Physics Flashcards
Give an overview of SI units.
Name and describe the seven base units
Name derived units common in anaesthesia and those with special symbols
Describe derived units in terms of the base units
Describe non-SI units related to anaesthesia
Describe the different units for pressure and how to convert them
The seven base units are second, mol, metre, ampere, candela, kelvin and kilogram. These can be remembered by the acronym SMMACKK
An understanding of some of the key derived units is essential
1 ATM = 1.013 Bar = 101.3 kPa = 760 mmHg = 1020 cmH20
You should have an understanding of the other non-SI units in common usage
Given that the joule is force x distance (which later becomes relevant to respiratory physiology and lung compliance), can you describe it in terms of base units?
Force is the newton, and is expressed as kg⋅m⋅s-2. The SI unit for distance is the metre, m. As joule is force x distance in SI units it must therefore be kg⋅m⋅s-2 x m.
This is simplified to kgm2/s2.
As discussed previously, the Pascal (Pa) is the SI unit for pressure, however it is clinically very small, so pressure is usually expressed in kiloPascals (kPa).
Are you able to match the appropriate units with the appropriate values? All of these values are representations of atmospheric pressure at sea level.
To make the maths slightly easier, assume that 1 atmosphere = 1 bar = 100 kPa.
The ones to remember and be able to inter-convert clinically are atm/bar, kPa, mmHg and cm H2O.
For example, if your monitor reads 35 mmHg ETCO2, divide by 7.6 = 4.6 kPa.
A CVP of 10 cmH2O is therefore 7.6 mmHg. These values depend on the physical properties of water and mercury. Mercury is heavier than water, therefore, it has a smaller increase in value for a given pressure.
PSI = Pounds per square inch, still sometimes used to measure tyre inflation pressures.
Regarding the base SI units:
A. The second is the time taken for light to travel a certain length in a vacuum
B. The base unit for weight is the kilogram
C. The base unit for temperature is the Celsius
D. The base unit for length is the metre
E. The base unit for electrical current is the volt
A. False. A second is the duration of a certain number of oscillations of a caesium-133 atom. The description of the metre is based on light travelling in a vacuum.
B. False. The kilogram is the base unit for Mass. Weight is not a base unit at all: it is force acting upon mass. Hence, people on the moon have less force, i.e. gravity, acting upon them so they weigh less, even though their mass is exactly the same.
C. False. The base unit for temperature is the Kelvin, which is Celsius minus 273.15.
D. True.
E. False. The base unit for electrical current is the Ampere.
Regarding derived and non-SI units:
A. Derived units must always be expressed as base units
B. Velocity and speed have the same base units
C. 101.3 kPa is equivalent to 760 cm H2O
D. The litre is the SI unit for volume
A. False. The Pascal, for example, can be expressed as base units (kg/m-1/s-2) or a combination of base and derived units (N/m2).
B. True. Speed is defined as m.s-1, Velocity is defined as m.s-1 but in a given direction.
C. False. It is equivalent to 760 mmHg. It would be equivalent to 1020 cmH2O.
D. False. The cubic centimeter is the approved unit, with 1 cm3 being equivalent to 1ml.
Regarding the units for pressure:
A. The kilopascal is the SI unit for pressure
B. 1 Bar is greater than 1 Atmosphere
C. An ETCO2 of 3.5 kPa is equivalent to 27 mmHg
D. 10 cmH2O is equivalent to 7.6 mmHg
A. True. Although the pascal is the actual unit, by using the approved kilo- prefix, it is still classed as an SI unit.
B. False. Although, for simplicity, both Bar and atmosphere are considered to have a value of 1, 1 atmosphere = 1.013 Bar.
C. True.
D. True.
In base units, the pascal is best described as:
A. kg x m2/s2
B. kg x m/s2
C. kg/m/s2
D. m/s
E. kg x m2/s3/A2
A. Incorrect. This is the joule.
B. Incorrect. This is the Newton.
C. Correct.
D. Incorrect. This is speed.
E. Incorrect. This is the ohm.
Give an overview of the kinetic theory of gases, and the relevance to anaesthesia.
List the assumptions of the kinetic theory of gases
Use the kinetic theory to describe the macroscopic properties of gases, such as pressure and temperature
Gases consist of large numbers of tiny particles
These particles are in constant random motion
Collisions of these particles with the walls of their container generates pressure
Temperature is a reflection of the average kinetic energy of these particles
The four postulates of an ideal gas are:
A. A real gas approximates to the behaviour of an ideal gas
B. Gases consist of a number of particles
C. Relative to their size, the particles are very close together
D. The particles move in a random pattern
E. The particles collide with one another
F. There are no attractive or repulsive forces between particles
A. False. Although this is a true statement in itself, it is not one of the four postulates.
B. True.
C. False. They are very far apart; the space between each particle is very much larger than the particle itself.
D. True.
E. True.
F. True.
Postulate 1
Gases consist of a large number of particles - either atoms or molecules.
These particles can be treated as point masses: in a gas they are very far apart, so that the space between each particle is very much larger than the particle itself. Therefore, the volume of the particles of a gas is negligible compared to the total volume of the gas.
Postulate 2
Individual particles are moving in random directions and at random speeds.
There is no general pattern governing the magnitude or direction of speed of the particles in a gas. As such, at any one time, they are moving in several different directions at different speeds.
Postulate 3
Individual particles travel in straight lines between abrupt collisions.
These collisions are with other particles, the walls of a container or other objects.
Collisions are perfectly elastic, which means that the total kinetic energy does not change during the collision.
Postulate 4
There are no attractive or repulsive forces between the particles.
If there were to be attractive forces, the particles would stick together and change phase, e.g. from a gas to a liquid, or liquid to solid.
This would involve a breach of Postulate 3, since it would involve a loss of kinetic energy.
The particles of a real gas:
A. Have a negligible volume
B. Are in constant random motion
C. Do not stick together or change phase
D. Have forces of attraction for each other
E. Deviate most from an ideal gas when the particles are far apart from one another
A. Incorrect. Real gases have particles that occupy a small but absolute volume.
B. Correct.
C. Incorrect. This may occur due to the finite forces of attraction between particles of a real gas.
D. Correct.
E. Incorrect. The deviation in behaviour of a real gas from an ideal gas is greatest when the particles are close together, such as at low temperatures or when under high pressure.
Real gases deviate slightly from the behaviour of the imaginary ideal gas because:
Real gas particles occupy a small but finite volume
The gas particles exhibit attractive forces for one another
These properties become increasingly important when particles are close together, for example at low temperatures or at high pressures.
Match each concept below to the appropriate description.
With regard to the postulates of the kinetic theory of gases:
A. The particles of a gas are in constant motion
B. All the particles within a gas, at a fixed temperature, will be moving at the same speed
C. The volume of gas particles relative to their container is small, but not negligible
D. Gas particles collide frequently with the walls of their container, but not with each other, as the attractive forces between molecules would result in them sticking to one another
E. Particles travel in straight lines between collisions
A. True.
B. False. Particles move at a range of speeds, governed by the Maxwell-Boltzmann distribution.
C. False. Although this statement is true for real gases, the kinetic theory describes an ideal gas with negligible molecular volume.
D. False. Particles collide in a perfectly elastic manner between each other and with the walls of their container. There are no attractive or repulsive forces between particles in the ideal gas described by the kinetic theory.
E. True.
The speed of gas particles:
A. Varies inversely with temperature
B. Is distributed normally
C. With greater mass, is faster on average than those with smaller mass at a given temperature
D. Has a distribution that shifts to the left as molecular weight increases
E. Is directly proportional to the kinetic energy of those particles
A. False. The speed and kinetic energy of gas particles will increase with temperature.
B. False. The speeds are distributed normally with a right skew.
C. False. Gases with a higher molecular weight have a lower average speed at a given temperature than a gas of lower molecular weight.
D. True. At a given temperature, a gas of higher molecular weight has particles moving at a lower average speed. The peak of the Maxwell-Boltzmann curve is therefore further to the left than the curve for a gas of a lower molecular weight.
E. False. Kinetic energy varies with the square of the speed and is governed by the relationship Ek = 1/2mv2.
The temperature of a gas:
A. Reflects the average potential energy of the gas particles
B. Falls as heat energy is removed from the gas
C. When rising, causes the distribution of the speed of the gas particles to shift to the left
D. Varies directly with the average speed of the gas particles
E. When rising, causes the rate of molecular collisions with the container walls to increase
A. False. Temperature is a reflection of the average kinetic energy of the gas particles.
B. True.
C. False. As the temperature rises, the distribution of the speeds of the gas particles shifts to the right.
D. False. Temperature reflects the average kinetic energy which varies directly with the square of the average speed of the particles.
E. True. As temperature rises the average speed of the particles will increase, resulting in more frequent collisions with the container walls and with other particles.
If heat energy is added to a gas, this manifests as an increase in the kinetic energy of the gas particles.
The average speed of the molecules of the gas therefore increases.
The peak of the distribution thus shifts towards the right, as shown in Fig 1. The area under the curve remains constant as this is a probability distribution and the total, therefore, is always sum to one.
Temperature is a reflection of the average kinetic energy of the particles of a gas.
The pressure of an ideal gas:
A. Increases in a fixed container, if the kinetic energy of the particles increases
B. Falls if the volume of the container is decreased, as the gas particles have a smaller surface area to collide with
C. Relies on the constant motion of its particles
D. Rises with temperature by virtue of a rise in the average kinetic energy of the gas particles, leading to more frequent collisions with the container walls
E. Remains constant as temperature rises, only if the container is not allowed to expand
A. True. As the kinetic energy increases, so too does the average speed of the gas particles. Therefore they collide more frequently with the sides of the container resulting in an increase in pressure.
B. False. As the volume of the container decreases, particles collide more frequently with the walls and the pressure therefore rises.
C. True. The constant motion of the gas particles leads to repeated collisions with the walls of the container. These collisions impart momentum which exerts a measurable force.
D. True. A rise in kinetic energy results in an increase in average speed of the gas particles. This leads to an increase in the frequency of collisions with the container walls and, therefore, an increase in pressure.
E. False. The container must be allowed to expand, otherwise the pressure increases with temperature due to more frequent collisions with the container walls.
Pressure may be explained by the kinetic theory as arising from the force exerted by gas particles impacting on the walls of their container.
In order to understand how this force is generated, consider a particle approaching a container wall at right angles at speed s.
The particle collides with the wall and, in a perfectly elastic collision, bounces off at the same speed but in exactly the opposite direction. Whilst its speed remains unchanged, its velocity has changed from +s to -s, in other words by 2 s.
Since acceleration is described as the rate of change of velocity, it follows that the particle undergoes acceleration as it bounces off the wall.
Force and pressure
Force = mass x acceleration
Hence the particle, which is of a fixed mass m, will exert a force as it bounces off the wall of the container.
Pressure is defined as the cumulative force generated, divided by the total area over which that force is applied.
Thus, an increase in pressure may be generated either:
By increasing the frequency of collisions of gas particles with the container walls, or
By reducing the area over which the collisions occur, for example, by reducing the volume of the container
Give an overview of Newtonian mechanics, and the relevance to anaesthesia.
Explain the concept of force
Define Newton’s three laws of motion
Explain the concepts of mass and inertia
Newton’s first law: In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity
Newton’s second law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass
Newton’s third law: If two objects interact, the force exerted by object A on object B will be equal but opposite to the direction of the force exerted by object B on object A, i.e., for every action there is an equal and opposite reaction
What is the relevance of Newtonian mechanics to the clinical practice of the anaesthetist?
The single most important application of the principles discussed in this session relates to the behaviour of gases. In an ideal gas, the molecules obey Newton’s laws of motion.
In relation to Newtonian mechanics:
A. The SI unit of mass is the kilogram weight
B. 1 N is the force required to give a mass of 1 kg an acceleration of 9.81 m/s2
C. Momentum is inversely proportional to mass
D. Inertia increases with the mass of an object
E. The acceleration of an object is inversely proportional to the force acting upon it
A. False. The SI unit of mass is the kilogram.
B. False. 1 N gives a mass of 1 kg an acceleration of 1 m/s2.
C. False. Momentum is proportional to mass.
D. True. The tendency of an object to resist changes in its state of motion is dependent upon mass. The greater the mass of an object, the more inertia it has – the more tendency it has to resist changes in its state of motion.
E. False. Acceleration is directly proportional to the force acting upon it.
If a mass (m) is acted on by net force (F):
A. The momentum of the mass would remain constant
B. The velocity of the mass would be inversely proportional to m
C. If m is doubled then F must also be doubled to achieve the same acceleration
D. The mass would not necessarily move
E. The velocity of the mass would remain unchanged
A. False. Momentum is the product of mass and velocity; the mass will accelerate, its velocity will increase and so too will its momentum.
B. False. The acceleration, not the velocity, will be inversely proportional to m (force = mass × acceleration).
C. True. Force = mass × acceleration.
D. False. The mass would not move if the net force is zero. Take the example of a book on a table acted on by the force of gravity.
E. False. The mass will accelerate therefore the velocity of the mass will increase.
In relation to Newtonian mechanics:
A. Mass and weight are equivalent
B. Acceleration is equal to the product of the mass of an object and the magnitude of the external force applied
C. Momentum varies directly with the velocity of an object
D. Weight is expressed in newtons
E. A stationary object has both momentum and inertia
A. False. Weight describes the force of gravity on a mass; a mass of 1 kg on the surface of the Earth weighs 9.81 N.
B. False. Acceleration = Force applied/mass.
C. True. Momentum is the product of mass and velocity.
D. True. Force is expressed in newtons, and weight is the force of gravity acting on a mass.
E. False. A stationary object has inertia but not momentum.
Give an overview of pressure, and the relevance to anaesthesia.
Define pressure
Explain the principles involved in measuring pressure
Describe key clinical applications of pressure and pressure measurement
Pressure = force/area
The SI unit of pressure is the pascal (Pa)
Absolute pressure = gauge pressure + atmospheric pressure
Atmospheric pressure = 1 atm = 101.3 kPa ≈ 1 bar
Pressure can be measured by manometers or gauges
A patient’s blood pressure is measured at 140/90 mmHg. What is this in kPa?
The gauge on a full oxygen cylinder reads 137 bar. What is the pressure expressed in terms of mmHg?
18.6/12 kPa
Since 1 kPa = 7.5 mmHg, the answer can be calculated very simply by dividing the pressure expressed in mmHg by a factor of 7.5.
It can be appreciated that the human pulse pressure (6.6 kPa in this case) is of much smaller magnitude than atmospheric pressure (101.3 kPa).
102 750 mmHg
As 1 bar = 100 kPa, then 137 bar = 13 700 kPa.
1 kPa = 7.5 mmHg, so 13 700 kPa = 102 750 mmHg.
Human pulse pressure is thus also several times smaller than the pressure in a full oxygen cylinder.
What is the absolute pressure in a full oxygen cylinder?
Absolute pressure = gauge pressure (137 bar) + atmospheric pressure (1 bar) = 138 bar, or 13 800 kPa
Is the human blood pressure reading of 140/90 mmHg referred to previously an absolute pressure or a gauge pressure?
Blood pressure readings are expressed relative to atmospheric pressure and, therefore, are gauge pressures.
Match each pressure scenario to the most appropriate absolute pressure value.
The key word in this question is absolute.
Remember that most pressures used in daily practice are in fact gauge pressures. Therefore, atmospheric pressure will need to be added to the gauge value to give the absolute value.
Barometric pressure on top of Everest = 33 kPa.
Intrapleural pressure during normal inspiration is a few cmH2O (approximately 10 cmH2O = 1 kPa) below atmospheric pressure = 101 kPa – 1 kPa = 100 kPa.
The pressure within an empty oxygen cylinder at sea level is approximately 101 kPa, or atmospheric pressure.
Human blood pressure is expressed as a gauge pressure, therefore, its absolute magnitude will be greater than atmospheric pressure. A blood pressure of 120 mmHg (16 kPa) will be equal to an absolute pressure of 101 kPa + 16 kPa = 117 kPa.
Pipeline gas supply pressure is a gauge pressure of 400 kPa therefore an absolute pressure of approximately 400 kPa + 100 kPa = 500 kPa.
The open-ended manometer consists of a vertical column of liquid within a tube. Mercury is frequently used, but other fluids such as water may be used. Manometers in their simplest form are commonly U-shaped. One end of the tube is open to air while the other end is exposed to the entrapped gas.
If the pressure of the gas (Pgas) is equal to atmospheric pressure (Patm), the mercury level in each arm of the tube is equal.
If the gas pressure exceeds atmospheric pressure, the gas pressure pushes the column of mercury to make its level in the open tube stand higher than its level nearest the gas. The difference in the heights represents the pressure difference (PHg), in mmHg.
If the gas pressure is lower than atmospheric pressure, the mercury column would be higher on the side of the gas, as the atmospheric pressure forces it up.
Question: What is a common clinical example of an open ended manometer?
There used to be a number of examples of open-ended manometers in clinical practice. Today, you are most likely to encounter the mercury sphygmomanometer, used to manually measure blood pressure.
The second type of manometer used is a closed-ended manometer (Fig 1).
This is particularly useful when the pressures to be measured are less than atmospheric pressure.
In a closed-ended manometer the arm furthest from the gas sample is sealed. The space above the mercury is a vacuum so that no atmospheric pressure exists above the mercury.
This is the principle of a mercury barometer.
Question: Does a closed-ended manometer measure absolute or gauge pressure?
A closed-ended manometer measures absolute pressure, in contrast to an open-ended manometer which measures gauge pressure.
Water has the advantage of being non-toxic and readily available.
Mercury manometers, however, are more useful than water where higher pressures are to be measured. This is due to the greater density of mercury. Mercury is 13.6 times as dense as water, so the force exerted by the weight of a column of mercury is proportionately greater than a column of water of similar height.
A pressure which supports a 7.5 mm column of mercury will support a 102 mm column of water.
Question: How high would the columns of mercury and water need to be in open-ended manometers able to measure up to 100 kPa?
Since 1 kPa = 7.5 mm Hg, the column heights would be:
Mercury: 750 mm or 75 cm
Water: 10 200 mm or 10.2 m
The required column height means that water based manometers are not practical for measuring such high pressures.
Various gauges may be used to measure pressure. One example used in anaesthesia is the Bourdon gauge. The Bourdon gauge was patented in 1849 by Eugene Bourdon, a French watchmaker and engineer.
This type of gauge may be used to measure pressures greater than 100 kPa, such as the high pressures in gas cylinders.
In the Bourdon gauge, a gas at high pressure enters a coiled tube (the Bourdon tube), causing it to uncoil; as the tube uncoils, the motion is transferred through a linkage to a gear train connected to a pointer, which moves over a scale on a dial (Fig 1).
Unlike manometers, there is no liquid in this type of gauge. They are sometimes called aneroid gauges, which means ‘without fluid’.
Question: The Bourdon gauge can also be used to measure temperature. How is this achieved?
The tube from a Bourdon gauge may be connected to a sensing element containing a gas (Fig 2). An increase in temperature will lead to a rise in pressure within the sensor by virtue of the third gas law, which states that at a constant volume the absolute pressure of a given mass varies directly with the absolute temperature.
This pressure change will cause the tube in the gauge to uncoil and move the pointer. The scale must be calibrated in units of temperature rather than pressure.
Which of the following are designed to give a reading of absolute pressure?
A. Mercury sphygmomanometer
B. Central venous pressure monitor
C. Theatre barometer
D. Pressure altimeter
E. Bourdon gauge on a gas cylinder
A. False. Clinical measurements are expressed as gauge pressures relative to atmospheric pressure.
B. False. See above.
C. True. A theatre barometer is designed to measure atmospheric pressure and therefore gives a reading of absolute pressure.
D. True. A pressure altimeter measures altitude by measuring the fall in atmospheric pressure with altitude. It therefore expresses absolute pressure.
E. False. Cylinder pressures are expressed as gauge pressures, such that when the cylinder is empty the gauge will read zero, despite being at atmospheric pressure.
Which of the following pressure readings is approximately equal in magnitude to 1 bar?
