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Hello world
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Hi all , this year I am trying to do my leading diver under CMAS *** Leading diver and last night myself and another candidate was at our clubhouse with our DO and Training officer going through a old exam paper , the thing is I am really struggling with gas laws and how to work out formulas , so I was wondering if anyone out there knew a way or any links to simplify this for me . I consider myself clever enough to learn but a more "Gas laws for dummies " route ! I did physics at school and was competent enough but this seems a whole new ball game . If anyone out there can help I would be much appreciated as the exam is 12th June and my nails have already gone!
 

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A short fat well off crap cave diver. Likes wrecks
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Hi all , this year I am trying to do my leading diver under CMAS *** Leading diver and last night myself and another candidate was at our clubhouse with our DO and Training officer going through a old exam paper , the thing is I am really struggling with gas laws and how to work out formulas , so I was wondering if anyone out there knew a way or any links to simplify this for me . I consider myself clever enough to learn but a more "Gas laws for dummies " route ! I did physics at school and was competent enough but this seems a whole new ball game . If anyone out there can help I would be much appreciated as the exam is 12th June and my nails have already gone!

The level of math required for diving is fairly basic so you shouldent have much of a problem.

Is there a particular question your strugling with?

ATB

Mark
 
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Nigel Hewitt
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All I can offer is this. I wrote it up because I saw so many people having trouble with formula that were too complex (do we add or subract 10 here?)
 

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Hello world
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Discussion Starter #4
I am out at moment but will post a question up later with answer and if you could help show the inbetween bit I would be much appreciated ! Thanks to both of you for the help and response
 

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Hello world
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Discussion Starter #5
Here is two of the questions from the 2008 exam
Q1 What volume of air in a lifting bag will hold a mooring with a volume of 20 litres and a mass of 100 kg neutrally bouyant in freshwater ?
The answer is 83 litres how is this calculated?
Q2 Your club RIB has a design burst pressure of 2 bar . You pump it to the recommended 1.75 bar pressure on a cool morning with a temperature of 7 c . The day turns out to be hot and sunny .How hot must the boat get to be at risk of bursting should the relief valves fail ?
the answer to this is 47 c also how is this calculated ? Help much appreciated. Thanks .
 

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One Team One Dream
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Here is two of the questions from the 2008 exam

Q2 Your club RIB has a design burst pressure of 2 bar . You pump it to the recommended 1.75 bar pressure on a cool morning with a temperature of 7 c . The day turns out to be hot and sunny .How hot must the boat get to be at risk of bursting should the relief valves fail ?
the answer to this is 47 c also how is this calculated ? Help much appreciated. Thanks .
Possibly the rise in temp 47c - 7c = 40 c
pressure required to burst 2.0 bar
Filled pressure 1.75bar

2.0 - 1.75 = 0.25

0.25bar / 40c = 0.00625 per 1degree C

So air heated by 1 degree increases its pressure by 0.00625bar

just a thought

its also called Boyles Law

http://www.scribd.com/doc/2280505/Expansion-of-Gas
 

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Out of his depth
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Possibly the rise in temp 47c - 7c = 40 c
pressure required to burst 2.0 bar
Filled pressure 1.75bar

2.0 - 1.75 = 0.25

0.25bar / 40c = 0.00625 per 1degree C

So air heated by 1 degree increases its pressure by 0.00625bar

just a thought

its also called Boyles Law

Expansion of Gas
Or to get that without knowing the 47 degree answer...

Pressure/Temp = a constant for a fixed volume of gas.

so, Pressure (morning)/Temp(Morning) = constant = Pressure (bursting)/Temp(Bursting temp)

Temp is in Kelvins (Normal temp + 273), so...

1.75/280 = 0.00625 so know we know the constant in this question, so to get bursting temp, we need to calculate

2/Burst Temp = 0.00625 (2 because that is the burst pressure). Re-arrange that, and it gives 2/0.00625=Burst temp
=320K or 47C (320-273)

I haven't got a scoobies about lift bags unfortunately.
 

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Q1:

To be neutrally buoyant you need the weight (downwards force) to be equal to the buoyancy (upwards). To do it properly you'd have to work in Newtons because Kg are not a unit of force but I presume they don't go to all that detail because it's not something we work with on a daily basis.

So you need a buoyancy equal to 100 Kg. Part of that buoyancy is already there and it comes from the volume of the mooring. Archimedes principle states that an object totally or partially immersed in a fluid will be acted upon by an upwards force (buoyancy) equal to the weight of the volume of liquid displaced by the object, meaning, it's immersed volume.

In this case, the mooring has a volume of 20 l. Freshwater has a density of 1 Kg/l. This means that the mooring will have an intrinsic buoyancy of 20 Kg.

But since its weight is 100 Kg, you still need to apply an extra 80 Kg of uplift to keep it neutral. That means 80 l.

This is the reasoning, which I don't think it's wrong although I get a different answer.
 

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About the questions with gases, they involve usually two equations, the Boyle's law (or Boyle-Mariotte) which states that:

pV = constant, for a fixed temperature and amount of gas (no gas lost or added)

and the Gay-Lussac's law:

P / T = constant, for a fixed volume and amount of gas.

Instead of memorising these two laws you can work with the Ideal Gas Law:

pV = nRT

n is related to the amount of gas, which for these exercises is always a constant and R is another constant, so you have:

pV = constant x T

From this you can work both the above laws, fixing either the temperature or the volume.
 

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The worlds slowest sailor.
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Q1:

To be neutrally buoyant you need the weight (downwards force) to be equal to the buoyancy (upwards). To do it properly you'd have to work in Newtons because Kg are not a unit of force but I presume they don't go to all that detail because it's not something we work with on a daily basis.

So you need a buoyancy equal to 100 Kg. Part of that buoyancy is already there and it comes from the volume of the mooring. Archimedes principle states that an object totally or partially immersed in a fluid will be acted upon by an upwards force (buoyancy) equal to the weight of the volume of liquid displaced by the object, meaning, it's immersed volume.

In this case, the mooring has a volume of 20 l. Freshwater has a density of 1 Kg/l. This means that the mooring will have an intrinsic buoyancy of 20 Kg.

But since its weight is 100 Kg, you still need to apply an extra 80 Kg of uplift to keep it neutral. That means 80 l.

This is the reasoning, which I don't think it's wrong although I get a different answer.
i agree.
so that means the lift bag and shackle needs 3lts of gas to also be neutral.
 

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Nigel Hewitt
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Q1 What volume of air in a lifting bag will hold a mooring with a volume of 20 litres and a mass of 100 kg neutrally bouyant in freshwater ?
The answer is 83 litres how is this calculated?
Eeek....
Well I can see 20L of displacement so in fresh water 20Kg of up-thrust so we need a further 80Kg of pull from the bag for neutral.
I can see that 80L of bag displacement won't quite work as 80L of air weighs 96gms so we need a teeny bit more but 3Kg more? Even a hundred meters down there's only a kilogram of gas in the bag.

If we were told something about the geometry of the bag and were quoting an answer in surface litres I can see it being more as the gas pressure in the whole bag is the absolute pressure at the gas/water interface which must be under the water level to get lift but we aren't.

I must have failed.
 
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