Air Physics Flashcards
What’s the absolute pressure at depth of 254m?
26,4 bar
What’s the absolute pressure at depth of 254fsw?
254/33= 7,69+1= 8,69ATA.
- A diver is working at 100 fsw for 30 minutes. What volume of gas will he use?
100/33 +1 = 4.03
Diver cons = 1,25 cfm
Time = 30min
4,03x1,25x30= 151,2 ft3
A diver is working at 120 msw for 4 hours. What volume of gas will he use?
120m= 13bar
Diver consumption = 35l/m
4 hours= 240min
13x35x240= 109.200 litres = 109.2m3
- Two divers are working out of the bell at 75 msw. What volume of gas will they use in 4 hours?
8,5x35x240 x 2 divers = 142.8m3
- An air diver is working at 60 fsw for 20 minutes. What volume of air will he use?
60/33+1 = 2,82 x 1,25 x 20= 70,45 ft3
- A 64 x 50 litre quad contains gas at a pressure of 100 bar. What is the total volume of gas in the quad?
- The quad in Question 7 will be changed over when the pressure falls to 40 bar. What volume of gas is available to the diver?
- In Question 8, the diver is working at 130 msw. How long can he work for before the quad is changed over?
64x50x100= 320m3
64x50x60= 192m3
130m = 14 bar
Diver cons= 35l/m
Total gas 192.000
14x35= 490l/m 192.000/490 = 391,83min / 60 = 6h 32min
A diver is working at 80 msw, breathing from a 16 x 50 litre quad at a pressure of 150 bar. How long could he work for? (Assume that the quad will be changed over at 40 bar.)
Total gas = 16x50x110= 88.000litres
80m = 9 bar x 35l/m = 315lm
88.000/315= 279,36 =
4h39min.
A 64 x 50 litre quad is at a pressure of 205 bar. It can be used until the pressure falls to 30 bar. How much gas is available for use?
205-30=175bar
64x50x175= 560m3
A 12 x 50 litre quad is at a pressure of 25 bar. What volume of gas will be needed to fill it to a pressure of 200 bar?
200-25=175bar
12x50x175= 105m3
- A bail-out bottle has a volume of 12 litres and contains gas at a pressure of 200 bar. If the diver is working at 100 msw, what volume of gas is available to him in an emergency?
200-10-11=179bar x 12 = 2148l or 2.15m3
A diver is working at 195 msw, breathing from a 16 x 50 litre quad at a pressure of 110 bar. The quad will be changed over at 40 bar. How long could he work for?
110-40=70bar
70x16x50= 56.000l
20,5x35= 717,5lm
56000/717,5= 1h18min
A bailout bottle has a FV of 6 litres charged to 190 bar, how long it last a diver at depth of 48 msw ? (Assume 15 bar losses)
a) 6min
b) 4.5min
c) 3min
d) 7min
b) 4.5min
190-15=175 x 6 = 1050 litres
5.8 x 40 = 232 lm
1050/232 = 4.5min
- How much gas would a diver consume in 80 min at depth of 42 msw?
a) 13.44m3
b) 11.76 m3
c) 145.6 m3
d) 14.56 m3
d) 14.56 m3
- What’s the apparent weight in the seawater, of a 1000 m3 pontoon that weights 1500 tonnes in air?
a) 470 tonnes
b) 500 tonnes
c) 1470 tonnes
d) 1030 tonnes
a) 470 tonnes
1000 x 1.03 = 1030
1500 - 1030 = 470 tonnes
1ft3 of sea water weights?
a) 1000kg
b) 62.5 lbs
c) 64.38 lbs
d) 1030 kg
c) 64.38 lbs - seawater
Remember : 62.5 lbs is fresh water
- What’s the PO2 in breathing air @ 43 msw?
a) 1.113 ba
b) 0.903 ba
c) 0.693 ba
d) 0.483 ba
a) 1.113 ba
5. 3 x 0.21 = 1.113 ba
- What the PO2 in breathing air at 97 fsw?
a) 1.025 ata
b) 0.827 ata
c) 0.609 ata
d) 0.468 ata
b) 0.827 ata
97/33 + 1 = 3.939 x 0.21 = 0.827
- What’s the PO2 of 32% nitrox breathed at 27 msw?
a) 1.184 ba
b) 1.458 ba
c) 0.119 ba
d) 0.109 ba
a) 1.184 ba
3. 7 x 0.32 = 1.184
- You require a PO2 of 1.4 ba @ 38 msw, what’s the suitable nitrox mix?