A. 1020 cm H2O
B. 100 N/m2
C. 1 atm
D. 75 cm Hg
E. 100 Pa
A. True. 1 kPa = 10.2 cm H20. Therefore, 1020 cm H20 = 100 kPa = 1 bar.
B. False. 1 N/m2 = 1 Pa. Therefore, 100 N/m2 = 100 Pa. However, 1 bar = 100 kPa.
C. True. 1 atm = 1 bar, which approximately equals atmospheric pressure.
D. True. 1 kPa = 7.5 mm Hg = 0.75 cm Hg. Therefore, 75 cm Hg = 100 kPa = 1 bar.
E. False. 100 kPa = 1 bar.
Which of the following statements regarding pressure are correct?
A. The SI unit for pressure is the bar
B. As a given force is spread over a greater area, pressure rises
C. A closed-ended manometer will give an absolute pressure reading
D. An example of an open-ended manometer is the theatre barometer
E. A manometer is an example of an aneroid gauge
A. False. The SI unit for pressure is the Pascal (Pa).
B. False. Pressure = force/area; therefore, as area increases, pressure will fall.
C. True. Closed-ended manometers have a vacuum as the reference pressure.
D. False. The theatre barometer is a closed-ended manometer.
E. False. A manometer is a liquid-filled tube; an aneroid gauge, by definition, contains no liquid.
Give an overview of the Gas Laws, with relevance to anaesthesia.
Define the gas laws
Explain what an ideal gas is
Explain some clinical applications of the gas laws
Boyle’s Law relates pressure and volume at a constant temperature
Charles’s Law relates volume and temperature at a constant pressure
The third gas law relates temperature and pressure at a constant volume
Ideal Gas Law: PV= nRT
Dalton’s Law of Partial Pressures: Ptot = P1 + P2 + P3
To demonstrate Boyle’s Law, assume that you have a syringe filled with air and you are obstructing the end of the syringe with your thumb.
Question: What happens to the volume of the air if you press the plunger down with the end obstructed? What is happening to the gas inside?
If you press the plunger down the volume decreases, you should also be aware of an increased pressure on both your thumb and your depressing finger. If the temperature is kept constant, the molecules would have the same energy of motion, but as they are now in a smaller volume, they would collide with the walls of the container more frequently. The greater the number of collisions, the greater the pressure of the gas in the container.
If you were obstructing the syringe with a pressure gauge and not your thumb you would see that if you halved the volume in the syringe, the pressure measured would double, if the temperature was kept constant.
Therefore V∝1/P or PV= constant (k1) (Fig 1)
Boyle’s Law states that at a constant temperature the volume of a given mass of gas varies inversely with the absolute pressure.
Using another example, Boyle’s Law can be applied to a cylinder of gas; if the piston is driven inwards to halve the volume then the pressure would double (Fig 2).
If you are transferring a patient you need to work out how much oxygen is in the cylinder before you start the transfer.
Question: If you have a 10 litre cylinder and the gauge pressure reads 137 bar, how much oxygen will it deliver?
To calculate how much oxygen is available in this cylinder for the transfer:
Step 1: Calculate the absolute pressure in the cylinder using Système International d’Unités (SI) units, which is gauge pressure + atmospheric pressure. So the absolute pressure is 13 700 kPa + 100 kPa, which is 13 800 kPa.
Step 2: Boyle’s Law states that:
P1 x V1 = P2 x V2
Where P1 is 13 800 kPa, V1 is 10 litres, and P2 is 100 kPa, as you are releasing the oxygen into atmospheric pressure.
Rearranging this equation:
V2 = P1 x V1/ P2
⇒ V2 = 13 800 x 10/ 100
Therefore, the volume of oxygen contained in the cylinder is 1380 litres.
Remember that 10 litres will remain in the oxygen cylinder, therefore only 1370 litres can actually be used on your transfer.
To define Charles’s Law, assume that you had the same syringe, half-filled with air and you obstructed the end with a pressure gauge, but allowed the plunger to move freely to keep the pressure constant.
Question: What would happen if you heated the syringe? What happens to the gas molecules as you heat them?
As the gas is warmed the mean velocity and hence kinetic energy of the gas molecules increases. The molecules collide with the walls more frequently and more vigorously. Under constant pressure the volume of the syringe must increase.
Therefore V∝T or V/T = constant (k2) (Fig 1)
Remember T is in Kelvins.
Charles’s Law states that at a constant pressure the volume of a given mass varies directly with the absolute temperature.
Take the example of gas in a sealed cylinder, but with a moveable piston loaded to apply a constant pressure. If the cylinder is then heated, the pressure remains constant but the volume increases as the gas expands (Fig 2).
Charles’s Law can be used to explain how a hot air balloon can be made to rise.
Question: Can you explain why?
The gas in the balloon is heated and so by virtue of Charles’s Law expands. As the gas expands it becomes less dense, and so rises taking the balloon and its basket with it.
To define the third gas law, or Gay-Lussac’s Law, assume you had the syringe half-filled with air and you obstructed the outlet with a pressure gauge, but prevented the plunger from moving.
Question: What would happen if you now heated the syringe?
The volume of the syringe will stay constant as you have prevented the plunger from moving. The molecules in the syringe gain kinetic energy as they are heated.
Therefore P∝T or P/T = constant (k3) (Fig 1)
The third gas law states that at a constant volume the absolute pressure of a given mass varies directly with the absolute temperature.
Again, take the example of gas in a sealed cylinder but with a constant volume. If heat is applied, the rise in pressure in the cylinder will be directly proportional to the rise in absolute temperature (Fig 2).
The combined gas law stated that:
PV/T = C
The constant C will be a constant only for a fixed quantity of gas. If a different quantity of gas is studied, the gas will still obey the combined gas law, but the resulting constant will have a different numerical value.
Question: What will be the value of the constant?
The value of the constant will vary with the number of moles of gas present, n.
To be more specific and look at the behavior of a single mole of gas, the constant can be given a new symbol R, where R is known as the universal gas constant.
Combining all of these statements, the ideal gas law is created:
PV/T = nR
or as it is usually expressed:
PV = nRT
It is worth noting that the same R also appears in the Nernst Equation. It has a value of 8.31 J K-1 mol-1.
Derivation of the ideal gas law
The ideal gas law is derived from the combined gas law and Avogadro’s number.
The combined gas law states the PV/T = C
The constant (C) is directly proportional to the amount of gas, n, which is Avogadro’s Law
The proportionality factor is the universal gas constant R therefore C = nR.
A full cylinder of nitrous oxide contains 3.4 kg of nitrous oxide, but how much gas are you able to get out of the cylinder?
Step 1: The molecular weight of nitrous oxide is 44, therefore 1 mole is 44 g.
Step 2: If you were using the cylinder at STP, you know that 1 mole will occupy 22.4 litres.
Step 3: If the cylinder contains 3400 g of nitrous oxide, you have 3400/44 moles, which is 77.2 moles. Therefore you will have 77.2 moles x 22.4 litres, which is 1730 litres of nitrous oxide at STP.
Don’t forget that as with the example used to illustrate Boyle’s Law, there will be a residual volume in the cylinder which will be unavailable.
If you had a mixture of gases in a container, the different molecules of the different gases would be interacting with the sides of the container. Pressure in the container is related to the frequency of the collisions and the mass and velocity of the colliding molecules. In the mixture all the molecules contribute to the pressure, but do so independently, therefore the pressure exerted by one molecule remains constant, regardless of whether other molecules are present.
Dalton’s Law of Partial Pressures states, that in a mixture of gases the pressure exerted by each gas is the same as that which it would exert if it alone occupied the container.
The ideal gas law becomes
PV = (n1 +n2 …)RT
where n1 equals the number of moles of gas 1, n2 the number of moles of gas 2 etc.
In a cylinder of air at atmospheric pressure, 20.9% of the cylinder is oxygen and 79% is nitrogen.
Question: How much pressure would each of the two gases exert?
The total pressure would be approximately 100 kPa (atmospheric pressure). Therefore the oxygen exerts a proportional pressure of 20.9 kPa and the nitrogen a pressure of 79 kPa.
Regarding the behaviour of a gas:
A. Behaviour deviates from that of an ideal gas as temperature and pressure fall
B. As temperature rises, if the volume is kept constant, pressure also rises
C. 32 g of nitrous oxide would occupy a volume of 22.4 litres at standard temperature and pressure
D. The pressure reading on the gauge of an oxygen cylinder varies directly with the volume of available oxygen remaining
E. One litre of an ideal gas at standard temperature and pressure would contain 22.4 moles
A. False. Behaviour deviates as temperature falls and pressure rises.
B. True. This is the third ideal gas law.
C. False. 44 g of nitrous oxide would occupy 22.4 litres. 32 g is equivalent to one mole of oxygen not nitrous oxide.
D. True. This is a refection of Boyle’s Law; PV = constant.
E. False. One mole of an ideal gas at standard temperature and pressure would occupy 22.4 litres.
Regarding oxygen:
A. The critical temperature of oxygen is -182°C
B. Oxygen cannot be liquefied at room temperature because its critical pressure cannot be reached
C. One mole of oxygen has a mass of 32 g
D. One mole of oxygen at 1 atmosphere pressure and room temperature would occupy 22.4 litres
E. The partial pressure of oxygen in room air would be approximately 101 kPa
A. False. The critical temperature of oxygen is -119°C.
B. False. Oxygen cannot be liquefied at room temperature regardless of pressure applied as it is above its critical temperature.
C. True.
D. False. One mole at standard temperature (273.15 K) and 1 atmosphere pressure would occupy this volume.
E. False. The partial pressure of oxygen in air would be approximately 21 kPa.
Regarding the ideal gas law PV = nRT:
A. The unit of the ideal gas constant is in moles
B. n must always be greater than 1
C. Temperature must always be expressed in Kelvins
D. Measurements must always be taken at standard temperature and pressure
E. The law becomes less accurate as P rises and T falls
A. False. The units are J K-1 mol-1.
B. False. n must be greater than or equal to zero.
C. True. The Kelvin scale must be used.
D. False. Temperature and pressure are variables.
E. True. As temperature falls and pressure rises the gas will start to liquefy.
Anaesthetists come across gases in solution in everyday practice. This session looks at some of the physical principles behind solubility, and the clinical applications of gases in solution.
Question: Consider the agents used for inhalational induction. Would anaesthesia be induced faster using sevoflurane or halothane?
Induction is faster with sevoflurane. It is less soluble than halothane and hence achieves equilibrium in the alveoli quicker than halothane.
This is one of the common clinical applications of the principles of solubility which this session examines.
Give an overview of gases in solution, and the relevance to anaesthesia.
Define the factors which affect solubility
Explain the ways of describing and comparing solubility
Identify the clinical applications of gases in solution relevant to the anaesthetist
There are a number of factors which affect solubility: temperature, pressure, type of gas and liquid
There are ways of describing and comparing solubility, such as the Ostwald and partition coefficients
The principles of gases in solution have relevant clinical applications for the anaesthetist
Henry’s law states that at a fixed temperature, the amount of a given gas dissolved in a given liquid is directly proportional to the partial pressure of the gas in equilibrium with the liquid.
In other words, the higher the pressure, the more the dissolved gas (Fig 1).
Question: How might Henry’s law be relevant to diving?
With rapid decompression, i.e. decrease in pressure, dissolved nitrogen comes out of solution and forms bubbles in the joints. This causes the ‘bends’.
Remember that Henry’s law applies at a constant temperature.
What happens if the temperature alters?
Question: To take an everyday example, why does a warm can of fizzy drink bubble up more than a cold one?
As the temperature increases, gas bubbles out of the liquid. Conversely, the solubility increases as the temperature decreases.
The other important factor to remember is that Henry’s law applies to a given liquid and a given gas.
Different gases and liquids have different solubilities. For example, nitrous oxide is more soluble than nitrogen.
Consider the following pnenomena, all of which are related to the clinical applications of gases in solution:
1) When a blood warmer is used, how does Henry’s law explain why bubbles form in the infusion line?
2) How does Henry’s law explain what would happen if a patient were taken out of a hyperbaric chamber too quickly?
3) How does the relative solubility of gases in liquids explain why sevoflurane induces anaesthesia faster than ether?
4) A patient with fractured ribs and a small pneumothorax is given inhaled Entonox for analgesia. How can a knowledge of gases in solution explain the risk doing this?
1) Henry’s law explains this by showing that as the blood warms, less gas is dissolved and air bubbles out.
2) Henry’s law explains what would happen by stating that as the pressure decreases, less gas is dissolved in liquid. Hence, nitrogen bubbles out of the blood leading to compression sickness.
3) Sevoflurane, which has a blood gas solubility of 0.6, induces anaesthesia faster than ether, which has a blood gas solubility of 12, because the less soluble agent achieves equilibrium in the alveoli faster than the more soluble agent.
4) The risk is that the pneumothorax contains nitrogen, which has a lower solubility than nitrous oxide i.e. nitrous oxide diffuses into the closed cavity faster than nitrogen diffuses out, thus increasing it’s volume.
Regarding the solubility of gases in solution:
A. SVP refers to the partial pressure of a vapour in equilibrium with its liquid
B. Henry’s law describes the relationship between amount of gas dissolved in solution and partial pressure
C. Henry’s law is independent of temperature
D. All gases have the same solubility
E. The liquid need not be specified when stating Henry’s law
A. True.
B. True.
C. False. Henry’s law is at a fixed temperature.
D. False. Solubility is dependent on the gas. For example, nitrous oxide is more soluble than nitrogen.
E. False. Henry’s law always specifies the gas and the liquid.
The partial pressure exerted by the vapour at equilibrium is called saturated vapour pressure (SVP).
SVP is the pressure exerted by molecules in the vapour component at the point of equilibrium. In confined evaporation, equilibrium occurs when dissolved gas molecules leave the liquid at the same rate at which others dissolve.
Regarding the Bunsen and Ostwald solubility coefficients:
A. The Bunsen and Ostwald are partition coefficients
B. The Bunsen coefficient is commonly used in anaesthetic practice
C. The Bunsen coefficient must be corrected to STP
D. Both Bunsen and Ostwald coefficients describe volumes of gas dissolved in unit volume of liquid
E. The Ostwald coefficient is independent of pressure
A. False. Bunsen and Ostwald are solubility coefficients.
B. False. The Ostwald coefficient is more commonly used.
C. True.
D. True.
E. False. The Ostwald coefficient is measured at a specified pressure.
Solutions of gases are described as volumes of gas dissolved in a volume of liquid.
Two ways of describing this are the Bunsen and the Ostwald solubility coefficients:
Bunsen: the volume of gas corrected to standard temperature and pressure (STP) dissolved in a unit volume of a liquid at the temperature concerned where the partial pressure of the gas above the liquid is one atmosphere
Ostwald: the volume of gas dissolved in a unit volume of liquid at the temperature concerned
As can be appreciated, the Ostwald is not corrected to STP and it is the one preferred by anaesthetists. It is measured at a known temperature and pressure.
Regarding the partition coefficient:
A. The partition coefficient refers to the ratio of a substance in two phases
B. The partition coefficient is similar to the Ostwald coefficient
C. The order of the phases need not be specified with the partition coefficient
D. One of the phases must be a gas
E. The temperature must be specified
A. True.
B. True.
C. False. The order needs to be specified. For example, the blood-gas partition coefficient is different from the gas-blood coefficient.
D. False. Both phases may be liquids.
E. True.
The term ‘partition coefficient’ describes the distribution of a substance in two phases, e.g. nitrous oxide in blood and gas (Fig 1).
The partition coefficient is the ratio of the amount of a given substance present in one phase compared with another, both phases being of equal volume and at equilibrium.
In some ways, the partition coefficient is similar to the Ostwald solubility coefficient, because the temperature and phases must be specified.
A key difference is that with the partition coefficient, the order of the phases must be specified. So, the blood-gas partition coefficient is different from the gas-blood coefficient. Another difference is that the partition coefficient can be applied to two liquids.
Give an overview of density and viscosity.
Define density and viscosity and explain their implications for the measurement of flow
Explain the significance of pressure and temperature for changes in density and viscosity
Describe how the physical characteristics of the common gases affect storage
Density is mass per unit volume and is an important determinant of turbulent flow
Viscosity is the ability of a fluid to resist flow and is an important determinant of laminar flow
The density and viscosity of a gas impact on its flow through a variable orifice device, such as a rotameter, so that each must be calibrated for a different gas
Oxygen is a commonly used gas which, for practical reasons, is stored as a liquid. Storage is complicated by a low boiling point
Nitrous oxide is a very soluble gas which increases the size of air filled spaces, causing serious clinical implications. It can be pressurized to a liquid at room temperature to increase storage mass within a cylinder
Helium has a low density which improves turbulent flow and proves advantageous in airway obstruction
What will happen to the flow in Fig 1 if the temperature is increased?
A. The flow will increase
B. The flow will decrease
C. The flow will remain the same
A. Correct.
B. Incorrect.
C. Incorrect.
What will happen to the flow in Fig 1 if the pressure gradient is increased?
A. The flow will increase
B. The flow will decrease
C. The flow will remain the same
A. Correct.
B. Incorrect.
C. Incorrect.
What will happen to the flow in Fig 1 if the viscosity is reduced?
A. The flow will increase
B. The flow will decrease
C. The flow will remain the same
A. Correct.
B. Incorrect.
C. Incorrect.
Laminar flow is affected by the factors in the Hagen-Poiseuille equation:
Laminar flow = ∆Pr4∏/8lη
Where ∆P is the pressure difference from the start of the flow to the end; r is the radius of the tube; l is the length of the tube and η (Greek symbol Eta) is the fluid viscosity.
Temperature reduces viscosity and hence increases flow. A pressure gradient increase describes a variation in pressure applied to the flow of fluid and is not the same as consistent pressure increase. Adding a pressure bag to a litre of fluid increases the rate of flow; this is an increased pressure gradient.
What will happen to the flow in Fig 1 if the temperature is increased?
A. The flow will increase
B. The flow will decrease
C. The flow will remain the same
A. Correct.
B. Incorrect.
C. Incorrect.
What will happen to the flow in Fig 1 if the pressure gradient is increased?
A. The flow will increase
B. The flow will decrease
C. The flow will remain the same
A. Correct.
B. Incorrect.
C. Incorrect.
What will happen to the flow in Fig 1 if the density is increased?
A. The flow will increase
B. The flow will decrease
C. The flow will remain the same
A. Incorrect.
B. Correct.
C. Incorrect.
Turbulent flow is proportional to r2 and √∆P and inversely proportional to length of the tube and density of the fluid.
Temperature reduces density and therefore increases flow. Increasing the pressure gradient increases turbulent flow but not as much as laminar flow, because the streamlining is lost.
Usually you exhale mainly air when using a PEFM. What would happen to the reading if you were exhaling mainly helium?
As you can see in the animation, because helium is less dense than air the flow would increase.
There are other flowmeters which are less commonly seen in practice. These include variable pressure, variable orifice flowmeters such as the watersight flowmeter. There are also other constant pressure, constant orifice flowmeters such as the bubble flowmeter.
Thermistor and ultrasonic flowmeters are also used.
Which of the following clinical devices measure flow, which measure volume and which measure pressure?
Since flow is volume per unit time, a volumeter with timing device could be used to measure flow. However, that is not the intended function of Wright’s Respirometer.
Oxygen cylinders reduce volume and pressure proportionally, so it is possible to measure either flow or remaining pressure on the gauge.
Regarding the density of a gaseous substance:
A. Density is mass per unit volume measured in kg.m
B. A decrease in pressure increases density
C. An increase in temperature increases the kinetic energy of the molecules within gas and therefore increases the density
D. Density has a more marked effect on turbulent flow than viscosity does
E. Density has a more marked effect than viscosity on the flow at the base of a rotameter
A. False. Density is measured in kg/m3.
B. False. Increasing pressure leads to an increase in density.
C. False. Decreasing temperature leads to an increase in density.
D. True.
E. False. The base of a rotameter has a narrow orifice around the bobbin causing laminar flow, which is more markedly affected by viscosity than density.
