Mathematical challenge wrt diving

[ClayJar]

1. Assuming a cylindrical chamber with hemispherical ends and nothing else inside (may as well simplify it all the way, eh?), that would be pi * (D * (L + D)) (with D being the diameter and L being the length of the cylindrical portion).

2. Well done.

3. If the chamber started at 100fsw (gauge), ended at 650fsw (gauge), and went from 72°F to 750°F, then the explosion must have stretched it such that the final internal volume was slightly more than 2.25 times the original volume.

 

[Jonas Isaksen]

Either there is additional information that was not presented here, or the given answer is incorrect.

Unfortunately there is no additional information, other than one should solve the exercise by using universal gas law. The given answer might be incorrect, but if it had been a typing error I should be able to calculate something that look like it not even 1750 degress F.

If anyone got that book the full name is:
Diving science
Essential physiology and medicine for divers
by Michael B. Strauss and
Igor V. Aksenov

 

[Blackwood]

I can get their answer if I assume there to be about 12.5 mol of gas in the chamber...

 

[Jonas Isaksen]

I can get their answer if I assume there to be about 12.5 mol of gas in the chamber...

Please show us how you did it, and please show how it fits in to the given information.

Thanks

 

[ClayJar]

I can get their answer if I assume there to be about 12.5 mol of gas in the chamber...

Assuming ideal gas, you have P₁V₁ = n₁RT₁ and P₂V₂=n₂RT₂. R is, by definition, constant, so you can write this as P₁V₁ / (n₁T₁) = P₂V₂ / (n₂T₂).

P₁, P₂, and T₁ are known. V₁ and V₂ are unknown, as are n₁ and n₂, but based on the problem statement's lack of any information regarding them, I considered both V and n to be constants (which then cancel). T₂ is, obviously, the value for which we're solving. Solving the equation for T₂, we get:

T₂ = T₁P₂ / P₁

When you say you assume there to be about 12.5 mol of gas in the chamber, how are you using that value? If it's a constant 12.5 mol, wouldn't that be irrelevant to the problem at hand?

 

[SmokinReefer]

3. If the chamber started at 100fsw (gauge), ended at 650fsw (gauge), and went from 72°F to 750°F, then the explosion must have stretched it such that the final internal volume was slightly more than 2.25 times the original volume.

I did one like this leaving volume constant and working back from the final temperature. The chamber temperature had to be way below zero before the explosion, -223F to be exact. That may explain why someone would be trying to light a fire in there in the first place.....

 

[ClayJar]

-224°F is more than cold enough to condense the xenon in your flashlight bulb. No wonder they were trying to warm it up.

If I convert 72°F to 22.2°C, forget to add 273 to make that an absolute temperature in Kelvins, and if I also use 650fsw and 100fsw, forgetting to add 33fsw to each to make them absolute pressures, then when I plug those three incorrect answers into the equation, I get a completely bogus answer of 144°. If I then add 273 to convert it to bogus "absolute temperature", then multiply it by 1.8 to get a bogus something-like-Rankine number, I get a value of 751.

Sure, it's wrong in so many places that it's hardly funny, but it gives me a closely matching answer.

 

[boat sju]

Odd, I came up with 468 degrees. The universal gas law says that pressure is proportional to temperature. The pressure increased by a factor of 6.5, i.e. from 100 to 650, so the temperature must also increase by a factor of 6.5.

I got the same answer. Assuming n, R and V are constant. I don't believe you need to convert pressures or temps as long as you're consistant, but I must be missing something here.

 

[OHGoDive]

Odd, I came up with 468 degrees. The universal gas law says that pressure is proportional to temperature. The pressure increased by a factor of 6.5, i.e. from 100 to 650, so the temperature must also increase by a factor of 6.5.

The ratio 100 to 650 is the change in effective depth (100fsw to 650fsw), not pressure. The pressure changed from a little over 4atm to a little over 20atm, a ratio of about 5.1.

The relative change in temperature would be in degrees K, not degrees F, right?

Is the original poster sure that there isn't some offhand description of the size of the chamber relative to the room (maybe half the room)? Because it seems that, all other things being equal, the temperature would change to about 1475 degrees F, which is 2 * "almost 750 degrees".

 

[Jonas Isaksen]

Is the original poster sure that there isn't some offhand description of the size of the chamber relative to the room (maybe half the room)? Because it seems that, all other things being equal, the temperature would change to about 1475 degrees F, which is 2 * "almost 750 degrees".

Unfortunately I have given all information that was provided with respect to this exercise. So far a lot of people have tried to solve it, no success, which means I as well start to look at it as an exercise where the necessary information where not given or the answer is completely wrong. By playing with numbers I have not found one that may look similar in any way.