a) 38/62
b) 40/60
c) 50/50
d) 28/72
d) 28/72
1. 4/4.8 = 0.291 x 100 = 29% O2
- What’s the maximum depth a diver could breathe 50/50 mixture with PO2 of 1.4 ba?
a) 12 msw
b) 1.8 msw
c) 18 msw
d) 15 msw
c) 18 msw
1. 4/0.5 = 2.8 bar = 18 msw
- What’s the maximum depth for breathing pure O2 when PO2 is 2.8 ba?
a) 12 msw
b) 1.8 msw
c) 18 msw
d) 15 msw
c) 18 msw
2. 8/1= 2.8 bar = 18 msw
- At what depth would air equal IMCA wet maximum PPO2?
a) 67 msw
b) 57 msw
c) 50 msw
d) 30 msw
b) 57 msw or 185 fsw
- 4/0.21=6.66-1x10= 56.66
- 66 x 3.28 = 185 fsw
- If a diver has 1 litre of gas in solution in his body at 1b, how much of the same gas would he have in solution at 50 msw after 10-12h?
a) 5 litres
b) 1 litre
c) 6 litres
d) 2 litres
c) 6 litres
- What is the apparent weight in seawater of a 1000 m3 pontoon that weighs 1500 tonnes in air?
1) 500 tonnes.
2) 1470 tonnes.
3) 470 tonnes.
4) 1030 tonnes.
3) 470 tonnes.
It’s in seawater remember, which is 1.03 tonnes per m3.
1000 x 1.03 = 1030 tonnes up thrust.
Weight of object in air minus the up thrust,
1500 - 1030 = 470 tonnes.
- A sunken pontoon 10m x 5m x 2m lies on a hard seabed and has an apparent weight of 4 tonnes, you de-ballast a compartment 2m x 2m x1m, what will happen?
1) It will be neutrally buoyant.
2) Nothing.
3) The apparent weight will be 8.12 tonnes.
4) It will raise to the surface.
4) It will raise to the surface.
2 x 2 x 1 = 4m3 of salt water up thrust which is 4.12 tonnes,
the object weighs only 4 tonnes in air so it will have a positive up thrust of .12 tonnes.
- An object has a density of 168 lbs / ft3 and is 50 feet long, 10 feet high and 8 feet wide. How heavy is it?
- An object has an area of 50 feet long, 10 feet high and 8 feet wide, what would be the up thrust in salt water? (Question before)
It weighs 300 tonnes.
50 x 10 x 8 = 4000 ft3
4000 x 168 lbs/ft3 = 672000 lbs
672000 / 2240 (tonne) = 300 tonnes.
It’s up thrust would be 257,520 lbs or 114.9 tonnes.
50 x 10 x 8 = 4000,
4000 x 64.38 lbs (ft3 of seawater) = 257,520 lbs
257,520 / 2240 (tonnes) = 114.9 tonnes.
- An object has an area of 50 feet long, 10 feet high and 8 feet wide, what the object appear to weight in salt water?
Density of 168ft/lbs
a) 185 tons
b) 288 tons
c) 3808 tons
d) 38.8 tons
a) 185 tons
50 X 10 X 8 = 4000 X 64.38 = 672000/2240 = 300tons
Weight of the object 300 tons
4000 X 64.38 = 257520/2240= 114.9tons
Upthrust 114.9 tons
300-114.9= 185 tons
- If an object 15m long x 4m wide x 3m high had a density of 3800 kg/m3, what would it appear to weigh in fresh water?
a) 860 tonnes
b) 684 tonnes
c) 86 tonnes
d) 504 tonnes
D. It would appear to weigh 504 tonnes.
15m x 4m x 3m = 180m3
3800 / 1000 = 3.8 tonnes x 180 = 684 tonnes.
684 tonnes - 180 tonnes (up thrust of object in fresh water) = 504 tonnes.
- 1) If an object 15m long x 4m wide x 3m high, floats in seawater with 1m submerged, what must the density be?
2) Also how much more weight could be placed on the object before it sank?
3) If equipment weighing 18.26 tonnes were loaded on to part one of the question, how much of the pontoon would be under water?
1) Its density is 343 kg/m3
15m x 4m x 3m = 180m3, only 1 meter is submerged,
so the object must weight the same as 15m x 4m x 1m = 61.8 tonnes seawater up thrust,
Divide up thrust by the whole volume to find it density, 61800kg / 180m3 = 343 kg/m3.
