chapter 1a gases

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Gases

"We live submerged at the bottom of an ocean of air - Torricelli, 1644"

Introduction

o Definitions: Pressure = force/area

o 2Units:

Pascal (Pa): 1 Pa = 1 Newton/m2

or 1N/m2

atmosphere (atm): one atm =101325 Pa = 14.7 lbs/in2

bar: 1 bar = 100000 Pascal

torr: 760 torr = 1 atm

mm Hg: 1 mm Hg = 1 torr

o Major properties associated with gas: pressure, volume, moles, andtemperature. Equations relating pressure, volume, temperature, and moles

for gas would be called an equation of state. The major properties of gases

are described by the relationship PV=nRT where R is the gas constant.

Lift Pump

note rod "C" that lifts movable piston which contains a valve "B". Also, note the valveA. When rod C is lifted, valve B closes and valve C opens. Then the water from below

the piston will flow to the chamber and the water above the piston will flow out



concurrently. Now when rod C is pushed down, valve A closes and valve B opens which

permits water to flow above the piston.

History: o Consider the lift pump above. Galileo was first to recognize or more

correctly record that a lift pump could only raise water is high as 10.4 m or 34 ft. The explanation was provided by Torricelli who suggested thatwater was being forced up the pipe as a consequence of the pressure

exerted by the atmosphere on the surface of the water at the bottom of the

pump. Carrying this idea further, since mercury (Hg) is about 14 times asdense as water, Torricelli predicted that atmospheric pressure would

support a column of mercury only 1/14 as high as the maximum water

column. See calculations below (reference 4).

In the Torricellian tube, the atmospheric pressure

supports, mercury 760 mm tall (Figure 12.2,

reference 4)



4The figure above illustrates the Torricellian tube which preceded today's

barometer. The glass tube is closed at one end, filled completely with mercury,

and then inverted into a bowl of mercury. As you know, the column of mercurywill drop and it continues to drop until it reaches the height of about 760 mm

above the lower surface. The pressure the atmosphere exerted on the mercury

surface in the bowl is equal to the pressure due to the weight of the mercury in the

tube. This condition is obtained at equilibrium. A vacuum exists at the top end of the tube and this vacuum has been designated the Torricellian vacuum.

o Continuing along this line of thinking, consider the following. The weight

of the mercury in the tube above is defined by its density * the volume*the acceleration. Density is represented by the symbol (rho).

o weight = *hA * g where h is the height and A is the cross-sectional area

of the mercury column. To obtain the equilibrium condition, thisdownward force must be balanced by an upward force that is equal. This



upward force comes from the atmospheric pressure which acts down on

the mercury in the bowl and is transmitted equally to all parts of the fluid,

in accord with Pascal's principle. Accordingly, the upward force on the bottom of the tube is, F = PA, where P is the atmospheric pressure.

o The condition of equilibrium requires that the two forces are equal,

therefore: *hA * g = PA and simplifying this expression we have P = *h

* g. Acceleration due to gravity is 9.81 m/s2, the density of

mercury is 13.6 x 103

kg/m3, and the height as we have noted

above, is760 mm or 0.760 m. [the expression 13.6 x 103

kg/m3

utilizes "scientific notation" and corresponds to 13.6 multiplied by

1000, which is represented as103

; therefore, the expression 13.6 x

103

is exactly equal to 13,600.]

Using the values and substituting in the equation, P = *h * g, weobtain P = 13.6 x 10

3kg/m

3* 9.81 m/s

2* 0.076m OR 101 x

103 N/m

2= 1.01 x 10

5Pa. This value, 1.01 x 10

5Pa, corresponds

atmospheric pressure. 1 atm is defined as the pressure exerted bythe earth's atmosphere at sea level-- the value to more significant

figures is 1.01425 x 105 N/m

2. {"The newton is the Standard

International Unit of force. In physics and engineering

documentation, the term newton(s) is usually abbreviated N . Onenewton is the force required to cause a mass of one kilogram to

accelerate at a rate of one meter per second squared in the absence

of other force-producing effects. In general, force ( F ) in newtons,mass (m) in kilograms, and acceleration (a) in meters per second

squared are related by a formula well known in physics: F =

ma"}5

Pressure:

o Before we speak specifically about application of gases in the context of anesthesia, we will first consider a number of preliminary aspects having

to do with the physics of gases, including the kinetic theory of gases.

