Basic Gas Laws, Pressures Flashcards

1
Q

What are the five principles of kinetic Gas laws?

A
  1. Gases consist of molecules. 2. Gas molecules randomly move and collide with one another and with walls of container. 3. Pressure is caused by collision of gas molecules with walls. 4. The pressure exerted by a gas or mixture of gases depends upon the number of collisions that take place. 5. The greater the number of gas molecules in a given space, the greater the number of collisions and hence, the higher the measured gas pressure.
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2
Q

Pressure is directly proportional too? Why?

A

Pressure is Directly Proportional to the Number of Molecules in a Container More molecules = More collisions = More pressure

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3
Q

Pressure equals? (mathematically)

A

Pressure= (number of gas particles/volume of container)x ideal gas constant x temp.

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4
Q

Heating a gas does what to its pressure and energy?

A

Heating the gas will increase the velocity (kinetic energy), producing more collisions per unit time and an increase in pressure (e.g. The pressure within an automobile tire increases with travel). Accordingly, the volume will increase.

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5
Q

In a constant temp system decreasing the volume does what to pressure?

A

it increases the pressure.

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6
Q

Pressure is directly proportional to?

A

the number of molecules in a container and their kinetic energy

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7
Q

Explain barometric pressure?

A

The atmosphere extends about 100 miles above sea level. These gas molecules push down on us due to earth’s gravitational pull, thus creating an atmospheric pressure. As we ascend higher the pressure is less due to the decrease in gas particles pushing down on us. but its important to note that the fractional composition of gasses is the same as we ascend.

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8
Q

What is Dalton’s Law?

A

Dalton’s Law (John Dalton 1766-1844) states that the total pressure (i.e., barometric pressure; PB) exerted by a mixture of gases, such as the earth’s atmosphere, is equal to the sum of the separate partial pressures each gas would exert if it occupied the entire volume (space) alone. That is, the total pressure exerted by a mixture of gases is equal to the sum of the individual partial pressures of the gases comprising the mixture. For the earth’s atmosphere, the total or barometric pressure (PB) is the sum of the individual partial pressures of the gases comprising the atmosphere.

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9
Q

What happens in regards to water vapor as we inhaled air?

A

Ambient air inhaled into the nasal passages and tracheobronchial tree is immediately warmed to body temperature and completely saturated with water vapor. The water vapor or water gas added to inspired air exerts a partial pressure just like the other gases comprising air. The ability and capacity of air to hold water vapor increases as the temperature of the air increases and is independent of the total air pressure. At body temperature (37q qC), air saturated with water vapor has a water vapor pressure (PH2O) of 47 Torr, provided PB exceeds PH2O. In a more practical sense, the PH2O in the airway of a person at sea level is the same as a person in Denver if their body temperatures are the same. Like the other gases present in air, PH2O also obeys Dalton’s Law. As a consequence, the addition of water vapor to inspired air reduces the partial pressure of other gases without changing the total gas pressure. For air in the tracheobronchial tree, PB is the sum of the partial pressure of atmospheric gases plus water vapor or PB = PN2 + PO2 + Par + PH2O + Pother gases. PO2 in airway= Fraction of O2 (21%) x [Pb (760 sea level 625 denver)-PH2O (47)]. At sea level, the PO2 of inspired air is reduced from 159 to 149 Torr in the airway due to the addition of water vapor. This is shown in the figure on the right. At higher altitudes, the addition of water vapor likewise decreases the PO2, along with the partial pressures of the other gases present.

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10
Q

How does a nasal cannula change the Fractionation of O2? how does this depend on the liters of O2? How is it different between a nasal cannula and a venturi mask?

A

Fraction of Inspired Oxygen (FiO2) for a nasal canula is shown in the table below. Specifically, we assume that the fraction of oxygen that is inspired (above the normal atmospheric level or 20%) increases by 4% for every additional liter of oxygen flow administered. Venturi Mask: 4L- .24-.28 6L- .31 8L-.35-.40 Nasal cannula: 0-.2 1- .24 2- .28 3- .32 4L- .36 5L- .4 6L- .44

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11
Q

The Alveolar gas equation allows us to calculate the Pressure of oxygen in the alveoli when what is changed? what else is the equation used for?