2) 123.7 tonnes.
61.8 tonnes per 1 meter high, the object is three meters high
so x 2 more loads of 61.8 tonnes (123.6 tonnes will make the object neutral)
So 123.7 tonnes will give the object negative buoyancy.
3) 1.29m underwater in total.
18.26 / 1.03 (seawater) = 17.72
17.72 / 60 (area of object) = 0.29
So the object would sink a further 0.29 meters,
which in total would be 1.29 meters underwater.
Remember : D=volume submerged x 1.03/ total volume
- A bell weighs 8 tonnes and displaces 300 ft3 of seawater. How much weight must be added to make it sink?
a) 0.63 tons
b) 0.53 tons
c) 57 lbs
d) 128 lbs
a) 0.63 tonnes would need to be added to make it sink.
300 x 64.38 lbs = 19,314 lbs / 2240 (tonne) = 8.622 tonnes
Bell weighs 8 tonnes with 8.622 tonnes up thrust,
So you would need 0.63 tonnes more weight to sink it.
- A twin set is filled to 200 bar in a surface temperature of 39°C, underwater it cools to 4°C, what is its new pressure?
a) 170 bar
b) 177 bar
c) 183 bar
d) 194 bar
It’s new pressure is 177 bar.
39°C + 273 (kelvin) = 312
4°C + 273 (kelvin) = 277
200 / 312 = 0.64
0.64 x 277 = 177 bar.
- A bailout bottle has a pressure of 2640 psi at 41°F, what was its filling pressure at 70°F?
Its original pressure was 2793 psi at filling.
41 + 460 (Rankin) = 501.
70 + 460 (Rankin) = 530.
So 2640 / 501 = 5.2694 x 530 = 2793 psi.
- A quad at a pressure of 198 bar at 27°C cools to 17°C, what will the new pressure be?
It’s new pressure will be 191.4 bar
27 + 273 (Kelvin) = 300.
17 + 273 (Kelvin) = 290.
So 198 / 300 = .66 x 290 = 191.4 bar.
- A pressurised chamber contains 92 m3 of gas at 68msw, if it cools from 31°C to 28°C, what will be the new depth?
It’s new depth will be 67.1 msw.
31 + 273 (Kelvin) = 304.
28 + 273 (Kelvin) = 301.
So 68 / 304 = .223 x 301 = 67.1 msw.
- An air reservoir has a MWP (max working pressure) of 200 psi, if it is at 180 psi, with a temperature of 16°C and stands in the sunshine until the temperature reaches 28°C, what will the new pressure become?
It’s new pressure will become 187.4 psi.
16 + 273 (Kelvin) = 289.
28 + 273 (Kelvin) = 301.
So 180 / 289 = .6228 x 301 = 187.4 psi.
- An object 11m x 7m x 7m in the seabed, weight 600t.
If pumped 50m3 of air. What will happen with the object?
IMCA EXAME!!!!!!
11x7x7 = 539 x 1.03 = 555.17t
600 - 555.17 = 44.83 t ( negative buoyant)
50 x 1.03 = 51.5 ton
ENOUGH TO BRING TO SURFACE POSITIVE BUOYANCY.
- A pontoon 12m x 7m x 3m floats with 2.5m submerged in seawater?
a) what’s the upthrust?
b) what’s the weight of the pontoon
c) how much weight could I place on the deck without it sinking?
a) 12 x 7 x 2.5 = 210 x 1.03 = 216.3 tonnes upthrust
b) as it floats the pontoon weights is equal to the up thrust
c) 12 x 7 x 3 = 252 x 1.03 = 259.56
259.56 - 216.3 = 43.26 tonnes
Or
12 x 7 x 0.5 x 1.03 = 43.26 tonnes
What is the pressure exerted at 1 atmosphere (Sea Level) ? a 14.7 psi b 14.5 psi c 0.445 psi d 29.2 psi
a
14.5 is gauge
Convert 6220 ppm to a %: a 0.0622% b 6.220% c 0.622% d 0.0062%
c
Ppm to % = divide by 10.000
% to ppm = times 10.000
Ppm to bar = move back 6 spaces
What is the absolute pressure exerted at a depth of 165 fsw? a 8.82 psi b 88.2 psi c 73.5 psi d 165 psi
b
165 +33 = 198/33 = 6x14.7 = 88.2
To convert degrees Celsius to absolute temperature you must add: a 273 b 460 c 32 d 100
a
+ 273 for celsius
+ 460 for kelvin
What is 32ºC expressed in FAHRENHEIT? a 305ºF b 57.6ºF c 89.6ºF d 92.4ºF
c
32x1.8+32 = 89.6
How much gas does a 10 litre flask, charged to 200 bar hold? a 2000 l b 200 l c 10 l d 2200 l
a
10x200= 2000l
Convert 1.000 litres into cubic metres: a 2 m3 b 1 m3 c 1.2 m3 d 28.32 m3
b
1000l = 1m3
How much gas would be required to pressurise a chamber with a floodable volume of 28 m3 to a depth of 37 msw?