Regarding the viscosity of a gaseous substance:
A. Viscosity is directly proportional to laminar flow
B. The viscosity of honey is greater than that of water
C. An increase in temperature increases viscosity
D. An increase in pressure leads to an increase in viscosity
E. A fluid with higher viscosity improves laminar flow, given fixed tube length and radius and pressure gradient
A. False. Viscosity is inversely proportional to laminar flow.
B. True.
C. False. Increasing temperature decreases viscosity, the ability of the substance to resist flow.
D. True.
E. False. Increasing viscosity reduces laminar flow.
Regarding flowmeters:
A. A pneumotachograph is an example of a fixed orifice, variable pressure device
B. A rotameter relies on the constant pressure drop supporting the weight of the bobbin
C. A peak expiratory flowmeter is an example of a constant pressure variable orifice device
D. A Wright’s Respirometer is used to measure flow
E. A Bourdon gauge can be used to calculate flow if a timing device is also used
A. True.
B. True.
C. True.
D. False. A Wright’s Respirometer is used to measure volume; only with the addition of a timing device can it measure flow.
E. False. A Bourdon aneroid gauge is used to measure pressure.
Constant orifice, variable pressure
Measurement of flow across a tube can be determined by the pressure difference.
A pneumotachograph creates laminar flow through multiple small diameter tubes or mesh, and uses the linear relationship between flow and pressure difference, as per the Hagen-Poiseuille equation. The small resistance offered by the tubes causes a pressure drop proportional to the flow and can be transduced to continuously monitor breathing flow. Calibration of the pressure difference is necessary by using a known flow rate, and must be repeated when a different gas is used or the change in viscosity and density will affect the results.
A simple pressure gauge can be used to calculate flow if upstream pressure is constant. An example is the pressure gauge of an oxygen cylinder, using a pressure gauge at the outlet of the oxygen cylinder.
Which of the following best describes the effect of changes on flow measurement near the top of a rotameter?
A. An increase in gas density causes the bobbin to rotate less
B. A decrease in viscosity leads to an increase in laminar flow
C. The pressure difference across a rotameter bobbin is constant
D. An increase in gas temperature decreases the density of the gas and increases flow
E. The flow measurement should be read from the top of the bobbin
A. Incorrect. An increase in density reduces flow and, therefore, causes the bobbin to rotate less. However, the rotation of the bobbin does not affect the measurement of flow.
B. Incorrect. Decreased viscosity causes an increase in laminar flow, but laminar flow is found mostly at the base of the rotameter where the orifice is small.
C. Incorrect. The pressure difference across the bobbin, with the weight of the bobbin pressing down and gas pressure pushing up, is constant and will not affect changes in flow measurement.
D. Correct. This statement is the best fit because it describes the effect of a change in gas density on the more turbulent flow found at the top of a rotameter.
E. Incorrect. The flow measurement should be read from the top of a flat-topped bobbin and from the middle of a spherical bobbin, neither of which affects changes in flow measurement at the top of a rotameter.
Constant pressure, variable orifice
A rotameter uses a bobbin of fixed size, which moves up and down in a tapered column of flowing gas, which is introduced via a needle valve.
Pressure from the gas flow pushes the bobbin up the tube until it is counteracted by the gravitational force pushing down on the bobbin, i.e. its weight, at which time the pressure across the bobbin is constant.
With the bobbin near to the base, flow is low and the bobbin is close to the sides and so the flowmeter resembles a tube. Higher up, the bobbin is further away from the sides to the extent that the diameter is greater than the length of the bobbin, creating an orifice relationship. So viscosity takes a more important role in determining flow at the base of the rotameter, and density a greater role at the top.
The flow measurement should be read from the top of a flat-topped bobbin and from the middle of a spherical bobbin, neither of which affects changes in flow measurement at the top of a rotameter
Individual gases have varying densities and viscosities, so the rotameter must be calibrated for each gas.
The boiling point of oxygen is:
Select one option from the list below.
Possible answers:
A. -183C
B. -88C
C. -269C
A. Correct.
B. Incorrect. -880C is the boiling point of nitrous oxide.
C. Incorrect. -2690C is the boiling point of helium.
This page describes the key features of gases which are essential to an understanding of their storage and use.
The low boiling point of oxygen requires either special provision for liquid storage or for its storage to be in gaseous form.
Nitrous oxide has a critical temperature above room temperature and so can be stored as a liquid.
Table 1 shows the key features of oxygen, nitrous oxide, carbon dioxide and helium.
The critical temperature of nitrous oxide is:
A. 720C
B. 300C
C. 36.50C
A. Incorrect. 72 bar is the critical pressure of nitrous oxide.
B. Incorrect. This is the critical temperature of carbon dioxide. It is similar to the critical temperature of nitrous oxide which is why they can both be condensed to liquids.
C. Correct.
This page describes the key features of gases which are essential to an understanding of their storage and use.
The low boiling point of oxygen requires either special provision for liquid storage or for its storage to be in gaseous form.
Nitrous oxide has a critical temperature above room temperature and so can be stored as a liquid.
Table 1 shows the key features of oxygen, nitrous oxide, carbon dioxide and helium.
Give an overview of gas storage, and the relevance to anaesthesia.
Define the concepts of critical temperature and pressure
Explain the relationship between gases, vapours and liquids
Define the ‘triple point’
Discuss the principles of gas storage, the different states of gases and their potential dangers
The critical temperature of a substance is the temperature above which the substance cannot be liquefied, however much pressure is applied
A gas is a substance existing above its critical temperature. A vapour is a substance in its gaseous state below its critical temperature
Phase diagrams give a great deal of information about the behaviour of substances. Fixed points such as the triple point are essential for calibrating scales such as temperature scales
Entonox® has unusual properties due to the Poynting effect. Below its pseudocritical temperature it may undergo dangerous liquefaction
Oxygen may be stored as a liquid or a gas. Liquid storage requires specific conditions within a VIE
Nitrous oxide is stored as a liquid in cylinders and its gauge pressure cannot be relied upon. Its content must be checked by weighing
What is the critical temperature and critical pressure of oxygen, nitrous oxide and carbon dioxide?
The critical temperature is the temperature above which a substance cannot be liquefied by pressure alone.
The critical pressure is the vapour pressure at the critical temperature.
An Entonox® cylinder that has been stored outside is about to be used. It is important to be aware that a dangerous separation of the nitrous oxide and oxygen may have occurred.
The oxygen rich gas, containing little N2O, is delivered initially and therefore provides little analgesia. As the cylinder is used, oxygen leaves the liquid N2O to establish equilibrium. As the cylinder nears empty it contains a hypoxic mixture of vapourized nitrous oxide that could be fatal to the patient.
Question: How can this potentially dangerous situation be averted?
This situation can be averted by ensuring that:
The temperature of the storage area is maintained
Cylinders are stored horizontally
A dip tube is in place to ensure the hypoxic mixture is not delivered
A dip tube reaches to the base of the cylinder and siphons off the liquid nitrous oxide first, ensuring that the lowest concentration of oxygen ever given to a patient is the 20% dissolved within the liquid from the start.
The following can be used to avoid the delivery of a hypoxic mixture from an Entonox® cylinder:
A. Storage area maintained above 10°C
B. Vertical storage of cylinders
C. Provision of Entonox® by pipeline supply
D. Intermittent use of Entonox®
E. Inclusion of a dip tube
F. Re-warming after Entonox® has been stored below its pseudocritical temperature
A. True.
B. False. Cylinders should be stored horizontally.
C. True.
D. False. Whilst intermittent use may reduce cooling from latent heat, the nature of Entonox® use makes this unlikely to be feasible.
E. True. A dip tube ensures that the initial gas drawn off has a minimum oxygen concentration of 20% that has remained mixed in the nitrous oxide. The mixture delivered later has excess oxygen, but lacks nitrous oxide as pain relief.
F. False. Re-warming would not redress the hypoxic mixture already potentially created, nor necessarily re-vaporize the nitrous oxide as Entonox®.
The critical temperature:
A. Of oxygen is -160°C
B. Of nitrous oxide is 36.5°C
C. Of Entonox® is 7°C at 137 bar
D. Of a substance increases with a drop in critical pressure
E. Of a gaseous substance describes the temperature below which a liquid cannot be produced by pressure alone
A. False. The critical temperature of oxygen is -118.6°C.
B. True.
C. False. The pseudocritical temperature of Entonox® is -7°C, not +7°C.
D. False. The critical temperature of a substance is fixed and does not vary with changes in pressure.
E. False. The critical temperature of a substance is a fixed temperature above which the substance cannot be liquefied no matter how much pressure is applied.
The critical temperature of a substance is the temperature above which it cannot be liquefied, however much pressure is applied.
Fig 1 shows that when a substance is above its critical temperature, a reduction in volume increases the pressure.
Below critical temperature, the substance would liquefy rather than causing an increase in pressure.
Critical pressure refers to the vapour pressure of a substance at its critical temperature and is thus the pressure required to liquefy the gaseous component at that temperature.
Regarding gas storage:
A. There is a global standard filling ratio for nitrous oxide
B. Liquid oxygen storage is costly, but convenient
C. A full oxygen cylinder has a gauge pressure of 137 bar when full
D. Oxygen cylinders are black with a continuous white collar
E. As high volumes of oxygen from a VIE are used, the supply pressure to the pipeline rises
A. False. The filling ratio describes the weight of fluid within a nitrous cylinder compared with the weight of the cylinder when filled completely with water. Filling ratio varies with geographical climate; in temperate climates it is 0.75, reduced to 0.67 in hotter climates because the increased vaporization leads to increased pressure within the cylinder.
B. False. Liquid oxygen storage in VIEs is space saving and cost saving.
C. True.
D. True.
E. False. High flow rates from a VIE could potentially lead to a fall in pressure due to latent heat of vaporization, however the pressure raising vaporizer system ensures that pressure is kept fairly constant.
To produce Entonox®, liquid nitrous oxide is carried into the gaseous phase via oxygen, which is bubbled through it, allowing a 50% mixture (Fig 1).
This is despite the nitrous oxide being below its critical temperature and above its critical pressure.
It should seemingly not be possible to produce a 50:50 mixture if the two are mixed as their gaseous constituents at 137 bar. This is made possible through a peculiar molecular interaction known as the ‘Poynting effect’.
The effect of combining the gases reduces the critical temperature of the mixture to –7°C at 137 bar.
If the temperature of an Entonox® cylinder is reduced to less than -7°C, liquefaction of N2O could take place. The nitrous oxide would liquefy and separate from the oxygen. This process is also called ‘lamination’ or ‘separation’.
After liquefaction, the nitrous oxide has approximately 20% oxygen dissolved in it.
This process is most likely to occur at a pressure of 117 bar. It is less likely at higher or lower pressures. For example, at a pipeline pressure of 4.1 bar, the pseudocritical temperature is lowered to -30°C.
Regarding nitrous oxide:
A. It can be mixed with air to form Entonox®, via the Poynting effect
B. It is generally stored as a gas
C. Gauge pressure can be a reliable measure of the cylinder content
D. It is stored in French blue cylinders
E. It should be stored in cylinders horizontally
A. False. Oxygen is bubbled through liquid nitrous oxide to form Entonox®. Air is not used.
B. False. A standard size E cylinder can store 1800 L of nitrous oxide as opposed to 680 L of gaseous oxygen. At a pressure of 52 bar nitrous oxide would liquefy anyway at room temperature.
C. True. The gauge pressure on a nitrous oxide cylinder can be an accurate measure of cylinder content once all liquid has vaporized, with a pressure below about 52 bar.
D. True.
E. False. Nitrous oxide is stored vertically. Entonox should be stored horizontally to prevent liquefaction.
Which of these statements explains the shape of the nitrous oxide isotherm at 20°C?
A. Nitrous oxide has a critical temperature of 46.5°C
B. There is a direct relationship between pressure and volume above the critical temperature of nitrous oxide
C. The critical pressure at 36.5°C is 132 bar
D. Latent heat of vaporization causes a temperature rise as the liquid vaporizes
E. The nitrous oxide steadily liquefies at a pressure of 52 bar and then demonstrates a clear inflexion point once all the vapour has turned to liquid
Isotherms are lines depicting the effect of temperature and pressure on the physical state of a substance. Each isotherm relates to a different temperature at, above or below the critical temperature.
A. False. Nitrous oxide has a critical temperature of 36.5°C. This does not relate to the isotherm at 20°C.
B. False. There is an indirect relationship between pressure and volume above the critical temperature of nitrous oxide.
C. False. The critical pressure at 36.5°C is 72 bar. This does not relate to the isotherm at 20°C.
D. False. The latent heat of vaporization causes a temperature drop as the liquid vaporizes, but the isotherm relates to the changes which occur at a fixed temperature, 20°C.
E. True.
Isotherms are a series of lines that describe the way in which temperature and pressure determine the physical state of a substance above and below the critical temperature.
Fig 1 shows the effects of compression on nitrous oxide at various temperatures:
Temperature 40°C: Nitrous oxide is above its critical temperature and exists as a gas. The inverse relationship between temperature and volume follows a hyperbolic curve which is explained by Boyle’s Law.
Temperature 36.5°C: Nitrous oxide is at its critical temperature. It exists as a vapour at low pressure and then liquefies at the critical pressure of 72 bar. Liquids are relatively incompressible, explaining the inflexion point in the curve and steep pressure rise for any reduction in volume.
Temperature 20°C: Nitrous oxide is below its critical temperature. At 20°C the vapour partly compresses to a liquid with only 52 bar. Further reduction in volume causes more vapour to condense with no change in pressure. A horizontal line is seen before the inflexion point at which all vapour has become liquid. This is characteristic of a nitrous oxide cylinder at room temperature.
The triple point is the temperature and pressure at which the solid, liquid and gaseous phases of a substance can exist in equilibrium. Fixed points such as this are essential in establishing a temperature scale.
Give an overview of gases and vapours,
Define the concept of vapour and saturated vapour pressure
Explain the principle of latent heat
Distinguish between adiabatic change and isothermic change
Describe how the concepts above relate to the anaesthetic vaporizer
SVP is the pressure exerted by the molecules in a vapour at the point of equilibrium. It is temperature dependent
Latent heat of vaporization is the energy required to convert a given mass of liquid into vapour (at the same temperature)
Adiabatic change describes the change in state of a substance which occurs without an exchange in heat energy with the environment
Isothermal change describes a thermodynamic process where heat energy lost is over a sufficient time to allow the surroundings to dissipate the heat and maintain the temperature of the substance
Vaporizers are designed to suit a specific agent. The required splitting ratio depends on the SVP of that agent
Vaporizers have a number of key features, such as temperature compensation which ensures a controlled and predictable concentration of anaesthetic agent reaches the common gas outlet
Try to match the gases with the temperature variance shown on the graph.
Why is it not possible to make a really good cup of tea in the high Himalayas?
A. SVP is lower at high altitude
B. Boiling point decreases with ambient pressure so the teabag will have to stew for longer to get the same flavour out
C. SVP is higher at high altitude
D. SVP cannot increase to atmospheric pressure
E. Boiling point is higher at high altitude so the temperature required cannot be attained with normal heating equipment
A. Incorrect.
B. Correct. The boiling point occurs when the vapour pressure of the liquid equals the ambient pressure above the liquid. Therefore, the boiling point varies with ambient pressure and is lower at high altitude. Molecules will escape as the liquid boils rather than the temperature rising further.
C. Incorrect.
D. Incorrect.
E. Incorrect.
Latent heat is important in anaesthesia in which of the following areas?
A. Liquid oxygen storage systems
B. Vaporizers
C. Heat and moisture exchange filters
D. Nitrous oxide cylinders
E. Wet drapes and head injured patients
F. Ethyl chloride spray
A. Correct. Vacuum insulated systems retain heat, cooling the remaining liquid as latent heat is expended.
B. Correct. Volatile agents lose heat as they are vaporized. The temperature of the remaining fluid falls, lowering its SVP, which reduces vaporizer output, requiring temperature compensation.
C. Correct. Heat and moisture exchange filters reduce water loss in the respiratory tract, reducing the requirement for further vaporization.
D. Correct. Latent heat is taken from the stored nitrous oxide as the gas is used and more liquid vaporizes.
E. Correct. Evaporation from wet drapes creates effective cooling. This cooling can improve outcome following head injury.
F. Correct. Ethyl chloride is stored as a liquid. When it is sprayed on skin it vaporizes, cooling the skin and sensory nerves providing brief but intense periods of local analgesia.
Not all molecules in a liquid have the same energy. The more vigorous molecules have a greater tendency to escape to the gaseous phase. This means the average energy of those left behind is lower.
As such, energy would have to be provided to enable those molecules left behind to continue to enter the gaseous phase at the same rate.
This energy is provided by heat taken from the liquid or the surrounding area and is known as latent heat.
Latent heat can be thought of as ‘hidden’ heat. It is the heat energy used to cause a phase change and it disappears into the surroundings without causing an increase in temperature.
The latent heat of vaporization is the heat required to convert a given mass of liquid into vapour whilst maintaining the same temperature.
The latent heat of fusion is the heat required to convert a given mass of solid into liquid at the same temperature.
Fig 1 describes the temperature of a substance when heated from the solid state.
It is vital to appreciate that heat is a form of energy, whilst temperature is a measurement of the hotness or coldness of a substance.
A solid such as deeply frozen ice will increase in temperature when heated until it reaches melting point. At that point, energy is required by the individual molecules to convert into the liquid state, and so the temperature overall does not increase during this stage. The same principle applies when the liquid water then boils and converts to a gas.
Latent heat of vaporization varies with the temperature of the liquid. Water closer to boiling point will require less latent heat to achieve complete vaporization than water at room temperature.
‘Specific latent heat’ is the heat required to convert one kilogram of a substance from one phase to another; it has units of J/kg and must be quoted with reference to a specific temperature.
For example, water at 100oC has a specific latent heat of 2.26 MJ/kg. At body temperature, this rises to 2.43 MJ/kg.
Fig 1 shows the change in latent heat of water with temperature. If the graph was extrapolated, the latent heat would eventually become zero.
What would be the danger of putting halothane in a sevoflurane vaporizer?
Select one option from the list below.
Possible answers:
A. Overdosing
B. Underdosing
A. Correct.
B. Incorrect.
Halothane has a higher SVP and a lower MAC and for both reasons potentially fatal overdosage may occur. Each vaporizer is agent specific in order to deliver the dialled percentage. There are many safety precautions taken to ensure that the only inhalational agent that can be put into the chamber is the one for which the vaporizer is calibrated.
The purpose of a vaporizer is to deliver a controlled and predictable concentration of anaesthetic agent in a carrier gas at the common gas outlet.
The saturated vapour pressure of volatile agents at room temperature is many times that required for anaesthesia. Vaporizers are designed to split the gas flow, fully saturate the chamber flow and then re-mix to provide the final, desired concentration. A dial controls the ratio of gas passing through each route and this is known as the splitting ratio.
The concentration of the anaesthetic vapour within the vaporizer chamber is known from its saturated vapour pressure. The dial on the front of the vaporizer ensures that when this vapour is mixed with the anaesthetic-free bypass gas, the concentration of anaesthetic in the gas leaving the vaporizer is known
Regarding saturated vapour pressure:
A. SVP is temperature dependent
B. SVP is pressure dependent
C. SVP is 23 kPa for Isoflurane at 20 oC
D. SVP leads to an increase in boiling point at high altitude
E. SVP determines the splitting ratio of an anaesthetic vaporizer
A. True.
B. False. SVP is unchanged with ambient pressure, only the proportion of total pressure occupied by the given vapour changes with ambient pressure.
C. False. SVP for Isoflurane at 20 oC is 32 kPa.
D. False. Liquid boils when SVP reaches ambient pressure, which is lower at high altitude. The SVP is unchanged but the boiling point is reduced.
E. True.
Saturated vapour pressure (SVP) is the pressure exerted by molecules in the vapour component at the point of equilibrium.
The higher the SVP, the more volatile a substance is and therefore the greater its tendency to vaporize. A volatile anaesthetic agent with a high SVP, e.g. isoflurane, will produce a higher concentration of agent in the carrier gas than an agent with a lower SVP, e.g. sevoflurane, when used at the same temperature.