Mostly we will be concerned with interrelationships of three parameters:

(1) pressure, (2) volume and (3) temperature. o We will begin by considering pressure which in terms of dimensions

would be force per area. (P = F/A) The standard international (SI) unitof pressure is the Pascal (Pa). In turn, 1 Pa = 1 Newton/meter

2. [a meter

will be abbreviated as m] From our earlier definition in which pressure

equals force / area, the Newton must be a measure of force. In fact, the

Newton is defined as a unit of force which is required to accelerate a massof 1 kg 1 m per second per second. Also note that the force of gravity

acting on a mass will give that mass an acceleration of 9.8 m/s2. If the

mass is 1 kg then the force of gravity on that mass would be 9.8 Newtons.

Therefore, 1 Pa = 1 Newton/m2

= kg/m s2. [simplified from (1kg

1m/s2)/m

2]

o Sometimes we discuss gas pressure in terms of different units, i.e. other

than Pascal units. For example, we may wish to describe gas pressure in



terms of "atmospheres" or mm Hg. The relationship between one

atmosphere and Pa units is: 1 atmosphere (1 atm) = 1.013 x 105

Pa

To determine the relationship between atm and mm Hg, we firstspecify a standard temperature of 25

oC, the density of mercury is

13.6 g/cm3

(water is defined as having a density of 1 g/cm3) and

that the acceleration due to gravity is 9.8 m/s

2

. We now need to determine, using mercury as the fluid, how tall a

column of mercury would be supported by a pressure of 1

atmosphere? To resolve this question we need to recognize that

there is a formal relationship between an applied pressure (P) andthe height of a fluid. This relationship is:

1. P = dgh where "d" is the density of the fluid (a

temperature-dependent value), "g" is the acceleration to

gravity and "h" would be the height of the column in, for example, cm. Again using the density of mercury as 13.6

g/cm3 we obtain the following taking care to appreciate the

units of dimension of the quantities. For consistency of units note that we represent Pa as kg/m s2

2. hHg = P/dHgg which is a rearrangement of the above

equation P = dgh, so

3. hHg = P/dHgg = [(1 atm) (1.013 x 105

Pa/atm) (kg/Pa m s2)]

/ [(13.6 g/cm3) (10

6cm

3/1 m

3) (1 kg/1000g) (9.8 m/s

2)]

OR

4. hHg = 0.76 m = 76 cm = 760 mm Hg

The above derivation is adapted from that of Professor Larry

Gladney, Ph.D. Dept of Physics, University of Pennsylvania



"The height of mercury is determined by the need to balance the weight of

mercury lifted against the weight of the air above it. The weight of air presses

down uniformly on everything on the surface of the earth, including the surface of the pool of mercury in the beaker. This pool transmits the pressure uniformly

through its volume and therefore maintains the height of mercury in the glass

tube. The 76.0 mm of mercury has a weight equal to the weight of a column of air

with the same cross-sectional area and a height of 150 km (roughly the height of our atmosphere)." Figure and legend by Professor Larry Gladney, Ph.D. Dept.

of Physics, University of Pennsylvania.

1Pressure-Volume Interrelationships: Let's begin this discussion by noting the equation P = F/A showing the

relationship between force, area and pressure. To use a familiar example, we willconsider to syringes of different sizes as shown below:

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The force exerted by the thumb on the plunger will be considered to be the same

in both cases. If the applied force is the same in both cases, then the pressure will

be determined in a manner dependent on the inverse of the area, P = F/A. Thearea of the smaller syringe plunger surface is A and the area of the larger syringe

plunger surface is 4A. If the smaller syringe were a 2 ml syringe, then the

plunger surface area would be about 5 x 10-5

m2

with the area of the larger

syringe (say a 20 mls syringe) would be about 2.5 x 10-4

m2

. Figure above: (©)

University of Kansas)

Consider the case in which the force applied by the thumb is set at 25 Newtons:

then substituting into our equation P = F/A we have 25 Newtons / 5 x 10-5 m2 which is 500 kPa (kiloPascals). Since the area is greater for the 20 ml syringe we

can calculate a new pressure, i.e. 25 Newtons / 2.5 x 10-4

m2

or 100 kPa. [100kPais approximately 1 atm]. Now 1 atm is about 760 mm Hg which is about 100

kPa. This means that if we take systolic blood pressure as 120 mm Hg (16

kPa), then our 100 kPa (1 atm) is about 6 times this systolic blood pressure. To better appreciate some of these pressures, let's consider the following clinical

correlation.

o In a type of regional anesthesia (Bier block), local anesthetic is injected

intravenously and the local anesthetic concentration is maintained by

pressure cuff which is inflated to above the systolic pressure.

o The clinical advantage of this procedure is that systemic toxicityassociated with local anesthetics, particularly CNS toxicity is less

likely to occur.