a 131.6 m3
b 1036 m3
c 103.6 m3
d 75.6 m3
c
3.7 absolute pressure x 28 = 103.6
A diver consumes 1.25 ft3/min at surface. How much gas would he require at depth of 100 fsw?
a 4.03 ft3
b 1.25 ft3
c 5.037 ft3
d 7.2 ft3
c
100+33/33=4.03 absolute pressure x 1.25= 5.037ft3
A compressor has a delivery of 7 ft3/min. To what depth would this support a
diver if his consumption is 1.25 ft3/min?
a 184.8 fsw
b 151.8 msw
c 151.8 fsw
d 51.8 fsw
c
7/1.25= 5.6 x 33= 184.8 -33 = 151.8fsw
A bail out bottle has a FV of 12 litres and is charged to 200 bar. How much gas is available to the diver at a depth of 42 msw? (Hat working pressure = 8 bar)
a 2400 l
b 2285.7 l
c 2253.6 l
d 2241.6 l
d
FV = 200-8-5.2= 186.8x12= 2.241.6l
A bank has 45 bottles, each have a FV of 50 litres. How much gas does it contain at a pressure of 185 Bar?
a 592 m3
b 450.2m3
c 416.25 m3
d 416.2 ft3
c
45x50x185= 416.250m3
A bailout bottle has a FV of 6 litres and is charged to 200 bar. How long would
it last a diver at a depth of 52 msw? (Use 40 l/min and 8 bar regulator pressure)
a 5 min 40 sec
b 4 min 30 sec
c 4 min 52 sec
d 4 min 41 sec
b 200-8-6.2= 185.8 x 6 = 1.114.8/248= 4.49min 6.2x40= 248 49x60=29sec 4m29sec
What is the pO2 at 47 msw if the breathing gas is air? a 0.37 bar b 1.2 bar c 0.99 bar d 0.21 bar
b
5.7x.21= 1.19
What is the partial pressure of Nitrogen in Air at a depth of 99 fsw? a 3.16 ATA b 2.16 ATA c 4.16 ATA d 0.79 ATA
a
99+33/33=4x .79= 3.16
If a diver has 1 litre of gas in solution in his body at Sea level, how much of the same gas would he have in solution at 50 msw?
a 5 litres
b 1 litre
c 6 litres
d 2 litres
c
50m 6 bar
Charles law
A diver weighs 300 lbs and displaces 4.5 ft3 of salt water. If he wanted to be 20lbs negative, how much extra weight would he require?
a 7 lbs
b 9.7 lbs
c 7.9 lbs
d 17.9 lbs
b
4.5x64.38= 289.71 + 20= 309.71 - 300 = 9.71lbs
A container measures 10 x 3 x 4 m, it draws 2 m draft. What is the upthrust.received?
a 60 t
b 68 t
c 61.8 t
d 16.8 t
c
10x3x2=60x 1.03= 61.8t ( If sea water )
10x3x2= 60x1 = 60t ( if fresh water)
A diving bell receives an upthrust of 14,250 lbs and weighs 6.5 ton. How much buoyancy is required to make it 500 lbs positive?
a 500 lbs
b 710 lbs
c 810 lbs
d 760 lbs
c
6.5x2240 = 14560 - 14250= 310 + 500= 810
During blowdown to a depth of 100 msw a chamber reached a temperature of 35ºC. What would be the reading by gauge after cooling to 25ºC?
a 96.75 msw
b 96.43 msw
c 106.75 msw
d 106.43 msw
b
11/35+273 = 0.3571x298= 10.64 -1x10= 96.428
During pressurisation a bottle reached a temp of 40ºC. If the start pressure was
195 bar, what would the pressure be, after cooling to 20ºC?