If isoflurane was delivered via a vaporizer designed for sevoflurane, a dangerously high concentration of agent would be administered.
If the SVP of two anaesthetic vapours is the same, then each could theoretically be given through the vaporizer designed for the other. Adjustment to the amount dialled for each would still be necessary because of the difference in Minimum Alveolar Concentration (MAC) value, but the percentage shown on the dial would be the same through either vaporizer.
If the kinetic energy of the molecules in a substance is increased by raising the temperature, then more molecules leave the liquid state. These molecules then exert a pressure. Once equilibrium is reached, a higher saturated vapour pressure is achieved. The relationship between SVP and temperature is non-linear (Fig 1).
When SVP is equal to atmospheric pressure, the liquid boils. Vapour concentration at the surface of the liquid is 100%.
SVP is unchanged with ambient pressure. Only the proportion of total pressure occupied by the given vapour changes with ambient pressure.
Regarding latent heat:
A. Latent heat of fusion explains the energy transfer when there is a change of state from liquid to solid, without a change in temperature
B. The ambient temperature must always be specified
C. Specific latent heat is measured in J/Kg/min
D. Latent heat of vaporization increases with increasing temperature of the liquid
E. Latent heat of vaporization is the temperature increase required to turn a liquid into a gas
A. False. Latent heat of fusion applies to the change of state from solid to liquid.
B. True.
C. False. Specific latent heat is measured in J/Kg and is the heat required to change a substance from one state into another.
D. False. Latent heat of vaporization decreases with increasing temperature because less energy is then required to convert the liquid into a gas.
E. False. Latent heat of vaporization describes the energy required rather than the increase in temperature and is measured in Joules.
Which one of the following safety features on the modern Tec vaporizer prevents the variation in concentration delivery from latent heat of vaporization?
A. The heat sink can be filled with water
B. Dial linked to the splitting ratio to determine concentration provided
C. The entrance to the vaporization chamber is controlled by a bimetallic strip
D. The entrance to the vaporizing chamber is partially occluded by bellows
E. A non-spill mechanism prevents excess agent entering the bypass channel
A. Incorrect. The heat sink in this vaporizer is metal only.
B. Incorrect. This is true but does not relate to the issue of latent heat.
C. Correct.
D. Incorrect. TEC vaporizers do not have bellows. These are features on the EMO and Ohio vaporizers.
E. Incorrect. This is true but does not relate to the issue of latent heat.
The temperature compensation methods used in modern vaporizers ensure that the correct concentration of volatile is produced despite the latent heat of vaporization.
Simple methods such as reducing gas flows can be employed by the anaesthetist, but the vaporizer itself must also have compensation devices.
Compensation methods may include the provision of heat energy from a store of high thermal conductivity encasing the vaporizer and/or a method that can create a mechanical change within the vaporizer to alter the splitting ratio with changing temperature.
Assuming fixed temperature, a change in altitude will affect delivery of anaesthetic agent in the following way:
A. The saturated vapour pressure changes
B. The proportion of total pressure occupied by a set concentration of vapour changes
C. The setting on the vaporizer should be adjusted in order to provide sufficient anaesthetic action
D. A vaporizer calibrated to give 1% isoflurane will give 2% when used at 5500 m above sea level
E. A vaporizer must be recalibrated for use at high altitude
A. False. Saturated vapour pressure is solely a function of temperature.
B. True.
C. False. The anaesthetic action relies on the partial pressure of anaesthetic vapour.
D. True. 5500 m above sea level corresponds to about half normal atmospheric pressure (101.325 kPa ?50.66 kPa). The vaporizer will give 2%, but the partial pressure will remain the same (1% of 101.325 = 2% of 50.66), and therefore the effect of the agent will be the same.
E. False. Dialling 1% isoflurane at sea level will give 1% isoflurane and the expected effects. Dialling 1% isoflurane at 5500 m will give 2%, but the partial pressure will remain the same and the effect the same. Therefore, the vaporizer should be used in the same way at any altitude.
If the state of a gas is altered without a change in heat energy to or from the gas, it is said to undergo adiabatic change.
If a compressed gas expands adiabatically, cooling occurs, e.g. as seen in a cryoprobe. Energy is required as the gas expands to overcome van der Waal’s forces. As no heat exchange occurs with the surroundings, the source of energy must be from the molecule’s own kinetic energy. As a result, the gas cools.
If a gas is rapidly compressed, its temperature rises. This is the Joule–Kelvin principle and it may be seen when a cylinder connected to an anaesthetic machine is turned on too quickly; the temperature rises and may, in the presence of oil or grease, lead to a fire or explosion.
If the compression or expansion occurs sufficiently slowly, heat can be conducted through the walls of the container, leading to no change in the temperature of the gas. This is known as an isothermal change
Give an overview of humidity, with relevance to anaesthesia.
Define relative humidity and absolute humidity
Define the implications of changes in temperature and pressure on both relative and absolute humidity
List the reasons for, and explain the importance of, humidification within the respiratory tract
Describe the principles behind the various methods of humidification
Explain the physical principles relating to the measurement of humidity
Relative humidity is the water vapour content of gas compared to the maximum amount of water vapour to fully saturate at the same temperature and pressure
Absolute humidity is the amount of water vapour per unit volume of gas at a given temperature and pressure
Dew point represents the temperature at which a gas is fully saturated and is seen as the point at which condensation occurs
Humidification is clinically essential to ensure normal production and function of respiratory secretions, cilia and epithelium, and to prevent heat loss
Humidification within the theatre environment must be strictly controlled: too high is uncomfortable for the staff, too low increases the risk of sparking
Humidifiers act by increasing the relative humidity of a gas and are most effective when an increase in temperature enables an increased mass of water vapour to be contained within the gas
Measuring humidity requires an understanding of the physical principles of humidity including condensation, evaporation and latent heat, dew point and saturated vapour pressure
If 2 m3 air, fully saturated with water vapour at 20°C and at sea level, is then heated to 37°C, what is the absolute humidity?
What is the relative humidity?
The absolute humidity is 17 g/m3 at any temperature.
The relative humidity becomes 17/44 (39%).
Not only does a change in temperature affect humidity, but a change in humidity also affects temperature via latent heat of vaporization. This can be utilized in the measurement of humidity using a wet and dry bulb thermometer (Fig 1). The greater the humidity of the atmosphere, the less likely the water to evaporate and cause a drop in the temperature reading.
Absolute humidity
Absolute humidity is the mass of water vapour present in a given volume of gas at a given temperature and pressure. It is expressed as g/m3 or mg/L. These quantities are numerically the same.
Relative humidity
Relative humidity describes the ratio of actual mass of water vapour in a gas compared to the maximum amount of water vapour that the gas could contain, at the same temperature and pressure, expressed as a percentage. Relative humidity can also be expressed in terms of vapour pressure:
Mass is proportional to the number of moles (n). The ideal gas law states:
n = PV / RT
where P is pressure, V is volume, R is the universal gas constant and T is temperature. Therefore, if temperature and volume are constant, the number of moles must be directly proportional to pressure.
Since this applies to both actual vapour pressure and the pressure at which full saturated vapour pressure (SVP) is reached, the ratio of these equals the relative humidity.
Relative humidity = Actual vapor pressure / SVP
Which of these are provided by, and which are prevented by, humidification?
Effective humidification has implications for respiratory support in anaesthesia and critical care. The majority of its benefits lie with preventing damage to the pulmonary epithelium and its products.
At 37°C, absolute humidity is 44 g/m3 and at room temperature (20°C) it is 17 g/m3. Air entering the upper trachea contains approximately 34 g/m3, i.e. almost fully saturated at a temperature of 34°C. Therefore, an intubated patient has no ability to humidify their gases without losing moisture from the respiratory mucosa (Fig 1).
In the theatre environment, staff health and safety must also be considered. High humidity is uncomfortable. Low humidity increases the risk of sparks and can cause cracked mucosal membranes, worsening of rhinitis and asthma and respiratory infections.
Regarding the measurement of humidity:
A. Dew point varies with ambient temperature
B. Hair length elongates with increasing partial pressure of water vapour
C. Ether is bubbled through water to obtain dew point on the silver casing of a Regnault’s hygrometer
D. Humidity is calculated from the temperature shown on the wet wick thermometer
E. Risk of sparking increases with increasing humidity
A. False. Dew point is dependent on humidity: it is achieved when the ambient temperature falls sufficiently for water to precipitate.
B. True. Hair length does increase with humidity and hence partial pressure of water vapour.
C. False. The Regnault’s hygrometer uses air bubbled through ether to find the dew point.
D. False. With wet and dry wicks, the humidity is calculated by the temperature difference between that shown on the wet and the dry wick.
E. False. Sparking risk is reduced with increasing humidity.
Increasing humidity increases the dew point. Increasing barometric pressure also increases the dew point, since relative humidity is increased.
The relationship between humidity and dew point can be used to calculate relative humidity when the dew point and saturated vapour pressure is known.
A Regnault’s hygrometer consists of a silver tube containing ether through which air is bubbled. This causes cooling, which leads to condensation forming on the outside of the tube. The temperature at which this occurs is the dew point, from which can be derived relative humidity.
This principle is illustrated on the graph of water in air against temperature. As shown by the moving red line, a drop in temperature leads to an increase in relative humidity. When relative humidity reaches 100%, there is full saturation and dew forms.
Regarding relative humidity:
A. It is measured in g/m3
B. It can be described as actual vapour pressure/saturated vapour pressure
C. It is increased with decreasing temperature
D. It is inversely proportional to pressure in an open space
E. It is always 100% at the dew point
A. False. Relative humidity is expressed as a percentage, absolute humidity is measured as g/m3.
B. True.
C. True.
D. False. Humidity is directly proportional to pressure in an open space.
E. True. Relative humidity must be 100% at the dew point for precipitation to occur.
Increasing temperature does not affect absolute humidity since the mass of water vapour is the same. However, the maximum possible absolute humidity to cause saturation increases with increasing temperature and, therefore, relative humidity decreases (Fig 1).
At room temperature (20°C) the mass of water vapour in fully saturated air at sea level is approximately 17 g/m3. At body temperature (37°C) the mass of water vapour in fully saturated air at sea level is approximately 44 g/m3.
At full saturation, the partial pressure of water vapour is the same as the saturated vapour pressure. SVP is unaffected by changes in pressure.
Where full saturation does not exist, the ideal gas law applies (Fig 1).
In an open space, relative humidity is directly proportional to barometric pressure until relative humidity reaches 100% and condensation occurs.
Dew point is the temperature to which a given volume of gas must be cooled in order for water vapour to condense out into water. If this point falls below freezing, it is called the frost point.
At the dew point, relative humidity is 100% and the air is fully saturated. As air temperature increases, the mass of water vapour that can be contained within that air increases too.
If air has low relative humidity then a greater temperature drop is required to reach dew point.
On a typical British summer day, with a temperature of 18°C, the dew point is about 8°C. In the early evening, when it is 12°C outside, the dew point is still 8°C. If the temperature reaches about 7°C overnight then the water vapour condenses out and forms dew on the ground.
Regarding the clinical implications of humidification:
A. Absolute humidity in the upper trachea is approximately 34 g/m3
B. Invasive ventilation increases respiratory tract mucus production
C. Low levels of humidification in the invasively ventilated patient are associated with increased risk of respiratory infection
D. A patient with acute asthma should not have humidified oxygen because of the risk of water droplets causing further obstruction to the airway
E. High humidity increases the risks of sparks in the theatre environment
A. True.
B. False. Intubation reduces humidification of the respiratory tract and so decreases mucus production and increases its viscosity.
C. True.
D. False. Though a patient with acute asthma should always have high flow oxygen initially, this is to enable as much oxygen delivery as possible. There is no justification to withhold humidification, which actually prevents further mucosal damage.
E. False. Humidification reduces sparking.
Effective humidification has implications for respiratory support in anaesthesia and critical care. The majority of its benefits lie with preventing damage to the pulmonary epithelium and its products.
At 37°C, absolute humidity is 44 g/m3 and at room temperature (20°C) it is 17 g/m3. Air entering the upper trachea contains approximately 34 g/m3, i.e. almost fully saturated at a temperature of 34°C. Therefore, an intubated patient has no ability to humidify their gases without losing moisture from the respiratory mucosa (Fig 1).
In the theatre environment, staff health and safety must also be considered. High humidity is uncomfortable. Low humidity increases the risk of sparks and can cause cracked mucosal membranes, worsening of rhinitis and asthma and respiratory infections.
Which one of the following best describes how temperature change affects relative humidity?
A. An increase in temperature does not affect the amount of water vapour measured in g/m3
B. A decrease in temperature of a gas causes formation of dew
C. Condensation occurs at 100% humidity
D. In an open space, relative humidity is directly proportional to barometric pressure
E. Increasing temperature increases the maximum amount of water vapour that can be contained within the atmosphere, so results in a decrease in relative humidity
A. Incorrect. Absolute humidity is measured in g/m3 and is not affected by temperature.
B. Incorrect. A decrease in temperature leads to formation of dew when saturation point is reached. However, this statement relates dew point to temperature rather than directly describing the change in relative humidity.
C. Incorrect. Condensation occurs when saturation is reached at a relative humidity of 100%. However, this statement does directly link temperature change to relative humidity.
D. Incorrect. This statement relates pressure change to humidity rather than temperature.
E. Correct. This is the best fit because it explains how relative humidity decreases with increasing temperature. Relative humidity is a comparison of the mass of water vapour present with the maximum amount that gas could contain, expressed as a percentage.
Give an overview of solvents and solutes, and the relevance to anaesthetics.
Define the terms solvent, solute and solution
Explain the difference between solutions, colloids and emulsions
Explain the efficacy of water as a solvent
Describe some physical mechanisms whereby solutions are created
A solution is a homogenous mixture of two or more substances
Solubility is a measure of the maximum amount of solute which can dissolve per amount of solvent under specified conditions of temperature and pressure. It is also affected by the polarity of the solvent and the solute: polar solvents will dissolve polar solutes; non-polar solvents will dissolve non-polar solutes
The term mixture describes any two or more substances that are dispersed through each other, but whose constituent parts retain their original identity
A suspension occurs where two or more substances are mixed together but where the heavier substance(s) will eventually settle out
A colloid is a type of mixture where a substance is dispersed evenly through another
An emulsion is a colloidal mixture of two or more immiscible liquids. Emulsions can be stabilized through the use of surfactants
Colligative properties are physical properties of a solution that vary by the number of dissolved particles rather than by the identity of the solute
Covalent bonds hold each hydrogen atom to the oxygen atoms via shared pairs of electrons, giving a non-linear molecular arrangement. This ‘bent’ structure of the water molecule makes it distinctly polar in nature: that is to say, the oxygen part of the molecule has a small negative charge whilst the hydrogen part retains a small net positive charge (Fig 1).
This structure is ideally suited to H-bond formation.
Question: How many H-bonds per molecule can water form?
Water can form four H-bonds per molecule, which is what endows water with its anomalously high boiling point, melting point, surface tension and heat of vaporization.
Fig 2 shows water molecule H bonding with four others.
Solubility describes the ability of a substance to dissolve in a solvent.
Question: What can affect the solubility of a substance?
Solubility can be affected by:
Pressure
Temperature
The nature of the solvent
The nature of the solute
The solubility of one substance in another is determined by the balance of intermolecular forces between solvent and solute. Factors such as temperature and pressure alter this balance and thus change solubility (Fig 1).
Solubility is a measure of the maximum amount of solute that can dissolve per amount of solvent under specified conditions of temperature and pressure.
Concentrations of solutions are commonly expressed in terms of mass of solute per volume of solvent, for example mg per ml.
However, as the volume of solvent and solution varies by temperature, the precise proportion of solute to solvent may change when measuring solubility in this way.
Question: How is it possible to express solubility in a way that is not susceptible to influence by fluctuation in temperature?
A more consistent unit of concentration would be mg per mg, which is a dimensionless ratio of masses. This is why local anaesthetic concentrations are traditionally expressed as percentage ratios.
For example, ‘lidocaine 1%’ means 1 gram in 100 grams of water, but this is usually interpreted as:
1 g in 100 ml, or
1000 mg in 100 ml, or
10 mg per ml
What are the defining characteristics of a suspension?
A suspension occurs where two or more substances are mixed together, but where the heavier substance/s will eventually settle out (Fig 2).
The settling process may take seconds or months, depending on the nature of the substances involved.
What are the defining characteristics of a colloid?
What term is used to describe liquid-in-liquid colloids?
A colloid is a type of mixture where a substance is dispersed evenly through another.
A colloidal system is made up of an:
Internal phase of small particles 1-1000 nm in diameter, i.e. the dispersed substance
External phase within which the internal phase is dispersed
The dispersed substance does not settle out under the influence of gravity.
Solids, liquids and gases may form colloidal systems. For example, smoke is a colloid of solid particles dispersed within gaseous air, and whipped cream consists of gas (air) dispersed within a liquid external phase.
Liquid-in-liquid colloidal systems are called emulsions.
Colloid suspensions are used in medicine as intravenous fluids (Fig 1). The large molecular weight of the suspended internal phase substances provides an increased oncotic pressure within the circulation. This helps to limit the degree of extravasation of administered fluid, and reduces the risk of pulmonary or cerebral oedema.
Consider these specific types:
Gelatin colloids
Starch colloids
Dextran colloids
They are given as intravenous fluids.
An emulsion is a colloidal mixture of two or more immiscible liquids.
Energy input in the form of stirring, shaking or spraying is required to form the initial emulsion, and more stable emulsions remain evenly dispersed for long periods of time.
However, emulsions tend to revert to their component phases. Fig 1a shows the yellow particles becoming emulsified through energy input and then starting to revert.
Question: What can be done to prevent this reversion to the component phases?
The use of emulsifiers stabilizes emulsions. Examples of emulsifiers include surfactants, i.e. surface active substances, such as egg yolk and soya lecithin.
When a surfactant is added to an emulsion, it binds electrostatically to the surface of the particles, enabling each to maintain its position in relation to the molecules in the other substance, again through the effect of electrostatic forces.
Propofol for injection is produced as an emulsion of lipid soluble propofol in a watery base (Fig 1). The stability of the emulsion is aided by egg and soya lecithin emulsifiers.
Regarding solvents, solutes and solutions:
A. A solution is a heterogeneous mixture of two or more substances
B. Alloys are a type of solution
C. Entonox is a solution
D. Warmer water dissolves more salt than cooler water
E. Warmer water dissolves more gas than cooler water
A. False. A solution is a homogenous mixture of two or more substances.
B. True. Though most solutions involve liquids, alloys do classify as a kind of solution.
C. False. Gases mixed together are just mixtures, although gases can dissolve in liquid.
D. True.
E. False. The bubbles seen in IV fluids that have been warmed show the dissolved gases coming out of solution because of the rise in temperature.
Regarding mixtures, colloids, suspensions and emulsions:
A. All colloids are mixtures, but not all mixtures are colloids
B. Where two immiscible liquids do not remain in their desired emulsified state, the only solution is to input energy and agitate the mix
C. Propofol for injection is a solution of propofol in water
D. Medical colloids are effective because the large molecular weight of the substances in them increases the oncotic pressure in the patient’s blood
E. An emulsion is a type of colloid
A. True. The term ‘mixture’ covers colloids, as well as emulsions and suspensions.
B. False. A surfactant emulsifier may in some cases be added to the mix to keep the emulsion stable.
C. False. Propofol is an emulsion of lipid-soluble propofol in a watery base.
D. True. The increased oncotic pressure can help prevent cerebral and pulmonary oedema.
E. True. An emulsion is a liquid-in-liquid colloid.
Regarding water as a solvent:
A. The water molecule is polar in nature
B. Water acts as a good solvent for polar and non-polar substances
C. The vapour pressure of water increases as a solute with a higher mole fraction is dissolved into it
D. The boiling point of water rises as a solute with a higher mole fraction is added to it
E. The solvent in lignocaine is water
A. True. The negatively charged oxygen atom and the positively charged hydrogen atoms give it its polarity.
B. False. Like dissolves like. Thus, water dissolves polar substances well.
C. False. The vapour pressure of water falls if a solute with a higher mole fraction is added to it.
D. True. Consistent with answer C, the boiling point of water rises as solvents are added.
E. True.
Regarding solubility:
A. Solubility is increased if pressure is increased
B. When gases dissolve in liquids, solubility increases if the temperature of the solvent increases
C. When liquids dissolve in liquids, solubility increases if the temperature of the solvent increases
D. Solubility is enhanced by the action of surfactants
E. Solubility is affected by intermolecular forces
A. True. Pressure increases solubility, especially when gases are dissolved in liquid.
B. False. Where gases dissolve in liquids, solubility falls as temperature rises.
C. True. Where liquids, or solids, dissolve in liquids, solubility increases with temperature.
D. False. Surfactants render emulsions stable by coating the particulate matter suspended within them.
E. True. Intermolecular forces cause solutes to disperse evenly throughout a solvent.
Give an overview of osmosis, with the relevance to anaesthetic practice.
