This document discusses the basic physics principles relevant to anesthesiology. It explains how anesthesiologists can calculate the amount of gas in a nitrous oxide cylinder using Avogadro's principle and the cylinder's weight. It also describes how pressure regulators work by balancing the pressure and surface area of two diaphragms. Additionally, it covers partial pressure gradients, diffusion of gases, the Joule-Thomson effect, ventilation principles, and how pulmonary surfactant prevents alveolar collapse.

1. BASIC PHYSICS FOR THE
ANAESTHESIOLOGISTS
RATHER APPLIED PHYSICS

2. WE HATE PHYSICS…
THAT’S WHY WE JOINED
MEDICINE….
ISA
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3. BUT… THERE IS A LITTLE BIT OF
PHYSICS IN ALLACTIVITIES OF
OUR DAY TO DAY LIFE….
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4. JUST NOW I’VE FITTED THE MACHINE
WITH A NEW N2O CYLINDER….
HOW LONG WILL IT LAST…?
CAN I TAKE A CASE WHICH MAY LAST
FOR 6 HOURS….?
How the physics knowledge helps the Anaesthesiologists?
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5. HOW TO CALCULATE THE AMOUNT OF
N2O GAS IN A CYLINDER..?
STORY OF O2 – IDEAL GAS
1360 L
3400 L 6900 L
1
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6. CONTENTS OF OXYGEN CYLINDER
The relationship between change of
pressure & volume remains linear…
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FILLING OF THE CYLINDER WITH N2O GAS
LIQUID
BUT THE STORY IS DIFFERENT
WITH N2O CYLINDER……
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8. Question 1
What is the temperature above which N2O can’t be
liquefied however much pressure is applied?
A.1000 C
B.360 C
C.520 C
D.180 C
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9. *******
*******
*******
Pressure Gauge may not indicate the
quantity of the contents of the N2O cylinder
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10. THEN HOW TO ASSESS THE QUANTITY OF N2O
IN A N2O CYLINDER..?
Let us take the help of
Avagadro’s principle..
One gram molecule of any gas
at NTP occupies 22.4 L.
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11. MOLECULAR WEIGHT=ADDITION OF ATOMIC WEIGHT
ATOMIC WEIGHT OF NITROGEN – 14, OXYGEN – 16
MOLECULAR WEIGHT OF N2O – 14+14+16 = 44
One gram molecule of any gas
at NTP occupies 22.4 L.
One gram molecule = Gram molecular weight = molecular weight
expressed in grams
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12. AVAGADRO HYPOTHESIS:
ONE GRAM MOLECULAR WEIGHT OF ALL GASES WILL CONTAIN THE
SAME NUMBER OF MOLECULES AND OCCUPY THE SAME VOLUME ( 22.4 L)
AT S.T.P
32 g OF O2 = 44 g OF N2O = 28 g OF N2 =22.4 LITERS
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13. AMOUNT OF N2O GAS IN A CYLINDER
TARE WEIGHT OF THE CYLINDER = 12.5 K.G
CYLINDER WEIGHT WITH N2O = 15 K.G
SO.. WEIGHT OF THE N2O = 2.5 K.G = 2500G
44G OF N2O = 22.4 L
THEREFORE 2500 G = 22.4/44 x 2500 = 1272 L
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14. COLOUR CODING
IDENTIFY THE CYLINDERS...
Question 2
Question 3
Question 4
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15. HOW DOES THE ALVEOLAR GAS
EXCHANGE OCCURS….?
O2 AND CO2 MOVE IN OPPOSITE
DIRECTION BECAUSE OF
PARTIAL PRESSURE GRADIENT.
2
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16. WHAT DO YOU MEAN BY
PARTIAL PRESSURE?
@ @
@
@
@
$ $
$
@ @
@
@
@ $
$
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
$
5 PSI 3 PSI 8 PSI
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17. DALTON’S LAW OF PARTIAL PRESSURE
THE PROPORTION OF THE PRESSURE EXERTED BY A GAS
IN THE TOTAL PRESSURE IS EQUATED WITH THE VOLUME
IT OCCUPIES
@ @
@
@
@
$
$
$
PRESSURE OF @ GAS = 5 PSI
PRESSURE OF $ GAS = 3 PSI
TOTAL PRESSURE = 8 PSI
What is the % contribution of pressure of @ gas
In the total pressure?
5/8 x 100 = 62.5%
So, gas @ will occupy
62.5% of the total volume
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18. THE % OF O2 IN THE ROOM AIR = 21% BY VOLUME
THAT MEANS…IT CONTRIBUTES 21% OF THE
TOTAL ATMOSPHERIC PRESSURE
21/100 X 760 = 159.6 mm of Hg. pO2 in room air = 159.6
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19. Gaseous
Component
Percentage in
Air
% x 760 (mm
Hg)
Partial
Pressure
exerted (mm
Hg)
Symbol
Nitrogen 79.04 0.7904 x 760 600.7 PN
Oxygen 20.93 0.2093 x 760 159.1 PO2
Carbon
Dioxide
0.03 0.0003 x 760 0.228 PCO2
Total 100 760
Partial Pressure of Atmospheric Gases Calculation
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20. 20/78