a 204.3 bar
b 190.2 bar
c 182.5 bar
d 179.4 bar
c
195/40+273=0.623x293= 182.53
What is the amount of gas required to blow down a chamber from 30 to 68 msw if the FV = 14 m3? a 53.2 m3 b 109.2 m3 c 67.2 m3 d 91.2 m3
a
3.8x14=53.2m3
How much gas would a diver consume over 4 hours at a depth of 110 msw if the consumption rate was 35 l/min? a 8.4 m3 b 100.8 m3 c 92.4 m3 d 25.2 m3
b
12x35x240= 100.8m3
Convert 0.37% into parts per million. a 37 ppm b 3700 ppm c 370 ppm d 2700 ppm
b
.37x10000=3700ppm
1 bar is equivalent to:
a. 760 mm Hg - FOR EXAM PURPOSES
b. 750 mm Hg
c. 770 mm Hg
d. 740 mm Hg
b
6 bar gauge is equivalent to:
a. 165 fsw
b. 175 fsw
c. 198 fsw
d. 231 fsw
c
6x33= 198fsw
A diver suffering from decompression sickness is to be given a total of 6 cycles of 25 minutes each. The consumption rate is 40 litres per minute at surface. Chamber depth is 70 msw. How much gas will be used?
a. 4.8 m3
b. 48 m3
c. 480 m3
d. 42 m3
b
8x40x25x6= 48m3
A DDC with an FV of 25 m3 is pressurised to 80 msw. The temperature on arriving at 80 msw is 35°C. When the temperature fall to 30°C what was the depth of the DDC?
a. 78.7 msw
b. 92.5 msw
c. 82.6 msw
d. 78.5 msw
a
80 x 303 / 308 = 78.7m
A liquid absorbs 1 litre of gas at 1 bar. How much will it absorb at 5 bar?
a. 1 litre
b. 2 litre
c. 3 litre
d. 5 litre
d
If you double the pressure on a gas, according to Boyle's law what will happen to the volume and density? a. Volume doubles Density doubles b. Density doubles Volume remains same c. Density is halved Volume is halved d. Volume is halved Density is doubled
d
What does the reference ± 0.25% marked on a 300 msw gauge, refer to in depth? a. 0.25 msw b. 0.5 msw c. 0.75 msw d. 0.1 msw
c
.25%x300= 0.75m
You have to pressurise a DDC with an FV of 15m3 to 122 msw. How much gas is required?
a. 55.39 m3
b. 74.4 m3
c. 183 m3
d. 198 m3
c
12.2x15= 183m3
A diver suffering from decompression sickness is given a total of 6 cycles of 25 minutes each. The consumption rate is 0.75ft ³at surface. Chamber depth is 100fsw. How much gas will be used? a 453ft³ b 425ft³ c 396ft³ d 341ft³
a
100+33/33=4.030x0.75x25x6= 453.4ft3
A chamber is pressurised to 200 fsw. The temperature on arriving at 200 fsw was 110°F. At what depth would the chamber be after cooling to 68°F a 211fsw b 182fsw c 152fsw d 179fsw
b
A liquid absorbs 1 cu.ft of gas at sea level. How much will it absorb at 6 atm? a 7 cu.ft b 12 cu.ft c 5 cu.ft d 6 cu.ft
d
Using a 500 msw gauge, what does the reference marked on the gauge ± 0.25% refer in depth? a 1 msw b 1.25 msw c 1.5 msw d 2.25 msw
b
Assumming that air contains 21 % oxygen and 79 % nitrogen, what is the PO² and PN² in air at 165 fsw ? a 1.56 atm and 4.02 atm b 1.05 atm and 3.95 atm c 1.26 atm and 4.74 atm d 1.32 atm and 4.25 atm
c
How much gas is required to pressurise a chamber system with a F.V. of 1400ft³ to a depth of 450 fsw a 19420ft³ b 18529ft³ c 19091ft³ d 17392ft³
c
What is the gas consumption of 2 divers at a depth of 138fsw for 6hours 27mins (1.25ft³/min consumption)
a 4009.3ft³
b 5202.7ft³
c 5011.7ft³
d 4044.2ft³
c
138+33/33=5.18x1.25= 6.47x2= 12.95x387= 5.011,65
A chamber loses 15m³ of gas per day. How many litres per hour does that relate to? a 62.5 ltrs b 652.5 ltrs c 52 ltrs d 625 ltrs
d
15000/24= 625l/h
A chamber system has a volume of 1200 ft³ . What volume of gas would it take to pressurise it to 500 fsw ? a 18763 ft³ b 18,182 ft³ c 19382.1 ft³ d 17846 ft³
b
500/33=15.15 x 1200=18.182ft3
- After filling a bailout to 3500 psi. The bottle temperature is 105°F What will the pressure in the bail-out be when the temperature drops back to 68°F
a 3271 psi
b 3745 psi
c 3714 psi
d 3299 psi
a
Convert F° in R°= +460
3500/ 565x528= 3.271psi
If a balloon has a volume of 5 litres on the surface, what will its volume be at 45 metres? a 2 litres b 1.5 litres c 4.5 litres d 1 litre
d
5/5.5= 0.9l
Boyles law
A 12 litre bailout bottle is at a pressure of 300 bar. What volume of gas is available to use (regulator working pressure 8 bar) dive depth is 140 msw. How long would this gas last the diver? Gas Available Time a 3324 l 5 mins 54 secs b 3324 l 5 mins 32 sec c 3324 l 6 mins 20 secs d 2794 l 4 mins 50 secs
b 300-15-8x12=3.324l 15x40=600l gas consumption ... So 3324/600= 5.54h .54x60= 32min 5h32min
- If you pressurise a 25m³ chamber to 75m from a 48 bottle quad, by how much would the quad pressure drop by?