Explain the principles of osmosis
Explain the van ‘t Hoff equation, and be able to compare osmotic pressure with the ideal gas laws
Describe colligative properties
Define osmolarity and osmolality
Describe the composition of some common intravenous crystalloid solutions
Osmosis is a physical process whereby a solvent moves by diffusion, across a semi-permeable membrane, from a solution of low concentration to a solution of high concentration
Osmotic pressure is a colligative property, meaning that it depends upon the molar concentration of solute rather than the identity of the solute
van ‘t Hoff showed that where temperature is 0°C and 22.4 litres of solution contains 1 mole of solute, the osmotic pressure is equal to 1 atmosphere; this is analogous to the ideal gas law
Intravenous fluids are manufactured to be approximately isotonic with plasma
Haemodialysis and peritoneal dialysis use osmosis and diffusion to remove toxins from the blood
Mannitol draws excess fluid by osmosis from injured brain cells, thus helping to reduce a raised intracranial pressure
What is it about the van ‘t Hoff equation that shows that osmotic pressure really is a colligative property?
n = iMRT
(n = osmotic pressure, i = factor, M = molarity, R = universal Gas constant, T = absolute temperature in K)
The van ‘t Hoff equation does not refer in any way to specific chemical elements. The nature of the solute has no effect on the osmotic pressure, just the molarity of the solution, M.
Osmotic pressure is a colligative property, meaning that it depends upon the molar concentration of solute rather than the identity of the solute.
Fig 1 shows unequal concentrations of potassium ions in plasma and intracellular fluid:
Plasma with a K+ concentration of 5 mmol/L, and
Intracellular fluid with a K+ concentration of 135 mmol/L
There is an osmotic gradient of potassium ions across the cellular membrane between the plasma and intracellular fluid compartments of about 4% or 1:25. This gradient determines the rate at which the fluid on both sides of the membrane becomes isotonic.
Question: In a living person this could never happen, so what is preventing the osmotic process from leading to isotonicity?
A rise of plasma potassium to 8 or 9 mmol/L would lead to cardiac arrest, so the isotonic state of 70 mmol/L could never occur.
In a living person, active pumps maintain the gradient of concentrations despite the potential effects of osmosis. Only after cell death and the loss of the ionic pumps could the solutions become equal in concentration.
An osmole is a unit of measurement that defines the number of moles of a chemical compound that contribute to a solution’s osmotic pressure. In calculating the number of osmoles present in a solution, salts are presumed to have separated into their component ions. For example, a mole of NaCl in solution gives two osmoles, i.e. one mole of Na+ and one mole of Cl-, while a mole of glucose dissolves to give one osmole.
Osmolarity Osmoles Per litre of solution
Osmolality Osmoles Per kilogram of solvent
Question: Why would one term be preferred over the other?
Osmolality is the preferred term because volume of solution alters with temperature. This means that a solution changes osmolarity, but not osmolality, as temperature varies.
Regarding the van ‘t Hoff equation, where π = osmotic pressure:
A. π = iMR/T
B. π = iMT/R
C. The equation takes no account of the nature of the solutes within the solution
D. π = iTR/M
E. 22.4 litres of a 1 molar solution at 0°C exerts 1 atmosphere of osmotic pressure
A. Incorrect. The elements in the van ‘t Hoff equation multiply to produce a figure for π (pressure).
B. Incorrect. See A.
C. Correct.
D. Incorrect. See A.
E. Incorrect. van ‘t Hoff explains that where:
the temperature is 0°C (273.15 K), and
22.4 litres of solution contains 1 mole of solute
then the osmotic pressure is 101.325 kPa, equivalent to 1 atmosphere (1 bar).
A 1 molar solution is not the same as a solution containing 1 mole of solute. 22.4 litres of a 1 molar solution would exert 22.4 atmospheres of osmotic pressure.
Plasma osmolality is a measure of the concentration of electrolytes and other solutes within the plasma, and is calculated in milliosmoles per kilogram of plasma solvent. Normal plasma osmolality is 280-303 milliosmoles per kilogram.
As cell membranes are generally freely permeable to water, changes in extracellular osmolality lead to changes in intracellular osmolality.
This is important, as alterations in intracellular osmolality affect cellular functioning and volume.
Question: Why is this important?
Alterations in intracellular osmolality affect cellular functioning and volume.
Therefore, extracellular osmolality is relevant in the manufacture of intravenous fluids.
Regarding osmolarity and osmolality:
A. Both osmolarity and osmolality describe the number of osmoles in a solution
B. Osmolarity is the preferred measure of osmotic activity as its calibration to the kilogram enables quick and accurate calculations
C. Two moles of glucose gives as many osmoles as two moles of NaCl
D. Salts in solution are assumed to have separated out into their component ions
E. Isotonic means exactly the same as isosmolal
A. True. Osmolarity, spelt with an r, is a measure of the number of osmoles per litre of solution, and osmolality, spelt with an l, is the number of osmoles per kilogram of solution.
B. False. Osmolality is the preferred term because the volume of solution alters with temperature. This means that a solution changes osmolarity, but not osmolality, as temperature varies.
C. False. Every mole of NaCl gives two osmoles, twice as many as a mole of glucose.
D. True.
E. False. Isotonic as a term places solutions with the same osmotic pressure in the context of a specific semi-permeable membrane, whereas isosmolal merely describes two or more solutions with the same osmotic pressure.
Regarding osmosis:
A. Osmosis is a process in which a solvent moves across a semi-permeable membrane from a hypertonic solution to a hypotonic solution
B. Erythrocytes exposed to a hypotonic solution take up water and swell
C. A semi-permeable membrane is defined as a membrane permeable to solute but not to solvent
D. Erythrocytes exposed to an isotonic environment neither take up nor lose water
E. Slugs thrive on a diet of table salt
Submit
A. False. Osmosis is the process in which a solvent moves across a semi-permeable membrane from a hypotonic solution to a hypertonic solution.
B. True.
C. False. A semi-permeable membrane is permeable to solvent but not to solute.
D. False. Erythrocytes in an isotonic environment both take up and lose water in equal measure. The net water content of the erythrocyte remains constant.
E. False. Slugs shrivel and die as their water content is lost to the powerfully hypertonic salt environment.
Regarding colligative properties:
A. Colligative properties include freezing point elevation and boiling point depression
B. Osmotic pressure is a colligative property
C. Colligative properties are employed in an osmometer
D. Colligative properties are influenced by the nature of the solute as well as the number of osmotically-charged particles
E. Dissolved molecules alter the physical properties of solutions
A. False. Freezing points are lowered and boiling points raised due to the colligative properties of solutes.
B. True.
C. True. An osmometer may measure the osmotic pressure of a solution indirectly by measuring the temperature at which it freezes.
D. False. Colligative properties are completely independent of the nature of the solute in a solution.
E. True. The osmotic activity of dissolved molecules alters the physical property of a solution.
Complete the table below so that it represents accurately the osmolality (in mmol/L) of some of the fluids that are commonly given intravenously.
Normal saline = 308 mmol/L
5 % glucose = 278 mmol/L
Plasma = 287.7 mmol/L
Intravenous fluids are usually manufactured to be practically isosmolal with plasma. Hypotonic fluids would cause cellular swelling and lysis, whereas hypertonic fluids would lead to cellular shrinkage.
Give an overview of heat and temperature.
Define common terms used in the physics of heat, energy and temperature
Explain why maintenance of normal body temperature is important
Name and describe the methods of heat transfer
Human physiology, biochemistry and pharmacology all rely on temperature-dependent enzymatic reactions
In the human body the optimal temperature of enzymatic reaction coincides with the normal human body temperature, around 37°C
Although an operating theatre has higher total heat energy than a human body, the human body transfers heat to the operating theatre because it is at a higher temperature than the theatre
Heat transfer is the passage of energy from a hot body to a cold body. The main methods of heat transfer are conduction, convection, radiation and evaporation
Radiation accounts for 40% of the body’s heat loss under anaesthesia, convection 30%, evaporation 20%. Losses through the respiratory tract account for the final 10%
The concept of specific heat capacity can be used to calculate the anticipated temperature drop of the body associated with the use of intravenous fluids
At what temperature do you think human enzyme systems function most efficiently?
In the human body the optimal temperature of enzymatic reaction coincides with the normal human body temperature, in other words around 37°C. An exception would be the enzymes in the testes which function optimally at a slightly lower temperature than normal body temperature.
Above or below this body temperature, enzymatic activity decreases markedly. As a consequence, body systems become less efficient. This is why maintaining body temperature is so important.
The transfer of heat in the operating theatre can have a significant impact on body temperature. The anaesthetist must be aware of the mechanisms involved and how these may be counteracted.
The relationship between heat energy and temperature can be illustrated by considering an iceberg and a pan of boiling water.
Question: Which has the greater heat energy, the iceberg or the boiling water?
An iceberg contains much more total heat energy than the pan of boiling water by virtue of its much greater mass. However, the boiling water is clearly at the higher temperature.
It is also the case that, despite its far lower heat energy, the boiling water would transfer heat to the iceberg causing it to begin to melt.
In other words, temperature is a measure of the likelihood that one substance gives heat to another, with heat being transferred from the substance with the higher temperature.
In more practical terms, this means that although an operating theatre has higher total heat energy than a human body, the human body transfers heat to the operating theatre because it is usually at a higher temperature than the theatre.
Take an average 70 kg patient and consider what happens when we transfuse 1 L of cold blood, i.e. approximately three units of packed red cells.
Assume the patient is initially normothermic at 37°C and the blood is at a temperature of 4°C.
The specific heat capacity of the patient is approximately 3.5 kJ/kg/°C and that of blood is 3.6 kJ/kg/°C.
The density of the blood is approximately 1.125 kg/L, therefore 1 L of blood weighs 1.125 kg.
When this blood is transfused into the patient, heat is transferred from the patient to the blood. The patient’s temperature therefore falls, whilst the blood is simultaneously warmed to the same, now somewhat lower body temperature.
Question: How would you use the information provided to calculate the body temperature after the transfusion?
Although the total heat of an object cannot be measured, the amount of heat that is transferred to an object of a certain mass in order to warm it by a specified amount can be calculated as follows:
The amount of heat transferred =
mass × change in temperature × specific heat capacity
Thus, the amount of heat lost by the patient:
= 70 kg × (37oC - T) × 3.5 kJ/kg/°C
Where T is the new body temperature. The amount of energy gained by the blood is:
= 1.125 kg × (T – 4oC) × 3.6 kJ/kg/°C
These two values must be equal, as the amount of heat energy transferred from the patient is equal to that taken up by the blood.
The equation can then be rearranged to calculate T, the new body temperature.
This is found to be 36.5°C. In other words, the transfusion of 1 L of unwarmed blood to this patient has resulted in a fall in body temperature of 0.5°C.
Heat transfer is the passage of energy from a warmer body to a cooler body.
It is important to understand how heat transfer can cause heat loss from a patient under anaesthesia.
The three main methods of heat transfer are:
Conduction
Convection
Radiation
Question: In what other way can heat loss occur in theatre?
Heat loss also occurs by evaporation, which is in fact heat loss by a combination of conduction, convection and radiation.
In order for a substance to change state, either from a solid to a liquid or from a liquid to a vapour, heat must be supplied.
The heat energy required is known as the latent heat of fusion and the latent heat of vaporization respectively.
The specific latent heat of vaporization is defined as the heat required to convert 1 kg of a substance from a liquid to a vapour at a given temperature. The unit of this quantity is J/kg.
For more information about latent heat, see session Gases and vapours (001-0829).
Evaporation causes heat loss from the body by taking the latent heat of vaporization needed to vaporize water or another liquid from the surrounding tissue.
Question: Why may a patient lose heat through evaporation in theatre?
Evaporative heat loss may occur when:
Sweat or antiseptic solution evaporates from the skin
Fluid evaporates from exposed moist internal body cavities
Evaporation is a mode of heat transfer that may usually be described in terms of the other three principal modes. For example, heat from the body may be transferred to sweat on the body surface by conduction, causing it to evaporate. This heat is then carried away by convection.
Evaporation accounts for about 20% of heat loss in theatre.
If two substances are at differing temperatures they have different average molecular kinetic energy. When these two substances are in contact with one another, heat transfers between them due to collision of the molecules of the two substances. (Fig 1).
When the molecules of the substance with the higher temperature collide with those of the substance with the lower temperature, its molecules’ average kinetic energy is reduced. This kinetic energy is transferred to the molecules of the substance with the lower temperature whose temperature, and heat energy, then increase.
Question: Is conduction an important method of heat loss in the operating theatre?
Conduction is not an important method of heat loss in the operating theatre because:
The air around the patient is a poor heat conductor
The table is well insulated
The area of contact between the patient and the table is relatively small
Regarding heat loss:
A. Conduction is the largest factor in patient heat loss
B. Convection is due to heating of the adjacent air layer, which is replaced by cooler air from the surroundings
C. Radiation accounts for about 20% of the body’s total heat loss in theatre
D. Heat lost in breathing dry gases is approximately 10% of total heat loss in the anaesthetized patient
E. The Stefan-Boltzmann law relates the radiated heat loss from an object to its temperature
A. False. Patient heat loss is mainly due to radiation (40%) and, to a lesser extent, convection and evaporation.
B. True.
C. False. Losses by radiation typically account for at least 40% of the total heat loss from the body in theatre.
D. True. About 10% of heat loss in the anaesthetized patient is via the respiratory tract. The main part of this is due to the latent heat needed to vaporize water to humidify the gas within the trachea.
E. True.
Radiation differs from conduction and convection in that it does not require matter to transfer heat. It is the means by which the sun heats the earth.
All objects with a temperature above 0 K emit radiation and the overall amount of radiation emitted compared to that absorbed is a function of the temperature of the object.
Radiation is emitted as electromagnetic waves. Those waves falling within the visual spectrum are seen as light, while those waves falling in the infrared spectrum are felt as heat (Fig 1).
The Stefan-Boltzmann law relates the total amount of radiation emitted by an object to its temperature.
E = sT^4
E = the total amount of radiation emitted per m2 of an object
s = a constant
T = the temperature in Kelvin of the object
Radiation accounts for about 40% of the body’s heat loss in theatre.
In summary, the contribution to heat loss under anaesthesia is as follows:
Radiation
40%
Convection
30%
Evaporation
20%
All of these modes of heat loss are exacerbated by increased exposure of the patient.
The final 10% of heat loss may be accounted for by losses through the respiratory tract.
This is due to humidification (8%) and warming (2%) of inspired gases and occurs via a combination of the other modes of heat transfer.
Regarding heat and temperature:
A. Heat is a form of energy
B. When measured in Kelvin, temperature is inversely proportional to the average kinetic energy of the molecules of a substance
C. Heat is always transferred from objects with a higher total heat energy to those with a lower total heat energy
D. Heat energy of a substance depends upon the mass of the substance
E. In the human body, enzymatic activity increases in proportion to increasing temperature
A. True.
B. False. Temperature in Kelvin is directly proportional to the average kinetic energy of the molecules of a substance.
C. False. Heat is transferred from an object with a higher temperature to an object with a lower temperature, regardless of the overall heat energy possessed by each object.
D. True. Heat energy is a measure of the total energy of molecular motion in a substance and, therefore, is dependent on the mass of the substance.
E. False. Enzymatic activity increases with increasing temperature but when the temperature increases above an optimum temperature, enzymatic activity falls.
Regarding heat and heat capacity:
A. Specific heat capacity is defined as the amount of heat energy required to raise the temperature of a given object by 1 K
B. The units of specific heat capacity are kJ/kg/K
C. The specific heat capacity governs the change in energy when a substance changes state from liquid to gas
D. During a transfusion of blood at room temperature, the amount of heat energy taken up by the blood is equal to that lost by the body
E. A change in temperature of 1°C is equal to a change in temperature of 1 K
A. False. This is the definition of heat capacity. Specific heat capacity is mass dependent.
B. True.
C. False. This is the latent heat of vaporization.
D. True. The overall energy must remain the same. Heat is transferred from the object with the higher temperature, i.e. the body, to that with the lower temperature, i.e. the blood.
E. True. However, temperature in Kelvin is 273 degrees higher than temperature measured in Celsius.
Give an overview of heat transfer, and the relevance to anaesthetics.
Describe the factors which affect heat loss in a theatre environment
List and describe methods available for reducing heat loss in theatre
There are four main factors that determine the degree of heat loss in theatre:
Environmental
Anaesthetic
Surgical
The patient
The fall in the body’s temperature under anaesthesia tends to follow a characteristic pattern, with an initial sharp decline, followed by a slower fall
Reduction in body temperature in theatre can be minimized by reducing heat loss and by active warming of the patient
How might general anaesthesia reduce heat production by the body?
General anaesthesia might reduce heat production in several ways.
Heat losses can become very significant during prolonged anaesthesia. The problem is further compounded by the administration of cold fluids and blood products. Most intravenous fluids are kept at room temperature and blood is stored at 4°C.
Regarding heat loss from the body:
A. On induction of anaesthesia, the patient undergoes a gradual drop in body temperature
B. The anaesthetized patient has impaired vasoconstriction and piloerection in response to hypothermia
C. Operating theatre humidity should be maintained at no more than 50%
D. Warmer ambient temperatures are often maintained in paediatric theatres
E. Heat loss through the respiratory tract is insignificant
A. False. Vasodilatation redistributes blood from the central compartment of the body to the peripheral compartment. This causes a rapid reduction in central temperature on induction.
B. True.
C. False. Theatre humidity should be at least 50% to reduce evaporative losses from the patient. Previously, when flammable anaesthetic vapours were used, static build-up associated with low humidity could lead to explosions.
D. True. Paediatric patients have an increased body surface area to weight ratio and, therefore, lose heat more quickly.
E. False. Heat losses through the respiratory tract can be significant, because the heat and moisture exchange devices used for the humidification and warming of inspired gases are only about 70% efficient.
A number of factors acting together cause the patient’s temperature to fall while under anaesthesia and undergoing surgery.
This fall in temperature tends to follow a characteristic pattern, with an initial sharp decline, followed by a slower fall (Fig 1a).
The initial sudden drop in temperature (Fig 1b) occurs on induction of anaesthesia and is due to vasodilatation. This results in the distribution of blood away from the central compartment to the peripheral compartment.
There is then a gradual, linear loss of heat (Fig 1c) until a temperature is reached which is the new set point for thermoregulation in the anaesthetized patient.
Thereafter, the temperature reaches equilibrium with the theatre environment (Fig 1d), when the heat generated by the patient’s metabolism equals that lost to the environment.
Blaise Pascal observed liquid being pushed out of a sphere with multiple holes in it.
Question: Look at Fig 1 and decide the angle at which the jet of fluid escapes from the sphere once it is punctured at point A.
Note that the piston (grey) is applying constant force.
What does this mean for subsequent perforations at points B, C and D?
Jet A escapes at an angle perpendicular to the surface of the sphere (Fig 2).
Jet A escapes at an angle perpendicular to the surface of the sphere (Fig 2).
