Boyle's Law question

[oszillodrom]

it should be P1/V1=P2/V2.

No, Boyle's Law is p1V1 = p2V2

If you look at the ideal gas law, pV = nRT, and nRT is constant (mass of air and temperature stay the same), you see that the product pV is always constant, i.e. if you halve the pressure, you double the volume. That is exactly what Boyle's Law says. Another expression of Boyle's Law is pV = const.

 

[mselenaous]

Boyle's Law = Breath Or Your Lungs Explode Stupid

 

[Scott]

Another problem: How many 80 cu ft scuba tanks can you fit in a room that is 8 ft high, 2 ft deep, and 5 feet wide?

Four, because the rest of the space is fill with other scuba stuff.

 

[NAM001]

You are correct in that those presures and water volumns provide the same stored gas n the cylendars. Wont work with the same or like tanks. The interpretaion assumes the same gas volumn in different sized vessels.

When using the same vessel then 1500= 40 and 3000=80 Your example says if you take the contents of one vessel and move it to another vessel twice the size it comtains the same amount of gas but under a different presure proportional to the difference in sizes of the containers.

No, Boyle's-Mariotte Law is P1V1 = P2V2, which means 1500 * 80 = 3000 * 40.

---------- Post added October 28th, 2014 at 06:09 PM ----------

yes if you half the pressure you have to double the amount of water volumn to have the same compressed volumn.

No, Boyle's Law is p1V1 = p2V2

If you look at the ideal gas law, pV = nRT, and nRT is constant (mass of air and temperature stay the same), you see that the product pV is always constant, i.e. if you halve the pressure, you double the volume. That is exactly what Boyle's Law says. Another expression of Boyle's Law is pV = const.

 

[agilis]

The bottom of the sea is littered with the drowned bodies of feckless divers who did not have the technical language and definitions of Boyle's law absolutely clear.

 

[Hickdive]

If only we had a thread on the superiority of the metric system when used in diving...

 

[oszillodrom]

The bottom of the sea is littered with the drowned bodies of feckless divers who did not have the technical language and definitions of Boyle's law absolutely clear.

God forbid people learn something outside of what they strictly need.

P.S. Understanding gas laws actually makes it easier, because then the only law you ever have to remember is the ideal gas law, pV = nRT.

 

[The Chairman]

No, Boyle's Law is p1V1 = p2V2

This is incorrect. You need to use the ideal Gas Law and assume that the temp is the same. p1/V1=p2/V2 or p1V2=p2V1 would be acceptable, but not p1V1=p2V2, which is an inverse relation.

1atm*0.39cf does not equal 204atm*80cf

Probably the most used iteration of this relationship is V2=p2/(p1/V2)

When you are dealing with a tank, the vessel is a constant size. We are dealing with variations of density and pressure rather than volume and pressure. Since we don't think in terms of Moles, I have reverted to using "atms".

 

[agilis]

God forbid people learn something outside of what they strictly need.

I actually think it's admirable and essential on a certain level to master the complex physics associated with using gas under pressure. My comment was meant only as a touch of levity. After all, this was in the 'Basic Scuba Discussions' forum, and forum appropriateness is an issue to which my attention has occasionally been called.

The OPs simple question asking for clarification in connection with gas volume under pressure and the water volume of the tank itself was completely answered by response #4. I think the technical discussion which followed was great fun and immensely valuable to the participants on that level.

 

[Storker]

This is incorrect. p1/V1=p2/V2 or p1V2=p2V1 would be acceptable, but not p1V1=p2V2, which is an inverse relation.

Sorry, you're wrong. It's P1V1 = P2V2, as you can see from the ideal gas law:

P1V1 = nRT
P2V2 = nRT

Since the number of atoms is constant, P1V1 = P2V2

Example: my 10L tank is rated to 300 bar. I fill it to the rated pressure, and I have 3000L at 1 bar (if I neglect the compressibility factor).
P1 = 10
V1 = 300
P2 = 1
V2 = 3000

10*300 = 1*3000, QED.

EDIT: I also have a 15L tank rated to 200 bar. I'll leave it up to the student to work out whether I'm carrying more gas, less gas or the same amount as if I'm using my 10L 300 bar tank (hint: PV=nRT). The student may neglect the effect of compressibility during this exercise.

1atm*0.39cf does not equal 204atm*80cf

No, you're right, it definitely does not. If you have 0.39cf at 1atm, you have 0.39*28.3L at 1atm, i.e. 0.39*28.3/22.4=0.5 mol of gas. If you have 80cf at 204atm, you have 80*28.3*204/22.4=20.6 mol of gas.

OTOH, if you have a tank with a water volume of 0.39cf and fill it to 204atm, that would equate 80cf at 1 atm. So 1atm*80cf = 204atm*0.39cf, P1V1 = P2V2

---------- Post added October 29th, 2014 at 02:42 PM ----------

If only we had a thread on the superiority of the metric system when used in diving...

My thoughts exactly