exceeds airway (atmospheric) pressure.
As a consequence, airway closure occurs in this region, and no gas enters with small
inspirations.

Question 25

Draw a graph comparing translung pressure (-10 to +30) to lung volume (0 to 100% TLC). On this graph show where on curve alveoli at the lung base are located. What is the significance for ventilation?

Where are alveoli at the apex located?

When is the effect of gravity less pronounced (in what position)? Why?

Answer 25

Slide 16.
Because of the difference in
alveolar volume at the apex and at the base of the lung, alveoli at the lung base are located along
the steep portion of the pressure-volume curve, and
they receive more of the ventilation (i.e., they have
greater compliance).

In contrast, the alveoli at the
apex are closer to the top of the pressure-volume
curve. They have lower compliance and thus receive proportionately less of the tidal volume.

The effect of
gravity is less pronounced when one is supine rather than upright, and it is less when one is supine rather than prone. This is because the diaphragm is pushed cephalad when one is supine, and it affects the size of
all of the alveoli.

Question 26

What are the two factors that the rate at which alveolar unit fills with air depends upon?

What is the time constant?

How will increased resistance affect time constant and filling?

How will decreased compliance affect time constant and filling?

Answer 26

The rate at which alveolar unit fills with air depends upon 2 factors: its Resistance and compliance

R x C product is the time constant for the unit,
the smaller the time constant, the faster the lung fills. (normal is .7 x..8=.56)

Increased resistance increases the time constant and slows the filling

Decreased compliance - filling is speeded but the unit is stiffer and only fills to half capacity.

Question 27

Given 3 scenarios, decreased compliance, normal, and increased compliance, graph these lines on a seconds vs volume change graph.

Answer 27

Slide 17.

Normal lung has time constant of .56 seconds. This unit reaches 97% of final equilibrium in 2 seconds (normal inspiratory time)

Unit with 2-fold increase in resistance will have its time constant doubled. Will fill more slowly and only reaches 80% equilibrium during a normal breath. The unit is underventilated.

The reduced compliance unit is stiff and its time constant is thus reduced. It fills faster than the normal unit but only receives half the ventilation of a normal unit.

Question 28

Describe nitrogen gas in lung (normal) and what happens when you exhale to RV then take a deep breath to TLC.

How does N2 dilution differ in apical vs basal alveoli? Why?

Answer 28

Slide 18.
While breathing atmospheric air, the lung is constantly filled with 79% nitrogen gas. If one exhales to RV, then takes a deep breath of 100% O2 gas to the TLC, the nitrogen in the lung gets diluted.

Since base of lung receives largest distribution of ventilation, the basal alveoli experience the highest N2 dilution (40%) whereas apical alveoli experience the lowest N2 dilution (70%)

The single-breath N2 washout curve is a
simple useful pulmonary function test of the regional distribution of ventilation. It clearly shows that not all lung units have equal
V/Q. The well-ventilated units (short time constant) empty faster
than less well ventilated units (long time constant). The portion
of the curve up to the first vertical dashed line represents the washout of dead space air mixed with alveolar gas. The long alveolar plateau rises slowly (

Question 29

Draw a graph exploring single breath nitrogen test to assess uniformity of ventilation. (Lung volume against %N2). Label TLC and RV, 100% O2 inhalation, washout, uniform ventilation, and slow emptying alveoli.

(The subject is asked to exhale slowly through an N2 meter at a constant flow rate from TLC to RV…N2% concentration then is plotted as a function of volume exhaled.)

Answer 29

Slide 18.

1) N2 % remains fixed at zero as the dead spaces filled with O2 empty
2) there is a rapid upswing in %N2 as alveolar regions start to empty
3) an alveolar plateau is reached where there is equal emptying of all lung zones from base to apex (%N2= 50%)
4) finally at the end of the plateau there is a second increase in %N2 due to slowly emptying alveoli

Question 30

If the dead space is 150 mL and tidal volume increases
from 500 to 600 mL for the same minute ventilation,
what is the effect on dead space ventilation?

Answer 30

Vt=500mL
V=150mL/500mL x V = .30x Ve

Vd= 150mL/600mL x Ve= .25 x Ve

As tidal volume increases, dead space ventilation decreases
for the same minute ventilation.

Question 31

Looking at a graph of the relationship between number of breaths to N2 concentration %, how would abnormal conditions (fast space and slow space) combine to shift the overall curve?

Answer 31

Slide 19.

Question 32

Define ventilation. What is it normally?

Describe dead/live space and which is filled first.

Write an equation for Vtotal, dead space ventilation, and alveolar ventilation.

Answer 32

Slide 20.
Ventilation = frequency, f (breath/min) x tidal volume (Vt)
= 15 x 500mL =7500mL/min

Lung can be partitioned into dead space (non-alveolated airways) and live space (alveolated airways), the tidal volume first fills the dead space (VD) and then the alveolar space (VA). So ventilation consists of dead space and “life space” ventilation.

Vtot= f x (VD +VA) = f x Vt

dead space ventilation= f x VD (does not contribute to gas exchange)

alveolar ventilation (effective for gas exchange): VA= f x VA = f x (Vt-VD)

Question 33

What is the first air to reach the deep alveoli during an inspiration?

Answer 33

“dirty” air from the previous exhalation

Slide 20

Question 34

Describe measuring dead space through Fowler’s method (single breath oxygen). Graph.

At the start of inspiration what is N2 concentration? start of expiration? alveolar plateau? Graph/sketch

Answer 34

Slide 21.

anatomical dead space is found on the single-breath nitrogen washout curve at the beginning of the first N2 upward inflection just below TLC.

(dead space is the volume up to the vertical dashed line, where areas A and B are equal)

(Dead space ventilation can also be measured by Fowler’s method. The patient takes a single breath of 100% O2 and then exhales into a tube that continuously measures the N2 concentration in the exhaled gas. As the patient exhales, the anatomic dead space
empties first. This volume contains 100% O2 and 0%
N2 because it has not participated in any gas exchange. As the alveoli begin to empty, O2 partial pressure falls and N2 partial pressure begins to rise. Finally, the partial
pressure of N2 is almost uniform, and it represents alveolar gas almost entirely. This phase of expired air exhalation is called the alveolar plateau. The volume with
initially 0% N2 plus half of the rising N2 volume is equal
to the anatomic dead space)

Question 35

What is physiological dead space?

How do you calculate it? (What is Bohr’s equation?)

What is a normal VD/Vt ratio?

Answer 35

total volume that does not participate in gas exchange
= anatomical dead space + alveoli that are ventilated but not perfused

See slide 22.
VD/VT= PaCO2 -Pexpiratory CO2/ PaCO2

Normal VD/Vt ratio is 0.2-0.35. In normal individuals, anatomical and physiological dead spaces are almost identical. In lung diseases, physiological dead space can be considerably larger.

Question 36

How does alveolar ventilation relate to pulmonary (tidal) ventilation?

If tidal volume is larger, how is dead space ventilation affected?

To increase VA would a change in respiratory rate or tidal volume be more effective?

Answer 36

Alveolar ventilation is less than pulmonary (tidal) ventilation

Exhaled minute volume (Vexhaled)= Vt x respiratory rate
= VD +VA

The larger the tidal volume, the smaller the dead space ventilation.

To increase VA, an increase in respiratory rate is less effective than increase in tidal volume.