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Help preparing for exam

Discussion in 'Advanced Scuba Discussions' started by Merryoceaner, Aug 28, 2019.

  1. Merryoceaner

    Merryoceaner Angel Fish

    # of Dives: 100 - 199
    Location: Saint Petersburg
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  2. Slate

    Slate Banned

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    I think in question 44, the water temperature is irrelevant and designed to fool you into thinking you need to account for it in the calculations.
     
  3. wnissen

    wnissen Barracuda

    # of Dives: 25 - 49
    Location: Livermore, Calif.
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    Have you learned about PV=NRT (pronounced "piv-nert")? Also, I do see an error that 85F should not be added to 273K.
    Well, that would be a good question for the OP. I would try working the two parts of the problem separately at first. If you held the pressure constant, what would the volume change due to temperature be? And vice versa, if you held the temperature constant, what would the volume change due to pressure. I worked this but got a slightly different answer so I don't want to potentially put anyone on the wrong path. I definitely recommend doing it in metric, though, much easier to calculate the pressure in meters.
     
    Slate likes this.
  4. Merryoceaner

    Merryoceaner Angel Fish

    # of Dives: 100 - 199
    Location: Saint Petersburg
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    Ahhha progress on #44 General gas law! P1 V1/ T1 = P2 V2/ T 2 and yes you need to convert to Rainken for absolute temperature so add 273.

    Onto the SAC ones! Thanks!
     
    The Chairman and Lorenzoid like this.
  5. wnissen

    wnissen Barracuda

    # of Dives: 25 - 49
    Location: Livermore, Calif.
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    I learned something, there is a Fahrenheit equivalent to Kelvin called Rankine. Who would have guessed?

    That's also one hell of a thermocline.
     
    mwerle likes this.
  6. Southside

    Southside Nassau Grouper

    # of Dives: 25 - 49
    Location: Chicago
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    You add 492 to Fahrenheit to convert to Rankine. The 273 is for converting from Celsius to Kelvin.

    As for the relevance to the question it is hard to say what they are looking for. They specify the water temperature at the different depths but say nothing about the temperature of the air inside the container. In the real world depending on the container the air inside would cool down as the water temperature decreased but it would not happen instantaneously, but that gets into thermodynamics and is beyond the scope of this problem.
     
  7. TMHeimer

    TMHeimer Divemaster

    # of Dives: 500 - 999
    Location: Dartmouth,NS,Canada(Eastern Passage-Atlantic)
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    For what course is this test if I may ask?
     
  8. dberry

    dberry Hydrophilic ScubaBoard Supporter

    # of Dives: 100 - 199
    Location: Philadelphia
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    Not to be a major jerk, but if the OP is so confused about temperature scales and how to use the gas laws, I think they need to spend more time studying the basics and less worrying about answering these specific questions. Just pounding through these to get answers won't be much help if/when they actually encounter gas planning problems in real life.
     
    ishmaelcat likes this.
  9. mwerle

    mwerle Angel Fish

    # of Dives: 200 - 499
    Location: Tokyo, Japan
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    We need to use the "complete gas equation": p1 * V1 / T1 = p2 * V2 / T2
    We want "v2", so rearrange : V2 = p1 * V1 * T2 / T1 * p2

    The only "gotcha" here is that we need the temperatures to be specified in absolute units (Kelvin), so add 273.16. Also the pressures need to be in kilopascal, but since that's just bar * 100, we can ignore that as we are dividing the two pressures with each other.

    Work out the pressures at depth (Fresh water has a density of 1m = 0.097bar - 3% lighter than salt water):
    p1 = 1 + 6.1 * 0.097 = 1.592
    p2 = 1 + 18.3 * 0.097 = 2.775

    Work out the absolute temperatures:
    T1 = 29.4 + 273.16 = 302.56
    T2 = 4.4 + 273.16 = 277.56

    Substitute in:
    v2 = 4 * 1.592 * 277.56 / 302.56 * 2.775
    = 1767.5 / 839.6
    = 2.10l (which is close, but not precisely one of the accepted answers... hmm)


    --> Ok, what did I do wrong here? I was expecting to get pretty much spot-on, unless the question was written in imperial and the conversions to metric are "iffy".

    -----

    The first step in Q40, is to determine the amount of usable air: 207-69 = 138bar

    Next is to work out the consumption rate at 27m - simply multiply the SAC by the pressure: 2.07 * 3.7 = 7.66bar/min.
    Last is dividing the usable air by the consumption rate to get the time - 138/7.66
    = 18min.

    You could also first convert SAC into litres/min as well as working out how many litres of air you have, but that's only useful if you want to plan another dive with a different sized tank.

    ----

    Q41 is a simple ratio calculation.
    2265 : 207 = ? : 152
    ? = 2265 / (207 / 152) = 1663.18 (~1665).

    Or, to make more "sense" for scuba diving, you could work out the size of the cylinder first:
    2265 / 207 = 10.942 (~11l)
    Then work out how many litres of air you have at the new pressure:
    10.942 * 152 = 1663.18 (~1665)

    Hmm, again, since these don't work out "exactly" I'm wondering how precise the exam-supplied metric numbers are..

    Do they work out better in imperial?
     
    Protondecay123 likes this.
  10. TrimixToo

    TrimixToo Regular of the Pub

    # of Dives: 200 - 499
    Location: New York State
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    PV=nRT. R and n are constants for these questions.
     

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