Return to Table of Contents

Regional anesthesia: Bier Block

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1In the diagram above (©) University of Kansas) note there are two forces to

consider.

o The first force is applied to inflate the pressure cuff (F, in blue), whereas a

second force (F, in red) is applied to the syringe plunger.

o From our calculations above using the 20 ml syringe example is possible to apply sufficient plunger force to cause pressures well in excess

of usual systolic blood pressures. Accordingly, if excessive pressure is applied to the syringe plunger

the local anesthetic can enter the vascular system distal to the cuff

at a pressure sufficient to overcome the "cuff" barrier and as aconsequence, the drug can gain access to the systemic circulation.

1Another clinical correlation example in which the relationship between pressure,

area, and force are highlighted would be that of bedsore formation. In one

example, if the patient weight over a localized region (10-2

m2) [which is equal to



10 cm x 10 cm] is set at 20 kg, then the force over this area can be found by

applying the relationship F = ma (force = mass times acceleration) which is

Newton's second law. Acceleration is due to gravity and is set at 9.8 m/s2

withthe mass being 20 kg in our example; therefore, the force = 20 kg x 9.8m/s

2= 196

Newtons.

o Now we can apply our equation P = F/A where P is the pressure and F isthe force with A representing area; 196 N/10

-2m

2= 19.6 kPa.

o The final step required to understand the clinical consequence is to

compare the pressure exerted on this region (19.6 kPa) with a systolic

blood pressure of 120 mmHg or about 16 kPa. Therefore the pressureexerted on the region is sufficient to cut off blood supply, thus promoting

bedsore development at the pressure point.

Pressure "Bed" sore secondary to loss of blood flow with subsequent tissue necrosis

Clinical Correlation

Peripheral Nerve Blocks: o 6

The Bier block is frequently referred to as intravenous regional anesthesia

and is most commonly performed on the upper extremity for hand or forearm surgery of limited duration. In order to ensure reliability, safety,

and expectable patient satisfaction, certain steps are required.



A small-gauge intravenous cannula is placed in the dorsum of the

hand vein, away from the surgical site, and distally. The needle

size will be about a 20-gauge needle.

A double tourniquet would be secured around the upper arm (not

all centers may use the double tourniquet method, although it is the

standard approach) -- see below

Bier block -- double tourniquet location

Bier block -- elastic bandage wrapping to

exsanguinate the arm



at

o The arm is exsanguinated by wrapping with a wide elastic band (esmarch). Theapproach might utilize a 10 cm Webril padding band.

o A pneumatic double cuff is placed and the proximal tourniquet inflated. Thelower end of the distal cuff will be secured using foam tape to prevent

movement.

o With the proximal cuff inflated, 40-50 mls (cc) of 0.5% lidocaine is injectedthrough the intravenous catheter -- subsequently (usually within minutes)



anesthesia is obtained and the operation may be initiated. Note that the distal cuff

remains in place, but uninflated

o When the patient begins to experience the discomfort from the effects of the proximal tourniquet, the distal cuff, under which anesthetic has been infused is

inflated -- and that point the proximal cuff may be deflated.

o Later, the distal cuff will also become painful, as a result of ischemia. Themaximal permissible time is really the maximal allowable ischemic time limitwhich limits the anesthesia period to about 90 minutes.

o A major concern in this anesthetic method is the possibility for local anesthetic

toxicity, which is a rare complication unless tourniquet time turns out to be <about 30 minutes. In that case repetitive deflation and re-inflation of the

tourniquet will tend to limit systemic local anesthetic toxicity. Communication

with the patient in the circumstances is required to detect early symptoms of

toxicity, which implies that excessive sedation should be avoided.

chapter 1a gases

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