21. WHAT WILL BE THE pO2
IN Mt EVEREST…?
A. 53 mm of Hg
B. 160 mm of Hg
C. 28 mm of Hg
D. 42 mm of Hg
Question 5
CLUE: Ht:8848 M BM:253 mm of Hg
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22. * *
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DIFFUSION OF GAS IN A LIQUID
 GAS GOES INTO SOLUTION
 SATURATION POINT
THE PRESSURE EXERTED BY THE GAS IN THE LIQUID
IS CALLED AS “TENSION”
3
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23. DIFFFUSION OF GAS IN A LIQUID
100 PSI 200 PSI
* *
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THE AMOUNT OF GAS DISSOLVED IN A GIVEN LIQUID
IS DIRECTLY PROPORTIONAL TO THE PARTIAL PRESSURE OF
THE GAS IN EQUILIBRIUM WITH THE LIQUID
HENRY’S LAW
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24. DIFFUSION OF GAS IN A LIQUID
PARTIAL PRESSURE OF THE GAS
TEMPERATURE
NATURE OF THE GAS
TYPE OF LIQUID
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25. PARTITION COEFFICIENT
BLOOD
N2O
0.47 L N2O
RATIO OF THE AMOUNT OF SUBSTANCE PRESENT IN ONE PHASE
COMPARED WITH ANOTHER,THE TWO PHASES BEING OF EQUAL VOLUME
AND IN EQUILIBRIUM
4
1 L N2O
SAME pN2O
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26. PARTITION CO-EFFICIENT
THE AMOUNT (VOLUME)OF DISSOLVED GAS TO RAISE THE SAME
PARTIAL PRESSURE AS IN THE CONTAINER MAY NOT BE
THE SAME……
5
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# #
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*******
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# # ###
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#
N2O
0.47
HALOTHANE
2.3
ETHER
12
SOLUBILITY OF ANAESTHETICS IN BLOOD
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28. THATS WHY, INDUCTION WITH ETHER IS SLOW……
Partial pressure in the alveoli builds up
very slowly as most of the ether goes into
solution in the blood.
Only when blood gets fully saturated, the
Partial pressure in the alveoli and then brain rises…..
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29. WHICH PART OF THE BOYLE’S MACHINE
UTILISES THE SAME PRINCIPLE AS THIS TOY…?
Question 6
A. Pressure regulator
B. Link-25
C. Bimetallic strip in vaporizer
D. Bourdon pressure gauge
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30. BOURDON PRESSURE GAUGE
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31. IN A CONTAINER,PRESSURE EXERTED BY A COMPRESSED GAS
EXERTS ITS EFFECT EQUALLY ON ALL SIDES OF THE CONTAINER.
IF THE CONTAINER IS MADE SUCH A WAY THAT IT CAN EXPAND OR
STRAIGHTEN OUT IN A DIRECTION, IT WILL DO SO AS LONG AS THE
PRESSURE IS MAINTAINED INSIDE.
ONCE THE PRESSURE IS RELEASED, THE CONTAINER WILL RETAIN ITS
ORIGINAL SHAPE
PRINCIPLE OF BOURDON PRESSURE GAUGE
6
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32. ------------------------
---------------------------------------
BASIC PRINCIPLE OF PRESSURE REGULATORS
TALL CONTAINER
PRESSURE = P
AREA OF ACTING
FORCE = a
SHORT,
WIDE
CONTAINER
Pressure = p
Area = A
P
a
p
A
HOW THE CYLINDER PRESSURE IS REDUCED
TO LOWER LEVEL IN THE BOYLE’S MACHINE…?
7
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33. BASIC PRINCIPLE OF PRESSURE REGULATOR
AT EQUILIBRIUM
FORCE ACTING IN THE TALL
CONTAINER
=
FORCE ACTING IN THE
SHORT,WIDE CONTAINER
P x a = p x A
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34. A LARGE FORCE ACTING ON A SMALL SURFACE AREA
CAN BE BALANCED BY
A SMALL FORCE ACTING ON A LARGE SURFACE AREA
BASIC PRINCIPLE OF
PRESSURE REGULATORS
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35. CYLINDER
PRESSURE = P
SMALL DIAPHRAGM
AREA = a
LARGE DIAPHRAGM
AREA = A
REDUCED PRESSURE = p
FLOW
METER
SCHEME OF A SIMPLE PRESSURE REGULATOR
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36. SIMPLE PRESSURE REGULATOR
FORCE ACTING TO OPEN
THE SMALL DIAPHRAGM = P x a
AT EQUILIBRIUM….
FORCE ACTING TO CLOSE
THE SMALL DIAPHRAGM = p x A
BALANCED BY
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37. EXAMPLE:
CYLINDER PRESSURE = 2000 psi
SURFACE AREA OF
SMALL DIAPHRAGM = 10 mm
PRESSURE OF THE GAS
IN THE REGULATOR = p
SURFACE AREA OF
LARGE DIAPHRAGM = 100 mm
2000 X 10 = p X 100
p = 2000 X 10 /100
p = 200 psi
SO, IF THE CYLINDER PRESSURE IS 2000, THE REGULATOR WILL REDUCE
IT TO 200 psi.
P x a = p x A
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38. 2000 X 10 = p X 100
p = 2000 X 10 /100
p = 200 psi
IN OTHER WORDS, THE RATIO OF THE SURFACE AREA
OF THE TWO DIAPHRAGMS DECIDE THE RATIO OF
PRESSURE REDUCTION
HERE IT IS 10 : 100 – SO, PRESSURE REDUCTION IS 1/10
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39. CYLINDER
PRESSURE = P
SMALL DIAPHRAGM
AREA = a
LARGE DIAPHRAGM
AREA = A
REDUCED PRESSURE = p
FLOW
METER
SCHEME OF A MODIFIED PRESSURE REGULATOR
S1
S2
P X a X S1 = p X A X S2
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40. MODERN REGULATOR:
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42. A pressure regulator being dismantled…
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43. Oil or grease application in any part of the
Boyle’s apparatus is not advised…
Why…?
8
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44. ADIABATIC COMPRESSION:
When a gas is subjected to sudden compression, heat energy
is produced rapidly. If there is no time for dissipation of this
heat, the temperature of the system rises enormously.
First law of thermodynamics
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45. ADIABATIC COMPRESSION OF GASES

