Reef Chemistry Question of the Day #275 Effect of Underwater Pressure on a Balloon

[Randy Holmes-Farley]

Reef Chemistry Question of the Day #275

Yesterday was a pressure question, so let's go with another...

Suppose that you diving on a reef and are 100 feet below the surface.

You fill an ordinary latex party balloon with a small amount of air from your tank, expanding it to far less volume than it would take to burst it on the surface, but it is stretched a little.

You release the balloon and watch it rise to the surface.

You have a keen eye and notice that the gas volume of the balloon increases as it rises.

Which of the following is the most accurate description of the change in the balloon volume as it rises?

A. The percentage increase in volume as it rises from 100 feet to 50 feet is the same as when it rises from 50 feet to 25 feet depth.

B. The percentage increase in volume as it rises from 100 feet to 50 feet is a bit less than when it rises from 50 feet to 25 feet depth.

C. The percentage increase in volume as it rises from 100 feet to 50 feet is a bit more than when it rises from 50 feet to 25 feet depth.

D. The percentage increase in volume as it rises from 100 feet to 50 feet is twice as much as when it rises from 50 feet to 25 feet depth.

E. The percentage increase in volume as it rises from 100 feet to 50 feet is half as much as when it rises from 50 feet to 25 feet depth.

Assume the water temperature is the same everywhere and the balloon does not pop.

Explain your answer if you want to!

Good luck!





































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[JimWelsh]

Static fluid pressure P = pgh, where p is the density of the liquid, g is a gravitational acceleration factor, and h is the depth of the fluid. In our ocean scenario, p and g are constant (unless you're counting the infinitesimal increase in g over the total 75 foot depth change), so P is a linear function of depth. For the gas in the balloon, (P * V) / T is constant, and since the question states that T is constant, then the volume of the balloon will just be the inverse of the pressure. Between 100 and 50 feet, the pressure halves, so the volume of the balloon doubles. Similarly, between 50 and 25 feet, pressure also halves, and volume again doubles. In both cases, the percentage volume increase is 100%. Answer is A (or else it is C, if you're counting the tiny, tiny, tiny change in g).

 

[Randy Holmes-Farley]

Static fluid pressure P = pgh, where p is the density of the liquid, g is a gravitational accelleration factor, and h is the depth of the fluid. In our ocean scenario, p and g are constant (unless you're counting the infinitesimal increase in g over the total 75 foot depth change), so P is a linear function of depth. For the gas in the balloon, (P * V) / T is constant, and since the question states that T is constant, then the volume of the balloon will simply be the inverse of the pressure. Between 100 and 50 feet, the pressure halves, so the volume of the balloon doubles. Similarly, between 50 and 25 feet, pressure also halves, and volume again doubles. In both cases, the percentage volume increase is 100%. Answer is A (or else it is C, if you're counting the tiny, tiny, tiny change in g).

Not looking at g, but I am looking at pressure more closely (not zero at surface, for example), and the balloon stretching also adds its own pressure.

 

[Sleepydoc]

Two relevant equations:
P = ⍴gh (P = pressure, ⍴ = density of the fluid, g = acceleration of gravity and h = depth)
PV=nRT (ideal gas law)

The first equation tells us that the hydrostatic pressure increases linearly with depth, so the decrease in pressure from 100 to 50 feet is twice as much as the decrease from 50 to 25 feet.

From the second equation, Assuming the temperature is constant, nRT will be constant. If we also assume the pressure exerted by the balloon on the gas is negligible, the volume will increase linearly with the decrease in pressure to keep PV constant. This means that the increase in volume from 100 to 50 feet will be twice as much as from 50 to 25 feet.

In terms of percentage, the pressure drop from 100 to 50 feet will be the same (50%), so the percentage increase in volume will also be the same. (A) If you want to take into account the pressure exerted by the balloon, it will exert more pressure on the gas at higher volumes (closer to the surface,) so the percentage increase from 100 to 50 will be slightly more (C)

 

[Alfrareef]

So, as a diver that already filled a plastic bottle with air while was bellow 100ft of water and observed what happened to the bottle as we return to the surface I’ll go with:
A
When you fill the ballon at 100ft you’ll use roughly 4 times more air that if it were at surface level so when you reach 50 ft the volume has doubled because the external pressure it’s half.
Here we must consider Arquimedes theory for water (weight) pressure versus the perfect gases... [emoji4]
What I don’t believe is that the ballon reach the surface without exploding....

 

[taricha]

Pressure at surface is not zero. It's one atmosphere, and to get another atmosphere of pressure you can go down about 10 meters under water...

 

[Sleepydoc]

Pressure at surface is not zero. It's one atmosphere, and to get another atmosphere of pressure you can go down about 10 meters under water...

How does this affect the question?

 

[taricha]

How does this affect the question?

It means that although P=pgh as stated above is correct for water pressure, the pressure 50 ft down is not half the pressure 100 ft down.
And the pressure 25 ft down is not half the pressure 50 ft down.
Water pressure, yes. Total pressure, no.

 

[Randy Holmes-Farley]

It means that although P=pgh as stated above is correct for water pressure, the pressure 50 ft down is not half the pressure 100 ft down.
And the pressure 25 ft down is not half the pressure 50 ft down.
Water pressure, yes. Total pressure, no.

Right.

 

[PatW]

At sea level, you have 1 atmosphere of pressure, 1 atm. At 32’, you get another atmosphere of pressure for a total of 2 atm. At 64’, you get a third atm and at 96’ you get your 4th atm. As it rises, the balloon will double in volume at 32’ and double again when it reaches the surface.