A

Clinically it is often important to calculate the mean PAO2 expected in a patient breathing room air or during ventilation with a gas mixture enriched with O2. The “alveolar gas equation,” along with a variety of abbreviated forms of this equation, are used to estimate the PAO2. This equation enables the computation of the expected PAO2 when either the inspired fraction of O2 is changed or the PB is altered. For example, if a patient is placed on 40% O2, the expected PAO2 could be calculated. In addition, this equation is used to specify the percentage of O2 necessary to yield a normal PAO2 for passengers in a pressurized aircraft or to calculate the expected PAO2 for non-pressurized airplanes at different altitudes.

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12
Q

What is the alveolar gas equation mathematically?

A

Notice that the first part of this equation [PO2 = FIO2 (PB- PH2O)] is the same equation used to calculate the PO2 of the airway. In addition, the alveolar gas equation subtracts the alveolar partial pressure of CO2 (PACO2). The latter is multiplied by an adjustment for a respiratory exchange ratio ( CO2/ O2), that can vary between 0.7 to 1.0. Basically, this equation states that the PAO2 is the PO2 in the airway minus the alveolar partial pressure of CO2 (PACO2). Since the arterial and alveolar PCO2 are nearly the same, the PACO2 is often estimated from the arterial PCO2 (PaCO2), obtained by arterial blood gas analysis

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13
Q

Explain bulk flow and how its used for us to breathe?

A

Air, like water or blood, flows from a region of higher to a region of lower pressure by bulk flow. When total gas pressure in the alveoli (PA) is the same as the atmospheric pressure (barometric pressure; PB), no air flow occurs because no pressure gradient exists. To initiate air flow into gas exchange sites of the lung, PA must decrease below PB, or PB must increase above PA, as occurs during mechanical ventilation. To expel gases from alveoli, PA must exceed PB. With normal breathing, PA needs to change because PB does not fluctuate on a minute-to-minute basis. When the inspiratory muscles contract, the thoracic cavity and lungs enlarge, which decreases PA. Air then moves into the alveoli by bulk flow until PA and PB equalize at end inspiration. The lung is stretched during inspiration, so when the inspiratory muscles relax, the lung recoils to compress the alveolar gas volume. This elevates PA above PB so air is expelled (expired) until PA again equals PB at end expiration.

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14
Q

local ambient pressure is often what in pulmonary medicine?

A

In physiology and pulmonary medicine: Local, ambient atmospheric pressure (PB) is often used as a reference point . . . That is, physiologistssay Patm= 0. E.g., PB= 760 mmHg becomes0. Pressureswithin the body are then measured with reference to this Patm.

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15
Q

difference between sucking and pushing? Lungs Vs. Vacuum

A

At this point it is important to note that pressure never sucks; pressure only pushes. To initiate air flow into gas exchange sites of the lung, PA must decrease below PB, or PB must increase above PA, as occurs during mechanical ventilation. When PA is below PB, the resulting pressure difference (‘P) between atmospheric and alveoluscreates the driving force to push air into the alveolus. In an analogous way, when you use a household vacuum cleaner, it is actually atmospheric pressure that is pushing air (and dirt) into the vacuum cleaner; not the vacuum “sucking” air and dirt in. When you inhale, it is atmospheric pressure “blowing” air into your lungs and not the lungs “sucking” in the air.

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16
Q

Explain the inspiration/expiration process with pressures.

A
17
Q

explain the cohesive forces of the lung?

A

Although the lung fills most of the thoracic cavity except for the space occupied by the heart and major blood vessels, no ligaments or other tissues hold or attach the outer lung surface to the inner chest wall. If air is allowed access to the “space” between the outer lung surface and inner chest wall, the lung will recoil away from the inner chest wall and collapse (figure). Collapse of the lung creates a condition known as pneumothorax. Thus, it is important to examine the forces that normally link the outer surface of the lung to the inner chest wall. These forces enable the lung and chest cage to function as a single unit.