46. Have you seen ice crystal formation over a cylinder valve
if N2O leaks through it?
Joule-Thomson effect
9
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47. Joule-Thomson effect:
When a compressed gas is allowed free expansion
through a narrow opening, the temperature of the
surrounding falls rapidly.
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48. Joule – Thomson effect
Anaesthetic significances
1. If the N2O cylinder contains water vapour, ice
crystals form inside the cylinder valve which
blocks the flow.
2. N2O cryoprobe
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49. If you want to transfuse blood rapidly…
What will you do?
A.Put a wider gauge cannula
B.Increase the drip stand height
C.Use a rapid infusion bag
D.Using a wider, filterless drip set
Question 7
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50. Hagen-Poiseuille Law:
When a fluid flows through a tube, then the quantity of the flow(Q) is
1. Proportional to the pressure difference between the two ends of the tube
2. Proportional to the fourth power of the radius of the tube
3. Inversely proportional to the viscosity of the fluid
4. Inversely proportional to the length of the tube
Q = π r4 (P1- P2)/ 8ηl
10
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51. What is the common factor among the following..?
1.Gas stove
2. Insect sprayer
3. Venti oxygen mask
4. Nebuliser
5. Ventilator
6. Bunsen burner
Question 8
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52. 52/78

53. Bernoulli’s theorem
How it works…?
53/78

54. Question 9
What is the stoichiometric concentration
of ether in air?
A.8.6%
B.12.2%
C.3.4%
D.1.5%
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55. WHY DO YOU REDUCE THE LUMEN SIZE OF THE PIPE
WHILE WATERING THE PLANTS?
Trying to understandthe principle of Venturi….
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56. 0
0
0
+
-
Clinical application of Bernoulli’s theorem:
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57. DIFFUSER ( VENTURI TUBE )
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58. 58/78