Jets here also escape at angles perpendicular to the surface of the sphere. The angle of escape bears no relation to the direction in which the pressure is applied
Because pressure is the force applied over an area, the pressure of a jet of water escaping from a hole in a container depends on:
The force being applied
The area of the hole in the container
Question: Assuming that 10 N of force is being applied (Fig 1), calculate the pressure at which a jet of water escapes from a hole with a surface area of 0.05 m2.
What would be the pressure of a jet that escaped through a hole of precisely twice the area?
What happens to the force exerted in the case of multiple perforations?
The pressure of the escaping jet would be halved to 100 Pa (Fig 3), because the force being exerted at the site of the hole is still 10 N, but the area is doubled.
Multiple perforations in a container do not alter the fact that at each point within the fluid a force of 10 N exists. For each perforation the pressure reduces as the total area of the perforations increases but the force remains constant, provided the pressure and supply of water is maintained
Fig 1 shows a diagram of a liquid in a closed container. Think about the topmost layer of molecules touching the top wall.
Question: In which direction is the force in this top layer exerted? Up or down? Or both equally?
The top layer of molecules must be pushing upwards (Fig 2).
The fluid, i.e. in this case the liquid, is in constant motion with its constituent molecules constantly colliding with container wall, thus exerting a pressure. See the reference to session Kinetic Theory of Gases in Related Sessions in Resources.
This means it is possible to imagine the container of fluid as being made up of multiple layers of molecules. Each layer exerts the same pressure for a given area and is balanced by an inward pressure arising from the vessel wall. A state of compression then arises when the two are in equilibrium.
Given that the topmost layer of fluid molecules in a closed container must be pushing upwards, think now about the layer of molecules immediately below the top layer (Fig 1).
Question: When compared with the upwards force exerted by the top layer, does the layer below exert a force on the topmost layer that is:
Greater?
Lesser?
The same?
The same pressure is exerted throughout the liquid (Fig 2).
For any layer of molecules, pressure equals force divided by area:
P = F/A
The final question relating to the forces at work within this situation focuses on the container.
Question: How would you describe the force that the container exerts on the topmost layer of molecules? The possibilities, of which only one is correct, are shown as blue arrows in the circles opposite (Fig 1).
And how does this force affect the subsequent layers of molecules?
Newton’s third law states that there must be an equal and opposite force exerted by the container on the topmost layer (Fig 2).
As the wall of the container is at rest, i.e. it is not a moving piston, then no net force can be acting and hence the reaction is equal and opposite.
This force is transmitted through all the layers, so that they transmit the reaction of the wall within the liquid.
This is the formula that describes the pressure at point a in a sealed container (Fig 1). It includes only the effects of gravity:
Pa = ρgh
In reality, however, pressures on the surface of the liquid in the container must also be considered.
Question: What is the formula to describe pressure at point a if a piston applies pressure b (Pb) to the top of the liquid (Fig 2)?
The pressure at point a is now expressed thus:
Pa = Pb + ρgh
What formula describes the pressure at point a if the piston is removed and atmospheric pressure (Patm) applies
The pressure at point a is expressed thus:
Pa = Patm + ρgh
So, the general formula is:
Pa = Pb + ρgh
bearing in mind that Pb may be atmospheric pressure.
The following principle applies to every vertical container with a flat shallow container at its base. ‘Shallow’ here should be taken specifically to mean having negligible height, and consequently no hydrostatic column effects.
Question: Look at points x, y and z as indicated in Fig 1 and decide at which point the pressure is greatest and at which point the pressure is least.
The pressure at all three points is equal.
This is Pascal’s principle: namely, that the change in pressure of an enclosed incompressible fluid is conveyed undiminished to every part of the fluid and to the surfaces of its container.
Fig 1a shows a column of liquid in an angled tube. Notice that there is a right-angled triangle with sides l (length), h (height) and b (base) created.
Question: Can you think of the possible direction of the forces in the liquid?
The purple arrows in Fig 1b indicate the direction of the forces in the liquid in the angled tube.
The pressure at point y (Py) is less where the column is angled than where the column is vertical.
With the tube in its fully vertical position, the pressure at point y could have been expressed as P = F/A. The angling of the tube reduces F, thereby reducing P (Fig 1). How does this occur?
Question: Thinking about what force actually consists of, why must the pressure be less at the base of the angled tube?
Pressure = force/area. Looking particularly at the force:
Force = density x acceleration due to gravity x height x area.
Everything remains the same apart from the vertical height, which reduces.
Therefore, the equations that you saw earlier:
F = ρghA
and
P = ρgh
always apply in angled columns of liquid, with P and F reducing as h reduces.
Consider the diagram opposite:
A. The pressure at point P1 = ρhgA
B. In a column of twice the height (2h) the pressure at point P1 = ρ2hg/A
C. The pressure at point P2= ρ(h/2)g
D. A column of twice the area (2A) would generate a pressure at point P1 of twice that generated by the column with area A
E. If the column were angled at 30° to the vertical the pressure at point P1 would remain the same
A. False. Pressure (P) = ρgh. The area of the column is not relevant to this calculation.
B. False. As P = ρgh, doubling the height makes the pressure greater, but the area of the column remains irrelevant.
C. True. P at point P2 = ρ(h/2)g.
D. False. P = ρgh. The height of the column, not its area, affects the pressure.
E. False. P1 is reduced. Measurement of the new vertical height would be required to perform the calculation and, as this would be less than h, P1 would be reduced.
Finding the Cardiac Output, Q
Upon what assumptions does this method of determining Q. rely and how plausible are these assumptions?
The method assumes that we can use the geometry of a box shape to work out the volume of the fluxoid - which we have measured independently by O2. However, the method will not deliver accurate results if, during the measurement period:
Q. varies, as in, for example, unstable cardiac output (Fig 1)
or CvO2 or CaO2 as in, for example, developing hypoxia, is not constant (Fig 2)
It is reasonable to expect that, in the few minutes the measurements take, the healthy subject at rest would have a fairly constant physiological state. In a sick patient, this is evidently not necessarily a secure assumption.
Without much effort, we can - by making appropriate adjustments - use the fluxoid representations to describe the calculations required to determine by the use of CO2 output. The determination of CO2 is quite straightforward by the use of expired gas collection - Douglas bags and the like. The same invasive lines are required.
This seems an equally promising approach at first sight.
Question: Why, even at a time when the O2 uptake method might have been used, did the equivalent method for CO2 remain unexploited? Why is it different?
We have spent quite a while examining how PaCO2 can be changed by altering the level of alveolar ventilation. This is the key to the problem here. While we may for a short period easily - by hyperventilation - exhale more CO2 than we are producing, we cannot change the value of O2 in the same way; it is determined by the metabolic rate.
Haemoglobin is normally working on the flat part of its sigmoid curve and even elevating PAO2 makes negligible difference to uptake. Uptake below what is required, rapidly manifests itself as hypoxia.
In other words, for PaCO2, the fluxoid ‘bricks’ may too easily not be brick-shaped at all and the method is too unreliable to use.
We can now proceed to derive the AGE. The details of the derivation may look a bit complicated because of the respiratory symbols involved. There is no need to remember the derivation itself. The exercise is to show that the AGE arises from a physiological model.
Understanding the origin of the AGE enables us to make a more intelligent interpretation of its predictions than if we just study the equation itself in isolation.
Since we have steady-state conditions, we can write an expression of the equality of inputs and outputs.
Task: Write down the steady-state input-output expression for O2 in the alveolar space.
The input-output relations in the case of O2 are a little more complicated than those for CO2 - but not much.
Task: Draw an input-output diagram, like the one for CO2, for the alveolar space and work out the equivalent input and output for O2.
We are only considering steady-states at this time:
In the absence of re-breathing we have a single input:
FIO2 . A (where FIO2 is the inspired O2 fraction) - the product of a flow and a concentration.
There are two outputs in this case compared with the single one for CO2:
The O2 uptake: O2
O2 in the expired gas: FAO2 . A
We already know that:
V.CO2 = V.A x FaCO2
Question: What is the relationship between V.O2 and CO2?
Use this to derive the alveolar gas equation.
The respiratory quotient, R, is the ratio of CO2 production to O2 consumption:
R = V.CO2 / V.O2
Since we now have expressions for both V.O2 and V.CO2 and know a relationship between them, we can derive the AGE.
The alveolar gas equation can now be completed:
As we have done before with equations containing a number of variables, we will examine the different solutions that arise if we fix certain variables. Firstly, allow PIO2 to vary but fix PaCO2 and R.
Conditions:
PIO2 variable
PaCO2 constant = 5 kPa
R constant = 0.8
Allow PIO2 to vary. Fix PaCO2 and R and we have an equation of the form:
y = mx + c
where x = PIO2, m = 1 and c = -[PaCO2]/R (Fig 1)
Question: Given this form, what features do we expect this graph to show?
The graph is linear with a gradient of unity and an intercept on the vertical axis at -6.25 kPa.
Stated in its most basic way, this graph answers the question:
What is the resulting PAO2 if we vary the partial pressure of oxygen breathed in, PIO2?
The answer is that PAO2 equals whatever the inspired pressure is minus 6.25 kPa.
If we increase the PIO2 by say 20 kPa, then PAO2 rises by the same amount.
We must however interpret the graph sensibly. Somewhere in the region PIO2 <8 kPa or thereabouts, the predicted relationship breaks down as we violate the steady-state condition upon which we constructed the model (Fig 1). We do not know exactly where this happens from our model.
We can test our model further by making PaCO2 the variable. Other terms remain constant.
Conditions:
PaCO2 is the variable
PIO2 is constant, say 20 kPa
R is constant; R = 0.8
Allow PaCO2 to vary. Fix PIO2 and R, we have an equation of the form:
y = mx +c
where x = PaCO2, m = -1/R and c = 20 kPa
This is another linear function.
Task:
Sketch the graph of this function and note the important features.
The graph is linear with a negative gradient of 1.25. The intercept on the vertical axis is at 20 kPa. Severe hypoventilation when breathing air leads to hypoxia. We know this but the graph shows us a quantitative prediction.
What are the two main physiological and clinical implications of this non-linear relationship between alveolar ventilation versus PAO2?
The effect of increasing A on the resulting PAO2 is rather small. Doubling the V.A from 4 to 8 L/min only increases PAO2 by 2.5 kPa. This is a feeble contribution compared with the ease of giving added oxygen - if you have it to hand, of course
The effect of hypoventilation when breathing air is a rapid decline in PAO2 as the steep part of the curve is encountered. This emphasizes the importance of oxygen therapy in the post-anaesthesia recovery room where hypoventilation (opioids, residual muscle relaxants) may easily occur
he model we have produced started simply and built upon what we have already learnt from the model of PaCO2 versus A in the prerequisite session Mathematical modelling and Input-Output Principle (IOP) (001-0839). There is a small and quite subtle flaw in it however. A clue lies in the graph (Fig 1).
Question: Can you spot the deliberate mistake - or rather the as yet unacknowledged approximation?
The input-output relations we used took no account of the fact that if the value of R is less than 1, as it typically is, then we extract more gas (O2) from the alveolar space than we return to it as CO2. Since the functional residual capacity (FRC) is a fixed volume, there must be a small amount, 50 mL at rest, of additional inflow. This is enabled by a small difference in pressure causing a passive, i.e. no ventilatory effort is required, influx of gas.
The effect of this extra passive influx is to cause a small increase in the value of PAO2. There is a more detailed version of AGE that has an additional term for predicting the amount by which PaO2 is increased by this process.
In practice, this makes so little difference that it can be ignored. All we need to note is the phenomenon that causes it and the fact that if the value of R = 1, the extra term disappears.
We have sidelined R for a while by setting its value to 1. Having obtained a good idea of how PAO2 varies with A we now reinstate R in its more typical value of 0.8.
Question: How do you expect division by a number less than 1 to affect the second term?
Division by a fraction is equivalent to multiplication by a number greater than 1: it amplifies the numerator. Writing 0.8 as a fraction = 4/5 and division by 4/5 is the same as multiplication by
5/4 = 1.25. This therefore increases the term we subtract from the constant PIO2 - it reduces the PAO2 further. The shape of the relationship is unchanged:
When R = 1
When R = 0.9
When R = 0.8
When R = 0.7
How does PACO2 relate to PaCO2 in apnoea?
There is no reason to change our earlier model assumption that PACO2 = PaCO2 in apnoea.
We are no longer in steady-state because the PaCO2 is rising, therefore less than the 200 ml of production must be crossing the alveolar membrane into the alveolar space. In fact, there is even more reason to treat it as negligible because the quantity of CO2 passing across the alveolar membrane is much less than the 200 ml of our normal, ventilated lung model. We are about to work out how much less this is.
Consider the following:
Do we expect the relationship to be linear?
Why is the relationship linear?
What does this imply?
What determines the gradient?
We have quoted the rate of rise of PaCO2 as typically 0.5 kPa/min. This is as a result of measuring it in real patients and discovering that it is plausibly linear. There are reasons when we might expect it to be non-linear; the graph relating blood PCO2 and CC’O2 - the CO2 dissociation curve - is not linear throughout after all.
We now consider a specific situation: we preoxygenate a normal, fit adult to FAO2 = 0.8, i.e. an alveolar concentration of approximately 80% before the onset of apnoea.
Question: What is the rate of rise of PaCO2 in apnoea?
Experiment shows it to be 0.3-0.7 kPa/min. We will take a middle value of 0.5 kPa/min (Fig 1).
What is the tolerable duration of apnoea in this circumstance, ignoring the contribution of tissue N2 - until PAO2 falls below 10 kPa? Remember we have preoxygenated to FAO2 = 0.8.
What will limit this procedure in fact?
If we have preoxygenated to a PAO2 of 80 kPa, and this falls at a rate of 0.5 kPa/min as we have calculated, we have about 140 min until the PAO2 is predicted to fall below 10 kPa (Fig 1). Before this can happen, the PaCO2 will rise to the point where the toxic effects and acidosis will cause fatal cardiac arrhythmias. This technique is limited by hypercapnia and not hypoxia provided we have furnished three things:
Adequate preoxygenation
An open airway
100% O2 in the airway and a fresh O2 flow of at least the amount being passively ventilated in (190 ml) by AMTO
Without doing any calculation, how much more rapidly do you think PAO2 would fall if there were only 50% O2 in the airway and all other factors were the same?
It is tempting to think that it would fall twice as fast since there is half as much available.
But if we have 50% O2 in the airway, then the volume discrepancy in O2 uptake and CO2 delivery is still 190 ml and this volume will flow into the alveolar space as before, but will only bring 95 ml O2 with it instead of the 190 ml before.
Instead of a mere 10 ml being missing, there is 105 ml (95 + 10 ml) more O2 being taken up than replaced. This means that the PAO2 will fall more than 10 times as fast as with 100% in the airway: over 5 kPa/min (Fig 1).
Work out approximately how rapidly the threshold PAO2 value of 10 kPa will be reached.
PAO2 is about 15 kPa maximum. PIO2 = 20 kPa. O2 is still 200 ml/min.
In the absence of preoxygenation, the O2 content of the FRC is 300 ml. 200 ml in 2 L FRC is PAO2 = 10 kPa. Consumption of 100 ml will take 30 seconds, and therefore in the absence of AMTO, e.g. when the airway is obstructed, this is how long it would take to reach our threshold.
Including AMTO, 190 ml of gas flowing into the alveolar space brings just 38 ml O2 with it, each minute.
The discrepancy between O2 input and output in the alveolar space is 162 ml negative each minute. It will take 100/162 = 0.62 min (approximately 35 seconds) to reach the threshold value of PAO2 = 10 kPa; 5 seconds or so extra. This is somewhat less than 140 min.
The importance of the open airway in the effectiveness of AMTO is obvious and should require no further emphasis.
Table 1 summarizes the factors involved in time to hypoxic threshold dependent upon starting value and rate of fall of PaO2.
There is a further twist to the tale of AMTO. Take a minute to recall the main features of the CO2 dissociation curve.
Question: What happens to the blood CO2 content in apnoea as it passes through the lung?
What happens to the CO2 attached to haemoglobin as it passes through the lung assuming adequate preoxygenation?
We have 10 ml CO2 passing out of the entire cardiac output – for example 5 L each minute - and crossing the alveolar membrane. This means 2 ml from each litre and hence 0.2 ml from each 100 ml - the conventional unit of volume in the CO2 dissociation curve diagram - instead of the typical 4 ml. This is 5% of the normal drop as we would expect since 10 ml is 5% of the 200 ml that crosses during normal ventilation.
Write down a formula for the proportional dilution after 30 seconds using n for the number of breaths. What have you noticed about this proportion as the number of breaths increases?
Remember that the total V.A remains the same whatever the number of breaths.
The proportional dilution is given by (n/(n+1))n ; 1000 breaths would be (1000/1001)1000 = 0.3681. The fraction falls as n increases but does so less steeply at higher respiratory rates than at low ones. 1 breath gives 0.5, and 16 breaths gives 0.3791. But 1000 breaths only reduces it to 0.3681.
Fig 1 demonstrates that as n (the number of breaths) increases, the fraction falls, but this fall flattens out quite quickly. The blue curve represents continuous, non-tidal flow.
We assume for convenience the starting values shown in Table 1 of breathing air before preoxygenation starts. FRC = 2 L. We are going to ignore any effects attributable to nitrogen wash-out from tissue stores. We also ignore water vapour. We will examine later what happens when we include these.
Question: Assuming perfect and complete preoxygenation what would the final values be?
We can calculate the approximate values for FAN2 and FAO2 after 60, 90 and 120 s of perfect preoxygenation (Table 1).
Question: How did we calculate these figures?
Firstly we know that the FAN2 will fall to about 37% of its starting value after 30 seconds. This in turn forms the starting value for the next 30 second period. After a further 30 seconds, i.e. a total of 1 minute of preoxygenation, the value will have fallen to 37% of 37% of the initial value.
This gives us values for FAN2 and we know that whatever the fall in FAN2 during any period, this will be added to the starting value of FAO2 for that period to give us the interim FAO2.
Figs 1 and 2 show the wash-in and wash-out curves.
A. The wash-in (O2) and wash-out (N2) processes take place at different rates
B. The curves are symmetrical about a horizontal line
C. N2 falls to half its starting value, i.e. to 0.4, at about 70% of the time constant
D. The proportional change in each curve is the same at equal times
E. The curves show an exact description of what happens in clinical preoxygenation
A. False. The wash-in and wash-out processes take place at the same rate because of molecule-for-molecule exchange.
B. False. The curves are not symmetrical because the O2 curve starts at a value of 0.15.
C. True.
D. True. They are taking place at the same rate.
E. False. Evidently untrue. It is an approximate model.
Fig 1 shows the structure of a covalent bond (CH4) visually contrasted with that of an ionic bond (NaCl).
Question: Which of the two bonding methods will produce the stronger bond - covalent or ionic?
Covalent bonds form the stronger and less reversible bond, as they share a pair of electrons. The structure of quartz and diamond consists of atoms held together by covalent bonds (Fig 2). These materials are known for their toughness.
The water-soluble salts created when ionic bonding takes place are, by contrast, capable of dissociating into ions.
o far the focus of the session has been on single bonds. That is to say, situations where two atoms share one pair of electrons, e.g. in an H2 molecule (Fig 1).
Question: What would be the effect on the strength and shape of the molecule if the atoms involved were to share two pairs of electrons?
Double-bonded molecules are stronger than single-bonded ones (and triple-bonded molecules even more so), as you might expect. As regards the shape of the molecules, single-bonded molecules can rotate easily, whereas double- and triple-bonded molecules become more rigid as the number of shared electron pairs increases. Free rotation becomes impossible and isomerism results. Fig 2 illustrates these bonds.
Because covalent bonds are so stable and strong, they are not really involved much with drug interactions.
Consider the paracetamol molecule, however, and you will notice the covalent bonds (Fig 1).
Question: Consider the process by which this molecule is eliminated from the body. What is it about the process that makes it (potentially) dangerous?