The outer surface of the lung is covered with a visceralpleura that is juxtaposed to the parietal pleura lining the inner chest wall. The “space” or potential space between the two pleura is termed the intrapleural space. The intrapleural space is occupied by a small amount of fluid. The presence of intrapleural fluid provides cohesive forces that help to join or link the visceral and parietal pleura together. The intrapleural cohesive forces resemble those present when a water droplet is placed between two glass slides. While the two glass slides move over one another very easily, they are difficult to separate perpendicular to their adjoining surfaces. Thus, as the chest wall expands during inspiration, the lung is obligated to follow, so the two structures expand as a single unit. When the respiratory muscles contract to increase thoracic volume, lung volume increases by a similar amount. To preserve the cohesive forces linking outer lung to the inner chest wall, the intrapleural “space” must be kept essentially free of air (gases) or excessive fluid accumulation, as would also be the case in the glass slide analogy. However, unlike the glass slides, the pleura membranes are modestly permeable to gases and water. So, forces must continually operate to maintain the intrapleural space essentially free of gases and fluid to preserve the cohesion between the lung and chest wall.

18
Q

explain negative intrapleural pressure during expiration and inhalation?

A

The total volume of fluid present in the intrapleural space is estimated to be only 2 to 10 ml. A small amount of protein is present in intrapleural fluid. At end expiration, mean intrapleural pressure (Ppl) is about 5 cm H2O below atmospheric pressure (-5cm H2O) and Ppl becomes more subatmospheric with inspiration (figure) such that at end inspiration of a typical tidal volume, Ppl may be -10 cm H2O. The magnitude of the subatmospheric or “negative” intrapleural pressure reflects the tendency of the lung to recoil or pull away from the inner chest cage. Because the intrapleural fluid is a liquid that cannot be expanded or contracted, it results in a “negative” intrapleural pressure that becomes more “negative” with lung inflation.

19
Q

Why doesn’t gas go into the intrapleural space? where do intrapleural gases get absorbed?

A

Because Ppl is below atmospheric pressure, a pressure gradient exists for gases to potentially diffuse into the intrapleural “space” to possibly disrupt the cohesive forces linking the visceral (lung) and parietal pleura (inner chest wall). However, this typically does not occur because the outer surface of the lung is not very permeable to gases and gases gaining access to the intrapleural space are rapidly absorbed into the parietal-side vessels (chest wall). Total gas pressure in the venous circulation of the parietal side, part of the systemic circulation, is about 72 cm H2O below atmospheric pressure (-72 cm H2O or -54 Torr). This low total gas pressure in parietal blood occurs because the decline in parietal blood PO2 (~78 cm H2O or 58 Torr) is not matched by a corresponding increase in PCO2 (~ 6 cm H2O or 5 Torr) after gas exchange occurs in the parietal capillaries. The difference between the large fall in the PO2 and small rise in the PCO2 yields a net “negative” pressure of about 72 cm H2O below atmospheric pressure (-72 cm H2O) in parietal veins. This subatmospheric pressure is a consequence of the different solubilities of O2 and CO2 and the shape of their respective dissociation curves (to be covered later). By contrast, the total gas pressure in visceral-side blood (i.e., pulmonary veins) is not different from atmospheric pressure because the blood has just equilibrated with alveolar gases. The most important site for absorption of intrapleural gases is the parietal-side veins because of their larger blood volume.

20
Q

Explain the starling forces determining the balance of fluid in the lungs?

A

The volume of fluid in the intrapleural “space” is small (2.0 to 10 ml), but closely regulated by Starling forces (factors familiar from Cardio). It is essential that intrapleural fluid volume remain small so that the cohesive forces linking the lung to the inner chest wall are preserved and thoracic cage enlargement results in lung expansion. A net Starling force of about 9 cm H2O favors fluid filtration from the parietal-side circulation into the intrapleural “space,” whereas a oncotic force of about 10 cm H2O favors the reabsorption of fluid from the intrapleural “space” into the visceral-side circulation. The visceral pleura is also more vascularized than the parietal pleura, so the net visceral reabsorptive force effectively operates over a greater area than parietal filtration to facilitate visceral fluid reabsorption. These interacting Starling forces function to keep the intrapleural space from excessive fluid accumulation. Simultaneously they maintain sufficient fluid volume for adequate lubrication of adjoining pleural surfaces, enabling them to slide over one another with minimal friction and to operate as a single unit during breathing. Nevertheless, a rather delicate balance exists between parietal filtration and visceral absorption that can be easily disturbed with some disease states. This is attested to by the frequency of pleural effusions (fluid accumulation in intrapleural space) in patients.
While the actual values and magnitude of these Starling forces are oversimplified and change constantly with the breathing and cardiac cycle, they are indicative of the forces that operate to keep the pleural space from accumulating excessive fluid.