59. 59/78

60. BASIC PRINCIPLE OF A NEBULISER

61. HOW ARE THE LUNG ALVEOLI ARE
KEPT INTACT WITHOUT GETTING
BURST / COLLAPSED?
11
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62. Laplace law:
The excess pressure inside a spherical gas-liquid
interface is equal to twice the co-efficient of surface
tension divided by the radius of the interface
ΔP= 2γ/ r
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63. 63/78

64. PRESSURE INSIDE THE ALVEOLUS WHICH
TRIES TO COLLAPSE IT=
2T/R
r
R
2T/2 2 X 10 / 2 = 10
2T/4 2X 10 / 4 = 5
4
2
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65. PRESSURE INSIDE THE SMALL ALVEOLUS
WILL BE HIGHER THAN THE
LARGER ALVEOLUS
SO-SMALLER ALVEOLI WILL EMPTY IN TO THE
LARGER ALVEOLI AND BURST
BUT..THIS DOESN’T HAPPEN
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66. FLUID-AIR
INTERFACE
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67. SURFACTANT
 PRESENCE OF SURFACTANT, LINING THE ALVEOLI
 SURFACTANT REDUCES THE SURFACE TENSION
 Dipalmitoylphosphatidylcholine
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68. r
R
2T/2 2 X 5 / 2 = 5
2T/4 2X 10 / 4 = 5
4
2
WHEN THE ALVEOLI SIZE DECREASES,THE SURFACTANT
PRESENT PER UNIT AREA OF ALVEOLAR WALL INCREASES
THEREBY REDUCING THE TENSION
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69.  BY REDUCING THE SURFACE TENSION,SURFACTANT
PREVENTS THE ALVEOLI FROM COMPLETELY
COLLAPSING ON EXHALATION
 IN ADDITION,THE DECREASED SURFACE TENSION
ALLOWS THE REOPENING OF THE AIR SPACE
WITH THE LOWER AMOUNT OF ENERGY
WHY SURFACE TENSION
HAS TO BE REDUCED?
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70. RECAP:
1.The contents of N2O cylinders can be arrived by weighing
it and using Avagadros’s law.
2. Pressure regulators work with a simple principle.
The ratio of the size of the diaphragm decide the extent of pressure
reduction.
3. Partial pressure gradient of gases decide the direction
Of gas diffusion. Not the volume % of gases.
4. Oil or Grease should not to be applied over the cylinder valve
To loosen it as it may cause explosion.
5. Venturi effect plays an important role in many gadgets
of anaesthesia.
5. Alveolar collapse is prevented by the presence of surfactant.
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71. Before I wind up….

72. THE BRILLIANCE OF ANCIENT CHOLA KING, RAJA RAJA CHOLAN
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73. IRRIGATION SYSTEM OF THE CAUVERY BASIN
WATER LET-OUT SYSTEM FROM THE LAKES
PILLAR
STONE BOX
WATER FLOW
STOPPING STONE
WATER OUTLET
MUD LET OUT OUTLET
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74. CLOSING STONE
MUD
LET-OUT
HOLE
WATER LET-OUT
HOLE
WATER
CHANNEL
BELOW THE BUND
BUND
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75. DANIEL BERNOULLI - 1738
GIOVANNI BATTISTA VENTURI - 1746 - 1822
RAJA RAJA CHOLAN - 985 - 1014
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76. THANK YOU
dr.r.selvakumar. m.d.d.a.dnb
professor of anaesthesiology,
k.a.p.viswanatham govt. medical college,
trichy
BETTER UNDERSTANDING OF THE PHYSICS
HELPS US TO NAVIGATE THE SHIP SAFELY
76/78

77. Here are the answers for the questions,
appeared during my lecture
Question number Answer
Question 1 B –Critical temp of N2O
360C
Question 2 CO2 CYLINDER-GREY
Question 3 ENTONOX- BLUE BODY
BLUE+WHITE SHOULDER
Question 4 BROWN -HELIUM
Question 5 A- Mt Everest-53 mm of Hg
Question 6 D- Bourdon Pressure Gauge
Question 7 A-Put a wider gauge cannula
Question 8 Venturi effect
Question 9 C -3.4%
77/78

78. thank you
dr.r.selvakumar.M.D.DA.DNB
professor & head,
institute of anaesthesiology,
madurai medical college,
madurai.
The end