Covalent bonds, including those in paracetamol, can be broken down with the help of enzymes, such as enzymic action during drug elimination in the liver. This can lead to intermediate molecules that are more reactive, such as N-acetyl-p-benzo-quinone imine (NAPQI) (Fig 2). This allows them to form new bonds at body temperature to small molecules that allow water solubility.
The intermediates can cause damage under certain circumstances, such as a paracetamol overdose.
For hydrogen bonding to take place there is usually a hydrogen atom (positively charged) covalently bound to another (negatively charged) atom.
This association of positive to negative within the molecule creates a strong dipolar attractive force. The H2O molecule illustrates this point (Fig 1a).
Question: Think about the orientation and alignment of a second water molecule in close proximity to the one shown opposite. Which way round will it present itself, given that it too will have the same positive and negative charges?
One of the hydrogen atoms will be attracted to the negative oxygen atom resulting in the alignment shown (Fig 1b). This is the ‘hydrogen bond’ and is the reason why water sticks together.
Hydrogen bonds are also found in other molecules, including DNA.
Hydrophobic bonding represents an interaction between non-polar molecules that come together and exclude water molecules from between them, as shown in the characteristic ‘tail’ of a soap molecule (Fig 1).
Question: What basic physical principle is at work here that should cause molecules to bond in this way?
The molecules that behave in this way are seeking to achieve a lower energy state.
Bonding of this type is usually non-specific. It is seen in phospholipid bilayer membranes and the inside of large protein molecules.
Does isoflurane (Fig 1a) have a chiral centre?
Yes, isoflurane has a chiral centre, as one of the carbon atoms is bonded to four different atoms or functional groups (Fig 1b).
This means that two enantiomers of isoflurane exist, with different physiochemical properties (Fig 1c).
In the past, enantiomers were known as optical isomers because they were defined by the direction of deflection of polarized light by a crystalline sample of the molecule. This gave rise to:
Left deflection, laevo or L forms
Right deflection, dextro or D forms
Optical isomers are now defined in a new way, based on the distribution of groups around the chiral centre, to give two forms:
Left-handed (S) configuration (Fig 2a)
Right-handed (R) configuration (Fig 2b)
Enantiomers typically exist as racemic mixtures, but can also be synthesized into a single enantiomer form.
Recent improvements in the understanding of chirality and the ability to synthesize molecules more accurately have led to the trend to produce drugs as a single enantiomer.
This can result in clinical benefits in terms of improved efficacy, more predictable pharmacokinetics or reduced toxicity.
Examples include:
(S) bupivacaine, confusingly called laevobupivacaine (less cardiotoxicity)
(S) ketamine (fewer psychotic emergence reactions)
What type of isomerism do these molecules exhibit?
A. Chirality
B. Structural isomerism
C. Stereoisomerism
D. Tautomerism
E. Cis-trans isomerism
A. Correct. The molecules exhibit chirality.
B. Incorrect. The molecules do not exhibit structural isomerism.
C. Correct. The molecules exhibit stereoisomerism.
D. Incorrect. The molecules do not exhibit tautomerism.
E. Incorrect. The molecules do not exhibit cis-trans isomerism.
The image shows the (R) and (S) stereoisomers of the amino acid alanine. They are enantiomers and therefore exhibit chirality.
Many drugs used in anaesthesia have isomers, which offer improved side-effect profiles and chemical properties.
How does the zeroth law of thermodynamics allow you to create a thermometer?
i.e. This law states that if two thermodynamic systems are separately in equilibrium with a third system, then they must also be in equilibrium with each other.
You can create a thermometer by calibrating the change in a thermal property such as the length of a column of mercury.
The thermometer is calibrated by placing it in thermal contact with a second physical system until thermal equilibrium is reached at a number of reference points. Such reference points commonly include the freezing and boiling points of pure water (Fig 1).
If the calibrated thermometer is then placed in contact with a third system (such as the human body), its temperature can then be determined by observation of the change in the thermal property of the thermometer, namely the length of the column of mercury (Fig 2).
An example more pertinent to anaesthesia is the Venturi principle which applies when fluid flows through a tube with a constriction in it. As the fluid passes through the constriction (from position A to B), its velocity (and therefore kinetic energy) must increase (Fig 1).
Due to the principle of conservation of energy, this increase in kinetic energy occurs at the expense of a fall in potential energy. This results in a fall in pressure at the constriction. This drop in pressure can be used to entrain a second fluid.
Question: Can you think of examples in clinical anaesthesia that make use of this phenomenon?
A good example is the function of a nebulizer, you may have thought of further examples.
In a nebulizer, fluid (in this case oxygen) passes along a tube in which a constriction lies. At this constriction, kinetic energy (and therefore fluid velocity) increases, at the expense of potential energy and pressure.
This fall in pressure may be used to entrain a second fluid, which in this case is room air. This entrained air bubbles through a reservoir containing the liquid to be nebulized, resulting in the creation of fine droplets which are carried forward in the stream of oxygen and air.
The first law of thermodynamics states that energy can neither be created nor destroyed. The total energy within a system remains constant, although energy may be converted from one form to another.
This law can be used to explain why a pendulum keeps swinging. The energy within the system remains constant, with energy being repeatedly converted from kinetic energy to gravitational potential energy and vice versa.
Question: With reference to Fig 1a, at which point on the pendulum’s trajectory will the kinetic energy be maximal?
Kinetic energy is maximal at the lowest point of the trajectory (position 3 in Fig 1b). At the extremes of the trajectory (positions 1 and 5), the kinetic energy is zero, having been converted into gravitational potential energy, which is maximal at these points.
At what pH do most body enzymes function optimally?
Are there any exceptions?
Most body enzymes function at normal intracellular or extracellular pHs, which have the values 7.0 and 7.4 respectively (Fig 1).
Notable exceptions are proteases in the stomach, which function best at lower pH values (1-2).
The hydrogen ion concentration also causes ionization of molecules in most metabolic pathways, such as phosphates, ammonium and carboxylic acid groups. This ionization confines these molecules to compartments in the body, especially the intracellular compartment, which enables these metabolic pathways to proceed.
Changes in pH can often have profound multi-system effects, including to the cardiovascular, respiratory and renal systems.
Can you derive the Henderson-Hasselbalch equation?
Buffers are solutions that are able to resist changes in the concentration of H+ and OH-. Buffers are commonly solutions of weak acids, and less commonly weak bases. Referring back to the dissociation equation of a weak acid:
HA ↔ H++ A-
Addition of acid or H+ to the solution tends to shift the dissociation equilibrium to the left, effectively ‘mopping up’ some of the additional H+ by combining them with acid anions to form HA, i.e. undissociated acid molecules. Thus, the change in H+ concentration is reduced or ‘buffered’ (Fig 1).
Similarly, the addition of alkali in the form of OH- ions shifts the dissociation equilibrium to the right; this is due to the combination of OH- with H+ to form H2O molecules, followed by the dissociation of further acid molecules to restore balance.
Question: What are the main buffering systems of the human body?
Given the importance of maintaining a normal body pH, the human body is equipped with three main buffering systems, namely:
Bicarbonate: CO2 + H2O ↔ HCO3- + H+
Phosphate: H2PO4- ↔ PO42- + H+
Proteins: HProt(n-1)- ↔ Protn-
Measurement of pH - combination glass electrode
In practice, a pH electrode is usually constructed such that both electrodes are contained concentrically within the same housing (Fig 1).
The H+ sensitive glass is fashioned into a bulb at the tip (1) and this is placed into the solution under test. The resulting potential difference is measured, as previously described, using a silver/silver chloride electrode (2) bathed in buffer solution and a reference electrode (3) bathed in potassium chloride (4). The circuit is completed by incorporating a semi-permeable membrane (5) into the outer wall of the device, which must also be in contact with the test solution.
The pH electrode must be calibrated before use using two reference solutions of known pH. The electrodes must be regularly cleaned and the semi-permeable membrane checked for holes.
When a pH electrode is used to measure the pH of human blood, it must be maintained at a constant temperature (usually 37°C) to ensure accurate measurements.
Question: Why is body temperature important in determining an accurate blood pH measurement?
Since the dissociation of acids and bases increases as temperature rises, if the body temperature of a patient is lower or higher than 37°C, i.e. the calibrated electrode temperature, then a correction must be made to determine the true pH of the blood sample at its actual temperature.
Also, temperature affects the amount of gas, such carbon dioxide, dissolved in solution, with more gas being dissolved in cooler solutions.
Place each category of EM radiation on to the appropriate wavelength.
Notice the inverse relationship between frequency and wavelength in each case.
Complete the sentence describing the relationship between wavelength and frequency.
Match each principle of light with its description.
Next, see how these principles relate to anaesthesia equipment, starting with the fibreoptic laryngoscope.
Identify a point where the ray of light undergoes total internal reflection in this glass fibre.
All the angles are points of total internal reflection, since the angle of incidence of light within a glass fibre is greater than the critical angle.
Identify these features on the diagram.
Light normally travels in a straight line until it encounters a surface. If the surface is shiny, such as the mirror in Fig 1a, then light bounces back. This is called reflection.
The angle at which light hits the surface is known as the angle of incidence (Fig 1b)
The angle at which light bounces off is known as the angle of reflection (Fig 1c)
Angle of incidence (θi) = Angle of reflection (θr)
Regarding infrared gas analysis, match each substance to the appropriate statement.
Next, see how these principles relate to anaesthesia equipment, starting with the fibreoptic laryngoscope.
What is the relationship between electrical charge and electrical current?
A. They are the same
B. Charge is rate of change of current
C. Charge is flow of current
D. Charge is electrons; current is positrons
E. Current is rate of change of charge
A. Incorrect. They are not the same.
B. Incorrect. Current is rate of change of charge.
C. Incorrect. Current is flow of charge.
D. Incorrect. Charge is the quantity of electrons; current is rate of flow of electrons.
E. Correct. Electrical current is the flow of charge, or the flow of electrons, or the rate of change of electrons per unit of time.
Charge
Charge (Q) is the presence (negative) or absence (positive) of electrons on a conducting or an insulating surface.
Unit: coulomb, C
A battery drives the electrons to a conducting plate (Fig 1). If they have nowhere else to go then they remain on the plate as negative charge.
Current
Current (I) is electron flow through a conductor or around a circuit (cf. fluid flow). It can be considered as rate of change of charge.
Unit: amp(ère), A
Fig 2 shows current flowing from high to low potential, like water flowing downhill.
It may be counterintuitive to think of electrons flowing from positive to negative, but this merely reflects the early failure to understand that current is due to electron flow; however this makes no difference to the way electrical circuits are analysed, so it is legitimate to describe current flowing from a high to a low potential.
What is the difference between voltage and electromotive force (EMF)?
A. EMF is a way of describing the driving voltage from a battery
B. EMF is with reference to the battery’s power; voltage is with reference to electrical earth
C. They have different units
D. EMF requires an inductance; voltage requires a resistance
E. There is no difference
A. Correct. EMF is a way of describing the driving voltage from a battery.
B. Incorrect. EMF and voltage are both electrical potentials with reference to earth.
C. Incorrect. Both are potentials, and have units of volts.
D. Incorrect. The definitions of voltage and EMF do not depend on which elements are in the circuit.
E. Incorrect. They are both a form of electrical potential, but EMF describes a driving potential, while voltage usually refers to a drop in potential.
Voltage
Voltage or potential is a ‘driving force’ to drive electrons around a circuit, like the potential that a water reservoir on a hilltop gives for that water to flow down the hill. The potential, V, of any point is the voltage of that point above zero (earth).
Unit: volt
A drop in voltage occurs as current flows through a circuit.
The potential difference between two points is the voltage difference between them.
Electromotive force
Electromotive force (EMF), E, from a battery or from the mains supply, is a voltage that drives current around the circuit (cf. reservoir or pump pressure) (Fig 1).
Unit: volt
In terms of the laws governing current flow and voltage distribution, what is the difference between a series circuit and a parallel circuit?
A. In a series circuit, the same current flows round the whole circuit
B. In a parallel circuit, the same current flows round the whole circuit
C. Uniquely in a series circuit, V = I/R
D. In a parallel circuit, the circuit resistance is the sum of the individual resistances
A. In a series circuit, the same current flows round the whole circuit
A circuit element that has resistance, R, impedes current flow I (cf. resistance to fluid flow in a pipe), causing a voltage drop (potential difference), V, across it (cf. pressure drop).
Unit: ohm, Ω
Ohm’s law can apply to individual elements in a circuit or to the whole circuit. It allows us to calculate V, I and R.
The following pages show how Ohm’s law can be applied to circuits containing resistances in series and in parallel.
What is the purpose of a Wheatstone bridge?
A. To provide a form of damping in an arterial waveform
B. To amplify a signal
C. To shunt mains currents to earth
D. To reject noisy signals common to both arms of the bridge
E. To act as a low pass filter
A. Incorrect.
B. Incorrect.
C. Incorrect.
D. Correct.
E. Incorrect.
A Wheatstone bridge is a double bridge circuit designed to allow signals common to both arms to remain undetected and to allow a possibly much smaller signal (such as an arterial waveform from a strain gauge transducer) to be detectable at a meter applied between the two arms of the bridge.
A Wheatstone bridge is a pair of voltage dividers which equally divides unwanted high voltage signals, e.g. mains, diathermy, common to a whole circuit, between two arms of a bridge so they cancel each other out.
The low-voltage signal to be measured, e.g. an arterial waveform from a transducer, appears in one arm of the bridge, and thus produces a measurable voltage output across the voltmeter (Fig 1).
R1, R2 and R4 are constant, and R3 varies with mechanical strain from a strain gauge diaphragm in an arterial transducer.
Fig 2 shows how the sensitivity of the bridge can be increased by having two strain gauge transducers on opposite sides of the transducer diaphragm and on opposite arms of the bridge. Compression of the strain gauge at R1 and tension of the gauge at R3 enhances the difference in signal between the two arms of the bridge.
A circuit contains two parallel resistances of value 300 Ω and 3 kΩ. What is the total value of the circuit resistance?
A. 3.3 kΩ
B. 272.7 Ω
C. 333 Ω
D. 0.273 kΩ
E. 27.3 Ω
A. Incorrect.
B. Correct.
C. Incorrect.
D. Correct.
E. Incorrect.
The two resistances, R1 and R2, are in parallel; therefore, the total resistance of the circuit R is given by the calculation
Which of the following is correct in relation to how a Wheatstone bridge works?
A. All four resistances must be equal
B. It is a pair of potential dividers connected in parallel
C. Only one resistance is ‘active’
D. With a means of detecting any potential difference between its two parallel arms, the effect of unwanted voltages common to both arms is eliminated
E. It helps to eliminate signals common to both arms
A. Incorrect.
B. Correct.
C. Incorrect.
D. Correct.
E. Correct.
A Wheatstone bridge is a pair of voltage dividers which equally divides unwanted high voltage signals, e.g. mains, diathermy, common to a whole circuit, between two arms of a bridge so they cancel each other out.
The low-voltage signal to be measured, e.g. an arterial waveform from a transducer, appears in one arm of the bridge, and thus produces a measurable voltage output across the voltmeter (Fig 1).
R1, R2 and R4 are constant, and R3 varies with mechanical strain from a strain gauge diaphragm in an arterial transducer.
Fig 2 shows how the sensitivity of the bridge can be increased by having two strain gauge transducers on opposite sides of the transducer diaphragm and on opposite arms of the bridge. Compression of the strain gauge at R1 and tension of the gauge at R3 enhances the difference in signal between the two arms of the bridge.
Which of the following statements apply to functions that RC and CR circuits perform?
A. An RC circuit acts as a mathematical differentiator
B. An RC circuit acts as a low-pass electronic filter
C. A CR circuit acts as a mathematical integrator
D. A CR circuit acts as a band-pass filter
E. A CR circuit acts as a high-pass filter
A. Incorrect. An RC circuit acts as a mathematical integrator.
B. Correct. An RC circuit acts as a mathematical integrator and a low-pass filter.
C. Incorrect. A CR circuit acts as a mathematical differentiator and a high-pass filter.
D. Incorrect. Neither an RC nor a CR circuit acts as a band-pass filter, but both circuits in series would.
E. Correct. A CR circuit acts as a mathematical differentiator and a high-pass filter.
An RC circuit looks a bit like this voltage divider (Fig 1).
A voltage divider is a device for producing an output voltage which is a modification or a proportion of the input voltage in this RC circuit (Fig 2) where:
(see figure)
If Vin is a square wave, then Vout has the output waveform shown, which represents an integration of Vin. The circuit acts as a mathematical integrator.
If f is high, then Vout/Vin is small. The circuit therefore tends to block high frequency signals and allow low frequency signals to pass. It acts as a ‘low-pass’ electronic filter.
Define inductance and the role of an inductor.
A. Inductance is the ability to induce current from electron flow
B. An inductor is a coil of wire carrying current, around which a magnetic field is induced
C. The magnetic field around the coil induces a voltage which increases the voltage drop across it
D. The time constant is inversely proportional to the inductance magnitude
E. An inductor is a device for inducing current from voltage
A. Incorrect. Inductance refers to the ability of any current-carrying conductor to induce a magnetic field around it. That magnetic field can in turn induce a current in a neighbouring conductor.
B. Correct.
C. Incorrect. An inductance coil (an inductor), by virtue of the coils of wire adjacent to each other, amplifies this effect, producing a strong magnetic field through the coil, and inducing a voltage which opposes the prevailing voltage drop and current flow.
D. Incorrect. The time constant for a circuit with R and L elements is L/R.
E. Incorrect. An inductor produces a strong magnetic field through the coil, and induces a voltage.
Inductance
Inductance (L) is the tendency of a conductor (carrying a varying current) to induce an electromotive force (EMF), or a voltage that opposes the voltage drop due to the circuit resistance. The unit of measurement for inductance is henry (H).
Inductor
Any wire carrying a current has an associated magnetic field around it that induces voltage and current in nearby (but unconnected) conductors. When such a wire is coiled, i.e. an inductor, this effect results in an induced voltage or EMF which opposes the main voltage drop. The induced (opposing) voltage is proportional to the rate of change of the current flowing through the coil. With DC, on closure of the switch, an initial high voltage drop across the inductor induces a magnetic field in the coil. This produces an opposing EMF that in turn reduces the voltage across it and limits the rate of change of the current flow.
Rate of change
The rate of change of current or voltage in an inductor (L) occurs, as in a capacitor, in an exponential fashion in response to a step (DC) change of input voltage or current, with an associated time constant (τ). (For a circuit containing R and L elements, τ = L/R.)
Define capacitance and the role of a capacitor.
A. Capacitance is the ability to store electrical charge
B. A capacitor discharges instantaneously
C. A capacitor can block alternating current
D. With DC, on switch closure, as the current increases, the voltage decreases
E. A capacitor stores equal and opposite charges on opposite plates
A. True. Capacitance means the ability to store charge.
B. False. As the time constant for a CR circuit is R.C., discharge does not occur instantaneously.
C. False. A capacitor functions by inducing equal and opposite charge on the distal plate as the charge on the proximal plate rises, but not when it has reached a steady state level; therefore a capacitor blocks DC, not AC.
D. False. On switch closure using a battery (DC), the voltage across the capacitor rises, and the electrons flow rapidly initially (high current), then less rapidly (decreasing current) as the capacitor becomes saturated with charge.
E. True. A capacitor functions by inducing equal and opposite charge on the distal plate as the charge on the proximal plate.
Capacitance (C) is the ability of an object to store electrical charge.
It is equal to the charge per unit voltage (C = Q / V, where C = capacitance, Q = charge and V = voltage).
Capacitance is measured using the farad (F) unit, but since this is a very large unit, most capacitance is measured in microfarads (μF).
Capacitor
A capacitor comprises a pair of conducting plates separated by a space filled with an insulator, e.g. air, on which charge is stored on the proximal plate and an equal and opposite charge is induced on the opposite plate.
With direct current (DC), such as from a battery, on switch closure, the potential on the proximal plate rises to maximum, as maximum charge is developed on it. While the charge on the proximal plate is rising, an equal and opposite charge is induced on the distal plate, thus ensuring current flow in the rest of the circuit.