21
Q
  1. From a vendor in the hospital lobby, you buy a helium balloon to take to your pediatric patient, whose room is on the 12th floor. Assuming no change in temperature, which of the following will increase as you climb the stairs to the 12th floor? 1. the amount (moles) of helium in the balloon 2. the volume of helium in the balloon 3. the pressure inside the balloon 4. the distending pressure of the balloon
A

2 and 4 are correct. As you climb the stairs (go up in altitude), atmospheric pressure decreases, so the distending pressure of the balloon increases (i.e. option 4 is correct). The increase in transmural pressure causes the balloon to expand a bit . . . i.e. its volume increases, so option 2 is correct. Option 1 is incorrect, because (assuming that the balloon is sealed) the following are all constant: the amount, mass, and moles of He (i.e., the number of He molecules) in the balloon. [Admittedly, He will escape very slowly from the balloon over several days, via diffusion through the rubber, but this will decrease,not increase, the amount of He in the balloon.] Option 3 is incorrect because, as the balloon expands, the pressure inside decreases; in a sealed container, pressure varies in inversely with volume (Ideal Gas Law).

22
Q

Your task, while assisting a trauma surgeon in a rural emergency medical clinic, is to aspirate (suction) blood from the surgical field so that the surgeon has a clearer view of the patient’s lacerated spleen. The “suction tip” you are holding (label A) is attached to a tube, and the other end of the tube (B) is inserted into a collection bottle (C). The “outlet” tube (D) of the collection bottle is attached to a “suction pump” (E).

Why does blood flow from the suction tip (A) into the collection bottle (C)? A. Gravity pulls downward on the blood in the tube, which creates a siphon effect, and that makes the blood flow into the bottle. B. The suction pump pulls blood into the bottle. C. The suction pump creates a vacuum in the bottle, which pulls blood into the bottle. D. The suction pump removes air from the bottle, so the bottle starts to collapse and the recoil of the bottle walls pulls blood into the bottle. E. The pump removes air from the bottle, so atmospheric pressure, which is higher than the pressure in the bottle, pushes blood through the tube and into the bottle.

A

E. Remember. Pressure never sucks . . . Pressure only pushes. The pump removes air from the bottle, which lowers the pressure in the bottle below atmospheric pressure. The resulting pressure difference (‘P) between atmospheric and bottle pressure creates the driving force to push fluid (whether air or blood) through the tube. In an analogous way, when you use a household vacuum cleaner, it is actually atmospheric pressure that is pushing air (and dirt) into the vacuum cleaner . . . . rather than the vacuum “sucking” air and dirt in. Think about this, could a vacuum cleaner”sweep up” dust on the moon? A “vacuum cleaner” would not work on the moon, and it’s true that a medical suction pump would not be able to “suck” fluid into its reservoir if atmospheric pressure were not there to provide the “push”. And it’s also true that when you inhale, it is atmospheric pressure “blowing” air into your lungs . . . and not the lungs “sucking” in the air.

23
Q

At a particular moment during breathing ….. barometric pressure (PB)= +740 mmHg alveolar pressure (PA) = -2 mmHg intrapleural pressure (PIP) = -8 mmHg. Is the subject inspiring or expiring at this moment? What is the transpulmonary pressure?

A

Transpulmonary pressure = Lung distending pressure = (PAlveolar -PIntrapleural ) = [(-2) (-8)] mmHg = +6 mmHg. That is, there is a net pressure of +6 mmHg acting to distend the lungs in this case. Remember that alveolar and intrapleural pressures are always referenced to atmospheric pressure, so saying that PA = -2 mmHg means that alveolar pressure is 2 mmHg less than atmospheric pressure (so, alveolar pressure is really 738 mmHg, in this example). Since alveolar pressure is less than atmospheric pressure, air would be flowing into the alveoli ….. i.e. inspiration is occurring (assuming that the airway is open, so as to permit air flow to occur).

