Mathematical challenge wrt diving

[Mad Scientist]

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.

PV=nRT, says that pressure and volume are inversely proportional, P=1/V. If you keep everything else constant than the chamber had to shrink using the assumption that everyone is making here. Since the Volume is constant (before the explosion) and pressure is increasing than some other argument is not constant. The pressure is coming from air being added to the chamber so n is not constant in this equation so you need to figure out the number of moles of gas in the chamber before and after the pressurizing occurred.

 

[boat sju]

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".

Depth X Density (of seawater which is constant here) gives you pressure in lbs/sq. ft.). Atm doesn't work in the equations (unit wise).

I think the book must have left something out.

 

[ClayJar]

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.

The pressures and temperatures can be expressed in any units you'd like, but they must be on absolute scales. You can do the problem in Kelvins or degrees Rankine, and you can use any units of pressure, as long as the values are absolute and not gauge.

For an example of non-absolute values giving completely preposterous results, take a tank from your LDS that is filled with air at 3000 psi when measured at 25°C (about 75°F). Now, drive over to dive at a local quarry and leave it in your freezing cold car overnight (0°C, 32°F). When you hook up your reg in the morning, does it show 0 psi? It would if you could do the math in Celsius. Does it show 1280 psi? It would if you could do the math in Fahrenheit. If you do the math in absolute temperature (Kelvin, Rankine, or any other absolute scale, it doesn't matter), you see it should actually show 2760 psi, which is what it will.

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.

We have indeed successfully solved the problem as written. That either the problem was poorly written (and incomplete) or the answer was completely wrong has no bearing on the successful solution of the straightforward problem presented.

PV=nRT, says that pressure and volume are inversely proportional, P=1/V. If you keep everything else constant than the chamber had to shrink using the assumption that everyone is making here. Since the Volume is constant (before the explosion) and pressure is increasing than some other argument is not constant. The pressure is coming from air being added to the chamber so n is not constant in this equation so you need to figure out the number of moles of gas in the chamber before and after the pressurizing occurred.

T is not constant, as per the problem definition. The problem statement requires assumptions for V and n, and as there was no information about either, the simplest assumption should be used, which is that both are constants. Assuming n is not constant would make the problem incompletely constrained, which would mean it would be, obviously, unsolvable.

Given the problem statement, the only logical course of action would be to solve for T₂, which is what the answer appears to be (albeit compeltely incorrectly).

Because it seems that, all other things being equal, the temperature would change to about 1475 degrees F, which is 2 * "almost 750 degrees".

1475°F is not roughly twice the temperature of 750°F. It is approximately 1.6 times the temperature of 750°F. When you deal with ratios, you must use absolutes. 1475°F is 1935°R, and 750°F is 1210°R. Dividing 1935°R by 1210°R gives 1.599.

Why must you use absolutes? Well, is 10°F "negative 1 times" -10°F?

 

[erparamedic]

My questions are:

1. "Can you calculate the area over which the guts are spread?"
and
2. "Do you like your meat rare, medium or well done?"

Now those are the REAL questions that should be asked!

 

[Mad Scientist]

T is not constant, as per the problem definition. The problem statement requires assumptions for V and n, and as there was no information about either, the simplest assumption should be used, which is that both are constants. Assuming n is not constant would make the problem incompletely constrained, which would mean it would be, obviously, unsolvable.

Given the problem statement, the only logical course of action would be to solve for T₂, which is what the answer appears to be (albeit compeltely incorrectly).

I never said T was constant. Using ratios to solve the problem will only work if we have a closed system, which it is obivous it is not, since the increase in pressure comes from outside the system not by changing the system itself. So n being constant can not be assumed. If you assume V and n being constant how did the pressure increase?

 

[ClayJar]

I never said T was constant. Using ratios to solve the problem will only work if we have a closed system, which it is obivous it is not, since the increase in pressure comes from outside the system not by changing the system itself. So n being constant can not be assumed. If you assume V and n being constant how did the pressure increase?

The pressure increased is due to the increase in temperature, of course! And the temperature increased because energy was added to the system.

The only numbers given are for initial temperature, initial pressure, and final pressure. The problem asks you to solve for the final temperature "using the universal gas law". Given the problem statement at the beginning of this thread, the problem boils down to a simple case of Amonton's Law.

Now, why they called the sudden addition of a considerable amount of energy an "explosion" is their problem. There is nothing in the problem that indicates a change in volume, and as there is no data, the standard assumption is to assume it is constant. There is also nothing regarding the change in quantity of gas (such as the addition of a large number of moles of reaction products as one would expect from a solid or liquid explosive being consumed), and so, while the problem statement's use of the word "explosion" may cause one to imagine reaction products being added to the chamber's contained gases, with no data, one must assume that the quantity of contained gas is also a constant.

It would be completely unscientific to make up numbers "out of thin air" (so to speak ). Assuming that an unspecified value is a constant is a standard assumption, given no data to the contrary.

 

[Blackwood]

Yah. I realized I made a stupid mistake after I left for lunch. Didn't know I could use _subscript on here. Thanks.



In any case, I solved for V given the initial conditions and then plugged that V back into the ideal gas law given the second conditions. I came up with a number much lower than theirs. Then I went: okay, what would my V have to be to come up with their answer? But I used n the first time I solved and and not the second time.

I agree that the given answer is a misprint. When instructed to use the ideal gas law, one can fairly assume that PV = constant for the process.

 

[Blackwood]

The pressures and temperatures can be expressed in any units you'd like, but they must be on absolute scales. You can do the problem in Kelvins or degrees Rankine, and you can use any units of pressure, as long as the values are absolute and not gauge.

If you are evaluating the ideal gas law, the units of volume, temperature and pressure required depend on what value you use for R (gas constant).

If you are using 8.314, you need m[super]3[/super], Kelvin and Pascals.

Darn. No superscript.

 

[ClayJar]

Okay, dropping the assumption that n is a constant, and replacing that with the assumption that T₂ is actually 750°F, we can turn it into a problem asking for the ratio of n₂ to n₁. So, if everything given in the problem is correct, and the volume of the chamber remained a constant (it's basically an infinitely strong chamber, and the explosive is basically infinitely dense pre-explosion), we find...

n₂/n₁ = 2.26

So, there's about two and a quarter times the moles of gas after the explosion as before. Of course, this is using the problem's alleged *answer* as part of the problem statement itself, and it's solving for something that was not at all specified in the original problem, but if you want to solve that completely irrelevant (but fun) problem, there you have it.

If you are evaluating the ideal gas law, the units of volume, temperature and pressure required depend on what value you use for R (gas constant).

If you are using 8.314, you need m³, Kelvin and Pascals.

Only if you're solving for a given set of conditions. If you're using the ideal gas law to compare two sets of conditions, the R disappears, so obviously you can use any units you want. Since diving physics problems almost always deal with comparisions, there's almost no need to deal with R. (Plus, if you were solving for, say, the number of moles of air in a tank, it'd probably not be ideal to use the ideal gas law anymore. )

Incidentally, [SUP]3[/SUP]

 

[Blackwood]

If PV = constant, R = constant, and T changes, n must too, right?