As the proximal plate becomes increasingly charged, the rate of rise of its potential decreases; therefore the current flow from the distal plate round the circuit decreases. Once the potential on the proximal plate has reached maximum, e.g. 9 V, no further charge is induced on the opposite plate, and no further current flows around the circuit.
The rate of change of current or voltage in a capacitor occurs in an exponential fashion in response to a step (DC) change of input voltage or current, with an associated time constant, τ. (For a circuit containing R and C elements, τ = R.C) It can be seen that current is proportional to rate of change of voltage.
Which of the following statements are true of RC circuits?
A. Vout/Vin is inversely proportional to capacitance and frequency
B. The higher the frequency, the higher is Vout/Vin
C. The output waveform is the integral of the input waveform
D. The higher the capacitance, the higher the output voltage
E. For DC circuits, Vout/ Vin = 1
A. Correct. Therefore Vout/ Vin is inversely proportional to both capacitance and frequency.
For an RC circuit:
Therefore Vout/Vin is inversely proportional to both f and C (and R). Therefore the higher the f or C, the lower is Vout/Vin. The circuit acts as an integrator.
B. Incorrect.
C. Correct.
D. Incorrect. See equation above.
E. Correct. See equation above, put f=0.
Which of the following are ‘tuned circuits’?
A. An RL circuit
B. A circuit with a frequency associated with maximum Vout/Vin
C. A circuit consisting of L, R and C elements
D. A circuit associated with a minimum output impedance
E. A CR circuit
A. False. A tuned circuit consists of an LCR circuit.
B. True. In a tuned circuit the expression for Vout/Vin can be shown to have a frequency-dependent maximum value, which is the same as having a minimum output impedance. It is ‘tuned’ to this frequency.
C. True.
D. True. In a tuned circuit the expression for Vout/Vin can be shown to have a frequency-dependent maximum value, which is the same as having a minimum output impedance. It is ‘tuned’ to this frequency.
E. False. A tuned circuit consists of an LCR circuit.
For this LCR circuit:
The top curves show the separate contribution to circuit resistance from the L and C elements when plotted against frequency (f). Note the L element’s resistance is directly to f, while the C element’s resistance is inversely proportional to f.
Frequency-dependent resistance is called ‘reactance’, and a circuit element which has components of R, L, and C has a combination of reactance and resistance called ‘impedance’.
At the ‘resonant’ frequency (r) for this circuit, the circuit reactance is at a minimum, and Vout/Vin is at a maximum as shown in the lower curve.
In converting an analogue signal to a digital one, the analogue signal is ‘sampled’ at intervals and digitized. Regarding the sampling rate:
A. The sampling rate is not important
B. The sampling rate must exactly match the analogue frequency
C. A sampling rate which is too low will inadequately represent the analogue signal
D. A sampling rate which is too high will also inadequately represent the analogue signal
E. An excessive sampling rate will distort the signal
A. Incorrect.
B. Incorrect.
C. Correct.
D. Incorrect.
E. Incorrect.
In digitizing an analogue signal, the sampling rate must not be too low or the resulting signal pattern will be an inadequate digital representation of the original analogue one. The sampling rate does not have to match any frequency relationship of the input signal, and a sampling rate which is high merely produces an unnecessarily accurately reproduced digital form of the original analogue signal.
Points on an analogue sine wave (Fig 1) are sampled at a chosen frequency. A digital value is assigned to each sample, giving a ‘staircase’ waveform. Low pass filtering smoothes this out to a facsimile of the original sine wave.
If the sampling rate is too low (Fig 2), then the digitized waveform will be a poor representation of the original, both in frequency and magnitude.
The following are true of common mode rejection ratio (CMRR):
A. It is the rejection of signals to a single input amplifier
B. It is the rejection of DC inputs
C. It is the acceptance of signals common to both inputs of a differential input amplifier
D. It is the rejection of signals common to both inputs of a differential input amplifier
E. A high CMRR is desirable
F. A CMRR of 10⁴ means a common mode signal of 10 000 mV is needed before 1 mV enters the op-amp for amplification
A. False. It is the rejection of signals common to both inputs of a differential input amplifier.
B. False. It is not related to the type of input signal.
C. False.
D. True. This is the definition of CMRR.
E. True. The ability to reject unwanted common mode signals is important for enhancing differential mode signals.
F. True.
If an amplifier has a single input it can, through capacitative coupling (the transfer of energy in a circuit via capacitance), amplify mains ‘hum’, in which an ECG signal may become swamped (Fig 1).
A differential input op-amp, with + and – inputs, allows noisy signals common to both inputs to cancel out, and the smaller differential signal to be amplified and appear at the amplifier output (Fig 2).
A good quality op-amp has a high ‘common mode rejection ratio’.
Monitoring requires a recording system to access and view the required potentials.
Question: What are the three main parts of a recording system?
The three main parts of a recording system (Fig 1) are the electrodes, the amplifier and the display unit. The latter may have a facility for the storage and printing of data.
Regarding the EEG:
A. It may be measured using scalp or intracranial electrodes
B. Measured potentials from scalp electrodes are usually in mV
C. Morphologically, it broadly resembles an ECG
D. Higher frequency activity is seen when patients are awake than when asleep
E. Evoked potentials only provide information about the integrity of the cortex
A. True.
B. False. They are heavily attenuated by meninges and the skull so they are in microvolts.
C. False. An EEG is much more complex and difficult to interpret than an ECG.
D. True.
E. False. Evoked potentials provide information about the stimulated pathway.
The EEG measures the summed electrical activity in the brain. As with the ECG, this is the result of ionic changes in response to stimulation. The EEG may be measured using scalp or intracranial electrodes (Fig 1).
The recording from scalp electrodes is attenuated by meninges, cerebrospinal fluid (CSF) and the skull. Measured potentials are usually in the region of 50 μV.
Depending on the area and the activity observed, there tends to be a predominant frequency. For example, delta waves are 3-Hz waves seen in slow-wave sleep. Beta waves are 12-25 Hz and are usually frontal, i.e. where a person is awake and concentrating.
The measured EEG frequencies are therefore variable and range from 3 Hz to 100 Hz.
The electrodes commonly used in anaesthetic practice rely on skin contact. This is for ease of use and patient comfort. However, non-invasive sensors are prone to certain unique problems.
Question: What problems may be caused by the use of non-invasive sensors?
The simplest issue is poor contact. This may be manifest as a complete loss of the signal display.
The biological signal can cause chemical changes in the recording electrode. This is called polarization and leads to altered performance of the sensor. Also, moisture trapped between the electrode and the patient’s skin can cause a battery effect (which results in the recording system generating a potential). This may interfere with the measured signal. A solution to these problems lies in the construction of modern electrodes. These contain silver (Ag) electrodes in contact with silver chloride (AgCl) and a conductive gel (Fig 1). The system is surrounded by an adhesive disc (Fig 2).
At a microscopic level ferromagnetic material is magnetized by:
A. Aligning it with the Earth’s magnetic field
B. Passing a small electric current through it
C. Aligning its unpaired electron spins in a magnetic field
D. Pairing up unpaired electron spins
E. Taking the material close to the equator
A. Incorrect. This is usually insufficient to permanently magnetize the material except near the poles where the magnetic flux density is high.
B. Incorrect. This is usually insufficient to permanently magnetize the material, though a large current might.
C. Correct. If unpaired electron spins within microscopic domains are aligned within a magnetic field of sufficient magnitude, then the material is magnetized.
D. Incorrect. Unpaired spins cannot be paired.
E. Incorrect. The magnetic flux density at the equator is low, so magnetization is unlikely.
As Fig 1 demonstrates, if an increasing magnetic field strength
H (amp. m-1) is applied to a ferromagnetic material, an increasing flux density B (tesla) develops as the magnetic domains become increasingly aligned with the applied field.
In Fig 2 region 1 represents a reversible pathway at low levels of H. Region 2 represents the onset of irreversibility of the magnetism. Region 3 is when the domains are irreversibly aligned (saturated) and the metal is a magnet. For iron this happens at flux densities of 1 to 2 tesla.
B = μ.H, where μ is the permeability of the material. The more permeable the material, the higher the degree of its magnetization that will respond linearly to an applied magnetic field. This is high for ferromagnetic materials, low for air:
What do you understand by ‘hysteresis’ in magnetic terms?
A. Magnetized material has a permanent associated magnetic field
B. A magnetic field can be reversibly applied to a material
C. Residual magnetism will remain on removal of the applied magnetic field
D. There is no energy loss in the magnetization – demagnetization process
E. It represents saturation of magnetic field strength
A. Incorrect. While this may be true of some materials, it does not define hysteresis.
B. Incorrect. Magnetization is seldom associated with reversibility, lack of which is a manifested as hysteresis.
C. Correct. Residual magnetism on removal of a magnetic field creates the irreversibility associated with hysteresis.
D. Incorrect. The fact that the magnetization – demagnetization process is irreversible, creates the hysteresis loop in the B – H curve, the area of which represents heat lost in the cyclic process.
E. Incorrect. Saturation in the process of magnetization is not a feature of hysteresis.
If the applied magnetic field (H) is removed, the magnetic flux density does not necessarily return to zero and the material may remain weakly magnetized. This failure to return to the earlier unmagnetized state is termed ‘hysteresis’. Magnetized ferromagnetic material therefore exhibits hysteresis.
0-1 The pathway taken on initial magnetization of the material
1-2 As the magnetic field H is removed, the magnetic flux density in the material does not fall to zero and there is some residual magnetism due to the partial irreversibility of the process
2-3 If the magnetic field H were reapplied from 2, the magnetic flux follows this path to re-establish the magnetism of the material nearly to its original level
2-4-5 A magnetizing force of reversed polarity is required to bring the flux density to zero (coercive force) (4), and a larger reversed magnetic field intensity produces saturation in the opposite direction (5)
5-1 Reversing the polarity of the magnetic field once again, the magnetic condition of the material follows this path, completing the hysteresis loop. If AC is used to induce the magnetic field, such a loop is traced out with each cycle. The area of the loop represents energy lost as heat in the magnetization-demagnetization process.
What feature does the voltage induced by the magnetic field around a coil have?
A. It enhances the magnetic field
B. It opposes the change in current flowing in the wire
C. It enhances the change in current flowing in the wire
D. It enhances the voltage drop along the wire
E. Its magnitude is maximum with DC
A. Incorrect. If anything the induced voltage opposes the driving voltage, which will attenuate the induced magnetic field.
B. Correct. This is the characteristic of such an induced voltage.
C. Incorrect. As the induced voltage opposes the driving voltage it will in fact reduce the change of current in the wire.
D. Incorrect. As the induced voltage opposes the voltage difference down the wire, it therefore reduces it.
E. Incorrect. Since the induced voltage is proportional to the rate of change of current in the coil, it will be zero when that current is DC.
f the wire is coiled around an iron core in a toroidal form, the flux density is further enhanced because of the high permeability of iron.
The high magnetic permeability μ of iron compared to that of air means that the flux density of such a coil will be greatly enhanced for a given magnetic field strength H since B=μH.
A coil of wire is more efficient than a straight wire in producing a magnetic field because:
A. Neighbouring coils produce mutually opposing magnetic flux
B. The magnetic permeability is increased
C. Neighbouring coils induce neighbouring reinforcing magnetic flux
D. An opposing voltage is induced in the coil
E. A larger current is induced
A. Incorrect. The neighbouring fluxes reinforce each other.
B. Incorrect. A coil does not affect permeability.
C. Correct. This is the reason a coil creates a stronger magnetic field.
D. Incorrect. While it is true that the voltage induced in the coil opposes the driving voltage, this is not a cause of a stronger magnetic field, but an effect.
E. Incorrect. The opposing voltage will reduce current change flowing in the coil, which will only partially offset the enhanced magnetic field.
If the current-carrying wire is coiled, a more intense magnetic field is created, because each wire in the coil which carries current induces a magnetic field around it. The close proximity of the coils to each other magnifies the flux density as many times as there are coils. If the single current-carrying wire, which induces flux Φ (weber), is coiled into N coil turns, then the flux thus induced is N.Φ (weber) A flat plane coil multiplies the flux density in its own plane, but an elongated coil ensures uniformity of the magnetic field along its length. The coil can have an air core or an iron core.
Henry discovered that a changing magnetic field induces a voltage in the coil proportional to the rate of change of the magnetic field, and Lenz said that the induced voltage opposes the change of magnetic flux which produced it.
Faraday deduced that the voltage V induced in a coil is proportional to the number of turns in the coil N, and the rate of change of magnetic flux dФ/dt.
∴V = N.(dФ/dt)
What happens when two electric coils are wound on the same core?
A. The core forms an electrical connection between the coils
B. The current in the first coil induces magnetic flux in the core
C. The flux in the coil induces an in-phase voltage in the second coil
D. The ratio of primary to secondary currents is the ratio of coil turns
E. The core is best made of a single piece of stainless steel
A. Incorrect. The function of the transformer does not depend on the core being an electrical connection.
B. Correct. This is the basis of a transformer’s function.
C. Incorrect. The current induced in the secondary coil opposes the flux, and thus the primary current, which induced it.
D. Incorrect. The ratio of voltages is the ratio of coil turns.
E. Incorrect. The core is best made of laminated steel.
In 1831 Faraday applied a voltage to one coil and observed a voltage on a second coil wound on the same iron core. This was the first transformer.
The transformer transfers electrical energy from one circuit to another by means of a magnetic field linking both circuits.
A changing voltage applied to the primary coil causes a changing current flow, which causes a changing magnetic flux in the core. Because of the changing flux, a voltage is induced in the secondary coil, which opposes the direction of the flux change in the core. No electrical connection between input and output is necessary. Because only changing currents (AC) are involved the output circuit can be isolated from a DC component at the input.
Isolated or floating circuits provide a circuit whereby a connection between the electrical source and the earth does not allow the current to flow. They are created by the use of an isolating transformer.
A transformer is used for:
A. Storing charge
B. Creating heat
C. Changing voltage
D. Creating magnetic flux
E. Transforming voltage to current
A. Incorrect. This is the purpose of a capacitor.
B. Incorrect. Any current carrying device creates heat, but this is not what a transformer is for.
C. Correct. The current induced in the secondary coil opposes the flux, and thus the primary current, which induced it.
D. Incorrect. Magnetic flux is created as an intermediate step, but it is not the purpose of a transformer.
E. Incorrect. This does not happen.
By adjusting the number of coils on the primary (N1) and the secondary winding (N2) it is possible to step up or step down the voltage between the input and the output.
V2/V1 = N2/N1. and N1.I1 = N2.I2
The coils are usually made of copper to reduce losses and the core is made of steel with a high permeability μ to minimize hysteresis loss, laminated to minimize eddy current loss.
Eddy currents are opposing flows of electrical current that can develop due to variations in the magnetic field or relative movement of the conductor. These flows induce their own magnetic field. This opposition results in resistive losses.
Laminated components are coated in high electrical resistance materials. This resistance minimizes the flow of eddy currents at the boundaries of the components. Thus the oppositional flow, magnetic field and resistance loss is reduced.
Regarding nuclear magnetism and MRI:
A. The nucleus of a hydrogen isotope has an equal number of protons and neutrons
B. Nuclear magnetism is the combination of positive electric charge and spin
C. Most hydrogen nuclei will orientate with a strong magnetic field
D. Net magnetization describes the sum of the magnetization of tissues
E. The Larmor frequency defines the magnetic field strength
F. Magnetic field strength is inversely proportional to precessional frequency
A. False. Hydrogen has a single proton in its nucleus.
B. True.
C. True.
D. True.
E. False. The Larmor frequency defines the precessional frequency.
F. False. Magnetic field strength is directly proportional to the precessional frequency. So, increasing field strength will increase the precessional frequency.
imagine that each hydrogen proton behaves like a tiny bar magnet.
Question: Each of the millions of hydrogen nuclei in the different tissues of the body has its own individual tiny magnetic field. Thus, when placed in the environment of a strong magnetic field, they react by orientating with the field. But how many react in this way: a few, most, or all of them?
The majority of hydrogen nuclei will orientate with the strong magnetic field, as demonstrated in the animation when you switch the field on and off.
Consider two tissues with different T1 relaxation times, one longer than the other.
Decide which statement describes the proportion of tissues returning to equilibrium (realigned with the main magnetic field), if TR is long (in the order of 2000 ms).
A. Strong likelihood that a large proportion of both tissues will have returned to equilibrium
B. Strong likelihood that only a small proportion of both tissues will have returned to equilibrium
C. Strong likelihood that there will be a difference between the two tissues, with one having largely returned to equilibrium and the other having only slightly returned to equilibrium
Submit
A. Correct. T1 relaxation times of tissue are exceeded, allowing tissues time to realign with the magnetic field.
B. Incorrect.
C. Incorrect.
Considering two tissues with different T1 relaxation times, decide what proportion of the tissues returned to equilibrium if the TR was short (in the order of 500 ms).
Select one option from the list below.
Possible answers:
A. Strong likelihood that a large proportion of both tissues will have returned to equilibrium
B. Strong likelihood that only a small proportion of both tissues will have returned to equilibrium
C. Strong likelihood that there will be a difference between the two tissues, with one having largely returned to equilibrium and the other having only slightly returned to equilibrium
A. Incorrect.
B. Incorrect.
C. Correct. T1 relaxation time of the tissue is longer than the TR.
As seen in the previous question, Fig 1 illustrates a way of differentiating between the T1 relaxation times of two tissues.
Remember, of course, that T2 relaxation occurs at the same time.
With this in mind, next consider what happens if the echo time is manipulated, but first:
Question: How much T2 relaxation time would you expect to have occurred when the echo time is relatively short – a large amount or a small amount?
A small amount. In this example, by using a short TR, you can differentiate the T1 relaxation times of grey and white matter. To maximize this T1 difference, an echo time must be selected which minimizes any other influence on the MR signal (T2 effects).
If T1 weighting is obtained using a pulse sequence with a short TR and a short TE, what would eradicate T1 information from such a pulse sequence?
A. Short TR, short TE
B. Short TR, long TE
C. Long TR, short TE
D. Long TR, long TE
Submit
A. Incorrect.
B. Incorrect.
C. Correct.
D. Correct.
By lengthening the TR, the differences in T1 relaxation are minimized, as shown next.
The MR signals coming from the tissues are converted into recognizable images by a process known as Fourier transformation. Here are some examples:
Short TE
Intermediate TE
Long TE
It is important to realize that these images bring out the differences in T1 relaxation of different tissues and are not a direct measure of T1 relaxation times. They are, therefore, designed to differentiate differences in T1 relaxation of tissues and are known as T1-weighted images.
The shorter the TR, the more heavily T1-weighted the image.
RF pulse sequences are set up with the scanning parameters manipulated to maximize contrast within an image. Clinicians refer to the images produced as:
A. T1 time
B. T1-weighted
C. T2 time
D. T2-weighted
A. False.
B. True.
C. False.
D. True.
These are the commonest scanning sequences used. It is possible to calculate relaxation times for different tissues but, in general, the pulse sequences have been developed to maximize the differences (contrast) in the relaxation times of different tissues.
MR images obtained from the pulse sequences require many hundreds (or even thousands) of measurements to allow the data acquired to be converted into recognizable MR images. The same section of anatomical tissue is, therefore, repeatedly exposed to the RF pulses throughout the scan. It is by manipulating the timing between RF pulses that T1 and T2 contrast is produced on MR images.
When an MR scanning sequence is being performed, the main factors that determine the T1 and T2 influence on the contrast of the images are:
The time between each of the RF pulse sequences. This is known as the repetition time and is given the symbol TR (Fig 1)
The time between the tissue being excited until a signal is detected. This is known as the echo time and is given the symbol TE (Fig 2)
The next page looks at what happens upon manipulation of the repetition time.
This image was acquired using a TR of 550 ms and a TE of 11ms. Based upon the signal intensity of the tissues indicated in the image, identify the following relative relaxation times.
This image was acquired using a TR of 2500ms and a TE of 100ms. Based upon the signal intensity of the tissues indicated in the image, identify the following relative relaxation times.