24
Q
  1. At a particular moment during quiet breathing, the following conditions exist for a patient: Barometric pressure (PB) = 750 mmHg Alveolar pressure (PA) = 1.2 cm H2O Intrapleural pressure (Pip) = 7.3 cm H2O The patient’s airways are open.
A

At this moment, the patient’s transpulmonary pressure is _______ and the patient is _______. A. 8.5 cm H2O … inspiring B. 6.1 cm H2O … inspiring C. 8.5 cm H2O … expiring D. 6.1 cmH2O … expiring E. 8.5 cm H2O … neither inspiring nor expiring

25
Q
  1. At the end of a normal tidal expiration: 1. transpulmonary pressure is negative 2. alveolar pressure is zero 3. lung volume equals functional residual capacity (FRC) 4. lung volume is less than it would be if there were a pneumothorax 5. lung volume is less than it would be if the respiratory muscles were paralyzed
A

2 and 3 are correct. At the end of a normal expiration, there is no air flowing through the airways, and alveolar pressure = atmospheric pressure = 0 mmHg. Intrapleural pressure is negative (sub-atmospheric), and transpulmonary pressure is greater than 0. The lung volume at the end of a normal expiration is called functional residual capacity. At the end of a normal expiration, the respiratory muscles are typically relaxed, so the lungs are at the same volume that would exist if the respiratory muscles were paralyzed. Under these conditions, the lungs are kept from collapsing by the positive transpulmonary pressure. After a pneumothorax, intrapleural pressure = atmospheric pressure = 0 mmHg. Under these conditions, transpulmonary pressure decreases to 0, and the natural, elastic recoil of the lung tissue causes the lungs to collapse to a very small volume (analogous to a balloon deflating

26
Q
  1. On a recent morning, the atmospheric pressure at MUC was 750 mmHg. A sample of the air was dried and found to contain 21% O2. Calculate the PO2of this dry air:
A

PO2= 0.21 u(750 mmHg) = 157.5 mmHg

27
Q
  1. [Continued from Question 6] When inspired by a medical student, this dry air became heated in the airways to body temperature (37qC) and saturated with water vapor. What was the partial pressure of water vapor in the saturated, inspired air? What fraction of the resulting, saturated mixture will be water vapor?
A

When air, or any other dry gas mixture, is exposed to water at 37qC, water evaporates into the gas mixture until the water vapor pressure in the mixture equals 47 mmHg. As soon as the water vapor pressure (PH2O ) equals 47 mmHg, water molecules in the gas mixture are forced back into the liquid water at the same rate that the water is evaporating. That is, there is an equilibrium between the liquid and gaseous phases of water. At a temperature of 37qC, water vapor pressure of any saturated gas mixture is 47 mmHg …… and it does not matter what the other gases are in the mixture! In this example, the total pressure is still 750 mmHg; the addition of water vapor to the air simply caused the air to expand a little bit, which diluted the O2 and other dry gases that were present. Fraction of H2O in the saturated air = 47 mmHg / 750 mmHg = 0.063 That is, water vapor makes up 6.3% of the saturated gas mixture.

28
Q
  1. [Continued from Question 7] This addition of water vapor to the dry, inspired air diluted the O2. Calculate the PO2of the saturated, inspired air:
A

O2still makes up 21% of the drygas in the saturated air, but the total pressure of the dry gases is reduced by 47 mmHg. Thus, PO2= 0.21 u(750 mmHg -47 mmHg) = 147.6 mmHg

29
Q
  1. One day last summer, the air temperature in Lansing reached 98.6 oF . The relative humidity was 80% (which means that the air was holding 80% of the maximum water vapor that it could at that temperature). What was the PH2O in the air?
A

As you know, at a temperature of 98.6oF and a relative humidity of 100% (saturated), the PH2O would be 47 mmHg. Thus, for air at that temperature with a relative humidity of 80%, the PH2O will be 0.8 47 mmHg = 37.6 mmHg.