 

[Alfrareef]

Yes, to keep it simple you’ve get 1atm at surface level (0 meters-sea level) and 1atm plus for every 10 meters you dive.
So it’s true that for this problem if you’re considering the variation it’s not important if you start from 0 or 1.
So I’ve returned to the answers and it’s D!

 

[Chuk]

C because p1v1=p2v2 and that equation uses absolute pressure not gauge

 

[Randy Holmes-Farley]

Don't forget the stretching balloon pressure.

 

[Alfrareef]

Considering the material stretching it should be C, but D it’s more didactic... [emoji3]

 

[Chuk]

Don't forget the stretching balloon pressure.

If I remember from mechanics of materials class way back when the tensile strength of something line a balloon ends up being a modified form of hooks law. I think it was due to the tangled polymer chains straightening out or something to that effect. The resultant pressure would be related to the balloon diameter and the spring constant of the rubber balloon I think it was something like P=k/v^2 or something like that (I tossed that book so I cant look it up quickly). Based on that half remembered half made up equation there would be a 25% difference in balloon stretching pressure from 100 ft to 50 ft and 50 ft to 25 ft. It should be like a 1 ft difference in pressure if the spring coefficient I found for a rubber band applies. That is almost negligible compared to the 58 ft total pressure at 25 ft depth and 83 ft total pressure at 50 ft depth. Now if my made up equation is wrong then you could be talking about some significant noticeable percentages but I wouldn't think you can see anything less than a 5 % change from you vantage point 75 ft away.

 

[Tautog]

Your answer is E.
Here’s why. Atmospheric pressure is 14.7 psi, 1 atmosphere, just under water you’re under 29.4 psi, or 2 atmospheres. For every 33 ft, or 1 atmosphere, 14.7 psi. At 50 ft, and under 34.1 psi and at 100 ft 63.5 psi. Due to that much pressure against the balloon and whatever volume of air in the balloon, as the balloon rises the air volume increases exponentially. Hence, why a diver can exhale and breathe at the same time while swimming back to the surface. Without exhaling, or assenting too fast, divers would would perish.

 

[Alfrareef]

Hi Randy!
It’s time for the correct answer...

 

[Randy Holmes-Farley]

oops

I'll post it later today.

 

[Alfrareef]

I’ve assumed you’re still going thru the numbers... Eheheheh!

 

[Randy Holmes-Farley]

OK, let's work through this in detail.

First, the volume of the balloon is a function of the pressure on the gas inside.

That pressure comes from three sources:
1. the weight of the water above it
2. the weight of the air above it
3. the elastic stretching of the balloon

Let's start with #2. At sea level, the pressure from the weight of the air is 1 atmosphere. It turns out that this pressure is equal to the pressure applied by about 33 feet of seawater:

https://www.scubatoys.com/education/introgas.asp

So here we calculate the pressure as the various depths. We'll take the easy route and calculate it in terms of feet of seawater (sort of like mm mercury; we discussed this type of "barometer" in an earlier question fo the day:

https://www.reef2reef.com/threads/reef-chemistry-question-of-the-day-216-seawater-barometer.324482/


Surface: 33 feet of seawater
25 feet: 25 + 33 feet of seawater = 58 feet of seawater
50 feet: 50 + 33 feet of seawater = 83 feet of seawater
100 feet; 100 + 33 feet of seawater = 133 feet of seawater


IF the water pressure was the only effect, and assuming the balloon gas follows the ideal gas law (it closely would), then the pressure changes we observe are:

100 to 50 feet: 133 - 83 = 50 feet of seawater (the pressure change)
consequently, if the volume started at 1 liter at 100 feet, the final volume at 50 feet would rise to 1 liter /(83/133) = 1.602 liters, or a rise of 60.2% from 100 to 50 feet

50 to 25 feet: 83-58 = 25 feet of seawater (the pressure change)
consequently, if the volume was 1.602 L at 50 feet, the final volume at 25 feet would rise to 1.602 liter /(58/83 L) = 2.293 liters, or a rise of 43.1%

Thus, we see that the percentage rise in volume is bigger from 100 feet to 50 feet than it is from 50 feet to 25 feet.

Much harder to calculate, but fortunately going in the same direction, is the fact that as the balloon stretches, it applies more and more of its own pressure on the gas in the balloon. That is, it gets harder and harder to blow it up bigger (at least when it is getting near its easily stretched limit) due to the pressure exerted by the stretching latex. One can see a detailed discussion of balloon inflation here:

http://members.tripod.com/the_common_loon/phyx.html

Fortunately, both of these effects (air plus water pressure and balloon stretching pressure) suggest that the balloon inflates less (as a percentage) from 50 feet to 25 feet deep than it does from 100 feet to 50 feet deep.

Answer is C:

Which of the following is the most accurate description of the change in the balloon volume as it rises?

A. The percentage increase in volume as it rises from 100 feet to 50 feet is the same as when it rises from 50 feet to 25 feet depth.

B. The percentage increase in volume as it rises from 100 feet to 50 feet is a bit less than when it rises from 50 feet to 25 feet depth.

C. The percentage increase in volume as it rises from 100 feet to 50 feet is a bit more than when it rises from 50 feet to 25 feet depth.

D. The percentage increase in volume as it rises from 100 feet to 50 feet is twice as much as when it rises from 50 feet to 25 feet depth.

E. The percentage increase in volume as it rises from 100 feet to 50 feet is half as much as when it rises from 50 feet to 25 feet depth.