a 88bar
b 78bar
c 75 bar
d 68 bar
b
7.5x25= 187.5m3
Assuming 200 bar std quad
48x50x200= 480m3
In a simple rule of 3
If 200 bar = 480m3
X = 187.5m3
X= 78.12 bar
Your company´s PO² during decompression is 500mb. At what depth would the O² % reach 21 % ? a 12msw b 14msw c 16msw d 18msw
b
0.5/.21= 2.38-1x10= 13.8m
30. A 64 x 50 litre bottle quad contains gas at a pressure of 195 bar. What is the total volume of gas in the quad ? a 656m³ b 592m³ c 468m³ d 624m³
d
64x50x195= 624m3
Convert 5200 PPM to a %. a 0.052% b 52% c 0.52% d 5.2%
c
5200/10000= 0.52%
To convert degrees fahrenheit to absolute temperature you must add? a 100 b 640 c 273 d 460
d
C to F + 273
F to R + 460
3. What is 65ºF expressed in celsius? a 52ºc b 18.3ºc c 31.2ºc d 20ºc
b
F to c = -32 / 1.8
65-32/ 1.8= 18.3
4. How much does a 12 litre flask hold when charged to 200 bar? a 240m3 b 25.20m3 c 4.8m3 d 2.4m3
d
12x200= 2400= 2.4m3
Convert 1500 litres into cubic metres. a 1.5m3 b 15m3 c 30m3 d 150m3
A
1m3=1000l
How much gas would be required to pressurise a chamber with a floodable volume of 20m3 to a depth of 65msw? a 150m3 b 1300m3 c 130m3 d 120m3
c
6.5x20= 130m3
A bailout bottle has a F.V of 12 litres and is charged to 300 bar. How much gas is available to the diver at a depth of 75m (hat working pressure of 8 bar)? a 3.5m3 b 3.4m3 c 3.6m3 d 4.0m3
b
300-8.5-8x12= 3402l
READY THE FUCKING QUESTION!!!!!!!
A 64 bottle quad, each have a floodable volume of 50 litres. How much gas does it contain at 195 bar? a 195m³ b 624m³ c 6.24m³ d 524m³
b
64x50x195=624m3
A bailout bottle has a F.V of 10 litres and is charged to 200 bar. How long would it last a diver at a depth of 50 msw (40 l/min and 8 bar reg pressure)? a 5mins b 4.6 mins c 6 mins d 7.7 mins
D
200-6-8=186x10=1860
6x40=240l sooo 1860/240= 7.75 min = 7min45sec
What is the partial pressure of O2 at 35 msw if the breathing gas is air? a 500 mb b 400 mb c 945 mb d 735 mb
c
4.5x.21=0.945 bar or 945mbar
If a diver has a litre of gas in solution in his body at 1 bar(A), how much of the same gas would he have in solution at 40 msw? a 1 ltr b 4 ltrs c 5 ltrs d 6 ltrs
c
A bell is blown down to a depth of 120 msw with a temperature at depth of 38ºC. What would the reading by gauge be after cooling to 28ºC? a 116 msw b 126 msw c 114 msw d 120 msw
a
13/ 38+273= 0.0418x 301=12.58-1x10= 115.81msw